Introductory Mathematics for Engineering Applications.pdf

Kuldip S.Rattan + Nathan W. Klingbeil Introductory Mathematics for Engineering Applications Introductory Mathematics for Engineering Applications Kuldip S. Rattan Wright State University Nathan W. Klingbeil Wright State University VP & Publisher: Editor: Editorial Assistant: Marketing Manager: Marketing Assistant: Designer: Associate Production Manager: Don Fowley Dan Sayre Jessica Knecht Chris Ruel Marissa Carroll Kenji Ngieng Joyce Poh This book was set by Aptara Corporation. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. 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ISBN 978-1-118-14180-9 (Paperback) Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 Contents 1 STRAIGHT LINES IN ENGINEERING 1 1.1 Vehicle during Braking 1 1.2 Voltage-Current Relationship in a Resistive Circuit 3 1.3 Force-Displacement in a Preloaded Tension Spring 6 1.4 Further Examples of Lines in Engineering 8 Problems 19 2 QUADRATIC EQUATIONS IN ENGINEERING 32 2.1 A Projectile in a Vertical Plane 32 2.2 Current in a Lamp 36 2.3 Equivalent Resistance 37 2.4 Further Examples of Quadratic Equations in Engineering 38 Problems 50 3 TRIGONOMETRY IN ENGINEERING 60 3.1 Introduction 60 3.2 One-Link Planar Robot 60 3.2.1 3.2.2 Kinematics of One-Link Robot 60 Inverse Kinematics of One-Link Robot 68 3.3 Two-Link Planar Robot 72 3.3.1 3.3.2 3.3.3 Direct Kinematics of Two-Link Robot 73 Inverse Kinematics of Two-Link Robot 75 Further Examples of Two-Link Planar Robot 79 3.4 Further Examples of Trigonometry in Engineering 89 Problems 97 4 TWO-DIMENSIONAL VECTORS IN ENGINEERING 106 4.1 Introduction 106 4.2 Position Vector in Rectangular Form 107 4.3 Position Vector in Polar Form 107 4.4 Vector Addition 110 4.4.1 Examples of Vector Addition in Engineering 111 Problems 123 5 COMPLEX NUMBERS IN ENGINEERING 132 5.1 Introduction 132 5.2 Position of One-Link Robot as a Complex Number 133 iii iv Contents 5.3 Impedance of R, L, and C as a Complex Number 134 5.3.1 5.3.2 5.3.3 Impedance of a Resistor R 134 Impedance of an Inductor L 134 Impedance of a Capacitor C 135 5.4 Impedance of a Series RLC Circuit 136 5.5 Impedance of R and L Connected in Parallel 137 5.6 Armature Current in a DC Motor 140 5.7 Further Examples of Complex Numbers in Electric Circuits 141 5.8 Complex Conjugate 145 Problems 147 6 SINUSOIDS IN ENGINEERING 157 6.1 One-Link Planar Robot as a Sinusoid 157 6.2 Angular Motion of the One-Link Planar Robot 159 6.2.1 Relations between Frequency and Period 160 6.3 Phase Angle, Phase Shift, and Time Shift 162 6.4 General Form of a Sinusoid 164 6.5 Addition of Sinusoids of the Same Frequency 166 Problems 173 7 SYSTEMS OF EQUATIONS IN ENGINEERING 184 7.1 Introduction 184 7.2 Solution of a Two-Loop Circuit 184 7.3 Tension in Cables 190 7.4 Further Examples of Systems of Equations in Engineering 193 Problems 206 8 DERIVATIVES IN ENGINEERING 218 8.1 Introduction 218 8.1.1 What Is a Derivative? 218 8.2 Maxima and Minima 221 8.3 Applications of Derivatives in Dynamics 225 8.3.1 Position, Velocity, and Acceleration 226 8.4 Applications of Derivatives in Electric Circuits 240 8.4.1 8.4.2 Current and Voltage in an Inductor 243 Current and Voltage in a Capacitor 247 8.5 Applications of Derivatives in Strength of Materials 250 8.5.1 Maximum Stress under Axial Loading 256 8.6 Further Examples of Derivatives in Engineering 261 Problems 266 9 INTEGRALS IN ENGINEERING 278 9.1 Introduction: The Asphalt Problem 278 9.2 Concept of Work 283 9.3 Application of Integrals in Statics 286 9.3.1 9.3.2 Center of Gravity (Centroid) 286 Alternate Definition of the Centroid 293 Contents 9.4 Distributed Loads 296 9.4.1 9.4.2 Hydrostatic Pressure on a Retaining Wall 296 Distributed Loads on Beams: Statically Equivalent Loading 298 9.5 Applications of Integrals in Dynamics 302 9.5.1 Graphical Interpretation 309 9.6 Applications of Integrals in Electric Circuits 314 9.6.1 Current, Voltage, and Energy Stored in a Capacitor 314 9.7 Current and Voltage in an Inductor 322 9.8 Further Examples of Integrals in Engineering 327 Problems 334 10 DIFFERENTIAL EQUATIONS IN ENGINEERING 345 10.1 Introduction: The Leaking Bucket 345 10.2 Differential Equations 346 10.3 Solution of Linear DEQ with Constant Coefficients 347 10.4 First-Order Differential Equations 348 10.5 Second-Order Differential Equations 374 10.5.1 Free Vibration of a Spring-Mass System 374 10.5.2 Forced Vibration of a Spring-Mass System 379 10.5.3 Second-Order LC Circuit 386 Problems 390 ANSWERS TO SELECTED PROBLEMS 401 INDEX 417 v Preface This book is intended to provide first-year engineering students with a comprehensive introduction to the application of mathematics in engineering. This includes math topics ranging from precalculus and trigonometry through calculus and differential equations, with all topics set in the context of an engineering application. Specific math topics include linear and quadratic equations, trigonometry, 2-D vectors, complex numbers, sinusoids and harmonic signals, systems of equations and matrices, derivatives, integrals, and differential equations. However, these topics are covered only to the extent that they are actually used in core first- and second-year engineering courses, including physics, statics, dynamics, strength of materials, and electric circuits, with occasional applications from upper-division courses. Additional motivation is provided by a wide range of worked examples and homework problems representing a variety of popular engineering disciplines. While this book provides a comprehensive introduction to both the math topics and their engineering applications, it provides comprehensive coverage of neither. As such, it is not intended to be a replacement for any traditional math or engineering textbook. It is more like an advertisement or movie trailer. Indeed, everything covered in this book will be covered again in either an engineering or mathematics classroom. This gives the instructor an enormous amount of freedom − the freedom to integrate math and physics by immersion. The freedom to leverage student intuition, and to introduce new physical contexts for math without the constraint of prerequisite knowledge. The freedom to let the physics help explain the math and the math to help explain the physics. The freedom to teach math to engineers the way it really ought to be taught − within a context, and for a reason. Ideally, this book would serve as the primary text for a first-year engineering mathematics course, which would replace traditional math prerequisite requirements for core sophomore-level engineering courses. This would allow students to advance through the first two years of their chosen degree programs without first completing the required calculus sequence. Such is the approach adopted by Wright State University and a growing number of institutions across the country, which are now enjoying significant increases not only in engineering student retention but also in engineering student performance in their first required calculus course. Alternatively, this book would make an ideal reference for any freshman engineering program. Its organization is highly compartmentalized, which allows instructors to pick and choose which math topics and engineering applications to cover. Thus, any institution wishing to increase engineering student preparation and vi For all of these students. this book would provide an outstanding resource for nontraditional students returning to school from the workplace. this book represents a onestop shop for how math is really used in engineering. or for upper-level high school students who are thinking about studying engineering in college. . for math and science teachers or education majors seeking physical contexts for their students. for students who are undecided or are considering a switch to engineering from another major. Finally.Preface vii motivation for the required calculus sequence could easily integrate selected topics into an existing freshman engineering course without having to find room in the curriculum for additional credit hours. The authors would also like to thank their many colleagues and collaborators who have joined in their nationwide quest to change the way math is taught to engineers.Acknowledgement The authors would like to thank all those who have contributed to the development of this text. Finally. This material is based upon work supported by the National Science Foundation under Grant Numbers EEC-0343214. and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation. Thanks also to Josh Deaton. This includes their outstanding staff of TA’s. Any opinions. whose unending patience and support have made this effort possible. but also played a critical role in the success of the first-year engineering math program at Wright State University. findings. DUE-0622466 and DUE0817332. viii . Werner Klingbeil and Scott Molitor. who has provided detailed solutions to all end-of-chapter problems. the authors would like to thank their wives and families. who have not only provided numerous suggestions and revisions. Special thanks goes to Jennifer Serres. who have contributed a variety of worked examples and homework problems from their own engineering disciplines. DUE-0618571. and the total stopping time on the graph.1) is in the slope-intercept form y = mx + b. and the relationship between force and displacement in a preloaded spring can all be represented by straight lines.5 v(t). as indicated in Fig. the velocity of a vehicle while braking. and b = vo . The equation of the velocity given by equation (1. It is assumed that the students are already familiar with this topic from their high school algebra course.85 Figure 1. This chapter will show. s 1. (b) Sketch the graph of the line v(t) and clearly label the initial velocity. with examples. The velocity satisfies the equation v(t) = at + vo (1.1. x = t.5 2. why this topic is so important for engineers.1 VEHICLE DURING BRAKING The velocity of a vehicle during braking is measured at two distinct points in time. t. where y = v(t). In this chapter. For example. the applications of straight lines in engineering are introduced. m/s 9. m = a.Straight Lines in Engineering CHAPTER 1 In this chapter. the voltagecurrent relationship in a resistive circuit. (a) Find the equation of the line v(t) and determine both the initial velocity vo and the acceleration a. 1 . 1.75 5. 1. the acceleration.1) where vo is the initial velocity in m/s and a is the acceleration in m/s2 .1 A vehicle while braking. The slope m is given by m= 𝚫y 𝚫x = y2 − y1 x2 − x1 . the equations of these lines will be obtained using both the slope-intercept and the point-slope forms. 2 shows the velocity of the vehicle after braking.5 The velocity of the vehicle can now be written in the slope-intercept form as v(t) = −3.5.6 m/s.85 = −3.9 m/s2 . which gives 0 = −3. v) = (1. Solving for vo gives vo = 15.2 Chapter 1 Straight Lines in Engineering Therefore.9 t + 15. s . Using the data point (t. The total stopping time (time required to reach v(t) = 0) can be found by equating v(t) = 0. The velocity of the vehicle can now be written as v(t) = −3.9 t + 15.6 m/s.6. The y-intercept b = vo can be determined using either one of the data points. yielding 5. Solving for t. Figure 1.5) + vo . 1.75 = = −3. v) = (2. Note that the stopping time t = 4.0 s. 4. 5.90 m/s2 Stopping time (x -intercept) 0 0 Figure 1.5.6 1 a 3.75 = −3. The y-intercept b = vo can also be determined using the other data point (t.9 t + vo .1 as a= v2 − v1 5. 9. m/s Initial velocity.9 (1. v0 (y -intercept) 15. the stopping time is found to be t = 4.9 (2.5 − 1.5) + vo .0 t.85 − 9. Solving for vo gives vo = 15.6 m/s.75) gives 9. the slope m = a can be calculated using the data in Fig.0 s and the initial velocity Velocity. t2 − t1 2.2 Velocity of the vehicle after braking.85). Also. The slope m is given by Δ y Δ Vs = .3. 1. 1.2 Voltage-Current Relationship in a Resistive Circuit 3 vo = 15.90 m/s2 is the acceleration of the vehicle during braking.1 Figure 1.4) The objective is to find the value of R and V when the current flowing through the circuit is known for two different voltage values given in Fig. A 0.0 Voltage drop. x = I. the relationship between the applied voltage Vs and the current I flowing through the circuit can be obtained using Kirchhoff’s voltage law (KVL) and Ohm’s law. I. and b = V.2) gives a linear relationship between the applied voltage Vs and the current I as Vs = I R + V. m = R.e. 1.4) is the equation of a straight line in the slope-intercept form y = mx + b..and y-intercepts of the line. 1. i.1 .1 1. VR = I R.2 VOLTAGE-CURRENT RELATIONSHIP IN A RESISTIVE CIRCUIT For the resistive circuit shown in Fig.1.3) Substituting equation (1. Applying KVL to the circuit of Fig. KVL states that the sum of the voltage rises is equal to the sum of the voltage drops. (1.6 m/s are the x.. Kirchhoff’s voltage law: ⇒ I ∑ Voltage rise = R VR VS V ∑ Vs . respectively. note that the slope of the line m = −3. 1. The voltage-current relationship given by equation (1. the slope R can be found as m=R= R= 20 − 10 = 10 Ω.3) into equation (1. Δx ΔI Using the data in Fig. V 10.3. For a closed-loop in an electric circuit.2) Ohm’s law states that the voltage drop across a resistor VR in volts (V) is equal to the current I in amperes (A) flowing through the resistor multiplied by the resistance R in ohms (Ω).1 − 0.3 gives Vs = VR + V.3 Voltage and current in a resistive circuit.e. i.0 20. (1. where y = Vs .3. (1. 1. 0 + 10. Note that the slope of the line m = 10 is the resistance R in Ω and the y-intercept b = 9 is the voltage V in volts.6) R R This is an equation of a straight line y = m x + b.1 9V 1. The slope and R I= VS . The y-intercept V can also be found by finding the equation of the straight line using the point-slope form of the straight line (y − y1 ) = m(x − x1 ) as Vs − 10 = 10(I − 0. (1. m = R1 is the slope.4 Voltage-current relationship for the data given in Fig. and b = − V is the y-intercept. the voltage-current relationship is given by Vs = 10 I + 9. The y-intercept b = V can be determined using either one of the data points. V = 9 V.4 Chapter 1 Straight Lines in Engineering Therefore.3.1) + V. (1. Figure 1.1) gives 10 = 10 (0. V 20 10 R 1 10 0 V 0 0. Solving for V gives V = 9 V. 1. Therefore. the current I can be written as a function of Vs as 1 V Vs − .1) ⇒ Vs = 10 I − 1. 0. I) = (10.5) Comparing equations (1.5). where x is the applied voltage Vs . From the voltage-current relationship Vs = I R + V.4 shows the graph of the source voltage Vs versus the current I. Using the data point (Vs . y is the current I.1 Figure 1. The values of R and V can also be determined by switching the interpretation of x and y (the independent and dependent variables). the values of R and V are given by R = 10 Ω. A . I. the source voltage can be written in slope-intercept form as Vs = 10 I + b.4) and (1. I) = (10.1 − 0. Solving for b gives b = −0.1 0. Therefore. the slope m can be found as 1.V y-intercept 0.1 Vs − 0.2 Voltage-Current Relationship in a Resistive Circuit 5 y-intercept can be found from the data given in Fig.9 A and the slope is = 0. 1.1) gives 0.3 using the slope-intercept method as Δy ΔI .5 Straight line with I as independent variable for the data given in Fig.1 = 0. R Figure 1.1 R ⇒ R = 10 Ω and V = −0.9.1.1 0. The y-intercept b can be determined using either one of the data points. Note that the y-intercept V 1 is − = − 0. = m= Δ x Δ Vs Using the data in Fig. A 1. 1. 20 − 10 Therefore.9.7) Comparing equations (1.1 (10) + b. Using the data point (Vs .7) gives 1 = 0. (1.1 = 0.3.9 ⇒ V = 0.1. 0.1 0 1 0 10 20 VS . the current I can be written in slope-intercept form as m= I = 0.3.1.1Vs − 0. R R − I. 1.9 Figure 1.6) and (1.1 Vs + b.9. the equation of the straight line can be written in the slope-intercept form as I = 0. .5 is the graph of the straight line I = 0.9 (10) = 9 V. 5N. 1.9) gives k = 5N∕m. the dependent variable y is the force f . N 1 5 y. 0. Using the data point (f .8 The equation of the force-displacement equation in the slope-intercept form can therefore be written as f = 5y + b.9) . if the values of the force and displacement are as given in Fig.1 0.5. 0.6 Chapter 1 Straight Lines in Engineering 1.9) + b. Comparing equations (1. and the y-intercept is the preload fo . the forcedisplacement relationship f = k y + fo is the equation of a straight line y = mx + b. The slope m can be calculated using the data given in Fig.6.9 − 0.6 as m= 4 5−1 = = 5. y f k f. The y-intercept b can be found using one of the data points. Method 1 Treating the displacement y as an independent variable. m 0.8) where f is the force in Newtons (N). y is the displacement in meters (m). 1. (1. the equation of the straight line can be written in slope-intercept form as f = 5y + 0.8) and (1.6 Force-displacement in a preloaded spring. Solving for b gives b = 0.5 N. y) = (5.1 0.9) gives 5 = 5 (0.9 Figure 1.3 FORCE-DISPLACEMENT IN A PRELOADED TENSION SPRING The force-displacement relationship for a spring with a preload fo is given by f = k y + fo . Therefore. where the independent variable x is the displacement y. (1. fo = 0. and k is the spring constant in N/m. The objective is to find the spring constant k and the preload fo . the slope m is the spring constant k. 1.8 = = 0. The slope m can be calculated using the data k given in Fig.5. Therefore. and the y-intercept is the negated preload k f divided by the spring constant − o . f ) = (0. This relationk k ship is the equation of a straight line y = mx + b. the forcef 1 displacement relationship f = ky + fo can be written as y = f − o . The y-intercept b can be found using one of the data points. .2. the equation of the straight line can be written in the slope-intercept form as y = 0.2f − 0. the force-displacement relationship for a preloaded spring given in Fig.1.9 − 0.2 (5) + b.2f + b.1.3 Force-Displacement in a Preloaded Tension Spring 7 Method 2 Now treating the force f as an independent variable.1 (5) = 0. where the independent variable x is the force f . Using the data point (y. 1. the dependent variable y is the displacement y. Solving for b gives b = −0.1 0.10) with the expression y = 1 = 0.1 k ⇒ fo = 0.6 is given by f = 5y + 0. 5−1 4 m= The equation of the displacement y as a function of force f can therefore be written in slope-intercept form as y = 0.10) f 1 f − o gives k k k = 5 N/m and − fo = −0. 5) gives 0. Therefore. the slope m is the 1 reciprocal of the spring constant . 1.9.9 = 0. Comparing equation (1.2 k ⇒ (1.6 as 0.5 N. 0 − 10 = −5 and the value of b can be calculated using the data 6−4 point (t. Therefore. It then cruises at a constant velocity of 10 m/s for 2 s before coming to rest at 6 s. Finally. between 2 and 4 s.8 Chapter 1 Straight Lines in Engineering 1. (b) 2 ≤ t ≤ 4: v = 10 m/s.7. between 4 and 6 s. 1. write the expression of v(t) for times between 0 and 2 s. Solution The velocity profile of the vehicle shown in Fig. s Figure 1. v(t)) = (6. 2−0 v(t) = 5t m/s. (c) 4 ≤ t ≤ 6: v(t) = mt + b. the third piece of the trajectory is a straight line starting at 4 s and ending at 6 s.7 Velocity profile of a vehicle. m/s 10 0 2 4 6 t. The first linear function is a straight line passing through the origin starting at time 0 sec and ending at time equal to 2 s. 0) as where m = 0 = −5 (6) + b ⇒ b = 0 + 30 = 30.7 is a piecewise linear function with three different equations. in other words. The equation of the piecewise linear function can be written as (a) 0 ≤ t ≤ 2: v(t) = mt + b where b = 0 and m = 10 − 0 = 5. The vehicle starts at rest (zero velocity) and reaches a maximum velocity of 10 m/s in 2 s. v(t). . The second linear function is a straight line with zero slope (cruise velocity of 10 m/s) starting at 2 s and ending at 4 s.4 Example 1-1 FURTHER EXAMPLES OF LINES IN ENGINEERING The velocity of a vehicle follows the trajectory shown in Fig. 1. Write the equation of the function v(t). and greater than 6 s. where v1 = 0 and t1 = 6. v(t) = −5(t − 6). 1. plot the acceleration of the vehicle. v − 0 = −5(t − 6). 3 6 9 t.4 Further Examples of Lines in Engineering The value of b can also be calculated using the point-slope formula for the straight line v − v1 = m(t − t1 ). or v(t) = −5t + 30 m/s. (d) t > 6: v(t) = 0 m/s. Thus. s 9 .1. (a) Determine the equation of v(t) for (i) 0 ≤ t ≤ 3 s (ii) 3 ≤ t ≤ 6 s (iii) 6 ≤ t ≤ 9 s (iv) t ≥ 9 (b) Knowing that the acceleration of the vehicle is the slope of velocity. m/s 24 12 0 Figure 1. Example 1-2 The velocity of a vehicle is given in Fig.8 Velocity profile of a vehicle. Therefore. v(t).8. 0) as where m = 0 = −4(9) + b. ⎩ 0≤t≤3s 3≤t≤6s 6≤t≤9s t>9s The plot of the acceleration is shown in Fig. v(t)) = (9. s . 0 − 12 = −4 m/s2 and b can be calculated in slope-intercept 9−6 form using point (t. ⎪ 0.10 Chapter 1 Straight Lines in Engineering Solution (a) The velocity of the vehicle for different intervals can be calculated as (i) 0 ≤ t ≤ 3 s: v(t) = mt + b. Therefore. m/s2 0 3 4 Figure 1. ⎪ ⎪ 0. b = 36 m/s and v(t) = −4t + 36 m/s.9. Therefore. 6 9 t. 1. the acceleration a in m/s2 is given by ⎧−4. (iii) 6 ≤ t ≤ 9 s: v(t) = mt + b. (iv) t > 9 s: v(t) = 0 m/s.8. 3−0 v(t) = −4t + 24 m/s. a=⎨ ⎪−4. where m = 12 − 24 = −4 m/s2 and b = 24 m/s. Acceleration. (b) Since the acceleration of the vehicle is the slope of the velocity in each interval. (ii) 3 ≤ t ≤ 6 s: v(t) = 12 m/s.9 Acceleration profile of the vehicle in Fig. 1. 25 and Fi = 400 lb.11). where C = 0. 800) gives 600 = 0.25. the force in the bolt Fb is related to the external load P as Fb = C P + Fi . (a) Determine the joint constant C and the preload Fi given the data in Fig. Solution (a) The force-load relationship Fb = CP + Fi is the equation of a straight line. and label C and Fi on the graph. Fb = 0. y = mx + b.11. which can be calculated as C= ΔFb 600 − 500 100 lb = = = 0. the y-intercept Fi can be calculated by substituting one of the data points into equation (1.25 P + Fi . 1. Substituting the second data point (Fb . where C is the joint constant and Fi is the preload in the bolt.25 × 800 + Fi . . The slope m is the joint constant C. Note that the joint constant C is dimensionless! (b) The plot of the force Fb in the bolt as a function of the external load P is shown in Fig.10.10 External force applied to a bolted connection. 1.1. P) = (600. ΔP 800 − 400 400 lb Therefore. (b) Plot the bolt force Fb as a function of the external load P. Fb (P) = 0.10. Solving for Fi yields Fi = 600 − 200 = 400 lb. 1. Therefore.4 Further Examples of Lines in Engineering Example 1-3 11 In a bolted connector shown in Fig.25 P + 400 is the equation of the straight line. P P (l(b) 400 800 Fb (l(b) 500 600 Figure 1.11) Now. (1. V = 10(I) + 10 I0 . Find the values of R and I0 if the voltage across the resistor V is known for the two different values of the current I as shown in Fig. Solving for I0 gives 10 I0 = 2. amp 0. For the electric circuit shown in Fig.2 − 2 = 0. 600) 600 C (400.2 V.12 Solution 800 R Io V I.2 2. 1.12 as R= ΔV 2.2 Circuit for Example 1-4.2. 1. where the slope m = R can be found from the data given in Fig.2. 0.12.12) The y-intercept b = 10 I0 can be found by substituting the second data point (2.2 1 volt = = = 10 Ω.2 = 100 × 0. ΔI 0.12) as 2.1 0. 500) 500 Fi 1 400 0 Figure 1. 1. I Figure 1.2 − 1.11 Example 1-4 400 P (lb) Plot of the bolt force Fb as a function of the external load P. The voltage-current relationship V = R I + R Io is the equation of a straight line y = mx + b.12. volt 1.2 + 10 I0 .2) in equation (1.1 0.1 amp Therefore. the relationship between the voltage V and the applied current I is given by V = (I + Io )R.12 Chapter 1 Straight Lines in Engineering Fb (lb) (800.2 − 0. . (1. Therefore. On the plot. V 5 10 vo . V = 10 I + 0.13 An Op–Amp circuit as a summing amplifier. (b) Plot the output voltage vo as a function of the input voltage vin . (a) Determine the value of R and vb . where the slope m = − can be found R from the data given in Fig. 100 kΩ R kΩ vin vo OP−AMP vb vin .13 as (a) The input-output relationship vo = − − 100 R ) Δvo −5 − 5 100 −10 = = = = −2.13 satisfies the relationship vo = − vin + 1 + vb . 1.4 Further Examples of Lines in Engineering 13 which gives I0 = 0. (1.1.13) as 5 = −2 × 5 + 3 vb . . vin ) = (5. ( Solution ( ) 100 vin + 1 + vb is the equaR 100 tion of a straight line. ( ( ) ) 100 100 vin + 1 + vb v0 = − 50 50 = −2 vin + 3 vb . 5) in equation (1.13 gives the values of the output voltage for two different values of the input voltage. Therefore. V 5 −5 Figure 1. Fig. 1.02 A. y = mx + b.13) The y-intercept b = 3 vb can be found by substituting the first data point (v0 .02 A. where R in R R kΩ is the unknown resistance and vb is the unknown voltage. clearly indicate the value of the output voltage when the input voltage is zero (y-intercept) and the value of the input voltage when the output voltage is zero (x-intercept).2. 1. R Δvin 10 − 5 5 Solving for R gives R = 50 Ω. Example 1-5 The output voltage vo of the operational (Op–Amp) ( amplifier ( ) )circuit shown in 100 100 Fig. and R = 10 Ω and I0 = 0. V 5 x−intercept 7.5 V. The force and voltage satisfy the linear relation F = kV.0 N? (c) Using the results of parts (a) and (b). R = 50 Ω. sketch the graph of F as a function of voltage V. 1. Therefore. The x-intercept can be found by substituting vo = 0 in the equation vo = −2 vin + 15 and finding the value of vin as 0 = −2 vin + 15. (a) Find the force produced by the actuator when supplied with V = 7.3 volts.5 N when supplied with V = 12 volts.14 An Op–Amp circuit as a summing amplifier.5 V Figure 1.14. (b) What voltage is needed to achieve a force of F = 6. Example 1-6 An actuator used in a prosthetic arm (Fig.15 Prosthetic arm. (b) The plot of the output voltage of the Op–Amp as a function of the input voltage if vb = 5 V is shown in Fig. Use the appropriate scales and clearly label the slope and the results of parts (a) and (b) on your graph.15) can produce a different amount of force by changing the voltage of the power supply. vo. and vb = 5 V. 1. where V is the voltage applied and F is the force produced by the prosthetic arm. which gives vin = 7. which gives vb = 5 V. vb 15 V 15 5 m 0 5 100 R 2 10 vin. Therefore. Figure 1. the x-intercept occurs at Vin = 7.5 V. The maximum force the arm can produce is F = 44. V y−intercept b 3.14 Chapter 1 Straight Lines in Engineering Solving for vb yields 3 vb = 5 + 10 = 15. vo = −2 vin + 15. . 14) as F = 3.71 N/V.16 1.3 volts is found by substituting V = 7.71 = 1. where the slope m = k can be found from the given data as k= 44.4 Further Examples of Lines in Engineering Solution 15 (a) The input-output relationship F = k V is the equation of a straight line y = m x.71 × 7. F.71 V 6. volt Plot of the actuator force verses the applied voltage. 12 Therefore.0 N in equation (1.0 N can be found by substituting F = 6.62 volts.14) Thus.3 12 V.15) (c) The plot of force F as a function of voltage V can now be drawn as shown in Fig.0 = 3.5 = 3.0 0 Figure 1. (1.62 7. 1.1.16.08 N.0 V= 3.71 V.14) as 6. (b) The voltage needed to achieve a force of 6. .3 = 27.3 in equation (1. the equation of the straight line representing the actuator force F as a function of applied voltage V is given by F = 3. N 44.71 6. the force produced by the actuator when supplied with 7. (1.5 27.1 m k 3. 0005. (a) Determine the value of m and b. F is the applied muscle force in N. 1.16 Chapter 1 Straight Lines in Engineering Example 1-7 The electrical activity of muscles can be monitored with an electromyogram (EMG).17) (b) The plot of the RMS amplitude as a function of the applied muscle force is shown in Fig. find the RMS value of the amplitude for a muscle force of 200 N. The following root mean square (RMS) value of the amplitude measurements of the EMG signal were taken when a woman was using her hand grip muscles to ensure a lid was tight on a jar.55 × 10−6 . N 110 275 Amplitude measurements of the EMG signal.55 × 10−6 (110) + b. ΔF 275 − 110 165 N The y-intercept b can be found by substituting the first data point (A.00075 = = = 4. 110) in equation (1. and m is the slope. Therefore. the equation of the straight line representing the RMS amplitude as a function of applied force is given by A = 4.17) as m= V ΔA 0. (1. where the slope m can be found from the EMG data given in the table (Fig.0005 = 4.0005 0. Solution (a) The input-output relationship A = mF + b is the equation of a straight line y = mx + b. EMG A.18. F) = (0.00125 − 0. . (b) Plot the RMS amplitude A as a function of the applied muscle force F.16) where A is the RMS value of the EMG amplitude in V.0005 0.55 × 10−6 F. V 0. Figure 1. 1.00125 F. (c) Using the equation of the line from part (a).17 The RMS amplitude of the EMG signal satisfies the linear equation A = mF + b (1.16) as 0. Solving for b yields b = 5 × 10−7 ≈ 0. 0015 0.4 476.0005 0 100 200 300 F. .19. Example 1-8 A civil engineer needs to establish the elevation of the cornerstone for a building located between two benchmarks. B1 and B2. of known elevations as shown in Fig. and m is the average slope of the grade. l is the distance from B1 along the grade. V 0. (a) Find the equation of the line E and determine the slope m of the grade.8 Sea level Figure 1.17) as A = 4. N Figure 1. find the elevation of the cornerstone E ∗ if it is located at a distance l = 565 m from B1.18 Plot of the RMS amplitude verses the applied muscle force. The elevation E along the grade satisfies the linear relationship E = m l + E1 (1.2 E (m) 428. (c) Sketch the graph of E as a function of l and clearly indicate both the slope m and elevation E1 of B1.1.18) where E1 is the elevation of B1.4 Further Examples of Lines in Engineering 17 A. 1.55 × 10−6 × (200) = 0. E2 l E1 E B1 E * B2 B1 B2 Building cornerstone l1 (m) 0 1001.001 m 4.91 × 10−3 V.19 Elevations along a uniform grade. (c) The RMS value of the amplitude for a muscle force of 200 N can be found by substituting F = 200 N in equation (1. (b) Using the equation of the line from part (a).55 10 6 0. 4 = = Δl 1001. 1. the equation of the straight line representing the elevation as a function of distance l is given by E = 0.19) (b) The elevation E ∗ of the cornerstone can be found by substituting l = 565 m in equation (1. m 476.7 Cornerstone E* y-intercept E1 428.8 E2 0. Therefore.4.18 Chapter 1 Straight Lines in Engineering Solution (a) The equation of elevation given by equation (1. 0) in equation (1.0483 m 1 455.18) is a straight line in the slopeintercept form y = mx + b.0483 × (565) + 428. l) = (428.2 l. (1.20 Elevation along a uniform grade. m . 1001. 1001.0483 l + 428.20.0483.18) as 428.4 m.19) as E ∗ = 0. (c) The plot of the elevation as a function of the length is shown in Fig. where the slope m can be found from the elevation data given in Fig.19 as m= ΔE 476.7 m.8 − 428.4 = 0. Elevation.2 − 0 48.4 m.4 0 565 Figure 1.2 The y-intercept E1 can be found by substituting the first data point (E.4 = 455. 1.0483 × (0) + E1 Solving for E1 yields E1 = 428.4 = 0. 2 m. The spring force F and displacement x for a close-wound tension spring are measured as shown in Fig. F (N) 135 222 x (cm) 25 50 Figure P1.4 Bolted connection for problem P1-4. 1-2.3. P1. In a bolted connection shown in Fig. P (l(b) 100 600 Fb (l(b) 200 400 Figure P1. (a) Using the given data.Problems 19 PROBLEMS 1-1. (a) Given the data in Fig. 1-3. 1-4.2.2 Close-wound tension spring for problem P1-2.5 3. P x (cm) 1.2 Figure P1. find the equation of the line for the spring force F as a function of the displacement x. P1. P1. (a) Using the given data in Fig. find the equation of the line for the spring force F as a function of the displacement x. F (N) 34. find the stiffness k of the spring. If the spring force and displacement satisfy the linear relation F = k x.4.0 Figure P1.4.5 57. Use appropriate axis scales and clearly label the preload Fi . and both given data points on your graph.1 Displacement of a spring in problem P1-1. determine the joint constant C and the preload Fi . the spring constant k. The spring force F and displacement x satisfy the linear equation F = k x + Fi . P1.0 The spring force F and displacement x satisfy the linear equation F = k x + Fi . . where k is the spring constant and Fi is the preload induced during manufacturing of the spring. A constant force F = 2 N is applied to a spring and the displacement x is measured as 0. P1. where k is the spring constant and Fi is the preload induced during manufacturing of the spring. and determine the values of the spring constant k and preload Fi . x k f F (N) 2 x (m) 0. and determine the values of the spring constant k and preload Fi . the force in the bolt Fb is given in terms of the external load P as Fb = C P + Fi . (b) Sketch the graph of F as a function of x.3 Close-wound tension spring for problem P1-3.2. (b) Sketch the graph of F as a function of x and clearly indicate both the spring constant k and preload Fi . The spring force F and displacement x for a close-wound tension spring are measured as shown in Fig. and determine both the initial velocity vo and the acceleration a.06 v(t) (m/s) 34.6 t (s) 0. (a) Given the data in Fig.7. A model rocket is fired in the vertical plane. find the equation of the line representing the velocity v(t) of the rocket.1 0. The velocity v(t) of a ball thrown upward satisfies the equation v(t) = vo + at.0 Figure P1.6 A ball thrown upward with an initial Figure P1. P1. (a) Given the data in Fig. P Fb (l(b) 300 660 P (l(b) 280 1000 1-7.0 3. and clearly indicate both the initial velocity and the acceleration on your graph.5. (b) Sketch the graph of the line v(t). P1. where vo is the initial velocity of the ball in ft/s and a is the acceleration in ft/s2 . The velocity v(t) of a ball thrown upward satisfies the equation v(t) = vo + at.8 A model rocket fired in the vertical velocity vo (t) in problem P1-6. . P1. P1.4 v(t) (m/s) 3.8. Repeat problem P1-4 for the data given in Fig.8.02 1.3 19. and determine both the initial velocity vo and the acceleration a.0 t (s) 0. The velocity satisfies the equation v(t) = vo + at. 1-8.5 Bolted connection for problem P1-5. v(t) v(t) (ft/s) 67. and determine both the initial velocity vo and the acceleration a. v(t) t (s) 1. Also determine the time at which the velocity is zero. The velocity v(t) is measured as shown in Fig. (a) Given the data in Fig.8 3.3 Figure P1. 1-6. v(t) Figure P1. and clearly indicate both the initial velocity and the acceleration on your graph. find the equation of the line representing the velocity v(t) of the ball. plane in problem P1-8.7 A ball thrown upward with an initial velocity vo (t) in problem P1-7. where vo is the initial velocity of the ball in m/s and a is the acceleration in m/s2 . P1. find the equation of the line representing the velocity v(t) of the ball. (b) Sketch the graph of the line v(t). Also determine the time at which the velocity is zero. where vo is the initial velocity of the rocket in m/s and a is the acceleration in m/s2 .5 2. 1-5.6.20 Chapter 1 Straight Lines in Engineering (b) Plot the bolt force Fb as a function of the load P and label C and Fi on the graph. where vo is the initial velocity in m/s and a is the acceleration in m/s2 . 1-11. Also determine the time at which the velocity is zero (i.12. Determine the equation for v(t) for (a) 0 ≤ t ≤ 2 s (b) 2 ≤ t ≤ 4 s (c) 4 ≤ t ≤ 6 s . and clearly label the initial velocity. P1. the time required to reach the maximum height). and determine both the initial velocity vo and the acceleration a.8 32.10 Velocity of a vehicle during braking in problem P1-10. The velocity satisfies the equation v(t) = vo + at.9 1-10.. and the total stopping time on the graph.0 Figure P1.5 t (s) 0.0 21 (a) Find the equation of the line v(t).e.0 4. The velocity of a vehicle is measured at two distinct points in time as shown in Fig. the acceleration. and determine both the initial velocity vo and the acceleration a.10. (b) Sketch the graph of the line v(t) for 0 ≤ t ≤ 10 seconds. (b) Sketch the graph of the line v(t). in problem P1-11.e. v(t) (m/s) 30 10 t (s) 1. The velocity satisfies the relationship v(t) = vo + at. and clearly indicate both the initial velocity and the acceleration on your graph.. 1-9. P1. (a) Find the equation of the line v(t). The velocity of a vehicle is measured at two distinct points in time as shown in Fig. Figure P1. v(t) v(t) (ft/s) 128. find the equation of the line representing the velocity v(t) of the rocket. v(t) (ft/s) 112. The velocity v(t) of a vehicle during braking is given in Fig.5 1. (b) Sketch the graph of the line v(t).5 37. P1.11. the time required to reach the maximum height). and clearly label the initial velocity. The velocity satisfies the relationship v(t) = vo + at. Also determine the time at which the velocity is zero (i.2 t (s) 1. P1. the acceleration. (a) Given the data in Fig.0 2. 1-12. A model rocket is fired in the vertical plane. and clearly indicate both the initial velocity and the acceleration on your graph. and determine both the initial velocity vo and the acceleration a.9. where vo is the initial velocity of the rocket in ft/s and a is the acceleration in ft/s2 . The velocity v(t) is measured as shown in Fig. where vo is the initial velocity in ft/s and a is the acceleration in ft/s2 .9. and the total stopping time on the graph.Problems (b) Sketch the graph of the line v(t) for 0 ≤ t ≤ 8 seconds. P1.11 Velocity of a vehicle during braking in problem P1-9.5 A model rocket fired in the vertical plane Figure P1. s 10 0 2 4 t. respectively (a) Find the equation of the line T(x).5 Figure P1.14. The acceleration of the linear trajectory of problem P1-13 is shown in Fig.13. P1.14 Acceleration of the robot trajectory. m/s2 v(t). and clearly label C1 and C2 on your graph. Determine the equation of a(t) for t. The temperature distribution in a wellinsulated axial rod varies linearly with respect to distance when the temperature at both ends is held constant as shown in Fig.75 ft). A linear trajectory is planned for a robot to pick up a part in a manufacturing process. The velocity of the trajectory of one of the joints is shown in Fig. .22 Chapter 1 Straight Lines in Engineering (a) 0 ≤ t ≤ 1 s (b) 1 ≤ t ≤ 3 s (c) 3 ≤ t ≤ 4 s a(t). P1. clearly indicate the temperature at the center of the rod (x = 0.0 1. Also. s 6 Figure P1. The temperature satisfies the equation of a line T(x) = C1 x + C2 .15. and determine both constants C1 and C2 . where C1 and C2 are constants of integration with units of ◦ F/ft and ◦ F. Figure P1.12 Velocity of a vehicle during braking in problem P1-12. s T(x) (◦ F) 30 70 x (ft) 0. (b) Sketch the graph of the line T(x) for 0 ≤ x ≤ 1. Determine the equation of v(t) for (a) 0 ≤ t ≤ 1 s (b) 1 ≤ t ≤ 3 s (c) 3 ≤ t ≤ 4 s v(t). x 0 1 3 4 Figure P1. 1-13. P1.15 Temperature distribution in a well-insulated axial rod in problem P1-15. m/s 10 30 0 15 1 3 4 t.13 Velocity of a robot trajectory. 1-14. m/s 10 1-15.5 ft. I R V VS I V + − Vs (volt) 9.16 Temperature distribution in a well-insulated axial rod in problem P1-16.17 Resistive circuit for problem P1-17.5 Figure P1. and R is the resistance of the resistor in ohms. The voltage-current relationship for the circuit shown in Fig.18. (a) Sketch the graph of I as a function of V if the resistance is 5 Ω. I is the current in amps. P1.0 R=5Ω I (amp) 2. (b) Sketch the graph of the line T(x) for 0 ≤ x ≤ 0.0 18.75 1. I (amp) 0. Also. 23 P1-19. 1-18.0 5. 1-19. respectively. Repeat problem P1-18 for the data shown in Fig.0 Figure P1. and clearly label C1 and C2 on your graph. clearly indicate the temperature at the center of the rod (x = 0. where V is the applied voltage in volts.17 is given by Ohm’s law as V = I R. and determine the values of R and V. (b) Sketch the graph of Vs as a function of I and clearly indicate the resistance R and voltage V on the graph. P1. P1. I is the current in amps.19 Single-loop circuit for problem Figure P1.0 0. .25 m). P1. The temperature distribution in a wellinsulated axial rod varies linearly with respect to distance when the temperature at both ends is held constant as shown in Fig. 1-17. P1. and determine both constants C1 and C2 . A voltage source Vs is used to apply two different voltages (12V and 18V) to the single-loop circuit shown in Fig. (a) Find the equation of the line T(x).0 Figure P1. I T(x) (◦ C) 0 20 R V VS x (m) 0. where C1 and C2 are constants of integration with units of ◦ C/m and ◦ C.18. (a) Using the data given in Fig.5 m.16. and Vs is the voltage in volts. The voltage and current satisfy the linear relation Vs = IR + V.18 Single-loop circuit for problem P1-18. The temperature satisfies the equation of a line T(x) = C1 x + C2 . find the equation of the line for Vs as a function of I.Problems 1-16. where R is the resistance in ohms.18.5 Vs (volt) 12. P1.0 18. The values of the measured current are shown in Fig. x (b) Find the current I if the applied voltage is 10 V.19. 22 Linear model of a diode for problem P1-22.035 0. (a) Find the equation of the line for Vs as a function of I and determine the resistance Rd and the voltage VON . and clearly indicate the resistance Rd and the voltage VON on the graph. .24 Chapter 1 Straight Lines in Engineering 1-20.20.21. Vs (volt) 5. Repeat problem P1-18 for the data shown in Fig.22.21 Linear model of a diode for problem Figure P1. Fig. 1-21.5 I I (amp) 24 32 Vs (volt) 2.0 VON I (amp) 0. P1.23) satisfies v in − the relationship vo = 1 + 100 R 2 ( ) 100 vb .23.086 0. where Rd is the forward resistance of the diode and VON is the voltage that turns the diode ON.5 11 Figure P1.5 2. V 5 10 Op−Amp 100 kΩ vo vo . The applied voltage and the measured current satisfy the linear equation Vs = I Rd + VON . P1. where R is the unknown resisR tance in kΩ and vb is the unknown voltage in volts. R V VS Vs (volt) 1. ( ) ( P1.135 Figure P1.21. The output voltage. 100 kΩ R kΩ I Rd VS Rd VS VON I (amp) 0.23 An Op–Amp circuit as a summing P1-21. 1-22. amplifier for problem P1. I (b) Sketch the graph of VS as a function of I. P1. P1.186 vb 100 kΩ vin vin . Repeat problem P1-21 for the data given in Fig. A linear model of a diode is shown in Fig.20 Single-loop circuit for problem P1-20.23 gives the values of the output voltage for two different values of the input voltage.0 6 Figure P1. 1-23. To determine the resistance Rd and voltage VON . vo . two voltage values are applied to the diode and the corresponding currents are measured. of the Op–Amp circuit shown in Fig. The applied voltage VS and the measured current I are given in Fig. P1.0 10. V 3. 24 An Op–Amp circuit for problem P1-24. A 0. two different values of the voltage ea are applied to the motor. and v2 is the unknown voltage. (b) Plot the output voltage vo as a function of the input voltage vin . V 2 4 ia . Figure P1. V 5 10 Gear ratio Load etach OP−AMP vo . clearly indicate the value of the output voltage when the input voltage is zero (y-intercept) and the value of the input voltage when the output voltage is zero (x-intercept). To maintain a constant speed. the relationship vo = − v2 + 100 R in where R is the unknown resistance in kΩ.24 satisfies ) v . where Ra is the resistance of the armature winding in ohms and eb is the back-emf in volts.25 gives the values of the current for two different values of the input voltage applied to the armature of the DC motor. vo . 1-24.26.25 Voltage-current data of a DC motor for problem P1-25. P1. of the Op– Amp circuit shown in Fig. (a) Find the equation of the line for ea as a function of ia and determine the values of Ra and eb .5 1. On the plot. Repeat problem P1-25 for the data shown in Fig.5 − 7. 1-26. (a) Find the equation of the line for vo as a function of vin and determine the values of R and v2 . The voltage ea and the current ia flowing through the armature winding of the motor satisfy the relationship ea = ia Ra + eb .24 gives the values of the output voltage for two different values of the input voltage vin . . 1-25.0 Figure P1.25.Problems (a) Determine the equation of the line for vo as a function of vin and find the values of R and vb . The output voltage.(P1. Clearly indicate the value of the output voltage when the input voltage is zero (y-intercept) and the value of the input voltage when the output voltage is zero (x-intercept). P1. A DC motor is driving an inertial load JL shown in Fig. (b) Plot the applied voltage ea as a function of the current ia . V − 4. (b) Plot the output voltage vo as a function of the input voltage vin . v2 vin i a Ra Tm eb Motor Jm TACH ea 20 Jtach JL 100 kΩ R kΩ vo vin .25 Figure P1. Clearly indicate the value of the back-emf eb and the winding resistance Ra . vin is the input voltage. ωm 100 kΩ 25 ea . Fig. P1. θm. clearly indicate the values of RD and VDD . 1-29. An actuator used in a prosthetic arm can produce different amounts of force RD by changing the voltage of the power supply. ea . Fig. On the plot.0 2.28.0 N when supplied with 10 V.0 V. Use the appropri10 2 ate scales and clearly label the slope Figure P1. V 0 5 iD . and the results of parts (a) and (b). Plot the output voltage vo as a function of the input drain current iD . The force and voltage satisfy the vo linear relation F = kV. sketch the graph of F as a func4 8 tion of voltage V. mA 10 5 Figure P1.27 gives VDD iD G S vo . P1.26 Chapter 1 Straight Lines in Engineering i a Ra Tm eb Motor Jm TACH ea Gear ratio 20 Jtach JL θm. where V is the voltage applied and F is the force provi G duced by the prosthetic arm. vi 1-27.27 satisfies the relationship vo = VDD − RD iD . S (a) Find the force produced by the actuator when supplied with 6. P1. 1-28. where RD is the unknown drain resistance and VD is the unknown drain voltage.0 10. V (b). ωm Load etach ia . the output voltage vo of the n-channel enhancementtype MOSFET (NMOS) circuit shown in Fig.28 NMOS for P1-28.0 N? (c) Using the results of parts (a) and iD . (b) What voltage is needed to achieve a force of 5. A 1.27 n-channel enhancement-type MOSFET.0 VDD iD RD Figure P1. In the active region. V 5. mA vo . The maximum force the arm can produce is 30.25 the values of the output voltage for two different values of the drain current. P1. .26 Voltage-current data of a DC motor vo in problem P1-26. Repeat problem P1-27 for the data given in Fig. and m is the rate of change of E with respect to l. The elevation E along the grade satisfies the linear relationship E = m l + E1 (1. where A is the RMS amplitude in volts. bpm) were recorded in an exercise physiology laboratory. and m is the slope of the line. R.32 Elevations along a uniform grade for exercise P1-32. The voltage across a thermocouple is calibrated using the boiling point of water (100◦ C) and the freezing point of Zinc (420◦ C). find the equation of the line for A(F). as shown in Fig.5 E-3 1. find the elevation of the cornerstone E ∗ if it is located at a distance l = 300 m from B1. B2 Building cornerstone B1 Sea level l1 (m) 0 500 R = mA + B where R is the heart rate in beats per minute and A is the age in years. (c) Using the relationship developed in part (a). V 0. which produces a voltage V proportional to the temperature at the junction of two dissimilar metals. 27 B1 B2 E (m) 500 600 Figure P1. P1. l is the taped distance from B1 along the grade. of known elevations. The electrical activity of muscles can be monitored with an electromyogram (EMG).25 E-3 F. A thermocouple is a temperature measurement device. The following two measurements of maximum heart rate R (in beats per minute. The RMS amplitude measurements of the EMG signal when a person is using the hand grip muscle to tighten the lid on a jar is given in the table below: A. F is the applied muscle force in N. years 30 45 The maximum heart rate R and age A satisfy the linear equation E1 E E2 l E* 1-31. (b) Sketch R as a function of A.20) where E1 is the elevation of B1. (a) Using the data provided.Problems 1-30. 1-33. P1. N 110 275 The RMS amplitude of the EMG signal satisfies the linear equation R = mA + B. (a) Find the equation of the line E and determine the slope m of the linear relationship. find the maximum heart rate of a 60-year-old person. B1 and B2. (a) Using the data provided in the table. bpm 183 169. find the RMS amplitude for a muscle force of 200 N. . A.5 1-32. (b) Sketch A as a function of F. (c) Using the relationship developed in part (a). (c) Sketch the graph of E as a function of l and clearly indicate both the slope m and elevation E1 of B1. (b) Using the equation of the line from part (a). A civil engineer needs to establish the elevation of the cornerstone for a building located between two benchmarks.32. find the equation of the line for R. as shown in Fig.33. where 𝛼 is the thermocouple sensitivity in mV/◦ K and TR is the reference temperature in ◦ K.8 Figure P1.34. 1-34. where 𝛼 is the thermocouple sensitivity in mV/◦ C and TR is the reference temperature in ◦ C.0 Figure P1. as shown in Fig. (b) Sketch the graph of T as a function of V and clearly indicate both the reference temperature TR and the sensitivity 𝛼 on the graph. (a) Find the equation of the line for the measured temperature T as a function of the voltage V and determine the value of the sensitivity 𝛼 and the reference temperature TR . and S𝜖 is .34. P1. (c) If the thermocouple is mounted in a chemical reactor and the voltage is observed to go from 10. find the equation of the line for the measured temperature T as a function of the voltage V and determine the value of the sensitivity 𝛼 and the reference temperature TR . The strain being measured (𝜖) and resistance of the sensor satisfy the linear equation R = Ro + Ro S𝜖 𝜖.33 Thermocouple to measure temperature in Celsius. 1-35.0 17. where Ro is the initial resistance (measured in ohms. It can be measured using a foil strain gauge shown in Fig. (a) Using the calibration data given in Fig.33. The junction temperature T and the voltage across the thermocouple V satisfy the linear equation T = 𝛼1 V + TR . P1. An iron-costantan thermocouple is calibrated by inserting its junction in boiling water (100◦ C) and measuring a voltage V = 5.0 70. The junction temperature T and the voltage across the thermocouple V satisfy the linear equation T = 𝛼1 V + TR . P1. find the equation of the line for the measured temperature T as a function of the voltage V and determine the value of the sensitivity 𝛼 and the reference temperature TR .0 mV to 13. and then inserting the juction in silver chloride at its melting point (455◦ C) and measuring V = 24.88 mV. P1.34 Thermocouple to measure temperature in Kelvin. Ω) of the sensor with no strain.28 Chapter 1 Straight Lines in Engineering Metal A Voltmeter Measuring junction Metal B T (◦ C) 100 420 V (mV) 5. Metal A Voltmeter Measuring junction T (◦ K) 373 1235 Metal B V (mV) 10. The voltage across a thermocouple is calibrated using the boiling point of water (373◦ K) and the freezing point of Silver (1235◦ K). Strain is a measure of the deformation of an object. (b) Sketch the graph of T as a function of V and clearly indicate both the reference temperature TR and the sensitivity 𝛼 on the graph. what is the change in temperature of the reactor? 1-36. (b) Sketch the graph of T as a function of V and clearly indicate both the reference temperature TR and the sensitivity 𝛼 on the graph.6 mV.36. (a) Using the calibration data given in Fig.27 mV. find the concentration of the sample if this sample had an absorption of 0.37.35 0. 1-39. find the equation of the line for the sensor’s resistance R as a function of the strain 𝜖. 1-38.02 Figure P1.Problems the gauge factor (a multiplier with NO units). (a) Find the equation of the line that describes the concentrationabsorbance relationship for this protein and determine the slope m of the linear relationship.578 The concentration-absorbance relationship for this protein satisfies a linear equation a = m c + ai .00419 𝜇g/ml.01 Figure P1. and ai is the y-intercept.38 Foil strain gauge to measure strain in problem P1-38. (c) If the sample is diluted to a concentration of 0. a graduate student used spectrophotometry to measure the absorbance given in the table below: c (𝜇g/ml) 3. End loops Grid Active grid length Alignment marks End loops Backing and encapsulation Solder tabs R. m is the the rate of change of absorbance a with respect to concentration c. 1-37. Ω 100 102 𝜖.342 0.50 8. (a) Using the given data. (b) Sketch the graph of R as a function of 𝜖. Ω 50 52 𝜖. Repeat problem P1-36 for the data shown in Fig. Repeat problem P1-38 if the absorptionconcentration data for another sample is obtained as c (𝜇g/ml) 4. the absorbance of the sample. and clearly indicate Ro on your graph. P1.00 a 0. (b) Using the equation of the line from part (a). what would you expect the absorption to be? Would this value be accurate? (d) Sketch the graph of absorbance a as a function of concentration c and clearly indicate both the slope m and the y-intercept.50 a 0. m/m 0 0. m/m 0 0. where c is the concentration of a purified protein.486.65 .00 8. a is 29 End loops Grid Active grid length Alignment marks End loops Backing and encapsulation Solder tabs R. and determine the values of the gauge factor S𝜖 and initial resistance Ro . To determine the concentration of a purified protein sample.36 Foil strain gauge to measure strain in problem P1-36. (a) Find the equation of the line that describes the temperature-volume relationship of the gas mixture and determine the slope m of the linear relationship.0 40.15 L. (b) Sketch the graph of T (◦ F) as a function of T (◦ C). (c) Calculate the thermostat setting needed to obtain a temperature of 320◦ F.08 1.24 The student knows that the gas volume directly depends on temperature. the freezing and boiling point of water is used as given in the table below: T (◦ F) 32 212 T (◦ C) 0 100 The relationship between the temperatures in Fahrenheit and Celsius scales satisfies the linear equation T (◦ F) = a T (◦ C) + b. (a) Using the given data. (d) Sketch the graph of the volumetemperature relationship for the gas mixture from −300◦ C to 100◦ C and clearly indicate both the slope m and the y-intercept.0 The relationship between the temperature T in Fahrenheit and the dial setting R satisfies the linear equation T (◦ F) = a R + b. (b) Sketch the graph of T (◦ F) as a function of R.43. What is the significance of the temperature when V = 0 L? 1-41. V(T) = m T + K. and clearly indicate a and b on your graph. forced cooling is provided by a pressurized lubricant flowing along the axial direction of the shaft (the x-direction) as shown in Fig. The lubricant . find the temperature interval in ◦ C if the temperature is between 20◦ F and 80◦ F. he could only obtain values at two temperatures as shown in the table below: T (◦ C) 50 98 V (L) 1. (a) Using the given data. that is. 1-42. In a pressure-fed journal bearing. the data for the temperature T (◦ F) versus the dial setting R was obtained as shown in the table below: T (◦ F) 110. find the temperature of the gas mixture if the volume is 1.0 R 20. K is the y-intercept in L. P1. (c) Using the equation of the line from part (a). find the equation of the line relating the temperature to the dial setting. To obtain the linear relationship between the Fahrenheit and Celsius temperature scales. where V is the volume in L. Due to technical difficulties. A thermostat control with dial marking from 0 to 100 is used to regulate the temperature of an oil bath. 1-43. and clearly indicate a and b on your graph. (c) Using the graph obtained in part (b).30 Chapter 1 Straight Lines in Engineering 1-40.0 40. (b) Using the equation of the line from part (a). find the volume of the gas if the temperature is 70◦ C. A chemistry student is performing an experiment to determine the temperature-volume behavior of a gas mixture at constant pressure and quantity. and m is the slope of the line in L/◦ C. T is the temperature in ◦ C. To calibrate the thermostat. find the equation of the line relating the Fahrenheit and Celsius scales. (c) Sketch the graph of the lubricant pressure p(x).0 in.43 Pressure-fed journal bearing. l where ps is the supply pressure and l is the length of the bearing.5 1. psi 24 12 x. in.5 Figure P1. P1. 0. (b) Calculate the lubricant pressure p(x) if x is 1. find the equation of the line for the lubricant pressure p(x) and determine the values of the supply pressure ps and the bearing length l. (a) Using the data given in the table (Fig. pressure satisfies the linear equation p p(x) = s x + ps .Problems 31 y Ps Bearing c l Groove x Journal p(x).43). . and clearly indicate both the supply pressure ps and the bearing length l on the graph. 2. In this chapter. the solution of quadratic equations will be obtained by three methods: factoring. 32 .1 A ball thrown upward to a height of h(t). 2. the applications of quadratic equations in engineering are introduced.CHAPTER 2 Quadratic Equations in Engineering In this chapter.1.1) is a quadratic equation of the form ax2 + bx + c = 0 and will be solved using three different methods. A quadratic equation is a second-order polynomial equation in one variable that occurs in many areas of engineering. Solution: h(t) = 96 t − 16 t2 = 80 or 16 t2 − 96 t + 80 = 0.1 A PROJECTILE IN A VERTICAL PLANE Suppose a ball thrown upward from the ground with an initial velocity of 96 ft/s reaches a height h(t) after time t s as shown in Fig. It is assumed that students are familiar with this topic from their high school algebra course. The height is expressed by the quadratic equation h(t) = 96 t − 16 t2 ft. and completing the square. the quadratic formula. For example.1) Equation (2. Find the time t in seconds when h(t) = 80 ft. the height of a ball thrown in the air can be represented by a quadratic equation. (2. h(t) = 96 t − 16 t2 Figure 2. then the quadratic formula to √ b2 − 4ac . rewrite the quadratic equation (2. t = 3 ± 2 or t = 1. To check if the answer is correct. substitute t = 1 and t = 5 into equation (2. 2 2 or Adding the square of t2 − 6 t + 9 = −5 + 9.3). −b ± First. . Hence. Therefore. Method 2: Quadratic Formula solve for x is given by If ax2 + bx + c = 0.3) x= 2a Using the quadratic formula in equation (2. t − 1 = 0 or t = 1 s and t − 5 = 0 or t = 5 s. the ball reaches the height of 80 ft at 1 s and 5 s. the ball reaches the height of Therefore.5) Equation (2. Hence.1 A Projectile in a Vertical Plane Method 1: Factoring 33 Dividing equation (2. 5 s. (2. the quadratic equation (2.1) by 16 yields t2 − 6 t + 5 = 0. Therefore.2) can be factored as (t − 1)(t − 5) = 0.2) can be solved as √ 6 ± 36 − 20 t= 2 6±4 . Substituting t = 1 s gives 16 × 1 − 96 × 1 + 80 = 0.1).4) gives ( )2 ( )2 −6 −6 2 t − 6t + = −5 + .2) as Method 3: Completing the Square ( ) t2 − 6 t = −5. (2. t = 2 2 80 ft at 1 s and 5 s. (2.2) Equation (2. = 2 6+4 6−4 = 1 s and t = = 5 s.2. (2.5) can now be written as √ (t − 3)2 = (± 4)2 or t − 3 = ±2.4) −6 (one-half the coefficient of the first-order term) to both 2 sides of equation (2. These three points (height at t = 1. substitute t = 5 s. Setting h(t) = 144 gives h(t) = 96 t − 16 t2 = 144.2 The height of the ball thrown upward with an initial velocity of 96 ft/s. Now. 96 t − 16 t2 = 0 6 t − t2 = 0 t (6 − t) = 0. To check this.34 Chapter 2 Quadratic Equations in Engineering which gives 0 = 0. . Therefore. The time when the ball hits the ground again can be calculated by equating h(t) = 0. It can be seen from Fig. Therefore. The ball is at 80 ft and going up at 1 s. This gives h(0) = 96(0) − 16(0)2 = 0 ft. Suppose now you wish to find the time t in seconds when the height of the ball reaches 144 ft.2. the maximum height can be found by substituting t = 3 s in h(t). The height of the ball at t = 0 is zero since the ball is thrown upward from the ground. 2.2 that the height of the ball is 80 ft at both 1 s and 5 s. Therefore. Therefore. Using these data points. t = 5 s is also the correct time when the ball reaches a height of 80 ft. which again gives 0 = 0. 2. the maximum height of ball must be halfway between 1 and 5 s. and it is at 80 ft and going down at 5 s. ft hmax 144 100 80 50 0 0 1 2 3 4 5 6 t. Therefore. However. t = 0 and 6 − t = 0 or t = 6 s. it will hit the ground again at t = 6 s. which is 1 + ((5 − 1)∕2) = 3 s. to plot the trajectory accurately. s Figure 2. additional data points can be added. and 5 s) can be used to plot the trajectory of the ball. the trajectory of the ball thrown upward with an initial velocity of 96 ft/s is shown in Fig. substitute t = 0 in h(t). Hence. 16 × 52 − 96 × 5 + 80 = 0. t = 1 s is the correct time when the ball reaches a height of 80 ft. which is h(3) = 96(3) − 16(3)2 = 144 ft. 3. h(t). Since the ball is thrown in the air from the ground (h(t) = 0) at t = 0. Setting h(t) = 160 gives h(t) = 96t − 16t2 = 160. (2.7) The quadratic equation given in equation (2.7) can be solved using the three methods as Factoring Quad Formula Completing the Square t2 − 6t + 10 = 0 t2 − 6t + 10 = 0 t2 − 6t + 10 = 0 cannot be factored using real integers √ 36 − 40 √2 6 ± −4 t= 2 √ t = 3 ± −1 t= 6± t =3±j t2 − 6t = −10 )2 )2 ( ( −6 −6 t2 − 6t + = −10 + 2 2 t2 − 6t + 9 = −1 (t − 3)2 = −1 √ t − 3 = ± −1 t =3±j . 3 t=3 s t − 3 = ±0 t = 3. 3 t=3 s Now suppose you wish to find the time t when the height of the ball reaches h(t) = 160 ft. 16t2 − 96t + 160 = 0 or t2 − 6t + 10 = 0.6) can also be solved using the three methods as Factoring Quad Formula Completing the Square t2 − 6t + 9 = 0 t2 − 6t + 9 = 0 2 t − 6t + 9 = 0 (t − 3)(t − 3) = 0 t−3=0 t=3 s t= t2 − 6t = −9 )2 )2 ( −6 −6 2 = −9 + t − 6t + 2 2 2 t − 6t + 9 = 9 − 9 ( 6± √ 36 − 36 2 t =3±0 =0 (t − 3)2 = 0 t = 3. 16 t2 − 96 t + 144 = 0 or t2 − 6 t + 9 = 0. Therefore. (2.2.6) The quadratic equation given in equation (2.1 A Projectile in a Vertical Plane 35 Therefore. 120 = VL + VR .2 CURRENT IN A LAMP A 100 W lamp and a 20 Ω resistor are connected in series to a 120 V power supply as shown in Fig. LAMP + VL + 120 V − 100 W − I + VR − 20 Ω Figure 2. 5 A √ 36 − 20 2 I =3±2 I= 6± I = 1. I 100 120 = + 20 I. 5 A . which gives VL = .10) can be solved using the three methods as Factoring Quad Formula I 2 − 6I + 5 = 0 2 I − 6I + 5 = 0 (I − 1)(I − 5) = 0 I = 1. From Ohm’s law. PL = VL I = 100 W.8) by I yields 120 I = 100 + 20 I 2 . Hence.3 A lamp and a resistor connected to a 120 V supply. since the power is the product of voltage and 100 current. the ball never reaches the height of 160 ft. therefore the roots of the quadratic equation are complex.36 Chapter 2 Quadratic Equations in Engineering √ In the above solution i = j = −1 is the imaginary number.9) by 20 and rearranging gives I 2 − 6 I + 5 = 0. Therefore. VR = 20 I. Also. The maximum height achieved is 144 ft at 3 s. The current I in amperes satisfies a quadratic equation as follows. (2.10) The quadratic equation given in equation (2. (2. 2. 2.8) I Multiplying both sides of equation (2.3. Using KVL.9) Dividing both sides of equation (2. (2. 5 A Completing the Square I 2 − 6I + 5 = 0 )2 )2 ( ( −6 −6 I 2 − 6I + = −5 + 2 2 I 2 − 6I + 9 = −5 + 9 (I − 3)2 = 4 I − 3 = ±2 I =3±2 I = 1. 2. 5 Case I corresponds to a lamp rated at 100 V. Case I: For I = 1 A. I 1 Case II: For I = 5 A. (2. resistance R = R1 + R2 R1 R2 R Figure 2.14) by 4 gives R22 − 100 R2 − 2500 = 0.000 = 0. 2. (2.4 Equivalent resistance of two resistors connected in parallel.14) Dividing both sides of equation by (2.13) gives 4 R22 − 400 R2 − 10. (2. The equivalent resistance of two resistors connected in parallel as shown in Fig.13) Simplifying equation (2. VL = 100 = 20 V. VL = 100 100 = = 100 V. 2. and Case II corresponds to a lamp rated at 20 V. find R1 and R2 .3 Equivalent Resistance 37 Note that the two solutions correspond to two lamp choices.4.3 EQUIVALENT RESISTANCE Suppose two resistors are connected in parallel. 5 R2 + 100 (4 R2 + 100) + R2 (2.12) Multiplying both sides of equation (2.12) by 5R2 + 100 yields 100 (5 R2 + 100) = 4 R22 + 100 R2 . R1 + R2 (2. as shown in Fig.2.4 is given by R1 R2 = 100 Ω.11) Substituting R1 = 4R2 + 100 Ω in equation (2.11) gives 4 R22 + 100 R2 (4 R2 + 100)(R2 ) 100 = = . If the equivalent R1 R2 = 100 Ω and R1 = 4R2 + 100 Ω.15) . 7) + 100 = 582.8 Ω. 2.5.16) (a) Find the time when h(t) = 245 m. Therefore.9 t2 Figure 2. 2.7 Ω. √ √ 100 ± 100 2 = 50 ± 50 2. R1 = 582.000 − 4(−2500) 100 ± 2(10. (c) Use the results of parts (a) and (b) to sketch h(t) and determine the maximum height.8 Ω and R2 = 120. Therefore. h(t) = 98 t − 4.9 t2 m. √ R2 = 50 + 50 2 = 120.7 Ω. (2.15) is a quadratic equation in R2 and cannot be factored with whole numbers.4 Example 2-1 FURTHER EXAMPLES OF QUADRATIC EQUATIONS IN ENGINEERING A model rocket is fired into the air from the ground with an initial velocity of 98 m/s as shown in Fig. The height h(t) satisfies the quadratic equation h(t) = 98 t − 4. (b) Find the time it takes the rocket to hit the ground. . R2 is solved using the quadratic formula as √ √ 100 ± 10.000) = . Substituting the value of R2 in R1 = 4R2 + 100 Ω yields R1 = 4(120.38 Chapter 2 Quadratic Equations in Engineering Equation (2. R2 = 2 2 Therefore. R2 = 2 Since R2 canot be negative.5 A rocket fired vertically in the air. 2 2 Substituting t = 10 s into equation (2.16).1 as Factoring Quad Formula t2 − 20t + 50 = 0 t2 − 20t + 50 = 0 20 ± t2 − 20t + 50 = 0 t2 − 20t = −50 √ 400 − 200 √ 2 t = 10 ± 50 t= can’t be factored with whole numbers Completing the Square t2 − 20t + 100 = −50 + 100 t = 10 ± 7.07 s. .18) The quadratic equation given in equation (2. crosses a height of 245 m at 2.93.93 and 17. At 10 seconds. t = 0 s and t = 20 s. 17.4 Further Examples of Quadratic Equations in Engineering Solution 39 (a) Substituting h(t) = 245 in equation (2.17) by −4. The plot of the rocket trajectory is shown in Fig. Since the rocket is fired from the ground at t = 0 s.07 s. it starts its downward descent and after crossing the height of 245 m again at 17.07 20 = = 10 s.9 t2 + 98 t − 245 = 0.9 t (20 − t) = 0. It can be seen from this figure that the rocket is fired from the ground at a height of zero at 0 s.07 t = 2.2. tmax = 2. it reaches the ground again at 20 s.9 gives t2 − 20 t + 50 = 0. Therefore.93 s.9(10)2 = 490 m.93 + 17. 17.6.07 t = 2. (2.07 s (b) Since the rocket hit the ground at h(t) = 0.17) Dividing both sides of equation (2.07 s (t − 10)2 = 50 √ t − 10 = ± 50 t = 10 ± 7.93. and continues moving up and reaches the maximum height of 490 m at 10 s. the quadratic equation is given by −4.9 t2 = 0 4. the rocket hits the ground again at t = 20 s.18) can be solved using the three methods used in Section 2. Therefore. h(t) = 98 t − 4.16) yields hmax = 98(10) − 4. (2. (c) The maximum height should occur halfway between 2. 2. (2.19) gives 8.07 20 t. (2. R1 + R2 (2. Example 2-2 The equivalent resistance R of two resistors R1 and R2 connected in parallel as shown in Fig.0 = 2 R21 + 4 R1 R1 (2 R1 + 4) . Also. or 24.19) (a) Suppose R2 = 2R1 + 4 Ω and the equivalent resistance R = 8. (b) Solve for R1 by each of the following methods: (i) Completing the square.20) by (3R1 + 4) yields 8. (ii) The quadratic formula.21) gives 2 R21 − 20 R1 − 32 = 0.0 in equation (2.22) .19) to obtain the following quadratic equation for R1 : 2R21 − 20R1 − 32 = 0. 2.6 The height of the rocket fired vertically in the air with an initial velocity of 98 m/s.21) Rearranging terms in equation (2.0 R1 + 32. = 3 R1 + 4 R1 + (2 R1 + 4) (2.20) Multiplying both sides of equation (2.0 = 2 R21 + 4 R1 .0 Ω. Solution (a) Substituting R2 = 2R1 + 4 and R = 8. determine the value of R2 corresponding to the only physical solution for R1 .40 Chapter 2 Quadratic Equations in Engineering h(t).4 is given by R= R1 R2 .0(3 R1 + 4) = 2 R21 + 4 R1 .93 4 8 12 16 17. m hmax 490 400 300 245 200 100 0 0 2. Substitute these values in equation (2. s Figure 2. 4 Further Examples of Quadratic Equations in Engineering 41 (b) The quadratic equation given in equation (2. k1 + k2 (2. 2. and the equivalent stiffness is k = 3.4)2 .4 = 11..22) can be now be solved to find the values of R1 .4 Ω in R2 = 2 R1 + 4 gives 20 ± R2 = 2(11.4 Ω and 5 − 6.24) Now.4 Ω. writing both sides of equation (2.4 = −1. R1 − 5 = ± 6.4. find k1 and k2 as follows: (a) Substitute the values of k and k2 into equation (2. (2. (ii) Method 2: Solving equation (2.8 Ω. which gives the values of R1 as 5 + 6. given by 2-3 k = k1 + k1 k2 .8 k1 − 14.8 Ω.25) to obtain the following quadratic equation for k1 : 5k21 − 2.4.4 Ω. (i) Method 1: Completing the square: Dividing both sides of equation (2.26) . (2.2.7 has an equivalent stiffness k.22) using the quadratic formula: √ (−20)2 − 4(2)(−32) R1 = 4 √ 20 ± 656 20 ± 25.4 Ω and R2 = 2R1 + 4 = 2(11.22) by 2 gives R21 − 10 R1 − 16 = 0.4 = 0.6 = = = 11. Since the value of R1 cannot be negative.6 lb/in. −1.23) ( Taking 16 on the other side of equation (2. R1 = 11. Substituting R1 = 11.23) and adding −10 2 )2 = 25 to both sides yields R21 − 10 R1 + 25 = 16 + 25. Example An assembly of springs shown in Fig. Therefore.24) as squares yields √ (R1 − 5)2 = (± 41)2 = (± 6.4) + 4 = 26. (2. R1 = 11.4.4) + 4 = 26.25) If k2 = 2k1 + 4 lb/in. 4 4 Since R1 cannot be negative. 2 10 = 2. (2.27) Multiplying both sides of equation (2.44.29) (b) The quadratic equation (2. k1 k1 k2 Figure 2.7 An assembly of three springs.0 in k2 = 2 k1 + 4 yields k2 = 2(2) + 4 = 8.0. Solution (a) Substituting k2 = 2 k1 + 4 and k = 3.25) yields 3.4 = 5 k21 + 8 k1 . solve equation (2.26) and determine the values of both k1 and k2 . = k1 + 3 k1 + 4 k1 + (2 k1 + 4) (2.6(3 k1 + 4) = k1 (3 k1 + 4) + 2 k21 + 4 k1 10.4 = 0.4 = 3 k21 + 4 k1 + 2k21 + 4 k1 10. Therefore. −1.28) gives 5 k21 − 2. (2.6 = k1 + 2 k21 + 4 k1 k1 (2 k1 + 4) .28) Rearranging terms in equation (2.42 Chapter 2 Quadratic Equations in Engineering (b) Using the method of your choice.0 lb/in. k2 = 8. .8 k1 + 14.6 in equation (2.8 ± (−2.8 k1 − 14. substituting k1 = 2. k1 = 2.0 lb/in.4) k1 = 10 = 2.8 k1 + 14. Since k1 cannot be negative.8 ± 17.29) can be solved using the quadratic formula as √ 2. Now.8)2 − 4(5)(−14.0.27) by (3k1 + 4) gives 3. 33) (b) The quadratic equation (2.25 F.30) Substituting L = 1.34) .0 = 𝜔(1) − 1 .31) Multiplying both sides of equation (2.2. (a) Suppose L = 1. (2.32) yields 𝜔2 − 3𝜔 − 4 = 0. 𝜔(0. (2. show that the angular frequency 𝜔 satisfies the quadratic equation 𝜔2 − 3𝜔 − 4 = 0. where 𝜔 is the angular fretotal reactance X in ohms is given by X = 𝜔L − 𝜔C quency in rad/s. 2.8 is given by X = 𝜔L − 1 .25) (2.0 Ω in equation (2.4 Further Examples of Quadratic Equations in Engineering Example 2-4 43 A capacitor C and an inductor L are connected in series as shown in Fig. (i) Method 1: Factoring: The quadratic equation (2. 2.33) can be solved by three different methods: factoring. completing the squares.8. which gives 𝜔 − 4 = 0 or 𝜔 + 1 = 0.25 F. and the quadratic formula.30) yields 3. completing the square.31) by 𝜔 gives 3𝜔 = 𝜔2 − 4. 𝜔 = 4 rad/s.33) can be factored as (𝜔 − 4)(𝜔 + 1) = 0. (ii) Method 2: Completing the squares: The quadratic equation (2. If the total reactance is X = 3. 𝜔C (2. and X = 3. The 1 . L C Figure 2. C = 0. Therefore. (b) Solve the quadratic equation for 𝜔 by each of the following methods: factoring.32) Rearranging terms in equation (2.0 H and C = 0. and the quadratic formula. Solution (a) The total reactance of the series combination of L and C shown in Fig.8 Series connection of L and C. (2.0 H.33) can be written as 𝜔2 − 3𝜔 = 4.0 Ω. Since 𝜔 cannot be negative. 𝜔 = 4 rad/s or 𝜔 = −1 rad/s. Since 𝜔 cannot be negative. and R = 8 Ω.44 Chapter 2 Quadratic Equations in Engineering ( Adding 3 − 2 )2 = 9 to both sides of equation (2. (2. 2 2 Therefore. 3 5 ± . Example 2-5 For the circuit shown in Fig. and the quadratic formula. . Since 𝜔 cannot 2 2 2 2 be negative. 𝜔− 2 2 (2.35) (2.3. 𝜔 = 4 rad/s. 𝜔 = 4 rad/s. −1. Show that the current I satisfies the quadratic equation I 2 + 4 I − 12 = 0. 𝜔2 − 3𝜔 + 4 4 Therefore. completing the square.37) can be written as 𝜔= 3± √ 2 25 = 3±5 .36) yields 𝜔− 5 3 =± . 2 which gives 𝜔 = 4. (iii) Method 3: Quadratic formula: Solving the quadratic equation (2.37) 𝜔= 2 𝜔= Equation (2.36) Taking the square root of both sides of equation (2. VL = 32 V.34) gives 4 ( ) ( ) 9 9 =4+ . (a) Suppose that P = 96 W. 𝜔2 − 3𝜔 + 9 25 = . 2. 4 4 Writing both sides of equation (2.35) as a square gives ( )2 ( )2 5 3 = ± . the power P delivered by the voltage source Vs is given by the equation P = I 2 R + I VL . (b) Solve the quadratic equation for I by each of the following methods: factoring. 2 2 3 5 3 5 which gives 𝜔 = + = 4 rad/s or 𝜔 = − = −1 rad/s.33) using the quadratic formula gives √ 3 ± (−3)2 − 4(1)(−4) . 44) can be written as √ −4 ± 64 −4 ± 8 = = −2 ± 4. (iii) Method 3: Quadratic formula: Solving the quadratic equation (2. (2.41) (2. I = −6 A or I = 2 A.42) Writing both sides of equation (2.40) can be solved by three different methods: factoring.43) gives I + 2 = ± 4. Therefore. ( )2 4 = 4 to both sides of equation (2.40) can be written as I 2 + 4 I = 12.38) by 8 gives 12 = I 2 + 4 I. Therefore.41).42) as a square yields (I + 2)2 = (± 4)2 . (2. . which gives I + 6 = 0 or I − 2 = 0.38) Dividing both sides of equation (2. (ii) Method 2: Completing the square: The quadratic equation (2.39) yields I 2 + 4 I − 12 = 0.4 Further Examples of Quadratic Equations in Engineering Solution 45 (a) Substituting P = 96 W. Adding 2 I 2 + 4 I + 4 = 12 + 4. (2. (2.40) can be factored as (I + 6)(I − 2) = 0.40) using the quadratic formula gives √ −4 ± (4)2 − 4(1)(−12) . and the quadratic formula.44) I= 2 Equation (2. and R = 8 Ω into the power delivered P = I 2 R + I VL yields 96 = I 2 (8) + I(32). (i) Method 1: Factoring: The quadratic equation (2. completing the squares. I= 2 2 which gives I = −2 − 4 = −6 A or I = −2 + 4 = 2 A. (2.40) (b) The quadratic equation given in equation (2.43) Taking the square root of both sides of equation (2. which gives I = −2 − 4 = −6 A or I = −2 + 4 = 2 A.2. (2. VL = 32 V.39) Rearranging terms in equation (2. I = −2 ± 4. 5 m above the water with an initial vertical velocity of 0. the power absorbed by the lamp is −6 × 32 = −192 W. (2. Use both the quadratic formula and completing the square.905 gives t2 − 0.6 m/s h(t) = 1.5 1.45) vo = 0. Solution (a) The time when the diver hits the water is found by setting h(t) = 0 in equation (2. (c) Use the results of parts (a) and (b) to sketch the height h(t) of the diver as a function of time. (2.6 t + 1.1223 t − 0.40). The diver’s height above the water is given by h(t) = −4. Example 2-6 A diver jumps off a diving board 1.6 m/s as shown in Fig.6 t + 1. The voltage across the resistor VR = 2 × 8 = 16 V and using KVL. I = −6 A is not one of the solutions of the quadratic equation given by (2. Vs = 16 + 32 = 48 V.9 Person jumping off a diving board.905 t2 + 0. for the applied power of 96 W (source voltage = 48 V). (b) Find the maximum height of the diver if it is known to occur at t = 0. Since the power absorbed by the lamp cannot be negative. Therefore.905 t2 + 0. the power absorbed by the lamp is 2 × 32 = 64 W and the power dissipated by the resistor is 96 − 64 = 32 W.9.5 m.5 = 0 (2.46) Dividing both sides of equation (2.40).45) as −4.47) .0612 s.46 Chapter 2 Quadratic Equations in Engineering Case I: For I = −6 A.3058 = 0. (a) Find the time in seconds when the diver hits the water. 2. Case II: For I = 2 A.5 m h(t) = 0 Figure 2.46) by −4. I = 2 A is the solution of the quadratic equation given by (2. 0612) = −4.0612 in equation (2.5563. Writing both sides of equation (2.51) Taking the square root of both sides gives t − 0.50) (2.0612)2 + 0.5563 = −0.1223 t = 0.5563. (2. . 2 (2.47) using the quadratic formula gives t= 0. (ii) Method 2: Completing the square: The quadratic equation (2. it takes 0.1223 ± 1. the diver’s height after jumping from the diving board is plotted as shown in Fig.48) Equation (2.49) )2 = 0.0612 = ± 0. it takes 0.45) as hmax = h(0.518 m.2382 = 0.0612 − 0.6(0.2.5 = 1.015 = 0.0612 ± 0. 2.0612 − 0.1223 t + 0.47) can be solved using the quadratic formula and completing the square as outlined below. ( Adding −0.1223 2 (2. t = 0.0612 ± 0.48) can be written as 0.0612 + 0.5563 = −0.0037.10. Since the time cannot be negative.3058 + 0. Therefore.5563 = 0.50) as a perfect square yields √ (t − 0.617 s.0037 to both sides gives t2 − 0.617 s.1127 = 0.5563.905(0.3058) .617 s for the diver to hit the water.0612)2 = (± 0. which gives t = 0. 2 which gives t = 0.1223 ± t= 2 √ 1.1223 ± √ (0. (b) The maximum height of the diver is found by substituting t = 0.3095)2 .47) can be written as t2 − 0.495 s or t = 0.0612 + 0.3058.617 s for the diver to hit the water.0612) + 1.5564 = 0.495 s or t = 0.4 Further Examples of Quadratic Equations in Engineering 47 The quadratic equation given in equation (2.1223)2 − 4(1)(−0. Since the time cannot be negative. (i) Method 1: Quadratic formula: Solving the quadratic equation (2. (c) Using the results of parts (a) and (b). 48 Chapter 2 Quadratic Equations in Engineering h(t).7 t. m Pipeline path through a parabolic hill.004 x2 + 0. s Height of the diver after jumping from the diving board. .10 Example 0.2 2-7 0.0612 0.11 Hill profile Tunnel entry A xA Pipeline path Tunnel exit B xB x. m 1.4 0.3 Pipeline Through Parabolic Hill: A level pipeline is required to pass through a hill having the parabolic profile y = − 0.5 0. m 2 0 Figure 2.3 x. (c) Find the length of the tunnel. y.5 0 0 0.617 0.11. 2. Figure 2.and y-coordinates is fixed at elevation zero near the base of the hill. (b) Solve the quadratic equation found in part (a) to determine the positions of the tunnel entry xA and exit xB using both the quadratic formula and completing the square.52) The origin of the x.518 1 0. (a) Write the quadratic equation for a pipeline elevation of y = 2 m. (2. as shown in Fig. 1 = 67. Therefore. the tunnel entry position is 7.53) can be written as x2 − 75 x = −500.5 ± 30.5 − 30.6 into equation (2.54) can be written as √ 75 ± 3625 75 ± 60.5 + 30.53).3 x which gives 0.5 = ± 30. which gives x = 37.4 and x = 67.6 m.56) as a perfect square yields √ (x − 37.25 = −500 + 1406. 2 2 which gives x = 37.53) (b) The quadratic equation given in equation (2.56) (2.25)2 . To check if the answer is correct.55) )2 = 1406.6 m.25 to both sides gives x2 − 75 x + 1406. ( Adding − 75 2 (2.5 − 30.5 + 30.004 x2 + 0.54) 2 Equation (2.4 m or x = 37. . x = 37.4 m and the tunnel exit position is 67.1 = 67.1.52) can be written as 2 = −0.3 x + 2 = 0 or x2 − 75 x + 500 = 0.1 = 7. (ii) Method 2: Completing the square: The quadratic equation (2. (i) Method 1: Quadratic formula: Solving the quadratic equation (2.1.5)2 = (± 906. Writing both sides of equation (2. Therefore.25.53) using the quadratic formula gives √ 75 ± (− 75)2 − 4(1)(500) x= .5 ± 30.1.004 x2 − 0.2. (2.4 Further Examples of Quadratic Equations in Engineering Solution 49 (a) Since the height y of the tunnel is 2 m.1 = 7.6 m.53) can be solved using the quadratic formula and completing the square as outlined below.57) Taking the square root of both sides gives x − 37. substitute x = 7.2 x= = = 37. (2. equation (2. (2.4 m or x = 37. Repeat problem P2-1 for the circuit shown in Fig. which yields the quadratic equation for the current I as 210 = 10 I 2 + 40 I. P2.4) + 500 = 0 55 − 555 + 500 = 0 0=0 Now.3 Resistive circuit for problem P2-3.50 Chapter 2 Quadratic Equations in Engineering Substituting x = 7.4 m and xB = 67. where I is in amps.6 − 7. which yields the quadratic equation for the current I as 3I 2 − 6 I = 45.1 yields the quadratic equation for the current I as 3I 2 + 6 I = 45.6) + 500 = 0 4570 − 5070 + 500 = 0 0=0 Therefore. 2-2.2.2 m PROBLEMS 2-1. xA = 7. where I is in amps.53) gives (67.4 = 60.6 m are the correct positions of the tunnel entry and exit.6 m into equation (2.4 m gives (7. where a and b are constants.42 ) − 75(7. (b) Solve the equation in part (a) by each of the following methods: factoring. and the quadratic formula. (c) The tunnel length can be found by subtracting the position xA from xB as Tunnel length = xB − xA = 67. P2. respectively. completing the square. substituting x = 67. P2. An analysis of a circuit shown in Fig. where I is in amps.3. 2-3. R=3Ω I Vs + − + R=3Ω I Vs + − − 6V + Figure P2. Vs + − + − 40 V Figure P2.2 Resistive circuit for problem P2-2.1 Resistive circuit for problem P2-1. . Repeat problem P2-1 for the circuit shown in Fig. (a) Rewrite the above equation in the form I 2 + aI + b = 0 . 6V − R = 10 Ω I Figure P2.62 ) − 75(67. Figure P2. (b) Using the method of your choice.5 Voltage across a capacitor.6. and simplify the resulting expression to obtain a single quadratic equation for R1 . find R1 and R2 as follows: (a) Substitute the values of R and R2 into equation (2. + − v(t) 51 If the total resistance of the circuit is R = 100 Ω and R2 = 2R1 + 100 Ω. and (b) i(t) = 84 A (use completing the square). Figure P2. The equivalent capacitance C of two capacitors connected in series as shown in Fig. Find t when (a) i(t) = 9 A (use the quadratic formula). P2. v(t) + − 2-7. solve the quadratic equation for R1 and compute the corresponding value of R2 .4 Current flowing through an inductor. R1 + R2 (2. the total resistance R of the circuit is given by R = R1 + R1 R2 . P2. The current flowing through the inductor shown in Fig.9 is given by C= C1 C2 C1 + C2 . P2.59) .2t − 0. In the purely resistive circuit shown in Fig. (2. P2.58).Problems 2-4.8 varies with time t in s according to the quadratic equation W = 3t2 + 6t. R1 L=1H i(t) R1 R2 R v(t) = 2t − 8 Figure P2. 2-5. and (b) v(t) = 27 V (use completing the square). The energy dissipated by a resistor shown in Fig.8 Resistive circuit for problem P2-8. Repeat problem P2-6 if R = 1600 Ω and R2 = R1 + 500 Ω. 2-8. The voltage across the capacitor shown in Fig.6 Series parallel combination of resistors. Find t when (a) v(t) = 16 V (use the quadratic formula).6) mA Figure P2. Solve for t if (a) W = 3 joules (b) W = 9 joules (c) W = 45 joules C = 100 μF i(t) i(t) v(t) + − R=3Ω i(t) = (0. 2-6.58) 2-9.5 is given by the quadratic equation v(t) = t2 − 6t. P2.4 is given by the quadratic equation i(t) = t2 − 8t. find L1 and L2 as follows: (a) Substitute the values of L and L1 in equation (2. (2. Also. completing the square. find C1 and C2 as follows: (a) Substitute the values of C and C2 in equation (2. and the quadratic formula. compute the corresponding values of L1 . L1 + L2 (2. C1 + C2 C1 L1 L2 L Figure P2. The equivalent capacitance C of three capacitors connected in series-parallel as shown in Fig.12 is given by L= L1 L2 L1 + L2 . Also. The equivalent inductance L of two inductors connected in parallel as shown in Fig. 2-10. 2-12. (2.61) and obtain the quadratic equation for L2 . (b) Solve the quadratic equation for C1 obtained in part (a) by completing the square and the quadratic formula.52 Chapter 2 Quadratic Equations in Engineering If the total capacitance is C = 120 𝜇F and C2 = C1 + 100 𝜇F.59) and obtain the quadratic equation for C1 .60) If the total capacitance is C = 125 𝜇F and C2 = C1 + 100 𝜇F. P2. 2-14. find C1 and C2 as follows: (a) Substitute the values of C and C2 in equation (2. The equivalent inductance L of three inductors connected in series-parallel as shown in Fig. compute the corresponding values of C2 . C1 C2 C 25 μF Figure P2. and the quadratic formula.61) If the total inductance L = 80 mH and L1 = L2 + 300 mH. C2 C Figure P2. 2-11. (b) Solve the quadratic equation for C1 obtained in part (a) by each of the following methods: factoring. P2. 2-13. compute the corresponding values of C2 .62) (a) Suppose L2 = L1 + 200 mH and that the equivalent inductance is L = 200 mH.60) and obtain the quadratic equation for C1 .14 is given by L = 125 + L1 L2 . Repeat problem P2-12 if L = 150 mH and L1 = L2 + 400 mH. (b) Solve the quadratic equation for L2 obtained in part (a) by completing the square. Also. Repeat problem P2-9 if C = 75 𝜇F and C2 = C1 + 200 𝜇F.12 Parallel combination of two inductors. Substitute these values in equation (2.9 Series combination of two capacitors.11 Series combination of two capacitors. P2.62) and obtain .11 is given by C C C = 25 + 1 2 . The ball falls according to the quadratic equation h(t) = 1000 − 4.17. A model rocket is launched in the vertical plane at time t = 0 s as shown in Fig.63) (b) Solve the quadratic equation (2. (d) Based on your solution to parts (a) through (c). P2. The height of the ball at time t is given by h(t) = 32 + 56t − 16t2 ft.15 A model rocket for Figure P2.16 is dropped from a height of 1000 meters. 000 = 0.Problems the following quadratic equation: L21 + 50 L1 − 15. determine the maximum height of the rocket and sketch the height h(t).08 m (c) 509. P2. (2. 2-17. from the top of a building.5 m (d) 0 m L 1000 m h(t) Figure P2.15. 125 mH L1 L2 53 2-16.905t2 . completing the square. as shown in Fig. Figure P2. The ball shown in Fig.14 Series-parallel combination of three inductors. (c) Find the time required for the rocket to hit the ground. a ball is thrown vertically from the top of the building at a speed of 56 ft/s. (a) Find the value(s) of the time t when h(t) = 48 ft. . P2. 2-15.16 A ball dropped from a height of 1000 m. (b) Find the value(s) of the time t when h(t) = 60 ft. At time t = 0. Find the time t in seconds for the ball to reach a height h(t) of (a) 921. and the quadratic formula.17 A ball thrown vertically problem P2-15.52 m (b) 686. v = 56 ft/s 32 ft h(t) h(t) = 64 t − 16 t2 Figure P2. The height of the rocket (in feet) satisfies the quadratic equation h(t) = 64 t − 16 t2 .63) for L1 by each of the following methods: factoring. 65) If k1 = k2 + 2 lb/in and the equivalent stiffness is k = 1.. An assembly of three springs connected in series as shown in Fig. find k1 and k2 as follows: (a) Substitute the values of keq and k2 in equation (2. find the values of k1 . (2. (b) Solve the quadratic equation for k1 obtained in part (a) by completing the square.67) (b) Solve equation (2. k1 + k2 obtained in part (a) and determine the values of k2 . and the equivalent stiffness is k = 2 lb/in. For each case. Two springs connected in series shown in Fig. k1 (2.19 is given by k k k = k2 + 1 2 . k3 = k1 + 8 lb/in. (b) Using the method of your choice.75 lb/in. Also.64) and obtain the quadratic equation for k1 .65) and obtain the quadratic equation for k2 . Also.67) for k1 by each of the following methods: (i) factoring. Substitute these values into equation (2.54 Chapter 2 Quadratic Equations in Engineering (a) Find the value(s) of the time t when h(t) = 32 ft. and sketch the height h(t) of the ball. find k1 and k2 as follows: (a) Substitute the values of k and k1 into equation (2. P2.2 N/m and k2 = 2k1 − 1 N/m. determine the value of k3 corresponding to the only physical solution for k1 . (ii) quadratic formula. k1 k2 Figure P2..20 has an equivalent stiffness k given by k= k1 k2 k3 . and (iii) completing the square. (2.18 Series combination of two springs. k2 k1 k2 Figure P2. solve the quadratic equation (2.66) to obtain the following quadratic equation: 4 k21 + 8 k1 − 96 = 0. The stiffness of the equivalent spring is given by keq = k1 k2 k1 + k2 . 2-18.19 Series-parallel combination of three springs. and the quadratic formula. (b) Find the time required for the ball to hit the ground. compute the corresponding values of k2 . The equivalent stiffness of a seriesparallel combination of three springs shown in Fig.20 Series combination of three springs. (c) Use the results to determine the maximum height. 2-19. k2 k3 + k1 k3 + k1 k2 (a) Suppose k2 = 6 lb/in. . If keq = 1.64) where k1 and k2 are the spring constants of the two springs. P2.18 can be represented by a single equivalent spring. P2.66) k k2 k3 Figure P2. 2-20. 2-22.0 mH and C = 1 F. R2 − 10 R − 200 = 0 (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula. Show that the angular frequency 𝜔 satisfies the quadratic equation 𝜔2 − 𝜔 − 1000 = 0. all measured in ohms. Rb = 3 R. show that the angular frequency − 20 3 𝜔 satisfies the quadratic equation 𝜔2 − 30 𝜔 − 400 = 0.Problems 2-21. 2-23. Substitute these values into equation (2.24. Ra = R. all measured in ohms.68) 2-25.24.24 Delta to Y conversion circuit. Consider a capacitor C and an inductor L connected in parallel. R2 − 300 R − 10. P2. 2-24.68) to obtain the following quadratic equation for R. (a) Suppose R1 = 6. If L = 0. P2. 000 = 0 (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula. and find the value of 𝜔 using both the quadratic formula and completing the square. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. find the value of 𝜔 using both the quadratic formula and completing the square. (a) Suppose R1 = 100. the resistance R1 is given by equation (2. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. If the total reactance is X = 1. P2. as shown in Fig.005 F. a R1 Ra L 55 C b R2 Rb R3 c Rc Figure P2. P2. Also. (b) Solve the quadratic equation for 𝜔 by the methods of completing the square and the quadratic formula.21. and the total reactance in problem P2-21 is X = Ω.68) to obtain the following quadratic equation for R.5 H. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. (a) Suppose L = 1. the resistance R2 is obtained as R2 = Ra Rc Ra + Rb + Rc (2. Assume that the total reactance in problem P2-21 is X = −1 Ω. the resistance R1 is obtained as R1 = Ra Rb Ra + Rb + Rc (2.0 Ω.69) .21 Parallel connection of L and C. 2-26.24.68). where 𝜔 is the given by X = 1 − 𝜔2 LC angular frequency in rad/s. Substitute these values into equation (2. and Rc = 100 + R. Figure P2. show that the angular frequency 𝜔 satisfies the quadratic equation 𝜔2 + 𝜔 − 1000 = 0. C = 0. Ra = Rb = R. The total reactance X in ohms is 𝜔L . and Rc = 100 + R. and Rc = 100 + R.56 Chapter 2 Quadratic Equations in Engineering (a) Suppose R2 = 12. L LC for problem P2-28. and damper system shown in Fig. (2. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. P2. The characteristic equation of a mass.70) (a) Suppose R3 = 36. Ra = R. Ra = R. spring.73) (a) If m = 1 kg.72) for the values of s by the methods of completing the square and the quadratic formula.28 is given as R 1 s+ = 0. (b) Repeat part (a) if R = 10 Ω. P2.28 Series RLC circuit (2. and C = 0. The characteristic equation of a series RLC circuit shown in Fig. Rb = 3 R.24. the resistance R3 is obtained as R3 = Rb Rc Ra + Rb + Rc (2. all measured in ohms. Substitute these values into equation (2. c = 2 Ns/m.29 Parallel RLC circuit for problem P2-29. R + 40 R − 1200 = 0 (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula. i(t) t=0 + R vc (0) C L − 2 R + 40 R − 1200 = 0 (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula. P2. and C = 25 + t=0 R L C v(t) − Figure P2. and C = 0.72) s2 + RC LC If R = 200 Ω. c = 3 Ns/m.30 is given by m s2 + c s + k = 0. L = 1 1 F. and k = 2 N/m. solve the quadratic equation (2.2 𝜇F.29 is given as 1 1 s+ = 0. (b) Repeat part (a) if m = 1 kg. P2. all measured in ohms.70) to obtain the following quadratic equation for R. L = 50 mH.71) (a) If R = 7 Ω. solve the quadratic equation (2. 2-30. (2. H. and Rc = 100 + R. 2-27. iL (0) 2 s2 + Figure P2. and k = 1 N/m.1 F. 2-29. . The characteristic equation of a parallel RLC circuit shown in Fig. solve the quadratic equation (2.73) for the values of s using the methods of completing the square and the quadratic formula.69) to obtain the following quadratic equation for R. 2-28. L = 1 H. Rb = 3 R.71) for the values of s (called the eigenvalues of the system) using the methods of completing the square and the quadratic formula. Substitute these values into equation (2. (2. 2-33. A diver jumps off a diving board 2.and y-coordinates is fixed at elevation zero near the base of the hill.981t + 2. The height h(t) above the water is given by h(t) = −4.981 m/s h(t) = 2. vo = 0. 2-32. A level pipeline is required to pass through a hill having a parabolic profile y = − 0. P2. (2.74) and obtain the quadratic equation for L. and damper system for problem P2-30. 57 (b) Solve the quadratic equation for L obtained in part (a) by completing the square.75) The origin of the x. If the perimeter P = 30 m and the area A = W × L = 36 m2 . (b) Find the maximum height of the diver if it is known to occur at t = 0.0 2. spring. Use both the quadratic formula and completing the square. P2. find the length L and width W as follows: (a) Substitute the values of P and A in equation (2. and the quadratic formula.32. (a) Find the time in seconds when the diver hits the water. (c) Use the results of parts (a) and (b) to sketch the height h(t) of the diver. The perimeter of a rectangle shown in Fig.30 Mass.33.0 m h(t) = 0 Figure P2.0 m.1 s.981 m/s as shown in Fig.74) P=2 L W Area = 36 m2 Perimeter = 30 m L Figure P2.905 t2 + 0. .32 Diver jumping off a diving board.005 x2 + 0.0 m above the water with an initial vertical velocity of 0.35 x. as shown in Fig.Problems x (displacement) k m f (force) c Figure P2.31 A rectangle of length L and width W. 2-31. Also. P2.31 is given by ) ( A +L . compute the corresponding values of W. 100 − x reaction that form the concentration of C = 2x are related to the equilibrium constant by the expression 4.9 P2 ). and the quadratic formula.000 mol . pores were generated that affect the strength of the material. Consider the following reaction having an equilibrium constant of 1.66 × 10−3 at a certain temperature: A(g) + B(g) ⇌ C(g) If 0. 2-36.33 Pipeline path through a parabolic hill. obtain the quadratic equation for volume fraction porosity P.300 − x and B = 0.9 t2 .76) (a) If a porous sample of silicon nitride has a modulus of elasticity of E = 209 GPa. (c) Find the equilibrium concentration A.300 − x) (0.e. The helmet attached to a weight is propelled downward with an initial velocity vi of 3 m/s from an initial height of 30 m. (0. and C. A research group is using a drop test to measure the force of attenuation of a helmet liner they designed to reduce the occurance of brain injuries for soldiers and athletes. the stiffer the material.58 Chapter 2 Quadratic Equations in Engineering y. 2-34. (b) Solve the quadratic equation found in part (a) to determine the positions of the tunnel entry xA and exit xB using both the quadratic formula and completing the square.00 at 1105◦ K temperature: CO(g) + H2 O(g) ⇌ CO2 (g) + H2 (g) Suppose the feed to a reactor contains 1. (b) Solve the quadratic equation for t obtained in part (a) by completing the square. the concentrations of A = 0.000 mol of CO and 2. m 3 0 Tunnel entry A xA fraction porosity P by Hill profile Tunnel exit Pipeline path B xB E = 304 (1 − 1.300 mol of A and 0. The behavior of the falling helmet is characterized by a quadratic equation h(t) = 30 − 3 t − 4. (b) Solve the quadratic equation found in part (a) using both the methods of completing the square and the quadratic formula.. 2-37. (c) Find the length of the tunnel. (c) Repeat parts (a) and (b) to find the volume fraction porosity of a silicon nitride sample with a modulus of elasticity of 25 GPa.100 − x) (a) Write the quadratic equation for x. m Figure P2.9 P + 0. The magnitude of elasticity is related to volume (2. During fabrication of a ceramic material from a powder form. h(t) = 0 m. 2-35. (a) Write the quadratic equation for a pipeline elevation of y = 3 m. i. The modulus of elasticity (E) is a measure of a material’s resistance to deformation. the larger the modulus. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula.100 mol of B are mixed in a container and allowed to reach equilibrium. B. Consider the following reaction having an equilibrium constant of 4.66 × 10−3 = (2 x)2 . x. (a) Write the quadratic equation for time t when the helmet and weight hit the ground. P2.000 − x and H2 O = 2. ft.000 − x) (2. (c) Find the equilibrium concentration of CO. Repeat problem P2-38 if the total area of the new asphalt needs to be 300 sq..38. H2 O. The area of the walkway is given by A = (50 + 5 x)(30 + 2 x) − 1500 (a) If A = 4500 sq. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula.77) 4x Walkway Walkway Figure P2. and the reaction mixture comes to equilibrium at 1105 K.000 − x reaction that form the concentration of CO2 = x and H2 = x are related to the equilibrium constant as 2x 1. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula. P2. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula. h New asphalt (a) If A = 200 sq. and H2 . 2-39. 2-38..40 Walkway around the swimming pool.Problems of H2 O(g).40. . ft. 2-40. Walkway The height h is required to be 10 ft. CO2 . 2 (2. The concentrations of CO = 1. obtain the quadratic equation for b. obtain the quadratic equation for x. An engineering co-op wants to hire an asphalt contractor to widen the truck entrance to the corporate headquarters as shown in Fig. and the total area of the new asphalt is given by 1 A = b(10 + b). more than the width b. The dimensions of the walkway along with the dimensions of the pool are shown in Fig.38 New asphalt dimensions of the 59 30 ft 30 + 2x x 50 ft 50 + 5x Figure P2.000 − x) (a) Write the quadratic equation for x. ft. A city wants to hire a contractor to build a walkway around the swimming pool in one of its parks. b x Walkway Swimming pool corporate headquarters driveway. (c) Find the dimensions of the width and height.00 = (1. 1) that is being rotated in the x-y plane by a motor mounted at the center of the table.2 ONE-LINK PLANAR ROBOT Consider a one-link planar robot of length l (Fig.2. In general. the joint displacement can be linear or rotational (angular). The direct or forward kinematics is the static geometric problem of determining the position and orientation of the end-effector (hand) of the robot from the knowledge of the joint displacement. 𝜃).1 Kinematics of One-Link Robot In Fig.2. But in this chapter. the tip of the robot moves on a circle of radius l (the length of the link of the robot) as shown in Fig. Assuming 60 . the inverse kinematics will determine the joint angle(s) from the Cartesian position of the tip. Going in the other direction. it has no end-effector or hand) and that only the position but not the orientation of the tip of the robot can be changed.2. Since no end-effector is considered in this chapter. The robot has a position sensor installed at the joint that gives the value of the angle 𝜃 of the robot measured from the positive x-axis. y) or in polar coordinates by the pair (l.CHAPTER 3 3. as the joint rotates from 0◦ to 180◦ and 0◦ to −180◦ . the point P (tip of the robot) can be represented in rectangular or Cartesian coordinates by a pair (x. which is also the location of the robot’s joint. the inverse or reverse kinematics is the problem of determining all possible joint variables (angles) that lead to the given Cartesian position and orientation of the end-effector. 3. Kinematics is the branch of mechanics that studies the motion of an object. Therefore.. the direct (forward) and inverse (reverse) kinematics of one-link and two-link planar robots are considered to explain the trigonometric functions and their identities.1 Trigonometry in Engineering INTRODUCTION In this chapter. 3. 3. Note that 180◦ = 𝜋 radians.e. 3. Furthermore. 3. it is assumed that the planar robot is wristless (i. only rotational motion is considered. The angle 𝜃 is positive in the counterclockwise direction (0◦ to 180◦ ) and it is negative in the clockwise direction (0◦ to −180◦ ). 2 Circular path of the one-link robot tip. y) Positive θ y (counterclockwise direction) 0 ≤ θ ≤ 180 A x Negative θ (clockwise direction) −180 ≤ θ ≤ 0 Adjacent side Figure 3. The position of the tip of the robot (x.2 One-Link Planar Robot 61 y P(x. This is known as the direct or forward kinematics of the robot. y) Link l θ y x x Motor Figure 3.2) y Opposite side Hypotenuse B y l O θ x P(x. a change in the angle 𝜃 of the robot changes the position of the tip of the robot. that the length of the link l is fixed.2 as cos(𝜽) = Adjacent side x = Hypotenuse l ⇒ x = l cos(𝜃) (3.1 One-link planar robot.3. . 3. y) in terms of l and 𝜃 can be found using the right-angle triangle OAP in Fig.1) sin(𝜽) = Opposite side y = Hypotenuse l ⇒ y = l sin(𝜃) (3. 62 Chapter 3 Trigonometry in Engineering Example 3-1 Use the one-link robot to find the values of cos(𝜃) and sin(𝜃) for 𝜃 = 0◦ . 90◦ . Solution Case I: 𝜃 = 0◦ y By inspection. x = l cos(−90◦ ) = 0 ⇒ cos(−90◦ ) = 0 y = l sin(−90◦ ) = −l ⇒ sin(−90◦ ) = −1 θ = −90° O x l P Case IV: 𝜃 = 180◦ y By inspection. −90◦ . l x = l cos(90◦ ) = 0 ⇒ cos(90◦ ) = 0 y = l sin(90◦ ) = l ⇒ sin(90◦ ) = 1 θ = 90° O Case III: 𝜃 = −90◦ x y By inspection. x = l cos(180◦ ) = −l ⇒ cos(180◦ ) = −1 ⇒ sin(180◦ ) = 0 y = l sin(180◦ ) = 0 θ = 180° P l O x . find the values of x and y. x = l cos(0◦ ) = l ⇒ cos(0◦ ) = 1 y = l sin(0◦ ) = 0 ⇒ sin(0◦ ) = 0 θ = 0° l P O Case II: 𝜃 = 90◦ x y P By inspection. and 180◦ . Also. y) l x = l cos(135 ) = − √ 1 ⇒ cos(135 ) = − √ 2 2 l 1 y = l sin(135◦ ) = √ ⇒ sin(135◦ ) = √ 2 2 ◦ ◦ y= l √⎯ 2 l 135° 45° x=− l O 2 √⎯ x Case IV: 𝜃 = −135◦ l x = l cos(−135◦ ) = − √ y 2 1 ⇒ cos(−135 ) = − √ 2 l y = l sin(−135◦ ) = − √ 2 1 ⇒ sin(−135◦ ) = − √ 2 ◦ x=− l y=− √⎯ 2 45° P(x. y) l 1 x = l cos(45◦ ) = √ ⇒ cos(45◦ ) = √ 2 2 l 1 y = l sin(45◦ ) = √ ⇒ sin(45◦ ) = √ 2 2 l y= 45° O Case II: 𝜃 = −45◦ x= l 2 √⎯ x= l 2 √⎯ l √⎯ 2 x y l 1 x = l cos(−45◦ ) = √ ⇒ cos(−45◦ ) = √ 2 2 l 1 ◦ ◦ y = l sin(−45 ) = − √ ⇒ sin(−45 ) = − √ 2 2 x −45° O y=− l l √⎯ 2 P(x. y) l l 2 √⎯ O −135° x . and −135◦ . −45◦ .2 One-Link Planar Robot Example 63 Find the position P(x. y) Case III: 𝜃 = 135◦ y P(x.3. 135◦ . y) of the robot for 𝜃 = 45◦ . 3-2 Solution Case I: 𝜃 = 45◦ y P(x. 64 Chapter 3 Trigonometry in Engineering Examples 3-1 and 3-2 show that in the first quadrant (0◦ < 𝜃 < 90◦ ). sin and csc are positive and all the rest of the trigonometric functions are negative. 45◦ .1.1 Values of sine and cosine functions for common angles. 60◦ .1. as shown in Fig. only the cos and sec are positive. 3. To remember this. 30◦ . In the second quadrant. for example) are functions of sin and cos functions. both the sin and cos functions are negative. Therefore.3. The values of sin and cos functions for 𝜃 = 0◦ . and 90◦ are given in Table 3.3 Trigonometric functions in the four quadrants. TABLE 3. both the sin and cos functions are positive. only the tan and cot are positive. sec = 1/cos. Finally. In the third quadrant. The values of sin and cos functions for many other angle can be found using Table 3. as explained in the following examples. and csc = 1/sin.” Another is “All Students Take Calculus. all the trigonometric functions are positive in the first quadrant. one of the phrases commonly used is “All Sin Tan Cos. Angle deg (rad) √ sin √ cos 0◦ (0) 0 4 4 4 =0 =1 30◦ ( 𝜋6 ) √ 1 = 12 4 √ √ 3 3 = 4 2 √ √ 45◦ ( 𝜋4 ) 2 4 2 4 = = 1 √ 2 1 √ 2 √ √ 60◦ ( 𝜋3 ) 3 4 = 1 4 = √ 3 2 1 2 90◦ ( 𝜋2 ) √ 4 =1 4 √ 0 =0 4 . in the fourth quadrant.” which is certainly true of engineering students! y sin +ve cos −ve 2nd Quadrant 1st Quadrant (sin is positive) sin cos +ve +ve (all trigonometric functions are positive) x sin cos −ve −ve (tan is positive) (cos is positive) 3rd Quadrant 4th Quadrant sin −ve cos +ve Figure 3. cot = 1/tan. Since the other trigonometric functions (tan = sin/cos. therefore.3.4 One-link planar robot with an angle of 120◦ . The reference angle is always positive. If the angle 𝜃 is in the second quadrant. However. if the angle 𝜃 is in the fourth quadrant. if the angle is in radians). find the position of the tip of the one-link robot for this angle. y P(x.2 One-Link Planar Robot 65 Example 3-3 Find sin 𝜃 and cos 𝜃 for 𝜃 = 120◦ . the reference angle is 180◦ − 𝜃 (𝜋 − 𝜃. the values of sin(120◦ ) and cos(120◦ ) can be written as l 2 √ 3 ◦ ◦ l. Note that the point P is in the second quadrant and. the reference angle is the absolute value of 𝜃.4. y = l sin(120 ) = l sin(60 ) = 2 x = l cos(120◦ ) = −l cos(60◦ ) = − √ Note that cos(120◦ ) = −cos(60◦ ) = − 12 and sin(120◦ ) = sin(60◦ ) = 23 . Also. Solution The position of the tip of the robot for 𝜃 = 120◦ is shown in Fig. the reference angle is 𝜃 + 180◦ . y) l y Reference angle 60° 120° x x Figure 3. which in this case is 60◦ . . Their values can be found using the reference angle of 𝜃 = 120◦ . 3. If the angle 𝜃 is in the third quadrant. and it is the acute angle formed between the x-axis and the terminal side of the angle (120◦ in this case). sin(120◦ ) should have a positive value and cos(120◦ ) should be negative. The values of sin(120◦ ) and cos(120◦ ) can also be found using the trigonometric identities sin(A ± B) = sin(A) cos(B) ± cos(A) sin(B) cos(A ± B) = cos(A) cos(B) ∓ sin(A) sin(B). the reference angle is the same as the angle 𝜃. If the angle 𝜃 is in the first quadrant. Therefore. sin(120◦ ) = sin(90◦ + 30◦ ) = sin(90◦ ) cos(30◦ ) + cos(90◦ ) sin(30◦ ) (√ ) ( ) 3 1 = (1) + (0) 2 2 √ 3 = 2 and cos(120◦ ) = cos(90◦ + 30◦ ) = cos(90◦ ) cos(30◦ ) − sin(90◦ ) sin(30◦ ) (√ ) ( ) 3 1 = (0) − (1) 2 2 1 =− . 2 Therefore. l x = l cos(−135◦ ) = −l cos(45◦ ) = − √ 2 l y = l sin(−135◦ ) = −sin(45◦ ) = − √ 2 ) ( l l . y 225° x O 45° y −135° l Reference angle P(x. 23 l .5 One-link planar robot with an angle of 225◦ . − √ 2 2 x . y) = −l2 . 3. position of the tip of the one-link robot if 𝜃 = 120◦ is given by ( the √ ) (x. y) Figure 3. The position of the tip of the robot for 𝜃 = 225◦ is shown in Fig. y) = − √ .66 Chapter 3 Trigonometry in Engineering Therefore.5. Example 3-4 Solution Find the position of the tip of the one-link robot for 𝜃 = 225◦ = −135◦ . (x. y) Figure 3.6 One-link planar robot with an angle of 390◦ .6.3. y) y l 390° y 30° Reference angle x x O Figure 3.7. 3. . The position of the tip of the robot for 𝜃 = 390◦ is shown in Fig. y) = .7 One-link planar robot with an angle of −510◦ . y Reference angle x x O 30° y l −510° P(x. P(x. 2 2 ◦ ◦ x = l cos(390 ) = l cos(30 ) = Example 3-6 Solution Find the position of the tip of the one-link robot for 𝜃 = −510◦ . 3.2 One-Link Planar Robot Example 3-5 Solution Find the position of the tip of the one-link robot for 𝜃 = 390◦ . 67 . The position of the tip of the robot for 𝜃 = −510◦ is shown in Fig. √ 3l 2 l y = l sin(390◦ ) = l sin(30◦ ) = 2 (√ ) 3l l (x. √ √ Therefore.2.4). Now dividing y in (3. y l sin𝜃 = = tan(𝜃). − . it is important to keep track of the signs of x and y. However.68 Chapter 3 Trigonometry in Engineering √ 3l x = l cos(−510 ) = −l cos(30 ) = − 2 ◦ ◦ y = l sin(−510◦ ) = −l sin(30◦ ) = − ( √ ) 3l l (x. x is either positive or negative. x2 + y2 = l2 . y) = − . y).4) 𝜃 = tan−1 x x In equation (3. This is called the inverse problem. l = x2 + y2 . the angle obtained from the atan function is between 0 and −90◦ x (fourth quadrant). This can be done by locating the point P in the proper quadrant or using the four-quadrant arctangent function (atan2) as explained in the following examples. .3) as (y) (y) = atan . Since the distance cannot be negative.2) give the relationship between the tip position and the angle 𝜃.1) and (3. if both x and y are negative (third quadrant). Using the trigonometric identity sin2 𝜽 + cos2 𝜽 = 1. the ◦ angle ( ) obtained from the atan function is between 0 and 90 (first quadrant) and if y is negative. 2 2 l 2 3. If x is positive. the angle 𝜃 can be determined from the position of the tip of the robot using equation (3. find the angle 𝜃 and length l. given x and y. y is divided by x before the inverse tangent (arctangent or atan) ( ) ( ) y y is calculated and therefore. the angles obtained from the atan function will be wrong since the angles should lie in the third or second quadrant. This is why the atan function is called the two-quadrant arctangent function.2) by x in (3. Equations (3. Squaring and adding x and y in these equations gives x2 + y2 = (l cos 𝜃)2 + (l sin 𝜃)2 = l2 (sin2 𝜃 + cos2 𝜃). it is required to find the joint angle 𝜃 by which the motor needs to move. or x is negative and y is positive (second quadrant). Therefore. for example. respectively.1).3) Therefore. (3. l = ± x2 + y2 . x l cos𝜃 (3.2 Inverse Kinematics of One-Link Robot In order to move the tip of the robot to a given position P(x. √ √ Also. y) = (1. l (0. √ √ Also. l = 1 and 𝜃 = −90◦ .2 One-Link Planar Robot Example 3-7 69 Find l and 𝜃 for the following points (x. ( ) 𝜃 = tan−1 01 = tan−1 (0) = 0◦ . l = x2 + y2 = (−1)2 + 02 = 1. 0): y By inspection. √ √ Also. y) = (0. l = x2 + y2 = 02 + 12 = 1. y) = (−1. ( ) 0 = tan−1 (−0) = 180◦ . l = x2 + y2 = 12 + 02 = 1. (0. ( ) 𝜃 = tan−1 10 = tan−1 (∞) = 90◦ . 0): y By inspection. 1): y By inspection. −1) Case IV: (x. y) = (0. 𝜃= ) tan−1 ( −1 0 = tan−1 (−∞) = x O −90◦ . l = 1 and 𝜃 = 0◦ . 0) x Case II: (x. √ √ Also. 𝜃 = tan−1 −1 l (−1. 1) l x O Case III: (x. −1): y By inspection. y): Case I: (x. l O (1. l = 1 and 𝜃 = 90◦ . l = 1 and 𝜃 = 180◦ . l = x2 + y2 = 02 + (−1)2 = 1. 0) O x .3. l = x2 + y2 = √ √ 2 2 Example 3-10 l √⎯ 2 ( √⎯2 . 𝜃 = tan−1 ⎜ 1 ⎟ ⎜ √ ⎟ ⎝ 2 ⎠ − l 2 √⎯ l l 45° 1 2 √⎯ ( √⎯2 . 2 − √1 2 . √ ⎜ ⎟ ⎝ 2⎠ ) 1 √ . y) = . 𝜃 = tan ⎜ 1 ⎟ ⎜−√ ⎟ 2⎠ ⎝ x θ l ( Find the values of l and 𝜃 if (x. √ 2 2 Find the values of l and 𝜃 if (x. l = x2 + y2 = √ − √ 2 2 ⎛ − √1 ⎞ 2⎟ −1 ⎜ = tan−1 (1) = 45◦ . 1 − 2 √⎯ O − 1 2 √⎯ l (− √⎯2 . − √⎯2 ) 1 1 θ 1 √⎯ 2 x . − √⎯2 ) ) − √1 . the calculator answer must be adjusted as explained in example 3-10. Example 3-8 ( ) 1 1 √ . y) = x l √⎯ 2 O ( ) l √⎯ 2 θ −1 ⎜ √ √( )2 ( )2 √ √ √ 1 1 + −√ = 1. y √ √( )2 ( )2 √ √ √ 1 1 2 2 l = x +y =√ √ + √ = 1.70 Chapter 3 Trigonometry in Engineering But a calculator will give an answer of 0◦ . 2 2 𝜃 = tan Example 3-9 l l ⎛ √1 ⎞ 2⎟ −1 ◦ ⎜ 1 ⎟ = tan (1) = 45 . For this case. y) = − √1 2 . y l √⎯ 2 O ⎛ − √1 ⎞ ⎜ 2⎟ = tan−1 (−1) = −45◦ . y √ √( )2 ( )2 √ √ √ 1 1 + −√ = 1. 2 Find the values of l and 𝜃 if (x. the correct answer is 𝜃 = 45◦ − 180◦ = −135◦ .2 One-Link Planar Robot 71 The answer 𝜃 = 45◦ obtained in example(3-10 is incorrect. Method 1: Obtain the reference angle and then subtract the reference angle from 180◦ .3.25)2 = 0. 3-11 l= √ √ x2 + y2 = (−0. The correct answer can also be obtained using the atan2(y. it is best to find the quadrant the point lies in and then correct the problem accordingly. can be obtained should lie between −90◦ and −180◦ . that the ) ( (Note y x atan2 function requires both x and y values separately instead of using MATLAB gives ( ) 1 1 atan2 − √ . Example Find the values of l and 𝜃 if (x.5) y (−0. ( ) y ◦ −1 . 0. The correct answer.25 𝜃 = tan−1 = tan−1 (−0.5 = 180◦ − 26. Therefore. x) function predefined in their libraries.5) = −26. the point lies in the third quadrant.5 .6) . 0. (3.57◦ .25 θ l 26. the angle therefore. (3. The atan2(y.559 Using your calculator.5. − 0.5.25). y) = (−0. in this case.25) 0. y) = √1 .57◦ obtained in example 3-11 is clearly incorrect.4◦ . The ( ) atan2(y. 𝜃 = 180◦ − reference angle ) ( 0. Most of the programming languages including MATLAB have the atan2(y.57◦ = 153.57° −0. − √ = −2. x) function computes the tan−1 y x function but uses the sign of both x and y to determine the quadrant in which the resulting angle lies.25 = 180◦ − tan−1 0. Since.) Therefore. and it is the same value ) obtained in example 3-8 where (x.57° x Reference angle The answer 𝜃 = −26.5 O −26. x) function. Remember that the calculator 2 2 ( ) y −1 always returns a value in the range of −90◦ ≤ 𝜃 ≤ 90◦ . ) ( 0.5)2 + (0. x) function is sometimes called a four-quadrant arctangent function and returns a value in the range −𝜋 ≤ 𝜃 ≤ 𝜋 (−180◦ ≤ 𝜃 ≤ 180◦ ).3562 rad 2 2 = −135◦ . To obtain function tan x the correct answer. The correct angle can be obtained using one of the following three methods. √1 . The other method is to obtain by subtracting 180 from angle obtain using tan x the reference angle and then add the reference angle to −180◦ . 4◦ .6779 rad) 180◦ 𝜋 rad ) = 153.5) = 2.7) Method 3: Use the atan2(y.6779 rad ( = (2. y) Upper arm l1 Shoulder motor (Joint 1) Figure 3.8 Two-link planar robot.8 shows a two-link planar robot moving in the x-y plane.8) TWO-LINK PLANAR ROBOT Figure 3. 3. and the lower arm of length l2 is rotated by the elbow motor. Position sensors are installed at the joints that give the value of the angle 𝜃1 measured from the positive real axis (x-axis) to the upper arm.4◦ .25. These angles are positive in the counterclockwise direction and negative in the clockwise y Lower arm P(x.3 (3.57◦ ) = 153. 𝜃 = 180◦ + tan−1 (y) x ) ( 0.72 Chapter 3 Trigonometry in Engineering Method 2: Use the tan−1 ( ) y x function and add 180◦ to the result.25 ◦ −1 = 180 + tan −0. x) function in MATLAB.5 = 180◦ + (−26. The upper arm of length l1 is rotated by the shoulder motor. −0. and the relative angle 𝜃2 measured from the upper arm to the lower arm of the robot. (3. 𝜃 = atan2(0. l2 θ1 Tip θ2 Elbow motor (Joint 2) x . 9. substituting equations (3. (3. x = x1 + x2 (3.11) y1 = l1 sin 𝜃1 .3 Two-Link Planar Robot 73 direction.13) y2 = l2 sin(𝜃1 + 𝜃2 ).9 Two-link planar robot. (3.10) yields y = l1 sin 𝜃1 + l2 sin(𝜃1 + 𝜃2 ).15) and (3. In this section.13) into equation (3.9) y = y1 + y2 . x1 = l1 cos 𝜃1 (3. (3.16) Equations (3.14) y P(x. .12) and (3.14) into equation (3.11) and (3. both the direct and inverse kinematics of the two-link robot are derived. y) y l2 y2 l1 y1 P1 B θ1 A O x1 θ2 θ1 + θ 2 θ1 x x2 x Figure 3. y) if the joint angles 𝜃1 and 𝜃2 are known. Substituting equations (3.1 Direct Kinematics of Two-Link Robot The direct kinematics of the two-link planar robot is the problem of finding the position of the tip of the robot P(x. (3.9) gives x = l1 cos 𝜃1 + l2 cos(𝜃1 + 𝜃2 ).16) give the position of the tip of the robot in terms of joint angles 𝜃1 and 𝜃2 .3. 3.15) Similarly.12) Similarly. x2 = l2 cos(𝜃1 + 𝜃2 ) (3.8 and 3.10) From the right-angle triangle OAP1 . (3. using right-angle triangle P1 BP.3. As illustrated in Figs. 3. 𝜃2 = 0◦ By inspection: x = −(l1 + l2 ) and y = 0.15) and (3. 2 −45° P(x. y). O x x = −(l1 + l2) Case III: 𝜃1 = 90◦ . 𝜃2 = −45◦ Using equations (3. Solution Case I: 𝜃1 = 𝜃2 = 0◦ By inspection: y x = l1 + l2 and y = 0.15) and (3. y Using equations (3.15) and (3. sin(90◦ − 90◦ ) −90° l2 l1 P(x. y Using equations (3. P(x.15) and (3.16): x = l1 cos(180◦ ) + l2 cos(180◦ + 0◦ ) = l1 (−1) + l2 (−1) = −(l1 + l2 ) P(x.16): x = l1 cos(90◦ ) + l2 cos(90◦ − 90◦ ) = l1 (0) + l2 (1) = l2 l1 sin(90◦ ) + l2 y= = l1 (1) + l2 (0) = l1 . y) 90° x O Case IV: 𝜃1 = 45◦ .74 Chapter 3 Trigonometry in Engineering Example 3-12 Find the position. of the tip of the robot for the following configurations. 𝜃2 = −90◦ By inspection: x = l2 and y = l1 . Also. y) l1 45° O l2 x .16): y ◦ ◦ ◦ x = l( 1 cos(45 ) ) + l2 cos(45 − 45 ) = l1 1 √ 2 + l2 (1) = l √1 2 + l2 ◦ ◦ ◦ y = l( 1 sin(45 ) ) + l2 sin(45 − 45 ) = l1 1 √ 2 + l2 (0) = l √1 . sketch the orientation of the robot in the x-y plane. y) l1 l2 y = l1 sin(180◦ ) + l2 sin(180◦ + 0◦ ) = l1 (0) + l2 (0) = 0. y) Using equations (3. l2 l1 O + l2 x x = l1 + l 2 Case II: 𝜃1 = 180◦ . P(x.16): x= y= l1 cos(0◦ ) + l2 cos(0◦ + 0◦ ) = l1 l1 sin(0◦ ) + l2 sin(0◦ + 0◦ ) = 0. 3. Assume l1 = l2 = 5 2. Solution In the algebraic solution. if the three sides of the triangle shown in Fig.17) . 6) 2 5 √⎯ 2 5 √⎯ 6 P1 θ1 O θ2 x l2 Figure 3. In this chapter. y P(12. the joint angles 𝜃1 and 𝜃2 are determined using the Pascal laws of cosines and sines. 3.3. (3. only the algebraic solution will be carried out.10.11 are known. This problem can be solved using a geometric solution or an algebraic solution. y) = (12.3 Two-Link Planar Robot 75 3. 6) as shown in Fig.11 A triangle with an unknown angle and three known sides. 3. The Pascal law of cosines can be used to find the unknown angles of a triangle if the three sides of the triangle are known. y) is known. 2bc c a γ α b Figure 3.2 Inverse Kinematics of Two-Link Robot The inverse kinematics of the two-link planar robot is the problem of finding the joint angles 𝜃1 and 𝜃2 if the position of the tip of the robot P(x. Example 3-13 Find the joint angles 𝜃1 and 𝜃2 if the position of the tip √ of the robot is given by P(x. For example.10 Two-link configuration to find 𝜃1 and 𝜃2 . the unknown angle 𝛾 can be found using the Law of Cosines as a2 = b2 + c2 − 2 b c cos 𝜸 or cos 𝛾 = b2 + c2 − a2 . 6) ⎯ ⎯8⎯0 √1 ⎯6 θ 2 θ2 ⎯⎯ 2⎯+ − ⎯2⎯ 180 1 √ P1 5 √2⎯ 12 5√ 2⎯ 76 2⎯ = O 6 Figure 3. three sides are known and one of the angles 180 − 𝜃2 is unknown. 𝜃2 lies either in the first or fourth quadrant based on the values of sin 𝜃2 as shown in Fig. (√ )2 ( √ )2 ( √ )2 ( √ )( √ ) 180 = 5 2 + 5 2 − 2 5 2 5 2 cos(180◦ − 𝜃2 ) 180 = 50 + 50 − 100 cos(180◦ − 𝜃2 ) 80 = −100 cos(180◦ − 𝜃2 ) −0. x . the unknown angle 𝛼 can be found using the Law of Sines as sin 𝜸 sin 𝜶 = c a or c sin 𝛾. equation (3. a Solution for 𝜃2 : In Fig. y θ 2 = 36. Applying the law of cosines to the triangle OP1 P gives sin 𝛼 = P(12. 3.87 0. the angle 𝜃2 can be obtained from triangle OPP1 formed by joining points O and P.87 −0. angle 𝜃2 is negative. 3.8 0. For the positive value of cos 𝜃2 . In this triangle (Fig.12). if sin 𝜃2 is negative.Chapter 3 Trigonometry in Engineering Similarly. If the value of sin 𝜃2 is positive.12 Using the law of cosines to find 𝜃2 .10.6 θ 2 = −36. 3.11 are known.13 Two solutions of 𝜃2 . However.8 = cos(180◦ − 𝜃2 ). (3. 3.18) can be written as cos 𝜃2 = 0. angle 𝜃2 is positive. if the two sides (a and c) and the angle (𝛾) of the triangle shown in Fig.6 Figure 3.8.18) cos(180◦ Since − 𝜃2 ) = −cos 𝜃2 .13. 6) Elbow up ° . the positive solution 𝜃2 = 36.87° Elbow up 6 O Figure 3.13◦ sin 𝛼 . 6) θ 2 = 36. The angle 𝛼 needed to find 𝜃1 in equation (3. Elbow-Up Solution for 𝜽1 : The angle 𝜃1 for the elbow-up solution is shown in Fig.19) x 12 Elbow-up configuration to find angle 𝜃1 .14.87◦ and 𝜃2 = −36. y P(12.13 α 143 Figure 3.87◦ is called the elbow-down solution. The angle 𝜃1 + 𝛼 can be obtained from Fig. . Elbow down θ 2 = −36.57◦ 𝜃1 = 26. √ = √ 5 2 180 .16. there are two possible solutions of 𝜃2 . 3.15 θ 2 = 36.87◦ is called the elbow-up solution and the negative solution 𝜃2 = −36.19) can be obtained using the law of sines or cosines from the triangle OP1 P shown in Fig. In Fig.3 Two-Link Planar Robot 77 Therefore. 3.57◦ − 𝛼.14 12 Elbow-up and elbow-down solutions of 𝜃2 . Using the law of sines gives sin 143. 3. 𝜃2 = 36.3. 3.15.87° 6 P1 θ1 O (3.15 as tan(𝜃1 + 𝛼) = 6 12 𝜃1 + 𝛼 = tan−1 ( 6 12 ) 𝜃1 + 𝛼 = 26.87◦ .87° P(12. (3.57 − 18.1 14 α 6 θ1 O 12 Figure 3.45◦ . √ 5 2 sin 𝛼 = √ sin 143.20) y Elbow down θ 2 = 36. Therefore.13◦ 180 = 0.3164. 3.57◦ 𝜃1 = 26. Therefore.45 = 8. the angle 𝛼 is positive.17.16 Elbow-up configuration to find angle 𝛼. The angle 𝜃1 − 𝛼 can be obtained from Fig. 6) when the elbow is up is given by 𝜃1 = 8.17 Elbow-down configuration to find angle 𝜃1 .19) yields 𝜃1 = 26.3164) 𝛼 = 18.Chapter 3 Trigonometry in Engineering Therefore. Since the robot is in the elbow-up configuration. the inverse kinematic solution for the tip position P(12.12◦ . Elbow-Down Solution for 𝜽1 : The angle 𝜃1 for the elbow-down solution is shown in Fig.87◦ .12◦ and 𝜃2 = 36. x . P ⎯ ⎯8⎯0 √1 2⎯ = ⎯ ⎯⎯ 6 13° 2 ⎯+ ⎯2⎯ 143. Substituting 𝛼 = 18. 𝛼 = sin−1 (0.57◦ + 𝛼.87° P(12. 3.45◦ into equation (3. 1 √ α P1 5 √2⎯ O 5√ 2⎯ 78 Figure 3.17 as 6 tan(𝜃1 − 𝛼) = 12 ( ) 6 𝜃1 − 𝛼 = tan−1 12 𝜃1 − 𝛼 = 26. 6) P1 3° 3. 3. the inverse kinematic solution for the tip position P(12.57 + 18. If the tip of the robot is located at the point P(x.9487 𝛼 = cos−1 (0.43◦ into equation (3. (a) Suppose 𝜃1 = 23𝜋 rad. 5 2 = 5 2 + Therefore.19. Therefore.3. determine the values of 𝜃1 and 𝜃2 ..43◦ .43 = 45◦ . 6) when the elbow is down is given by 𝜃1 = 45◦ and 𝜃2 = −36. y) = (12. ( √ ) (√ ) 0 = 180 − 2 5 2 180 cos 𝛼 cos 𝛼 = √ 180 √ 2 × 180 × 5 2 cos 𝛼 = 0. 3.20) can be obtained using either the law of sines or cosines for the triangle OP1 P shown in Fig. 5 √2⎯ P1 3° 5√ 2⎯ 3. 3.3. y).87◦ . (b) Suppose now that the same robot is located in the first quadrant and oriented in the elbow-up position. and determine the x and y coordinates of point P(x. . Substituting 𝛼 = 18. l1 = 10 in. Using the law of cosines gives ( √ ) (√ ) ( √ )2 ( √ )2 (√ )2 180 − 2 5 2 180 cos 𝛼.18 Elbow-down configuration to find angle 𝛼. 𝜃2 = 56𝜋 rad. 3.19. 16). and l2 = 12 in.1 α 14 2 ⎯6 ⎯+⎯ ⎯ ⎯2⎯ √1 2⎯ P 0⎯ 1⎯8⎯ =√ O Figure 3.9487) 𝛼 = 18. Sketch the orientation of the robot in the x-y plane. with positive orientations of 𝜃1 and 𝜃2 as shown in Fig.20) yields 𝜃1 = 26. as shown in Fig.18.3 Two-Link Planar Robot 79 The angle 𝛼 needed to find 𝜃1 in equation (3.3 Further Examples of Two-Link Planar Robot Example 3-14 Consider a two-link planar robot. The x and y coordinates of the tip position are given by x = l1 cos 𝜃1 + l2 cos(𝜃1 + 𝜃2 ) = 10 cos(120◦ ) + 12 cos(270◦ ) ( ) 1 + 12 (0) = 10 − 2 = −5 in. y) l2 l1 θ2 P1 θ1 x O Figure 3. in.. l1 = 10 in. 3.34 in.20 Orientation of the two-link planar robot for 𝜃1 = 120◦ and 𝜃2 = 150◦ . in.20. (3. −3. y) = (−5 in. P1 θ 2 = 150° 10 12 θ 1 = 120° O x. is shown in Fig. P(x. −3. y) = (−5.21) Therefore. and l2 = 12 in. 𝜃2 = 56𝜋 rad = 150◦ . y.34) Figure 3.34 in.).80 Chapter 3 Trigonometry in Engineering y P(x.19 Two-link planar robot for example 3-14. Solution (a) The orientation of the two-link robot for 𝜃1 = 23𝜋 rad = 120◦ . P(x. . y = l1 sin 𝜃1 + l2 sin(𝜃1 + 𝜃2 ) = 10 sin(120◦ ) + 12 sin(270◦ ) (√ ) 3 = 10 + 12 (−1) 2 = −3. 22) Since the robot is in the elbow-up configuration. the angle 𝜃1 + 𝛼 can be determined using the right-angled triangle OAP as ( ) 16 −1 16 ⇒ 𝜃1 = tan − 𝛼.23) yields 𝜃1 = 53.65) = 49.13◦ in equation (3.19. .13◦ . 12 sin(180◦ − 49. 3.13◦ − 𝛼.13 − 27. The angle 𝛼 can be found from the triangle OP1 P using either the law of cosines or the law of sines.3. 𝜃1 = 53. The unknown angle 180◦ − 𝜃2 and the three sides of the triangle OP1 P are shown in Fig. the angle 𝜃2 is positive and is given by 𝜃2 = cos−1 (0. y) = (12. (3.4560) 𝛼 = 27.21.23) 20 2⎯ = ⎯ 6 ⎯1⎯ θ2 2⎯⎯+ − 0 1⎯2⎯ √ 18 P1 α 10 θ1 O Figure 3. tan(𝜃1 + 𝛼) = 12 12 Therefore. Using the law of cosines gives 202 = 102 + 122 − 2(10)(12) cos(180◦ − 𝜃2 ) 400 = 244 + 240 cos 𝜃2 156 = 240 cos 𝜃2 ⇒ cos 𝜃2 = 0. 3. 16) 16 A 12 The triangle OP1 P to find the angles 𝜃1 and 𝜃2 . 3. 3.21. from Fig. (3.46◦ .46◦ ) 20 = 0.4560 𝛼 = sin−1 (0.13 = 26. sin 𝛼 = Substituting 𝛼 = 27.65.3 Two-Link Planar Robot 81 (b) For the two-link robot located in the first-quadrant as shown in Fig. Also. the angle 𝜃2 can be found using the law of cosines on the triangle OP1 P shown in Fig. 12 20 Therefore.21 12 P(x.21. Using the law of sines gives sin(180◦ − 𝜃2 ) sin 𝛼 = .0◦ . y) = (−4 cm.82 Chapter 3 Trigonometry in Engineering Example 3-15 Consider a two-link planar robot with positive orientations of 𝜃1 and 𝜃2 as shown in Fig. Solution (a) The orientation of the two-link robot for 𝜃1 = 120◦ . Suppose 𝜃1 = 120◦ . (b) Determine the x and y coordinates of point P(x. l1 = 8 cm. (c) Determine the distance from point P to the origin. l1 = 8 cm. . y. 10. 𝜃2 = −30◦ . y) 30° 4 P1 8 120° O x.93 cm. and l2 = 4 cm is shown in Fig. 𝜃2 = −30◦ . cm Figure 3. and l2 = 4 cm. cm P(x. (b) The x. y = l1 sin 𝜃1 + l2 sin(𝜃1 + 𝜃2 ) = 8 sin(120◦ ) + 4 sin(90◦ ) (√ ) 3 + 4 (1) =8 2 = 10.93 cm). P(x. 3. y).22. (a) Sketch the orientation of the robot in the x-y plane.19.22 Orientation of the two-link planar robot for 𝜃1 = 120◦ and 𝜃2 = −30◦ .and y-coordinates of the tip position are given by x = l1 cos 𝜃1 + l2 cos(𝜃1 + 𝜃2 ) = 8 cos(120◦ ) + 4 cos(90◦ ) ( ) 1 + 4 (0) =8 − 2 = −4 cm. Therefore. 3. 24. 3.. determine the values of 𝜃1 and 𝜃2 .64 cm. 3.07 10 θ1 −17. y) = (−17.23 Solution (a) The orientation of the two-link robot for 𝜃1 = 135◦ .3.07 O x.).3 Two-Link Planar Robot 83 (c) The distance from the tip P(x. The x and y coordinates of the tip position are given by x = l1 cos 𝜃1 + l2 cos(𝜃1 + 𝜃2 ) = 10 cos 135◦ + 10 cos 180◦ ( √ ) 2 = 10 − + 10 (−1) 2 = −17.07 in.07 in. and l1 = l2 = 10 in. the distance from the tip of the robot to the origin is 11.. 7.93)2 = 11. y) 10 θ2 P1 7. Two-link planar robot in the elbow-up position with P(x. y). Therefore. 𝜃2 = 45◦ and l1 = l2 = 10 in. (a) Suppose 𝜃1 = 135◦ . 𝜃2 = 45◦ . (3. P(x. Figure 3. (b) Suppose now that the tip of the same robot is located in the second quadrant and oriented in the elbow-up position as shown in Fig. is shown in Fig. y.07 in. and determine the x and y coordinates of point P(x.64 cm. in. Sketch the orientation of the robot in the x-y plane. y) to the origin is given by √ d = x2 + y2 √ = (−4)2 + (10.19. y) = (−17. 3.24) . y = l1 sin 𝜃1 + l2 sin(𝜃1 + 𝜃2 ) = 10 sin 135◦ + 10 sin 180◦ (√ ) 2 = 10 + 10(0) 2 = 7. 7.07 in.). in.07 in.23.07 in. Example 3-16 Consider a two-link planar robot with positve orientations of 𝜃1 and 𝜃2 as shown in Fig. If the tip of the robot is located at the point P(x. in.).5◦ .47 A 7 −17.0◦ . in.9239 = cos 𝛼. −17. 3. y) P1 10 10 135° O x.07 θ1 O x. Using the law of cosines gives 102 = (18. and 𝜃2 = 45◦ . P(x. The unknown angle 180◦ − 𝜃2 and the three sides of the triangle OP1 P are shown in Fig.48) cos 𝛼 −341. y. Since the robot is in the elbow-up configuration.4 = 200 cos 𝜃2 ⇒ cos 𝜃2 = 0.48)2 + 102 − 2(10)(18.4 = −369.25. 7. The angle 𝛼 can be found from the triangle OP1 P using either the law of cosines or the law of sines.25.⎯0⎯7⎯2 α =1 8. Using the law of cosines gives (18. 𝜃1 = 157. . 45° P(x.5◦ − 𝛼. 3. Figure 3. in..707. y) = (−17.54 cos 𝛼 0. the angle 𝜃1 + 𝛼 can be determined using the right-angled triangle OAP as 𝜃1 + 𝛼 = atan2(7. P(x. (3.4 = 200 + 200 cos 𝜃2 141.07 in.25 The triangle OP1 P to find the angles 𝜃1 and 𝜃2 .24 The orientation of the two-link planar robot with 𝜃1 = 135◦ . from Fig.⎯0⎯7 2 ⎯⎯+ 10 2 ⎯⎯7⎯ . (b) The angle 𝜃2 can be found using the law of cosines on the triangle OP1 P in Fig. 3.48)2 = 102 + 102 − 2(10)(10) cos(180◦ − 𝜃2 ) 341. angle 𝜃2 is positive and is given by 𝜃2 = cos−1 (0.07. in. y) 10 θ2 P1 180 √1⎯ −θ 7⎯.84 Chapter 3 Trigonometry in Engineering Therefore. Also.707) = 45.25.07 7. Therefore.07 in.25) y. Figure 3.07) 𝜃1 + 𝛼 = 157. 07 Two-link planar robot in the elbow-down position with P(x. y) 10 θ 2 P1 O θ1 x. 3. (b) Suppose now that the tip of same robot is located in the third quadrant and oriented in the elbow-down position (clockwise direction).. −7. −7. and l1 = l2 = 10 in.5◦ = 135. (a) Suppose 𝜃1 = −135◦ . as shown in Fig.07 in.).25) yields 𝜃1 = 157.5◦ in equation (3. . 3.0◦ .. y) = (−17. Sketch the orientation of the robot in the x-y plane.and y-coordinates of point P(x.).5 − 22. 3. 𝛼 = 22. Figure 3. Therefore. 𝜃2 = −45◦ . −7.and y-coordinates of the tip position are given by x = l1 cos 𝜃1 + l2 cos(𝜃1 + 𝜃2 ) = 10 cos(−135◦ ) + 10 cos(−180◦ ) ( √ ) 2 + 10 (−1) = 10 − 2 = −17.). and determine the x.07 in.3. and l1 = l2 = 10 in.26 Solution (a) The orientation of the two-link robot for 𝜃1 = −135◦ . y. Example 3-17 Consider a two-link planar robot with positive orientations of 𝜃1 and 𝜃2 as shown in Fig. The x.07 in.5◦ . y) = (−17. y = l1 sin 𝜃1 + l2 sin(𝜃1 + 𝜃2 ) = 10 sin(−135◦ ) + 10 sin(−180◦ ) ( √ ) 2 + 10(0) = 10 − 2 = −7. If the tip of the robot is located at point P(x. P(x.07 in.3 Two-Link Planar Robot 85 Therefore.07 in. y) = (−17. in. −17.07 in. determine the values of 𝜃1 and 𝜃2 .26.07 in.07 10 P(x. is shown in Fig.. y). Substituting 𝛼 = 22.27.07 in. 𝜃2 = −45◦ .19. −7. in. 17.07) 𝜃1 + 𝛼 = −157. (3.4 α O θ1 x.707) = 45. in.0⎯7⎯) ⎯⎯1⎯7 √(⎯− P(x.4 = 200 cos 𝜃2 ⇒ cos 𝜃2 = 0. .48)2 = 102 + 102 − 2(10)(10) cos(180◦ − 𝜃2 ) 341.28 The triangle OP1 P used to find the angles 𝜃1 and 𝜃2 . 10 P(x.07 Figure 3.5◦ − 𝛼. y) 10 θ2 P1 77 18.86 Chapter 3 Trigonometry in Engineering y.0◦ or −45◦ . from Fig. in. (b) The angle 𝜃2 can be found using the law of cosines on the triangle OP1 P shown in Fig.5◦ . Also.707.4 = 200 + 200 cos 𝜃2 141.28. 3. 𝜃2 = cos−1 (0. y) −45° P1 Figure 3. −17. −θ 2 180 ⎯⎯( 2⎯⎯+ ⎯. in. 𝜃2 = −45◦ .27 The orientation of the two-link planar robot with 𝜃1 = −135◦ and 𝜃2 = −45◦ . Since the angle 𝜃2 is in the clockwise direction. the angle 𝜃1 + 𝛼 can be determined using the right-angled triangle OAP as 𝜃1 + 𝛼 = atan2(−7. 3.07. 3. in. 𝜃1 = −157.28.07 A 2⎯ = ⎯0⎯7⎯) ⎯−⎯⎯7. Therefore. Therefore. O −135° 10 x. 10 −7. The unknown angle 180◦ − 𝜃2 and the three sides of the triangle OP1 P are shown in Fig. Using the law of cosines gives (18.26) y.28. 477)2 + 102 − 2(10)(18.. Since angle 𝛼 is in the clockwise direction.and y-coordinates of the tip position are given by x = l1 cos 𝜃1 + l2 cos(𝜃1 + 𝜃2 ) = 10 cos(−45◦ ) + 10 cos(0◦ ) (√ ) 2 = 10 + 10(1) 2 = 17. Example 3-18 Consider a two-link planar robot with positive orientations of 𝜃1 and 𝜃2 as shown in Fig. y. 𝜃2 = 45◦ . and determine the x. y) Two-link planar robot in the elbow-up position with P(x. .3 Two-Link Planar Robot 87 The angle 𝛼 can be found from the triangle OP1 P using either the law of cosines or the law of sines.07 in. and l1 = l2 = 10 in.29 Solution (a) The orientation of the two-link robot for 𝜃1 = −45◦ .477) cos 𝛼 −341.07 in.3.07 O θ1 x.) Figure 3. 3.5◦ ) = −135.9239 = cos 𝛼. The x.19. Using the law of cosines yields 102 = (18.5◦ − (−22. Sketch the orientation of the robot in the x-y plane. −7. determine the values of 𝜃1 and 𝜃2 . 3. y) = (17. in. Substituting 𝛼 = −22. 17. y).4 = −369.).29.07 in. and l1 = l2 = 10 in.and y-coordinates of point P(x.5◦ . 10 −7.54 cos 𝛼 0. 𝜃2 = 45◦ . as shown in Fig..07 in. −7.30. 3. Therefore. (a) Suppose 𝜃1 = −45◦ . y) = (17. 𝛼 = −22.07 P1 θ2 10 P(x. (b) Suppose now that the tip of same robot is located in the fourth quadrant and oriented in the elbow-up position (counterclockwise direction). If the tip of the robot is located at the point P(x. is shown in Fig. in. 𝜃1 = −135◦ .0◦ .5◦ in equation (3.26) gives 𝜃1 = −157.07 in. y. P(x.707. in. O −7.707) = 45◦ .0⎯7⎯)⎯2 ⎯+ ⎯⎯(⎯ α − ⎯⎯7⎯.5◦ . Also.07 17.07 in. The unknown angle 180◦ − 𝜃2 and the three sides of the triangle OP1 P are shown in Fig. 0⎯7⎯)⎯2 10 ⎯= 18. 3. y) Figure 3.48)2 = 102 + 102 − 2(10)(10) cos(180◦ − 𝜃2 ) 341.31.30 The orientation of the two-link planar robot with 𝜃1 = −45◦ and 𝜃2 = 45◦ . in. Since the robot is in the elbow-up configuration.07. Therefore.07 θ1 √ A (⎯1⎯7⎯ .07 in.. Using the law of cosines gives (18. x O x. 17.88 Chapter 3 Trigonometry in Engineering y = l1 sin 𝜃1 + l2 sin(𝜃1 + 𝜃2 ) = 10 sin(−45◦ ) + 10 sin(0◦ ) ( √ ) 2 + 10(0) = 10 − 2 = −7.4 = 200 cos 𝜃2 ⇒ cos 𝜃2 = 0. . y. angle 𝜃2 = cos−1 (0. in.31. 3.07 in. 3. −7. y) Figure 3.07) 𝜃1 − 𝛼 = −22.31. from Fig.). x. (b) The angle 𝜃2 can be found using the law of cosines on the triangle OP1 P shown in Fig.4 180 − θ 2 8 P1 θ2 10 P(x.4 = 200 + 200 cos 𝜃2 141.31 The triangle OP1 P to find the angles 𝜃1 and 𝜃2 . −45° 10 10 y P1 45° P(x. in. y) = (17. the angle 𝜃1 − 𝛼 can be determined using the right-angled triangle OAP as 𝜃1 − 𝛼 = atan2(−7. y) of the tip is given by x = 1. 𝜃1 = −45◦ . the position P(x.5 + (−22.9239 = cos 𝛼. It can be seen from this figure that the tip is located in the third quadrant and the reference angle is 15◦ .48)2 + 102 − 2(10)(18.32.9659 = −1.32 1.5 m is moving in the x-y plane.27) The angle 𝛼 can be found from the triangle OP1 P using either the law of cosines or the law of sines.4 Further Examples of Trigonometry in Engineering 89 Therefore. Therefore. Using the law of cosines yields 102 = (18.5 cos(15◦ ) = −1. locate the tip P(x.4 = −369. y) of the robot in the x-y plane. The tip of the one-link robot for 𝜃 = −165◦ is shown in Fig.5 sin(15◦ ) = −1. Since angle 𝛼 is in the clockwise direction.4 Example 3-19 Solution FURTHER EXAMPLES OF TRIGONOMETRY IN ENGINEERING A one-link planar robot of length l = 1. 3.5 m −165° One-link planar robot for Example 3-19.5 × 0.5)◦ = −45. y) Figure 3.449 m y = 1.2588 = −0.5 cos 𝛼 0.0◦ . Substituting 𝛼 = −22. 𝛼 = −22.5 × 0. If the joint angle 𝜃 = −165◦ . .3.5 cos(165◦ ) = −1. Since both the sin and cos functions are negative in the third quadrant.5 sin(165◦ ) = −1.388 m y x P(x.5◦ in equation (3.27) gives 𝜃1 = −22.48) cos 𝛼 −341.5◦ + 𝛼. 3. 𝜃1 = −22. (3.5◦ . the angle 𝜃 is given by ( ) 5 ◦ 𝜃 = 180 − atan 10 = 180◦ − atan(0. the civil engineer calculated the elevation of the building cornerstone located between two benchmarks shown in Fig.90 Chapter 3 Trigonometry in Engineering Example 3-20 The x. −10) = 2.5) = 180◦ − 26. as shown in Fig.6779 rad) ( 180◦ 𝜋 rad ) = 153. 3. find the length l of the link and the angle 𝜃.4◦ . 3. Also. Using the atan2(y. respectively. y P(x.4◦ . Solution The tip of the one-link robot with x = −10 cm and y = 5 cm is shown in Fig. Since the tip of the robot is in the second quadrant. x) function in MATLAB.57◦ = 153.33. The length l is given by √ l= (−10)2 + (5)2 = √ √ 100 + 25 = 125 = 11.34. y) 5 cm l θ −10.and y-components of the tip of a one-link planar robot are given as −10 cm and 5 cm.19. the angle 𝜃 is given by 𝜃 = atan2(5.0 cm x O Figure 3.18 cm. 1. Example 3-21 In Example 1-8.6779 rad = (2.33 One-link planar robot for example 3-20. . Locate the tip of the robot in the x-y plane. The same engineer is now required to calculate the angle of inclination and horizontal distance between the two benchmarks. (c) Check the results of part (b) using the Pythagorean theorem.34 Elevation between the two benchmarks.768◦ .2 m. The distance L between the two benchmarks B1 and B2 is 1001. Also. (a) Find the angle of inclination 𝜃 of the grade. the angle of inclination 𝜃 = sin−1 (0. 3. .4 B1 x.4 Further Examples of Trigonometry in Engineering 91 y.8 − 428. respectively.8 B2 L Δy θ 428.8 m. Δx (b) Calculate the horizontal distance between the two benchmarks.0483) = 2.2 = 0. Therefore.84%. The percent grade can now be calculated as Percent grade = Δy × 100 Δx = 100 × tan 𝜃 = 100 × tan(2. m Δx Figure 3. Δy × 100. calculate the percent grade.34 as sin 𝜃 = Δy L = E2 − E1 L = 476.768◦ ) = 4.3. and their elevations are 428.4 m and 476.4 1001.0483. m 476. Solution (a) The angle of inclination 𝜃 can be determined from the right-angle triangle shown in Fig. x 5 θ2 .. 𝜃2 = 45◦ . in.36 Ankle-up position. in. y).. l1 = 20 in.35 x.).28) (c) The horizontal distance can also be calculated using the Pythagorean theorem as √ Δx = L2 − Δy2 √ = (1001. 2.2) cos(2. 3.36.2)2 − (48. l1 l1 P(x.0 m. If the end of the toes are located at P(x. y) = (19.468◦ ) = (1001.92 Chapter 3 Trigonometry in Engineering (b) The horizontal distance between the two benchmarks can be calculated as Δx = L cos𝜃 = (1001. as shown in Fig.5 in. Determine the x.35. 3.5 in. and l2 = 5 in.and y-coordinates of the position of the toes P(x. y P(19.4)2 = 1000 m.2) (0.99883) = 1000. 3-22 y. (b) Now suppose that the same leg is positioned such that the tip of the toes is located in the first quadrant and oriented in the ankle-up position (counterclockwise direction) as shown Fig. determine the values of 𝜃1 and 𝜃2 . Example Consider the position of the toes of a person sitting in a chair. (a) Suppose 𝜃1 = −30◦ .5) θ1 O 20 Figure 3. 2. θ1 Toes position of a person sitting on a chair. y) θ2 l2 Figure 3. (3.5. 71 in.6 2 ⎯⎯(2 ⎯⎯.1◦ . from Fig.1925) = 78. Using the law of sines yields sin(𝜃1 + 𝛼) sin(180◦ − 𝜃2 ) = .3. (180◦ − 𝜃2 ) = cos−1 (0. (b) The angle 𝜃2 can be found using the law of cosines on the triangle OP1 P shown in Fig.9◦ .66)2 = 202 + 52 − 2(20)(5) cos(180◦ − 𝜃2 ) 386. 5 19.5) A α 5 θ1 θ2 20 180 − θ 2 Figure 3.37.15 in.4 Further Examples of Trigonometry in Engineering Solution 93 (a) The x. Therefore.37 x P1 The triangle OP1 P to find the angles 𝜃1 and 𝜃2 . y) = (22. 3. The unknown angle 180◦ − 𝜃2 and the three sides of the triangle OP1 P are shown in Fig.1925 y O 6 ⎯⎯)2⎯ = 19. 2. P(x. the angle 𝜃1 + 𝛼 can be determined using from the triangle OP1 P using either the law of cosines or the law of sines.9659) 2 = 22. Since the leg is in the ankle-up configuration. Using the law of cosines gives (19. y = l1 sin 𝜃1 + l2 sin(𝜃1 + 𝜃2 ) = 20 sin(−30◦ ) + 5 sin(−30◦ + 45◦ ) ( ) 1 = 20 − + 5(0.37.).. 3. −18.and y-coordinates of the toes position can be calculated as x = l1 cos 𝜃1 + l2 cos(𝜃1 + 𝜃2 ) = 20 cos(−30◦ ) + 5 cos(−30◦ + 45◦ ) (√ ) 3 = 20 + 5(0. Also.66 .37. Therefore.15 in.5 = 425 − 200 cos(180◦ − 𝜃2 ) −38.5 = −200 cos(180◦ − 𝜃2 ) ⇒ cos(180◦ − 𝜃2 ) = 0. 𝜃2 = 101.2588) 2 = −18. 3.5.5⎯)⎯⎯+ √(1 P(19.5 ⎯⎯9⎯.71 in. 3.45 − 𝛼 = 14. The length of the foot segment (from ankle to toe) is 7.. determine the values of 𝜃1 and 𝜃2 .94 Chapter 3 Trigonometry in Engineering Therefore.5.45◦ . sin(𝜃1 + 𝛼) = (3.38. (a) Given the angles shown in Fig. find the position of the knee if the runner’s toes touch the ground at the point x = y = 0. Example 3-23 In a motion capture study of a runner.1 in. 3. 5 sin 78. one frame shows the subject supporting her weight on one leg.4° Ankle 37. 3.45◦ − 7.66 𝜃1 + 𝛼 = 14.38. y) = (−4 in. as shown in Fig.14◦ . 19.29) gives 𝜃1 = 14.38 Knee x Position of the runner’s leg during motion capture study. Substituting the value of 𝛼 in equation (3.5) = 7.31◦ or 𝜃1 = 7.2° Toes Figure 3. 24 in.).9◦ 19.39.9 in.29) The angle 𝛼 can be found from the right-triangle OAP shown in Fig. y 82. and the length of the lower leg (from ankle to knee) is 17. 3. If the end of the toes are located at P(x. .31◦ . (b) Now suppose that the same leg is positioned such that the knee is located in the second quadrant and oriented in the knee-down position (clockwise direction) as shown Fig.37 as 𝛼 = atan2(2. 9 in. in. 24) Knee θ2 Ankle θ1 x.3. Angle 𝜃1 and 𝜃2 to find the position of the knee. (−4.1 θ 2 = −97.7046) = 5. y) 17. .7096) = 16.2° θ 1 = 142. P(x. y = l1 sin 𝜃1 + l2 sin(𝜃1 + 𝜃2 ) = 7.76 in.1(0.1(0.6◦ ) = 7. Knee P(x.6046) + 17.39 Solution Position of the runner’s leg to find 𝜃1 and 𝜃2 .6◦ ) = 7.. in.8◦ ) + 17. in.40.and y-coordinates of the knee position are calculated as x = l1 cos 𝜃1 + l2 cos(𝜃1 + 𝜃2 ) = 7.9 sin(142. y) = (5.40 x.76 in.9 (−0. the x.9 in.9 (0.8◦ ) + 17.4 Further Examples of Trigonometry in Engineering 95 y. 3. Therefore.4° Ankle 7. Toes Figure 3.9 cos(142.7965) + 17.8° Toes Figure 3. 16.8◦ − 97.8◦ − 97.1 cos(142.1 sin(142. y.6° 82.9 37. (a) Using the angles 𝜃1 and 𝜃2 shown in Fig.). in. The angle 𝛽 can be found from the right-triangle TKP shown in Fig.8◦ .96 Chapter 3 Trigonometry in Engineering (b) The angle 𝜃2 can be found using the law of cosines on the triangle TAK shown in Fig.41.33 which gives 𝛼 = 33◦ . since the robot is in the knee-down configuration.33)2 = 7. Angle 𝜃1 can now be found by adding angles 𝛼 and 𝛽 as shown in Fig. 3. 24) y. 17. 3. 24 17.41.41 The triangle TAK to find the angles 𝜃2 . Also.92 + 17. Therefore.33 180 − θ 2 A 7. the angle 𝛼 can be determined from the triangle TAK using either the law of cosines or the law of sines. Also from Fig. .28◦ .9 θ1 α P −4 β x.41 as 𝜃1 = 𝛼 + 𝛽 = 51.41.82◦ . The unknown angle 180◦ − 𝜃2 and the three sides of the triangle TAK are shown in Fig.33 Therefore. sin(𝛼) = 17.41 as 𝛽 = atan2(24. Using the law of cosines gives (24.6181 ⇒ 𝜃2 = 51.9)(17. T Figure 3.18 cos(180◦ − 𝜃2 ) 167 = 270. 𝜃1 = 151. in. 3.46◦ = 151.1 θ2 24. 3.2◦ 24.1 24.82◦ + 99. 3. −4) ⇒ 𝛽 = 99.1 sin 129. in.46◦ . the angle 𝜃2 = −51. Using the law of sines yields sin(𝛼) sin(180◦ − 𝜃2 ) = .12 − 2(7.82◦ .1) cos(180◦ − 𝜃2 ) 592 = 425 − 270.18 cos(𝜃2 ) ⇒ cos(𝜃2 ) = 0. K(−4. coop student calculates the area of lot 1 shown in Fig. . P3. A laser range finder records the distance from the laser to the base and from the laser to the top of a building as shown in Fig. If the height of the basketball hoop is 10 ft from the floor. from the floor. Is this the correct answer? If not. Figure P3. as shown in Fig.6 Laser beam for problem P3-6.73 3-3.73 m. The same coop student calculates the area of lot 2 shown in Fig. find the correct answer. player are 82 in. 3-2. The origin of the beam is 5 m from the circle as shown in Fig. The eyes of a 7 ft 4 in. Repeat problem P3-2 if the player’s height is 72 in. 156 in. a city hires a coop student to determine the area of different lots in a new subdivision.1 Using a range finder to find the height of a building.5 as 50. P3. 10 ft 82 in. P3. 3-4. 30° 200 m 0m Lot 1 h 10 θ 300 m Figure P3.4 Dimension of lot 1 in the new subdivision. Find the angle 𝜃 and the height of the building. find the correct answer.000 m2 .1. To calculate the property tax. P3.5 Dimension of lot 2 in the new θ subdivision. Is this the correct answer? If not.Problems 97 PROBLEMS 3-1. P3.640 m2 .6.2 A basketball player in front of the basketball hoop. A laser beam is directed through a small hole in the center of a circle of radius 1. 75 m Range finder Figure P3.2. find the distance l and angle 𝜃 from the player’s eye to the hoop. 3-5. 3-6. 120° 200 m Lot 2 300 m l Figure P3. 1. The Beam θ 5m Laser Figure P3.4 as 94. If the control tower is 15 m tall. For the given l and 𝜃. 100 cm Figure P3. 80 θ cm 3-11. P(x. cm θ cm 80 80 80 100 cm 45° x Figure P3.7. A one-link planar robot moves in the x-y plane as shown in Fig.10 A one-link planar robot for problem 100 cm cm m P3-10. P3. y Figure P3. 5 1.98 Chapter 3 Trigonometry in Engineering y What should be the angle 𝜃 of the beam for the beam to go through the hole? Use the law of sines. 3-9.12.10. determine the height h of the rocket from the ground when it is located at a distance d = 575 m from the top of the control tower. find the position P(x. P3. find the position P(x.11 A one-link planar robot for problem P3-11. Also. .5 3-7.12 A one-link planar robot for problem P3-12. y) 1. determine the angle 𝜃. y) of the tip.7 Truss structure for problem P3-7. x −45° m 3-10. P(x. A one-link planar robot moves in the x-y plane as shown in Fig. Repeat problem P3-8 if l = 300 m and d = 500 m. Rocket d Control tower h θ l 15 m Launch Pad 3-12.8. P(x. y) of the tip. For the given l and 𝜃. y 3-8. For the given l and 𝜃. A truss structure consists of three isosceles triangles as shown in Fig. y) 1. Determine the angle 𝜃 using the laws of cosines or sines. P3.8 A rocket taking off from a launch pad for problem P3-8. A rocket takes off from a launch pad located l = 500 m from the control tower as shown in Fig. 5 m 135° x Figure P3. find the position P(x. A one-link planar robot moves in the x-y plane as shown in Fig. y) Figure P3. P3. P3.11. y) of the tip. Repeat problem P3-16 if (a) P(x. y x −135° 1. P3. 𝜃2 = −45◦ . y). and (a) 𝜃 = 150◦ 2𝜋 (b) 𝜃 = − rad 3 (c) 𝜃 = 420◦ 9𝜋 (d) 𝜃 = − rad 4 O P1 x Figure P3. y) = (4. 4) in. A one-link planar robot moves in the x-y plane as shown in Fig. y P(x. find the position P(x. P3. y) y l2 x θ2 l1 θ1 Figure P3. −3) cm (d) P(x. l1 = l2 = (d) 𝜃1 = 4 2 5 cm (e) 𝜃1 = −30◦ .14 A one-link planar robot for problem P3-14. y) = (3. Consider the one-link planar robot shown in Fig.14. Sketch the orientation of the robot and determine the (x. l1 = l2 = (c) 𝜃1 = 4 2 5 cm 3𝜋 𝜋 rad. Consider again the one-link planar robot shown in Fig. (b) P(x. y) Figure P3. y) = (−5. 3-18.5 m P(x.13 A one-link planar robot for problem P3-13. y) y l θ O x P(x. P3. y) coordinates of position P for 𝜋 (a) 𝜃 = rad 4 4𝜋 rad (b) 𝜃 = 4 (c) 𝜃 = −135◦ 𝜋 (d) 𝜃 = − rad 4 3-15. 𝜃2 = rad. sketch the position of the tip of the robot and determine the (x.14. y) = (−2. y) = (6. 3-14. If l = 5 cm. Determine the length l and angle 𝜃 if the tip of the robot is located at the following points P(x. −7. y 99 3-16.5) in. 𝜃2 = 45◦ . Consider the two-link planar robot shown in Fig. l1 = l2 = 5 cm (b) 𝜃1 = 30◦ .18 A two-link planar robot for problem P3-18. 𝜃2 = 45◦ . 4) cm (b) P(x. y) = (−3.Problems 3-13. y) of the tip. −4) cm 3-17. 𝜃2 = − rad. P3. −6) in. (c) P(x. 2) in. l1 = l2 = 5 cm . Repeat problem P3-14 if l = 10 in. For the given l and 𝜃. (d) P(x. y) = (−4. 3) cm (c) P(x. y) coordinates of point P for (a) 𝜃1 = 30◦ . l1 = l2 = 5 cm 3𝜋 𝜋 rad. (a) P(x. y) = (5.18.13. 10 θ2 O x. 9). y. y. in. 3-21. 𝜃2 = − rad. If the tip of the robot is located at the point P(x.18 is located in the first quadrant and is oriented in the elbowup position. Assume l1 = 6 in. P3.73 Figure P3.21.23 A two-link planar robot for problem P3-23. −1). −8. determine the values of 𝜃1 and 𝜃2 . l1 = l2 = 5 cm 3𝜋 𝜋 rad.22. determine the values of 𝜃1 and 𝜃2 . P3. 5). Consider a two-link planar robot with l1 = l2 = 10 cm and oriented in the elbow-up position as shown in Fig.83. Consider the two-link planar robot with l1 = l2 = 5 in. θ1 P(7.100 Chapter 3 Trigonometry in Engineering (f) 𝜃1 = −30◦ . determine the values of 𝜃1 and 𝜃2 .22 A two-link planar robot for problem P3-22. . y) = (−17. as shown in Fig. cm θ1 10 P −16. determine the values of 𝜃1 and 𝜃2 . y) = (10.5. l1 = (h) 𝜃1 = − 4 2 l2 = 5 cm 3-19. and oriented in the elbowup position. and l2 = 8 in.83. and l2 = 8 in. If the tip of the robot is located at the point P(x. 3-22. l1 = l2 = (g) 𝜃1 = − 4 2 5 cm 3𝜋 𝜋 rad.24.36). −2. in. Assume l1 = 6 in. 𝜃2 = −45◦ . If the tip of the robot is located at the point P(x. O x. in.8). −2.5. 3-20. P3. Consider the two-link planar robot with l1 = l2 = 5 in. cm −1 θ1 P 10 10 θ2 Figure P3. P3. cm −4.36) Figure P3. 𝜃2 = rad. y) = (4. O x. P3. −8.24 A two-link planar robot for problem P3-24. cm −17 O x.5 Figure P3. Consider a two-link planar robot with l1 = l2 = 10 cm and oriented in the elbow-down position as shown in Fig. in.8) 5 5 3-24. θ1 5 5 θ2 P(4. determine the values of 𝜃1 and 𝜃2 . and oriented in the elbow-down position as shown in Fig. If the tip of the robot is located at the point P(x. Suppose that the two-link planar robot shown in Fig. Suppose that the two-link planar robot shown in Fig. y) = (9.18 is located in the first quadrant and is oriented in the elbowdown position. If the tip of the robot is located at the point P(x. 3-23.23.21 A two-link planar robot for problem P3-21. If the tip of the robot is located θ2 y. y. P3. y) = (7. 25 A two-link planar robot for problem P3-25. 15). and l2 = 8 in. The river flows due east at a speed of 4 mph. P3.29. Assume l1 = 10 in. P3.28. −3.66 cm.27. Find the magnitude of the velocity V and the angle 𝜃 of the plane relative to the ground using the laws of sines and cosines. 3-28. y 13.5. 3-27. The wind is blowing at 30◦ southwest at a speed of 50 mph.66 cm P(x. Find the magnitude of the velocity V and the angle 𝜃 of the barge using the laws of sines and cosines.66 cm). If the tip of the robot is located at the point P(x. determine the values of 𝜃1 and 𝜃2 .25. x Figure P3. 10 θ1 O x −1 −13 Figure P3. P3.73). Assume l1 = 10 in. y) = (−1. y) = (−4. A large barge is crossing a river at a heading of 30◦ northwest with a speed of 12 mph against the water as shown in Fig.27 A two-link planar robot for problem P3-27. and l2 = 8 in.Problems values of 𝜃1 and 𝜃2 using the laws of cosines and sines.66 cm 12 θ2 8 x θ1 −3. −16. Consider a two-link planar robot oriented in the elbow-up position as shown in Fig.26. Consider a two-link planar robot oriented in the elbow-down position as shown in Fig. y) = (13. If the tip of the robot is located at the point P(x. 12). . y) = (−13. at the point P(x. P3. Consider a two-link planar robot oriented in the elbow-up position as shown in Fig. If the tip of the robot is located at the point P(x. determine the values of 𝜃1 and 𝜃2 . y 30° y 50 E W S 50 P N 0 15 V θ2 8 θ 60° 10 Figure P3. y) 10 cm 10 cm θ2 y P 101 Figure P3. 3-25.28 Velocity of an airplane for problem θ1 −1 x O P3-28. P3.26 A two-link planar robot for problem P3-26. determine the values of 𝜃1 and 𝜃2 . determine the 3-29. 3-26. An airplane travels at a heading of 60◦ northwest with an air speed of 500 mph as shown in Fig. 33 Impedance triangle to find the capacitive reactance. where VR is R = 100 Ω θ XC = 30 Ω the voltage across the resistor. VL is the Z XC = 30 Ω voltage across the inductor. The impedance triangle of a resistor (R) and an inductor (L) connected in series in an AC circuit is shown in Fig. The impedance triangle of a resistor (R) and a capacitor (C) connected in series in an AC circuit is shown in Fig.005 KΩ is the total impedance of the circuit. P3. P3.30 A series AC circuit containing R and L. Find the capacitive reactance XC and the phase angle 𝜃. P3. 3-30. where R = 100 Ω is the resistance of the resistor and XC = 30 Ω is the capacitive reactance of the capacitor.29 A barge crossing a river against the water current.33. Find the impedance Z and the phase angle 𝜃. P3. where R = 1000 Ω is the resistance of the resistor and Z = 1.118 KΩ is the XL = 30 Ω total impedance of the circuit.30. phase angle 𝜃. N E θ x S V = Vbw + V Figure P3. Find the impedance Z and the phase angle 𝜃. Find the total voltage V and the Figure P3.102 Chapter 3 Trigonometry in Engineering y Vw = 4 Vb w =1 2 W V 30° 3-32.34. The impedance triangle of a resistor (R) and a capacitor (C) connected in series in an AC circuit is shown in Fig. R = 100 Ω .32. Find the inductive reactance XL and the phase angle 𝜃.31 A series AC circuit containing R and C.31. Figure P3. Z θ R = 100 Ω XL = 30 Ω Ω 1005 Z= θ R = 1000 Ω XL Z θ R XL Figure P3.32 Impedance triangle to find the inductive reactance. and V is the AC voltage applied to the RL circuit in volts. The impedance triangle of a resistor (R) and an inductor (L) connected in series in an AC circuit is shown in Fig. where R = 100 Ω is the resistance of the resistor and XL = 30 Ω is the inductive reactance of the inductor. 3-34. P3. The phasor diagram of a series RL circuit is shown in Fig. 3-31. 3-33. where R = 1000 Ω is the resistance of R = 100 Ω the resistor and Z = 1. R = 1000 Ω θ Z= 1118 Ω XC Z θ R XC Figure P3. R VC = 5 V V C Figure P3. 3-38. respectively. P3.0 3-35. 103 3-37.. m Δx θ Δy + b a Vab Figure P3. B2 L Δy θ 500.39.e. (b) Calculate the horizontal distance between the benchmarks. P3.35. VR = 20 V θ V (a) Find the angle of inclination 𝜃 of the grade. P3. The distance L between the benchmarks B1 and B2 is 500 m. Find the total voltage V and the phase angle 𝜃. where VR is the voltage across the resistor.36 Three-phase AC system. Find the voltage between phase a and b (i. 3-39. and their elevations are 500 m and 600 m.0 x. Repeat problem P3-37 if the distance L between the two benchmarks is 200 m. The phasor diagram of a series RC circuit is shown in Fig.37 Elevation between the two benchmarks. find Vab ).0 B1 n 100 V + − 120° − L 100 V x. Also calculate the percent grade.35 Phasor diagram of an RC circuit. The distance L between the benchmarks B1 and B2 is 200 m. Consider the elevation between the two benchmarks shown in Fig.34 Phasor diagram of an RL circuit. and V is the AC voltage applied to the RL circuit in volts. y. A three-phase AC system has a trigonal planar configuration as shown in Fig. respectively. (c) Check the results in part (b) using the Pythagorean theorem. B2 . c y.39 Elevation between the two benchmarks for problem P3-39. m 600. P3.36. 400. m + − 500. 3-36.0 Figure P3. Consider the elevation between the two benchmarks shown in Fig. The voltage of each phase is 100 V and the angle between the adjacent phase is 120◦ . VC is the voltage across the capacitor.Problems R V VL = 5 V L θ VR = 10 V Figure P3.37. and their elevations are 500 m and 400 m. m Δx B1 Figure P3. 42 Planar configuration of boron trifluoride. as indicated in the Fig. (b) What is the angle the patellar forms with the horizontal.43? FQuadriceps θ FResultant FPatellar Figure P3. P3.104 Chapter 3 Trigonometry in Engineering (a) Find the angle of inclination 𝜃 of the grade. To find the height of a building.41 Survey set up to find the height of a building. Knee O x. Repeat problem P3-39 if the distance L between the two benchmarks is 100 m. −19.3. a surveyor measures the angle of the building from two different points A and B as shown in Fig. Repeat P3-43 if the force exerted by the quadriceps is 30 N and the force by the patellar is 40 N. 𝜃2 = 30◦ .45 Ankle-up position of the leg. 3-43.42. Find the distance between adjacent fluoride molecules.3 Angstrom.41. 60° 10 m B Figure P3. (a) Suppose 𝜃1 = −45◦ . (a) Find the magnitude of the resultant force. P3. Find the height h of the building. (b) Calculate the horizontal distance between the benchmarks. Also calculate the percent grade.. The gas boron trifluoride (BF3 ) has a trigonal planar configuration as shown in Fig. y. P3. and l2 = 5 in. 3-45. h 40° A 3-44. l1 = 20 in.43. (c) Check the results in part (b) using the Pythagorean theorem. in.82) Toes 5 120° B F F Figure P3. P3. 3-42. The force exerted by the quadriceps is 38 N and the force by the patellar is 52 N. Adjacent fluoride molecules form a 120◦ angle. 3-41. in. θ1 20 F P(14. The B-F bond length is 1.43 Forces exerted on the knee. Consider the position of the toes of a person sitting in a chair as shown in Fig. 3-40.45. The distance between the two points is 10 m. P3. The force exerted on the kneecap by the quadriceps and patellar tendons are shown in Fig. Ankle θ2 Figure P3. Determine the . If the end of the toes are located at P(x. (a) Given the angles shown in Fig. −19. y) = (14. x. 105 Toes (b) To find 𝜃1 and 𝜃2 . Figure P3. . P3. (b) Now suppose that the leg is positioned such that the tip of the toes is located in the fourth quadrant and oriented in the ankle-up position (counterclockwise direction) as shown Fig.25. y) = (−6. one frame shows the subject supporting her weight on one leg.33 in. y. determine the values of 𝜃1 and 𝜃2 .25 in.46.46(a). and the length of the lower leg (from ankle to knee) is 18 in.). If the toes are located at P(x. y).). as shown in Fig. as shown in Fig. The length of the foot segment (from ankle to toe) is 8 in. in.Problems x and y coordinates of the position of the toes P(x. find the position of the knee if the runner’s toes touch the ground at the point x = y = 0... In a motion capture study of a runner. 3-46.82 in. P3. P3. 25) Knee y Knee θ2 Ankle 60° Ankle 30° Toes θ1 x (a) During a motion capture study. (b) Now suppose that the same leg is positioned such that the knee is located in the second quadrant and oriented in the knee-down position (clockwise direction). 25 in.46. (−6. in.46 Position of the runner’s leg.45. P3. determine the values of 𝜃1 and 𝜃2 . The concepts of unit vectors. and momentum have not only a magnitude but also a direction associated with them. acceleration. an automobile traveling north at 65 mph can be represented by a two-dimensional vector in polar coordinates with a magnitude (speed) of 65 mph and a direction along the positive y-axis. 4. velocity. a vector OP be drawn as shown in Fig. both the distance and direction are needed. 4. and direction of a vector will be introduced. A vector is a convenient way to represent both magnitude and direction and can be described in either a Cartesian or a polar coordinate system (rectangular or polar forms).1. The tip of the one-link and two-links planar robots introduced in Chapter 3 will be represented in this chapter using vectors both in Cartesian and polar coordinates. For example. magnitude. To describe the displacement of an object from its initial point.1 A representation of a vector. It can also be represented by a vector in Cartesian coordinates with an x-component of zero and a y-component of 65 mph. forces. The quantities such as displacement (position). The magnitude of the vector is the distance between points O and P (magnitude = P) and the direction is given by the direction of the y P e= P ud it agn M θ O Figure 4. electric and magnetic fields.1 INTRODUCTION ⃖⃖⃖⃖⃖⃗ or simply P⃗ with the initial point O and the final point P can Graphically.CHAPTER 4 Two-Dimensional Vectors in Engineering The applications of two-dimensional vectors in engineering are introduced in this chapter. 106 x . Vectors play a very important role in engineering. of the vector P⃗ are given by Px = P cos 𝜃 Py = P sin 𝜃. 4. 4. Note that the magnitude of the unit vectors is equal to 1. can be obtained from the Cartesian components Px and Py as P= √ Px2 + Py2 𝜃 = atan2(Py . .2) can be written in rectangular form as P⃗ = Px î + Py j. the vectors are also written as a boldface P.2 One-link planar robot as a position vector in Cartesian coordinates.3 Position Vector in Polar Form 107 arrow or the angle 𝜃 in the counterclockwise direction from the positive x-axis as shown in Fig. Px ).2 POSITION VECTOR IN RECTANGULAR FORM The position of the tip of a one-link robot represented as a 2-D vector P⃗ (Fig. Px and Py . y P Py P ˆj θ Px ˆi x Figure 4. 4. 4. ⃗ and where P is the magnitude and 𝜃 is the angle or direction of the position vector P. The arrow above P indicates that P is a vector. The x.2.4.and y-components. 4.̂ where î is the unit vector in the x-direction and ĵ is the unit vector in the y-direction as shown in Fig. In many engineering books.1.3 POSITION VECTOR IN POLAR FORM The position of the tip of a one-link robot represented as a 2-D vector P⃗ (Fig. 4.2) can be also be written in polar form as P⃗ = P∠𝜃. 2 Therefore.0 ĵ m. Find the x.2 is given as P = 0.5 sin 30◦ ( ) 1 = 0. Find the magnitude (length) and direction of the 4 ⃗ robot represented as a position vector P. 4.0 m 2 √ √ Py = 2 sin 135◦ = 2 sin 45◦ ( ) √ 1 = 2 √ = 1. the position of the tip of the one-link robot can be written in vector form as P⃗ = 0. The x.and y-components Px and Py and write P⃗ in vector notation. . Example 4-2 Solution √ The length of a one-link robot shown in Fig.5 = 0. 2 Therefore. Solution The x.25 ĵ m.and y-components Px and Py are given by √ √ Px = 2 cos 135◦ = − 2 cos 45◦ ( ) √ 1 =− 2 √ = −1. Example 4-3 √ 3 The x. Find the x.433 î + 0.0 m.5 = 0.and y-components Px and Py are given by Px = 0. as shown in Fig.25 m.433 m 2 Py = 0.and y-components of the one-link robot are given as Px = m and 4 1 Py = m. 4.5 cos 30◦ (√ ) 3 = 0.3.0 î + 1. the position of the tip of the one-link robot can be written in vector form as P⃗ = −1.5 m and the direction is 𝜃 = 30◦ .2 is given as P = 2 m and the direction is 𝜃 = 135◦ .108 Chapter 4 Two-Dimensional Vectors in Engineering Example 4-1 The length (magnitude) of a one-link robot shown in Fig. 4.and y-components Px and Py and write P⃗ in rectangular vector notation. 5 m and the direction 𝜃 is given by ( 𝜃 = atan2 1 .5∠30◦ m. Solution The length (magnitude) of the one-link robot is given by √ P = Px2 + Py2 √ √( √ ) 2 ( )2 √ √ 3 1 =√ + 4 4 = 0. The position of the tip can also be written in Cartesian coordinates as √ 3 ̂ 1 ̂ i + j m. the position of the one-link robot P⃗ can be written in polar form as P⃗ = 0. 4 √ ) 3 4 = 30◦ .4.3 Position Vector in Polar Form 109 y P P 1 4 θ x 3 √⎯ 4 Figure 4. Therefore. P⃗ = 4 4 Example 4-4 A person pushes down on a vacuum cleaner with a force of F = 20 lb at an angle of −40◦ relative to ground. as shown in Fig. . Determine the horizontal and vertical components of the force.3 One-link planar robot for Example 4-3. 4.4. 1) can be carried out by adding the x. the addition of two vectors can be obtained by placing the initial point of P⃗2 on the final point of P⃗1 and then drawing a line from the initial point of P⃗1 to the final point of P⃗2 . 4.5. Solution The x. Graphically. P2 P1 P O Figure 4.32 lb Fy = F sin (−40◦ ) = −20 sin 40◦ = −12.110 Chapter 4 Two-Dimensional Vectors in Engineering Fx F = 40 20 lb y Fy x Figure 4. Algebraically. F⃗ = 15.86 ĵ lb.32 î − 12.4 A person pushing a vacuum cleaner. (4. Vectors P⃗1 and P⃗2 can be written in Cartesian form as P⃗1 = Px1 î + Py1 ĵ (4. the addition of two vectors given in equation (4.86 lb. 4.5 Graphical addition of two vectors. forming a triangle as shown in Fig.1) Vectors can be added graphically or algebraically. Therefore.and y-components of the force are given by Fx = F cos (−40◦ ) = 20 cos 40◦ = 15.4 VECTOR ADDITION The sum of two vectors P⃗1 and P⃗2 is a vector P⃗ written as P⃗ = P⃗1 + P⃗2 .and y-components of the two vectors.3) .̂ (4.2) P⃗2 = Px2 î + Py2 j. 4 Vector Addition 111 Substituting equations (4.1 Examples of Vector Addition in Engineering Example 4-5 A two-link planar robot is shown in Fig.4.6 Position of two-link robot using vector addition.5 m.5 m.4.3) into equation (4. 4. Solution The x.and y-components. 𝜃1 = 45◦ .1) gives P⃗ = (Px1 î + Py1 ĵ ) + (Px2 î + Py2 ĵ ) = (Px1 + Px2 ) î + (Py1 + Py2 ) ĵ = Px î + Py ĵ where Px = Px1 + Px2 and Py = Py1 + Py2 .6. . Therefore. In other words. the length 2 of the second link P2 = 0. and 𝜃2 = −15◦ . addition of vectors algebraically amounts to adding their x.and y-components of the first link of the planar robot P⃗1 can be written as Px1 = P1 cos 45◦ ( )( ) 1 1 = √ √ 2 2 = 0. write P⃗ in polar coordinates. 4. Find the magnitude and angle of the 1 position of the tip of the robot if the length of the first link P1 = √ m.2) and (4. y P2 15° P2 P Py2 P1 P1 P Py1 θ 45° x O Px1 Px2 Figure 4.5 m Py1 = P1 sin 45◦ )( ) ( 1 1 = √ √ 2 2 = 0. P⃗1 = 0. 0. The voltage phasors (vector used to represent voltages and currents in AC circuits) across the resistor R = 20 Ω and inductor L = 100 mH are given as V⃗ R = 2 ∠0◦ V and V⃗ L = 3.25 ĵ ) = 0.197∠38.433 m Py2 = P2 sin 30◦ ( ) 1 = 0. VL − .̂ The magnitude and direction of the vector P⃗ are given by √ P = (0. find the magnitude and phase (angle) of V.7 Sum of voltage phasors in an RL circuit. P⃗ = 1.25 m.77 ∠90◦ V. Therefore.433 î + 0.433 î + 0.7.933) = 38.75 j. If the total voltage phasor V⃗ across R and L is V⃗ = V⃗ R + ⃗ V⃗ L .933)2 + (0. P⃗ = (0. P⃗2 = 0.79◦ .5 î + 0.197 m.5 ĵ m. i = 100 sin(120 πt) mA + R = 20 Ω V + VR − + L = 100 mH − Figure 4.5 2 = 0. 4.25 ĵ m.75)2 = 1.112 Chapter 4 Two-Dimensional Vectors in Engineering Therefore. Finally.75. respectively. Similarly.and y-components of the second link P⃗2 can be written as Px2 = P2 cos 30◦ (√ ) 3 = 0. 𝜃 = atan2(0. Example 4-6 A sinusoidal current is flowing through the RL circuit shown in Fig. since P⃗ = P⃗1 + P⃗2 . the x.79◦ m.933 î + 0. Therefore.5 ĵ ) + (0.5 î + 0.5 2 = 0. Similarly.0) = 62. N W E y S P2 = 300 = P1 45° 20 0 P θ 0 Figure 4.77.77 V. V⃗ = 4.77 ĵ V. then 300 miles due east as shown in Fig.and y-components of the voltage phasor V⃗ R are given by VRx = 2 cos 0◦ = 2. Therefore. the magnitude and phase of the total voltage phasor V⃗ are given by √ V = (2.27 V 𝜃 = atan2(3.77 ĵ V. since V⃗ = V⃗ R + V⃗ L . 2. 4. V⃗ = (2. Finally.05◦ V.0 î + 0 ĵ ) + (0 î + 3. Find the resulting position of the ship.05◦ .77 cos 90◦ VLy =0 V = 3. x .0 î + 3.0 V VRy = 2 sin 0◦ = 0 V. Therefore. the x.27∠62.and y-components of the voltage phasor V⃗ L are given by VLx = 3. Therefore.4 Vector Addition Solution 113 The x.8 Resulting position of the ship after travel.77 sin 90◦ = 3.8.0)2 + (3. V⃗ L = 0 î + 3.77 ĵ ) = 2. Thus. V⃗ R = 2.0 î + 0 ĵ V.77)2 = 4.4. Example 4-7 A ship travels 200 miles at 45◦ northeast. Similarly.9.4 mi.4 î + 141.and y-components of the position vector P⃗1 are given by Px1 = P1 cos 45◦ ( ) 1 = 200 √ 2 = 141. Example 4-8 Relative Velocity: An airplane is flying at an air speed of 100 mph at a heading of 30◦ southeast. P⃗1 = 141.5 miles.and y-components of the position vector P⃗2 are given by Px2 = P2 cos 0◦ = 300 (1) = 300 mi Py2 = P2 sin 0◦ = 300 (0) = 0 mi. . the ship is now located at 463.114 Chapter 4 Two-Dimensional Vectors in Engineering Solution The x.5∠17.4 ĵ mi.4 ĵ mi.4)2 = 463. Therefore. In other words. determine the resultant velocity of the plane with respect to the ground.4 î + 141.4 mi Py1 = P1 sin 45◦ ) ( 1 = 200 √ 2 = 141.4) = 17. If the velocity of the wind is 20 mph due west.4 ĵ ) + (300 î + 0 ĵ ) = 441. P⃗2 = 300 î + 0 ĵ mi. Thus. 4. since P⃗ = P⃗1 + P⃗2 .4 î + 141.5 mi 𝜃 = atan2(141.4.76◦ northeast from its original location. Finally.76◦ miles. P⃗ = (141. the x.4)2 + (141. 441. the distance and direction of the ship after traveling 200 miles northeast and then 300 miles east are given by √ P = (441. P⃗ = 463.76◦ . as shown in Fig. Therefore. Therefore. 17. 4. Similarly. V⃗ ag = −20 î + 0 ĵ mph. are given by Vxpa = Vpa cos (−30◦ ) (√ ) 3 = 100 2 = 86.6 î − 50 ĵ mph.and y-components of the velocity of air (wind) relative to ground V⃗ ag are given by Vxag = Vag cos (180◦ ) = 20 (−1) = −20 mph Vyag = Vag sin (180◦ ) = 20 (0) = 0 mph.0 ĵ mph. Therefore. .6 mph Vypa = Vpa sin (−30◦ ) ( ) 1 = −100 2 = −50.4 Vector Addition 115 N W E y S x 30° θ Vpg V pa = 100 Vag = 20 Figure 4. Solution The x. Finally.0 mph. V⃗ pa .6 î − 50.6 î − 50 ĵ ) + (−20 î + 0 ĵ ) = 66. Therefore.9 Velocity of the plane relative to ground. the velocity of the plane relative to ground V⃗ pg = V⃗ pa + V⃗ ag is given by V⃗ pg = (86. V⃗ pa = 86. the x.and y-components of the velocity of the plane relative to air. Therefore.116 Chapter 4 Two-Dimensional Vectors in Engineering Thus.10 The triangle to determine the speed and direction of the plane. . The direction of the Vpg velocity of the airplane relative to ground can now be found as 𝜃 = 30 + 𝛼 = 36. the speed and direction of the airplane relative to ground are given by √ Vpg = (66.10. 4. it becomes unwieldy when adding three or more vector quantities. the speed of the airplane relative to ground can be determined using the law of cosines as 2 = 202 + 1002 − 2 (20) (100) cos(30◦ ) Vpg = 6936 y x 30° α Vpg 100 30° 20 Figure 4.9◦ . In such cases. Vpg 20 20 sin 30◦ = 0. therefore.6) = −36. Also.28 mph. Therefore. the algebraic approach is preferable.3 mph 𝜃 = atan2(−50.9◦ mph.6)2 + (−50)2 = 83. 66. Vpg = 83. Using the triangle shown in Fig.89◦ . using the law of sines.3∠−36.12 and the value of 𝛼 = 6.9◦ mph. V⃗ pg = 83. given as Therefore.896◦ . Note that while this geometric approach works fine when adding two vectors.3∠−36. Note: The velocity of the airplane relative to ground can also be found using the laws of cosines and sines discussed in Chapter 3. sin 𝛼 = V⃗ pg = 83. The velocity of the plane relative to ground is. the angle 𝛼 can be found as sin 𝛼 sin 30◦ = . 4. Determine the tension in each cable.and y-components of the tension T Tx1 = −T1 cos 45◦ ( ) 1 = −T1 √ N 2 Ty1 = T1 sin 45◦ ( ) 1 = T1 √ N. 45° 45° 100 kg g = 9. 4. ∑ F⃗ = 0 or ⃗1 + T ⃗2 + W ⃗ = 0.11. i.11 An object hanging from two cables. 4.11 can be drawn as shown in Fig. T1 T2 T2 sin 45° T1 sin 45° 45° T1 cos 45° 45° T2 cos 45° y 100 kg x W = mg = (100)(9.4) . 4. Solution The free-body diagram (FBD) of the system shown in Fig. and therefore the sum of all the forces are equal to zero (Newton’s first law). 2 (4.12.12 Free-body diagram of the system shown in Fig. It is assumed that the system shown in Fig. 4.4.81) = 981 N Figure 4.e.81 m/sec2 Figure 4. T ⃗ 1 are given by The x.11 is in static equilibrium (not accelerating)..11.4 Vector Addition Example 4-9 117 Static Equilibrium: A 100 kg object is hanging from two cables of equal length as shown in Fig. Therefore. from 2 equation (4. T T ⃗ = 0 î + −981 ĵ N. Similarly.7 N. that is. . both cables have the same tension. Substituting T⃗ 1 .5) is zero.and y-components of the ten2 2 ⃗ are given by sion T⃗ 2 and weight W Tx2 = T2 cos 45◦ ( ) 1 = T2 √ N 2 Ty2 = T2 sin 45◦ ) ( 1 N = T2 √ 2 Wx = W cos (−90◦ ) =0 N Wy = W sin (−90◦ ) = −981 N. T1 = T2 = 693.7) 2 2 2T Adding equations (4. the sum of forces in the x.5) −√ + √ î + √ + √ − 981 ĵ = 0. Therefore.6) and (4. ( ) ∑ T2 T1 Fx = − √ + √ =0 (4.and y-directions are zero. ( ) ∑ T2 T1 Fy = √ + √ − 981 = 0. (4.6). the x-component is the sum of the forces in the x-direction and the y-component is the sum of forces in the y-direction. or ∑ ∑ Fx = 0 and Fy = 0.7 N.5). Since the right-hand side of equation (4.6) 2 2 and. 2 2 2 2 In equation (4. T2 = T1 . T⃗ 1 = − √1 î + √1 ĵ N. T⃗ 2 = √2 î + √2 ĵ N and W 2 2 ⃗ into equation (4. Also.7) gives √ 2 = 981 or T2 = 693.118 Chapter 4 Two-Dimensional Vectors in Engineering T T Therefore. Therefore.4) gives and W ( ) ( ) T1 T1 T2 T2 − √ î + √ ĵ + √ î + √ ĵ + (0î − 981ĵ ) = 0 2 2 2 2 ) ( ) ( T2 T2 T1 T1 (4. T⃗ 2 . the x. 4.4 Vector Addition Example 4-10 119 Static Equilibrium: A 100 kg television is loaded onto a truck using a ramp at a 30◦ angle. Find the normal and frictional forces on the TV if it is left sitting on the ramp. TV g 30° Figure 4.13 Loading a TV onto the truck using a ramp. Solution The free-body diagram (FBD) of the TV sitting on the ramp as shown in Fig. 4.13 is given in Fig. 4.14 where W = 100 × 9.81 = 981 Newton. y x N F 30° W = (9.81)(100) = 981 N Figure 4.14 Free-body diagram of a TV on the 30◦ ramp. Note that we are using the rotated axes to simplify the computation below. It is assumed that the system shown in Fig. 4.13 is in static equilibrium; therefore, the sum of all the forces is equal to zero (Newton’s first law): ∑ F⃗ = 0 ⃗ = 0. F⃗ + N⃗ + W ⃗ are given by The x- and y-components of the TV weight W Wx = −W sin 30◦ ( ) 1 = −981 2 = −490.5 N Wy = −W cos 30◦ (√ ) 3 = −981 2 = −849.6 N. (4.8) 120 Chapter 4 Two-Dimensional Vectors in Engineering ⃗ = −490.5 î − 849.6 ĵ N. Similarly, the x- and y-components of the Therefore, W frictional force F⃗ and normal force N⃗ are given by Fx = F cos 0◦ =F N Fy = F sin 0◦ =0 N Nx = N cos (90◦ ) =0 N Ny = N sin (90◦ ) = N N. ⃗ N, ⃗ and W ⃗ into Therefore, F⃗ = F î + 0 ĵ N and N⃗ = 0 î + N ĵ N. Substituting F, equation (4.8) gives (F î + 0 ĵ ) + (0 î + N ĵ ) + (−490.5 î − 849.6 ĵ ) = 0 (F + 0 − 490.5) î + (0 + N − 849.6) ĵ = 0. (4.9) Equating the x- and y-components in equation (4.9) to zero yields Example 4-11 F − 490.5 = 0 ⇒ F = 490.5 N N − 849.6 = 0 ⇒ N = 849.6 N. A waiter extends his arm to hand a plate of food to his customer. The free-body diagram is shown in Fig. 4.15, where Fm = 400 N is the force in the deltoid muscle, Wa = 40 N is the weight of the arm, Wp = 20 N is the weight of the plate of food, and Rx and Ry are the x- and y-components of the reaction forces at the shoulder. (a ) Using the x-y coordinate system shown in Fig. 4.15, write the muscle force F⃗ m , ⃗ p in standard vector ⃗ a , and the weight of the plate W the weight of the arm W notation (i.e., using unit vectors i and j). (b) Determine the values of Rx and Ry required for static equilibrium: R⃗ + F⃗ m + ⃗a+W ⃗ p = 0. Also compute the magnitude and direction of R. ⃗ W Deltoid muscle Plate of food Shoulder joint Fm y Rx x (a) Figure 4.15 Waiter handing a plate to a customer. 30° Ry Wa (b) Wp 4.4 Vector Addition Solution 121 (a) The x- and y-components of the deltoid muscle force F⃗ m are given by Fm,x = Fm cos 150◦ = −400 cos 30◦ = −400 × 0.866 = −346.4 N Fm,y = Fm sin 150◦ = 400 sin 30◦ = 400 × 0.5 = 200 N. Therefore, F⃗ m = −346.4 î + 200 ĵ N. Similarly, the x- and y-components of ⃗ a and weight of the plate Wp are given by the weight of the arm W Wa,x = Wa cos (−90)◦ =0 N Wa,y = Wa sin (−90)◦ = −40 N Wp,x = Wp cos (−90)◦ =0 N Wp,y = Wp sin (−90)◦ = −20 N. ⃗ p = 0 î − 20 ĵ N. ⃗ a = 0 î − 40 ĵ N and W Therefore, W (b) It is assumed that the system shown in Fig. 4.15 is in static equilibrium; therefore, the sum of all the forces is equal to zero (Newton’s first law): ⃗a+W ⃗ p + R⃗ = 0. F⃗ m + W (−346.4 î + 200 ĵ ) + (0 î − 40 ĵ ) + (0 î − 20ĵ ) + (Rx î + Ry ĵ ) = 0 (Rx − 346.4 + 0 + 0) î + (Ry + 200 − 40 − 20) ĵ = 0 (4.10) Equating the x- and y-components in equation (4.10) to zero yields ∑ Fx = 0 : ⇒ Rx − 346.4 = 0 ⇒ Rx = 346.4 N ∑ Fy = 0 : ⇒ Ry + 200 − 40 − 20 = 0 ⇒ Ry = −140 N. Therefore, R⃗ = 346.4 ⃗i − 140 ⃗j N. The magnitude and direction of R⃗ can be obtained as √ R⃗ = (346.4)2 + (−140)2 ∠atan2(−140, 346.4) = 373.6 ∠−22◦ N (4.11) 122 Chapter 4 Two-Dimensional Vectors in Engineering Example 4-12 Using motion capture, the positions of each arm segment are measured while a person throws a ball. The length from shoulder to elbow (P1 ) is 12 in. and the length from elbow to hand (P2 ) is 18 in. The angle 𝜃1 is 45◦ and 𝜃2 is 20◦ . y P2 P Ball θ2 P 1 Elbow Shoulder Figure 4.16 θ1 x Position of the arm throwing a ball. (a) Using the x-y coordinate system shown in Fig. 4.16, write the position of the ball P⃗ = P⃗1 + P⃗2 in the standard vector notation. ⃗ (b) Find the magnitude and direction of P. Solution (a) The x- and y-components of the position P⃗1 are given by P1x = P1 cos 45◦ √ 2 = 12 × 2 = 8.49 in. P1y = P1 sin 45◦ √ 2 = 12 × 2 = 8.49 in. Therefore, P̂ 1 = 8.484 î + 8.484 ĵ in. Similarly, the x- and y-components of the position P⃗2 are given by P2x = P2 cos (45◦ + 20◦ ) = 18 × (0.4226) = 7.61 in. P2y = P2 sin (45◦ + 20◦ ) = 18 × (0.9063) = 16.31 in. Problems 123 Therefore, P⃗2 = 7.61 î + 16.31 ĵ in. The position of the arm P⃗ can now be found in standard vector notation by adding vectors P̂ 1 and P̂ 2 as P⃗ = P⃗1 + P⃗2 = 8.49 î + 8.49 ĵ + 7.61 î + 16.31 ĵ = 16.1 î + 25.80 ĵ in. (4.12) (b) The magnitude of the position P⃗ is given by √ P = Px2 + Py2 √ = 16.12 + 25.82 = 29.6 in. The direction of position P⃗ is given by 𝜃 = atan2(Py , Px ) = atan2(25.8, 16.1) = 58◦ . Therefore, the vector P⃗ can be written in the polar form as P⃗ = 29.6 ∠58◦ in. PROBLEMS 4-1. Locate the tip of a one-link robot of 12 in. length as a 2-D position vector with a direction of 60◦ . Draw the position vector and find its x- and ycomponents. Also, write P⃗ in both its rectangular and polar forms. 4-2. Locate the tip of a one-link robot of 1.5 ft length as a 2-D position vector with a direction of −30◦ . Draw the position vector and find its x- and y-components. Also, write P⃗ in both its rectangular and polar forms. 4-3. Locate the tip of a one-link robot of 1 m length as a 2-D position vector with a direction of 120◦ . Draw the position vector and find its x- and y-components. Also, write P⃗ in both its rectangular and polar forms. 4-4. Locate the tip of a one-link robot of 2 m length as a 2-D position vector with a direction of −135◦ . Draw the position vector and find its x- and y-components. Also, write P⃗ in both its rectangular and polar forms. 4-5. The tip of a one-link robot is represented as a position vector P⃗ as shown in Fig. P4.5. Find the x- and y-components of the vector if the length of the link is P = 10 in. and 𝜃 = 30◦ . Also, write the vector P⃗ in rectangular and polar form. y P Py P θ Px x Figure P4.5 A one-link robot represented in polar coordinates. 124 Chapter 4 Two-Dimensional Vectors in Engineering 4-6. Repeat problem P4-5 if P = 14.42 cm and 𝜃 = 123.7◦ . 4-7. Repeat problem P4-5 if P = 15 cm and 𝜃 = −120◦ . y P Py P θ O 2 P 4-8. Repeat problem P4-5 if P = 6 in. and 𝜃 = −60◦ . 4-9. The x- and y-components of a vector P⃗ shown in Fig. P4.9 are given as Px = 2 cm and Py = 3 cm. Find the magnitude and direction, and write the vector P⃗ in its rectangular and polar forms. y, cm P θ x, cm −1.5 Figure P4.11 A position vector for problem P4-11. 4-12. The x- and y-components of a vector P⃗ shown in Fig. P4.12 are given as Px = −1 in. and Py = −2 in. Find the magnitude and direction, and write the vector P⃗ in its rectangular and polar forms. x Px Figure P4.9 A position vector for problem P4-9. y, in. −1 x, in. θ 4-10. The x- and y-components of a vector P⃗ shown in Fig. P4.10 are given as Px = 3 in. and Py = −4 in. Find the magnitude and direction, and write the vector P⃗ in its rectangular and polar forms. P −2 P Figure P4.12 A position vector for problem P4-12. y x θ 3 P 4 P 4-13. A state trooper investigating an accident pushes a wheel (shown in Fig. 4.13) to measure skid marks. If a trooper applies a force of 20 lb at an angle of 𝜃 = 45◦ , find the horizontal and vertical forces acting on the wheel. Figure P4.10 A position vector for problem P4-10. 4-11. The x- and y-components of a vector P⃗ shown in Fig. P4.11 are given as Px = −1.5 cm and Py = 2 cm. Find the magnitude and direction, and write the vector P⃗ in its rectangular and polar forms. θ Figure P4.13 A wheel to measure skid marks. Problems 4-14. Repeat problem P4-13 if the trooper is applying a force of 10 lb at an angle of 𝜃 = 60◦ . 4-15. In a RL circuit, the voltage across the inductor VL leads the voltage across the resistor by 90◦ as shown in Fig. P4.15. If VR =10 V and VL = 5 V, find the total voltage V⃗ = V⃗ R + V⃗ L . + + V VL = 5 V θ V R VL L − − VR = 10 V VR − + 125 Find the sum of the two voltages; in other words, find V⃗ = V⃗ 1 + V⃗ 2 . 4-20. In an electrical circuit, voltage V⃗ 2 leads voltage V⃗ 1 by 60◦ as shown in Fig. P4.20. Find the sum of the two voltages; in other words, find V⃗ = V⃗ 1 + V⃗ 2 . V2 10 V 60° V1 20 V Figure P4.15 Voltage phasor diagram of RL circuit. ⃗ 1 and V⃗ 2 for problem Figure P4.20 Voltages V P4-20. 4-16. Repeat problem P4-15 if VR =10 V and VL = 15 V. 4-17. In an RC circuit, the voltage across the capacitor VC lags the voltage across the resistor by −90◦ as shown in Fig. P4.17. If VR = 20 V and VC = 5 V, find the total voltage V⃗ = V⃗ R + V⃗ C . + VR = 20 V θ V VR R VC C − + VC = 5 V V − 4-21. An airplane travels at a heading of 45◦ northeast with an air speed of 300 mph. The wind is blowing at 30◦ southeast at a speed of 40 mph as shown in Fig. P4.21. Find the speed (magnitude of the ⃗ and the direction 𝜃 of the velocity V) plane relative to the ground using vector addition. Check your answer by finding the magnitude and direction using the laws of sines and cosines. y Figure P4.17 Voltage phasor diagram of RC circuit. 4-18. Repeat problem P4-17 if VR = 10 V and VL = 20 V. 4-19. In an electrical circuit, voltage V⃗ 2 lags voltage V⃗ 1 by 30◦ as shown in Fig. P4.19. V1 15 V 30° 10 V V2 ⃗ 1 and V⃗ 2 for problem Figure P4.19 Voltages V P4-19. 40 0 30 45° θ V N 30° W x E S Figure P4.21 Velocity of airplane for problem P4-21. 4-22. An airplane travels at a heading of −60◦ with an air speed of 500 mph. The wind is blowing at 30◦ at a speed of 50 mph as shown in Fig. P4.22. Find the speed ⃗ and the (magnitude of the velocity V) 126 Chapter 4 Two-Dimensional Vectors in Engineering direction 𝜃 of the plane relative to the ground using vector addition. Check your answer by finding the magnitude and direction using the laws of sines and cosines. y (c) Repeat part (b) using the laws of sines and cosines. 4-24. A ship is crossing a river at a heading of −150◦ with a speed of VSW = 30 mph against the water as shown in Fig. P4.24. The river is flowing in the direction of 135◦ with a speed of VW = 10 mph. x 60° y θ V = VSW + VW 500 V x VS −150° −135° 50 Figure P4.22 Velocity of airplane for problem P4-22. current. (a) Calculate the resultant velocity V⃗ of the ship using vector addition (use the î and ĵ notation). (b) Determine the magnitude and ⃗ direction of V. (c) Repeat part (b) using the laws of sines and cosines. y N Vw = 5 w θ Figure P4.24 A ship crossing a river against the 4-23. A large barge is crossing a river at a heading of 30◦ northwest with a speed of 15 mph against the water as shown in Fig. P4.23. The river flows due east at a speed of 5 mph. Vb VSW VW 30° =1 5 W V θ 30° S x 4-25. A boat moves diagonally across a river at a heading of 45◦ southwest and at a E water speed of Vbw = 12 mph as shown in Fig 4.25. The river flows due east at a speed of Vw = 4 mph. V = Vbw + Vw y Figure P4.23 A barge crossing a river against the current. (a) Calculate the resultant velocity V⃗ of the barge using vector addition (use the î and ĵ notation). (b) Determine the magnitude and ⃗ direction of V. Vbw Vw x 0 45° θ V N W E S Figure P4.25 A boat moving diagonally across a river. 29.e. In so doing. (b) Determine the magnitude and direction of the position of the robot tip.28 A two-link robot for problem P4-28.27. (a) Calculate the position of the tip P⃗ of the planar robot using vector addition (use the î and ĵ notation). express all vectors using standard vector notation (i..25 P1 P θ 8 60° x. A two-link planar robot is shown in Fig. write vector P⃗ in the polar form.14 P 4-27. 4-26. 8 θ O O 120° θ x Figure P4. 14. A two-link planar robot is shown in Fig. (a) Calculate the position of the tip P⃗ of the planar robot using vector addition (use the î and ĵ notation). P P2 150° 7. A two-link planar robot is shown in Fig. x. (a) Calculate the position of the tip P⃗ of the planar robot using vector addition (use the î and ĵ notation). P4. in. in. (c) Repeat part (b) using the laws of sines and cosines. P4. determine the magnitude V and direction 𝜃. In other words. which is given by V⃗ = V⃗ bw + V⃗ w .27 A two-link robot for problem P4-27. in.28. A two-link planar robot is shown in Fig.26. (b) Determine the magnitude and direction of the position of the robot tip. (c) Repeat part (b) using the laws of sines and cosines. P4. . (c) Repeat part (b) using the laws of sines and cosines. using unit vectors î ⃗ also and ĵ ). 4-28. write vector P⃗ in the polar form. y. In other words.Problems (a) Determine the resultant velocity V⃗ of the boat.21 P1 P −135° P4-26. y. −45° P2 Figure P4. O Figure P4. in. (b) Repeat part (a) using the laws of sines and cosines. 4-29. P4.26 A two-link robot located for problem 21. y P2 P 4 P1 P 127 (b) Determine the magnitude and ⃗ direction of P. Given your result for V. ⃗ ⃗ frictional force F. (c) Determine the values of F and N required for equilibrium (i. and the weight W in rectangular vector notation. cm O Figure P4.28 12 P1 P2 30° P Figure P4. (a) Determine the angle 𝜃.31 A weight suspended from two cables x. . T⃗ 1 + T⃗ 2 + ⃗ = 0).81 m/sec2 y.32 A vehicle parked on an inclined driveway for problem P4-32. W 4-31.32. P4. A 200 lb weight is suspended by two cables as shown in Fig.31.30 A weight suspended from two cables for problem P4-30. W W = 10 kN 4m F θ N g Figure P4.29 A two-link robot for problem P4-29.. T1 y 5f t t 10 f x 10 m T2 α 45° y α 45° x 200 lb 200 lb Figure P4. Assuming that the weight is ⃗ = 0). 4-30. Fig..e. as shown in Fig. P4. (b) Express T⃗ 1 and T⃗ 2 in rectangular vector notation and determine the values of T1 and T2 required for static equilibrium (i. cm for problem P4-31. ⃗ the (b) Express the normal force N.128 Chapter 4 Two-Dimensional Vectors in Engineering (a) Calculate the position of the tip P⃗ of the planar robot using vector addition (use the î and ĵ notation).30. (a) Determine the angle 𝛼. A weight of 100 kg is suspended from the ceiling by cables that make 30◦ and 60◦ angles with the ceiling as shown in 4-32.e. −135° θ P 28. find the not moving (T⃗ 1 + T⃗ 2 + W tensions T⃗ 1 and T⃗ 2 . P4. (b) Determine the magnitude and ⃗ direction of P. A vehicle weighing 10 kN is parked on an inclined driveway. F⃗ + N⃗ + ⃗ = 0). 30° 60° T1 T2 100 kg g = 9. (c) Repeat part (b) using the laws of sines and cosines. (b) Determine the values of F and N required for static equilibrium.35.34.e. P4. (c) Determine the values of F and N required for equilibrium (i. write the friction force F⃗ and the normal force N⃗ in rectangular vector notation (i.32 ft Figure P4. in terms of unit vectors î and ĵ ). A vehicle weighing 2000 lb is parked on an inclined driveway.e. .36. N⃗ + W W = 2000 lb y 129 4-35.66 ft x 17. (a) Determine the angle 𝜃. as shown in Fig. as shown in Fig. find the values of F and ⃗ = 0.e. F⃗ 2 .e.35.e. A two-bar truss supports a weight of W = 750 lb as shown in Fig. in other words. P4..34 A crate resting on a ramp for problem P4-34.34. in terms of unit vectors î and ĵ ). P4.33. F⃗ 2 .33 A vehicle parked on an inclined 15 ft driveway for problem P4-33. A crate of weight W = 100 lb sits on a ramp oriented at 27 degrees relative to ground. in terms of unit vectors î and ĵ ). as shown in Fig. (a) Using the x-y coordinate system shown in Fig. in other words. (c) Determine the values of F and N required for static equilibrium. such that F⃗ 1 + F⃗ 2 + W 4-37.. W 100 lb x 27° F N Figure P4.37. write ⃗ in the forces F⃗ 1 . as shown in Fig.7◦ . (a) Determine the angle 𝜃. P4.35 A crate resting on an inclined ramp 4-34. The free-body diagram showing the external forces is also shown in Fig. ⃗ ⃗ frictional force F. (a) Using the positive x-y coordinate system shown in Fig.35. and weight W rectangular vector notation (i. The free-body diagram showing the external forces is also shown in Fig. F N Figure P4. ⃗ the (b) Express the normal force N. N if F⃗ + N⃗ + W W x y 10 ft N F 500 N θ 8. P4. A two-bar truss is subjected to a vertical load P = 346 lb. (b) Determine the values of F1 and F2 ⃗ = 0.37.34. P4.Problems 4-33. P4. and P⃗ in rectangular vector notation (i. write the friction force F⃗ and the normal force N⃗ in rectangular vector notation (i.. P4. (b) Determine the values of F1 and F2 such that F⃗ 1 + F⃗ 2 + P⃗ = 0. The truss is constructed such that 𝜃 = 38. and the weight W in rectangular vector notation.4◦ .36. P4.. write the forces F⃗ 1 . in terms of unit vectors î and ĵ ).6◦ and 𝜃2 = 62.. N if F⃗ + N y θ for problem P4-35. F⃗ + ⃗ = 0). (b) Using the x-y coordinate system shown in Fig. P4. A 500 N television sits on an inclined ramp. P4. find the values of F and ⃗ +W ⃗ = 0. (a) Using the positive x-y coordinate system shown in Fig. 4-36. The truss is constructed such that 𝜃1 = 39. Rx and Ry are the 2 F y F2 FBD: (Joint 2) F 59. Express ⃗ F⃗ 1 .37 A two-bar truss subjected to a y vertical load. rectangular vector notation (i. The freebody diagram is√shown in Fig. and F⃗ 2 in terms of unit vecforces F. F⃗ 2 .40 where Fm = 250 2 N is the force in the deltoid muscle. 3 F1 F2 . 4-39. Wa = 35 N is the weight of the arm.39 A weight resting on a hinged shelf. (a) Using the positive x-y coordinate system shown in Fig. A force F = 100 N is applied to a twobar truss as shown in Fig.39. A block of weight W = 125 lb rests on a hinged shelf as shown in Fig.38 Force applied to joint 2 of a truss. The shelf is constructed such that 𝜃 = 41. A waiter extends his arm to hand a plate of food to his customer. P4. and weight W 4-40.3° x 1 Figure P4. ̂ tors i and ĵ and determine the values of F1 and F2 such that F⃗ 1 + F⃗ 2 + F⃗ = 0. P4. write ⃗ in the forces F⃗ 1 . Wp = 15 N is the weight of the plate of food.e. Figure P4.36 A weight supported by a two-bar truss. in terms of unit vectors î and ĵ ). such that F⃗ 1 + F⃗ 2 + W P P y θ1 θ2 F1 3 1 W F2 x W Figure P4. F1 θ θ x 4-38. P4.38.. P4.7◦ .130 Chapter 4 Two-Dimensional Vectors in Engineering F1 y θ x θ F2 W W Figure P4. (b) Determine the values of F1 and F2 ⃗ = 0.0° 56.39. (a) Using the x-y coordinate system shown in Fig.e. 4-43. Wp = 3 lb. The angle 𝜃1 is 60◦ and 𝜃2 is 65◦ .5 ft. x. and 𝜃2 is 60◦ . Wa = 7 lb. write the position of the ball P⃗ = P⃗1 + P⃗2 in standard vector notation. and 𝜃 = 45◦ . Also comR⃗ + F⃗ m + W pute the magnitude and direction ⃗ of R. 𝜃1 = 45◦ . Repeat problem P4-42 if P1 = 1 ft.and y-components of the reaction forces at the shoulder. Using motion capture. P4. the positions P⃗1 and P⃗2 of each arm segment are measured while a person throws a ball. P2 = 1. The length from shoulder to elbow (P1 ) is 10 in. the weight of ⃗ a .Problems 131 Deltoid muscle Plate of food Shoulder joint Fm y θ Rx x Ry Wa Wp Figure P4. and the weight of the the arm W ⃗ plate W p in the standard vector notation (i.42 Position of the arm throwing a ball. P4. (b) Determine the values of Rx and Ry required for static equilibrium: ⃗a+W ⃗ p = 0. 4-41. and the length from elbow to the hand holding the ball (P2 ) is 13 in.42..40. Ball P y P2 θ2 Elbow P1 Shoulder θ1 x Figure P4. (b) Determine both the magnitude and ⃗ direction of P. and 𝜃 = 30◦ 4-42.40 Waiter handing a plate to a customer. . write the deltoid muscle force F⃗ m . (a) Using the x-y coordinate system shown in Fig. Repeat problem P4-40 if Fm = 50 lb. using unit vectors i and j). the imaginary part of the complex number is measured by the vertical axis (on the Cartesian plane. the two parts of a complex number cannot be combined. and a good understanding of this topic is necessary. in engineering and physics they are in fact used to represent physical quantities such as impedance of RL.CHAPTER 5 5. The imaginary number −1 is represented by the letter i by mathematicians and by almost all the engineering disciplines except electrical engineering.1 Complex Numbers in Engineering INTRODUCTION Complex numbers play a significant role in all engineering disciplines. Although imaginary numbers are not commonly used in daily life. a + bj. On this plane. The square root of a negative real number is said to be imaginary because there is no real number √ that gives a negative number after it has been squared. Complex numbers are numbers that consist of two parts. RC. Note that the example of one-link planar robot is used only to show similarities between the twodimensional vector and the complex number. this is the y-axis) and the real number part goes on the horizontal axis (x-axis on the Cartesian plane). The position of the tip of the robot is generally not described by a complex number. it is especially important for the electrical engineer to master this topic. operations with complex numbers follow the same rules as those for real numbers. In general. The complex number is useful for representing two-dimensional variables where both dimensions are physically significant and are represented on a complex number plane (which looks very similar to the Cartesian plane discussed in Chapter 4). or RLC circuit. In many ways. However. The one-link robot discussed in Chapter 3 could be described using a complex number where the real part would be its component in the x-direction and the imaginary part would be its component in the y-direction. where a is the real part (real number) and bj is the imaginary part (real number times the imaginary unit j). Electrical engineers use the letter j to represent imaginary number because the letter √ i is used in electrical engineering to represent current. one real and one imaginary. imaginary numbers are used in combination with a real number to form a complex number. However. 132 . An imaginary number is the square root of a negative real number (−1). −1 will be represented by the letter j throughout this chapter. To remove this confusion. 5. Therefore.1 Representation of a one-link planar robot as a complex number.1 can be represented by a complex number as P = Px + j Py or P = Px + Py j.2 Py θ 0 Px x Figure 5. Px = Re(P) = l cos(𝜃) is the real part.2 Position of One-Link Robot as a Complex Number 133 Even though the parts are joined by a plus sign. √ where j = −1 is the imaginary number. Similarly. The expression must be left as an indicated sum.5. The numbers Px and Py are like the components of P in the x. the one-link planar robot can be represented in polar form as P = |P| ∠𝜃. P = Px + j Py = |P| cos 𝜃 + j (|P| sin 𝜃) = |P| (cos 𝜃 + j sin 𝜃) = |P| e j 𝜃 (5.1) .and y-directions (analogous to a 2-D vector). y ∣= l P ∣P 5. and Py = Im(P) = l sin(𝜃) is the imaginary part of the complex number P. where |P| = l = √ Px2 + Py2 is the magnitude and 𝜃 = atan2(Py . POSITION OF ONE-LINK ROBOT AS A COMPLEX NUMBER The one-link planar robot shown in Fig. the addition cannot be performed. Px ) is the angle of the complex number P. 3. 5. 5. The cosine function is the real part of the exponential function and the sine function as the imaginary part of the exponential function.134 Chapter 5 Complex Numbers in Engineering where cos 𝜃 + j sin 𝜃 = e j𝜃 is Euler’s formula. If R = 100 Ω. AND C AS A COMPLEX NUMBER 5.3.2 can be written as ZR = R Ω.2 A resistor. L. 5. a point P in the rectangular plane (x-y plane) can be described as a vector or a complex number as P⃗ = Px i⃗ + Py j⃗⋯ ⋯ vector form P = Px + j Py ⋅ ⋅ ⋯ ⋯ complex number in rectangular form P = |P| ∠𝜃 ⋅ ⋅ ⋅ ⋅ ⋯ ⋯ complex number in polar form P = |P| e j 𝜃 ⋅ ⋅ ⋅ ⋯ ⋯ complex number in exponential form 5. If L = 25 mH and f = 60 Hz. where R is the resistance in ohms (Ω). where L is the inductance in henry (H) and 𝜔 = 2𝜋f is the angular frequency in rad/s (f is the linear frequency or frequency in hertz (Hz)).1 Impedance of a Resistor R The impedance of the resistor shown in Fig.. R Figure 5. The complex number written as in equation (5. i.2 Impedance of an Inductor L The impedance of the inductor shown in Fig.3 IMPEDANCE OF R. the impedance of the inductor can be written as a complex number in rectangular and .1) is known as the exponential form of the complex number and can be used to express the cosine and sine functions.e. the impedance of the resistor can be written as a complex number in rectangular and polar forms as ZR = R Ω = 100 Ω = 100 + j 0 Ω = 100 ∠0◦ Ω ◦ = 100 e j 0 Ω. cos𝜃 = Re(e j 𝜃 ) and sin𝜃 = Im(e j 𝜃 ) In summary.3 can be written as ZL = j𝜔L Ω. 426 e j 90 Ω. 5.3 An inductor.3 Impedance of a Capacitor C 1 Ω. The impedance of the capacitor can also be written in polar and exponential form as ZC = 132. .025) Ω = 0 + j 9.6 Ω j = 0+ 132.4 A capacitor. C Figure 5.3.426 Ω = 9.3 Impedance of R.4 can be written as ZC = ZC = 1 Ω j𝜔C = 0+ 1 Ω j (120 𝜋) (20 ∗ 10−6 ) = 0+ 132.6 ∠−90◦ Ω ◦ = 132.426 ∠90◦ Ω ◦ = 9. 5.6 j ( )Ω j j = 0+ 132. L Figure 5. If C = 20 𝜇F and f = 60 Hz.5.6 j Ω √ where j2 = ( −1)2 = −1. the impedance of the capacitor can be written as a complex number in rectangular form as The impedance of the capacitor shown in Fig. and C as a Complex Number 135 polar forms as ZL = j 𝜔 L Ω = j (2 𝜋 60) (0.6 e −j 90 Ω. L.6 j Ω j2 = 0 − 132. j𝜔 C where C is the capacitance in farad (F) and 𝜔 is the angular frequency in rad/s. 3 as ZR = 100 + j 0 Ω ZL = 0 + j 9.4 Chapter 5 Complex Numbers in Engineering IMPEDANCE OF A SERIES RLC CIRCUIT The total impedance of the series RLC circuit shown in Fig.6 Ω. C = 20 𝜇F.5 A series RLC circuit. the impedance of R. the total impedance of the series RLC circuit can be calculated as ZT = (100 + j 0) + (0 + j 9.426) + (0 − j 132.7 ∠−50. Since ZT = ZR + ZL + ZC . 5. 5.174)2 = 158. j𝜔C R L C Figure 5. ZL = j𝜔L Ω.174 Ω. L.426 + (−132. L = 25 mH. 100) = −50. where √ |ZT | = 1002 + (−123. The polar and exponential forms of the total impedance can be calculated from the rectangular form as Polar Form: ZT = |ZT | ∠𝜃. If the complex numbers are given in the polar or exponential form. Exponential Form: ◦ ZT = |ZT | e j 𝜃 = 158.426 Ω ZC = 0 − j 132.93 Ω. For the values of R = 100 Ω.93◦ Ω. they . the total impedance of the series RLC circuit shown in Fig.7 Ω 𝜃 = atan2(−123. Note: The addition and subtraction of complex numbers is best done in the rectangular form.93◦ .2) 1 Ω. and 𝜔 = 120𝜋 rad/s. and C were calculated in Section 5. Therefore.6)) Ω = 100 − j 123.5 in rectangular form is ZT = 100 − j 123. Therefore.174.7 e −j 50.5 is given by ZT = ZR + ZL + ZC where ZR = R Ω.174 Ω.6) Ω = (100 + 0 + 0) + j (0 + 9.136 5. and ZC = (5. ZT = 158. 5 Impedance of R and L Connected in Parallel 137 should be converted to rectangular form to carry out the addition or subtraction of these complex numbers. In summary. of the subtraction of two complex numbers Z1 and Z2 (Z1 − Z2 ) is obtained by subtracting the real part a2 of complex numbers Z2 from the real part a1 of complex number Z1 . the real part. of the addition of complex numbers. 5. respectively. Z. if the result is needed in the polar or exponential form. c. the real part. Therefore. a. the imaginary part of the subtraction of the two complex numbers Z1 − Z2 is obtained by subtracting the imaginary part b2 of complex number Z2 from the imaginary part b1 of complex number Z1 .6 is given by Z= where ZR = R Ω and ZL = j𝜔L Ω. 5. ZR ZL ZR + ZL . However. Addition of Two Complex Numbers: The addition of two complex numbers Z1 = a1 + j b1 and Z2 = a2 + j b2 can be obtained as Z = Z1 + Z2 = (a1 + j b1 ) + (a2 + j b2 ) = (a1 + a2 ) + j (b1 + b2 ) = a + j b. Similarly. and the imaginary part of the addition of the two complex numbers is obtained by adding the imaginary parts of complex numbers being added. the conversion from the rectangular to polar or exponential forms is carried out as a last step. the subtraction of the two complex numbers Z1 and Z2 can be obtained as Z = Z1 − Z2 = (a1 + j b1 ) − (a2 + j b2 ) = (a1 − a2 ) + j (b1 − b2 ) = c + j d. Therefore. the addition and subtraction of two complex numbers can be carried out using the following steps. Subtraction of Two Complex Numbers: Similarly. is obtained by adding the real parts of the complex numbers being added.5 IMPEDANCE OF R AND L CONNECTED IN PARALLEL The total impedance Z of a resistor R connected in parallel with an inductor L as shown in Fig.5. 3 as ZR = 100 + j 0 Ω ◦ = 100 e j 0 Ω ZL = 0 + j 9. it will be shown in Section 5. the total impedance of the R connected in parallel with L can ZR + ZL be calculated as Since Z = ◦ ◦ (100 e j 0 )(9.88 + j 9. Therefore. If the complex numbers are given in rectangular form.62◦ ) + j 9.88 + j 9.7 that it is best to carry out these operations in polar form. ZR ZL .426 e j 90 ) ZT = (100 + j 0) + (0 + j 9.384 cos(84.62◦ .384 e j 84.34 Ω.384 sin(84.138 Chapter 5 Complex Numbers in Engineering ZL Z ZR R L Figure 5.62◦ ) = 0. However.384 ∠84.426 Ω ◦ = 9.6 e j 90 100.384 ∠84.426) ◦ = 942.6 A parallel RL circuit. they should be converted to polar form and the result of the multiplication or division is then obtained in the polar form. However.6 e j 90 100 + j 9. the impedances of R and L were calculated in Section 5. For R = 100 Ω.384◦ ◦ = 9. Therefore. Z = 0. if the result is needed in rectangular form.443 e j 5.62◦ Ω.426 e j 90 Ω.426 = 942. the total impedance of a 100 Ω resistor connected in parallel with a 25 mH inductor in the polar form is Z = 9. The rectangular form of the total impedance can be calculated from the polar form as Z = 9.62 ◦ = 9. L = 25 mH and 𝜔 = 120 𝜋 rad/s. the conversion from polar to rectangular form is carried .34 Ω. Note that the multiplication and division of complex numbers can be carried out in rectangular or polar forms. the product of all the dividend and the divisor complex numbers . The angle ∠𝜃 of the resultant is obtained by subtracting the angle of the divisor complex numbers Z2 from the angle of the dividend complex number Z1 .5 Impedance of R and L Connected in Parallel 139 out as a last step. Therefore. Note that this procedure is not restricted to multiplication of two numbers only. Note that if the dividend or the divisor are the product of complex numbers. it can be used for multiplying any number of complex numbers. the magnitude |Z| of the division of two complex numbers given in polar form is obtained by dividing the magnitudes of dividend complex number Z1 by the magnitude of divisor complex number Z2 . Division of Complex Number in Polar Form: The division of the two complex numbers Z1 and Z2 in the polar form can be carried out as Z= = = Z1 Z2 M1 ∠𝜃1 M2 ∠𝜃2 M1 e j ∠𝜃1 M2 e j ∠𝜃2 = M1 j (∠(𝜃 −𝜃 )) 1 2 e M2 = M1 ∠(𝜃1 − 𝜃2 ) M2 = |Z| ∠𝜃. Multiplication of Complex Number in Polar Form: The multiplication of the two complex numbers Z1 = M1 ∠𝜃1 and Z2 = M2 ∠𝜃2 can be carried out as Z = Z1 ∗ Z2 = M1 ∠𝜃1 ∗ M2 ∠𝜃2 = M1 e j ∠𝜃1 ∗ M2 e j ∠𝜃2 = (M1 M2 ) e j ∠(𝜃1 +𝜃2 ) = (M1 M2 ) ∠(𝜃1 + 𝜃2 ) = |Z| ∠𝜃. the magnitude |Z| of the multiplication of complex numbers given in polar form is obtained by multiplying the magnitudes of complex numbers being multiplied and the angle ∠𝜃 of the resultant is the sum of the angles of the complex numbers being multiplied.5. The steps to carry out the multiplication and division in the polar form are explained next. Therefore. 5.3 13.74 e j 43.10) ◦ = 110 e j0 13. Converting the numerator and denominator in equation (5.6 ARMATURE CURRENT IN A DC MOTOR The winding of an electric motor shown in Fig.3) Since it is easier to multiply and divide in exponential or polar forms. find the current I = V Z where Z = ZR + ZL and V = 110 V.01 e −j 43.3◦ = 110 e j(0 −43.426 = 110 + j 0 A.01 ∠−43. The total impedance of the winding of the motor is given as Z = ZR + ZL .140 Chapter 5 Complex Numbers in Engineering should be obtained first. 5.4) . Therefore.3◦ A.426 Ω.3 = 8. Z = (10 + j 0) + (0 + j 9. I R 110 V L 60 Hz Motor Figure 5. 60 Hz voltage A flowing through the winding of the motor. If the motor is connected to a 110 V.426. 10 + j 9. source as shown. and the current flowing through the winding of the motor is V I= Z 110 = 10 + j 9. the current in equation (5. (5. and then the division of the two complex numbers should be carried out.4262 e j atan2(9.426) = 10 + j 9.426 Ω.3) will be calculated in the polar/exponential form.7 has a resistance of R = 10 Ω and an inductance of L = 25 mH.7 Voltage applied to a motor.74 ◦ ◦) ◦ = 8. where ZR = R = 10 + j 0 Ω and ZL = j 𝜔 L = 0 + j 9.426 (5.3) to exponential form yields ◦ 110 e j0 I= √ 102 + 9. 1 ∠30◦ A.1 ∠30◦ )(100 + j 30) V will be calculated by multiplying the two complex numbers using their rectangular forms as well as their polar/exponential forms to show that it is much easier to multiply complex numbers using the polar/exponential forms. The voltage V = I Z = (0. I.8 that the current is lagging (negative angle) the voltage by 43. the current flowing through the winding of the motor is 8. 5. It can be seen from Fig. The polar form of the current given in equation (5. 5.3◦ − j sin 43.8. Find V if I = 0.49 A. The phasor diagram (vector diagram) showing the voltage and current vectors is shown in Fig. converting to radians is unnecessary for the purpose of multiplying and dividing complex numbers.01 (cos 43.3 = 8.9 produces a voltage. V = I Z. However. 43.7 Further Examples of Complex Numbers in Electric Circuits 141 Therefore. 5. flowing through the RL circuit shown in Fig.3° A I Figure 5.3◦ ) = 5. 110 V V 01 8.7 Example 5-1 FURTHER EXAMPLES OF COMPLEX NUMBERS IN ELECTRIC CIRCUITS A current. Solution The impedance Z of the RL circuit can be calculated as Z = R + j XL Ω = 100 + j 30 Ω. Note: In general e ± 𝜃 = cos 𝜃 ± j sin 𝜃 requires 𝜃 in radians.3◦ . 5.01 e −j 43. where Z = R + j XL .5.8 The current and voltage vector.83 − j 5. + R = 100 Ω I ↑ V XL = 30 Ω − Figure 5. .01 A.4) can be converted to rectangular form as ◦ I = 8.9 Current flowing through an RL circuit. where j = −1.66 + j 2. Compute the voltage V1 given by V1 = Z2 V.66 + j 2.10.4 e j 16.05)(100 + j 30) = (0.7◦ = 104. all measured in ohms: (a) Express the impedance Z1 and Z2 in both rectangular and polar forms.0866 + j 0.7◦ ) = 7.598 + j 5 + 1.1 ∠30◦ = 0.598 = 10.7 Ω Z = 100 + j 30 Ω V = IZ V = IZ ◦ ◦ = (0.44 ∠46.4 ∠16.598 + j 5 + 1.10 Solution Voltage divider circuit for Example 5-2.05 A ◦ = 104.5 j 2 = (10.7 ) = 8. and XC = 20. XL = 10.1 e j 30 )(104. (5. Suppose R = 10.44) ∠46.7◦ V Example 5-2 In the voltage divider circuit shown in Fig.7◦ V ◦ +16. 100) = 0. the impedance of the resistor is given by Z1 = R.1 (cos 30◦ + j sin 30◦ ) Z = 100 + j 30 √ = 1002 + 302 ∠atan2 (30.1 e j 30 A = 0.1 ∠30◦ I = 0.4 e j 16.5 (−1) = (10.142 Chapter 5 Complex Numbers in Engineering Rectangular Form Polar/Exponential Form ◦ I = 0.44) e j(30 = 8. Z1 + Z2 R V L C + V1 − Figure 5. 5. (a) The impedance Z1 can be written in rectangular form as Z1 = R = 10 + j 0 Ω. √ (b) Suppose the source voltage is V = 100 2 ∠45◦ V.5) .0866 + j 0. The total impedance√of the inductor and capacitor in series is given by Z2 = j XL + 1j XC .16 + j 7. 10) = 10 ∠0◦ Ω. (b) Z2 V Z1 + Z2 ( ) √ 10 ∠−90◦ = (100 2 ∠45◦ ) (10 + j 0) + (0 − j 10) ( ) √ 10 ∠−90◦ = (100 2 ∠45◦ ) 10 − j 10 ) ( √ 10 ∠−90◦ (100 2 ∠45◦ ) = √ 10 2 ∠−45◦ ( ) √ 1 = √ ∠−45◦ (100 2 ∠45◦ ) 2 V1 = = 100 ∠0◦ V = 100 + j 0V. The impedance Z2 can be written in polar form as √ Z2 = 02 + (−10)2 ∠atan2(−10. and XC = 10. and ZC = 1j XC . Example 5-3 In the circuit shown in Fig. the impedance√of the various components are ZR = R. The impedance Z2 can be written in rectangular form as 1 X j C ( ) j 1 = j 10 + 20 j j = j 10 − j 20 = −j 10 Z2 = j XL + = 0 − j 10 Ω. ZR ZL in both rectangular and polar ZR + ZL . XL = 10. ZL = j XL . (a) Express the total impedance Z = ZC + forms. 5. where j = −1. Suppose R = 10. 0) = 10 ∠−90◦ Ω.5. all measured in ohms.11.7 Further Examples of Complex Numbers in Electric Circuits 143 The impedance Z1 can be written in polar form as √ Z1 = 102 + 02 ∠atan2(0. 11 Total impedance of the circuit for Example 5-3.144 Chapter 5 Complex Numbers in Engineering √ (b) Suppose a voltage V = 50 2 ∠45◦ V is applied to the circuit shown in Fig. Find the current I flowing through the circuit if I given by I= Solution V Z (5. The impedance ZL can be written in rectangular and polar forms as ZL = j XL = 0 + j 10 Ω = 10 ∠90◦ Ω. 5. ZC = The total impedance can now be calculated as Z = ZC + ZR ZL ZR + ZL = 0 − j 10 + (10 ∠0◦ ) (10 ∠90◦ ) (10 + j 0) + (0 + j 10) = 0 − j 10 + 100 ∠90◦ (10 + j 10) . The impedance ZC can be written in rectangular and polar forms as 1 X j C = −j XC = 0 − j 10 Ω = 10 ∠−90◦ Ω.11.6) (a) The impedance ZR can be written in rectangular and polar forms as ZR = R = 10 + j 0 Ω = 10 ∠0◦ Ω. C R Z L Figure 5. Example 5-5 A sinusoidal current source is = 100 sin(120 𝜋 t + 60◦ ) mA is applied to a parallel RL circuit. Solution Since sin(𝜃) = Im(e j 𝜃 ). . therefore. (b) I= V Z √ 50 2 ∠45◦ = √ 5 2 ∠−45◦ = 10 ∠90◦ A = 0 + j 10 A.8 COMPLEX CONJUGATE The complex conjugate of a complex number z = a + j b is z∗ = a − j b. Write the current source is in the exponential form.5. Solution Since cos(𝜃) = Re(e j 𝜃 ). 5. the voltage source vs can be written in the exponential form as ◦ vs = Re(100 e j(100 t+45 ) )V. Example 5-4 A sinusoidal voltage source vs = 100 cos(100 t + 45◦ ) V is applied to a series RLC circuit. Write the voltage source vs in the exponential form. the current source is can be written in the exponential form as ◦ is = Im(100 e j(120 𝜋 t+60 ) )mA. therefore.8 Complex Conjugate 145 100 ∠90◦ √ 10 2 ∠45◦ √ = 0 − j 10 + 5 2 ∠45◦ = 0 − j 10 + = 0 − j 10 + 5 + j 5 = 5−j5 Ω √ = 5 2 ∠−45◦ Ω. Also. Therefore. √ |z|2 = ( a2 + b2 )2 = a2 + b2 . Solution The conjugate of the complex number z = 3 + j 4 is given by z∗ = 3 − j 4. √ = 25. 3) z∗ = 5 ∠−53. Example 5-6 If z = 3 + j 4.1◦ z∗ = 3 − j 4 √ z∗ = 32 + (−4)2 ∠atan2(−4. z z∗ = |z|2 = a2 + b2 . calculating z z∗ using the polar form.146 Chapter 5 Complex Numbers in Engineering The multiplication of a complex number by its conjugate results in a real number that is the square of the magnitude of the complex number: z z∗ = (a + j (b)(a − j (b) = a2 − j a b + j a b − j2 b2 = a2 − (−1) b2 = a2 + b2 .1◦ . we have Therefore. Calculating z z∗ using the rectangular form yields z z∗ = (3 + j 4)(3 − j 4) = 32 − j (3)(4) + j (3)(4) − j2 42 = 32 − (−1) 42 = 32 + 42 = 25. z z∗ z = 3+j4 √ z = 32 + 42 ∠atan2(4. Now. 3) z = 5 ∠53. Note: |z|2 = ( 32 + 42 )2 = 25. find z z∗ using the rectangular and polar forms. 1◦ )(5 ∠−53. P5.e.1◦ ) = (5)(5)∠(53. (c) Write the real and imaginary parts of V. Assume VR = 1 ∠0◦ V and VL = 1 ∠90◦ V. if the angle of VR is 0◦ . (a) Write IR and IL in rectangular form. but the angle of the complex conjugate is the negative of the angle of the complex number.4 RC circuit for problem P5-4. voltage VL leads voltage VR by 90◦ (i. PROBLEMS 5-1.. if the angle of VR is 0◦ .1◦ − 53. P5.32 ∠−30◦ V. 5-7.6◦ V and VL = 4. (b) Determine V = VR + VC in both its rectangular and polar forms.7. Note that the complex conjugate of a complex number in polar form has the same magnitude as the complex number. Repeat problem P5-4 if VR = 10 ∠60◦ V and VC = 17. Therefore.e.5 ∠63. 5-2. Assume iR = 100 ∠0◦ mA and iL = 200 ∠−90◦ mA.44◦ V and VC = 3. In the parallel RL circuit shown in Fig. Repeat problem P5-4 if VR = 9. In the series RL circuit shown in Fig. P5. (a) Write VR and VL in rectangular form. the angle of 5-5. Repeat problem P5-1 if 10 ∠−45◦ V and VL = 5 ∠45◦ V. the angle of VL is 90◦ ). (a) Write VR and VC in rectangular form.5 ∠18. (c) Write the real and imaginary parts of V.1.4◦ V. VR = 5-3. voltage VC lags voltage VR by 90◦ (i. z z∗ = 25.. Repeat problem P5-1 if VR = 9 ∠−26.4.56◦ V. R V VR − + C VC − VL Figure P5. Assume VR = 1 ∠0◦ V and VC = 1 ∠−90◦ V. (b) Determine V = VR + VL in both its rectangular and polar forms. 5-4. the total current I is the sum of the currents flowing through the resistor (IR ) and the inductor (IL ). In the RC circuit shown in Fig. 5-6. .Problems 147 z z∗ = (5 ∠53.1 A series RL circuit for problem P5-1. − Figure P5. + R V + VR − + L VC is −90◦ ).16 ∠−71.1◦ ) = 25 ∠0◦ = 25. R L C IC I ↑ C IR Figure P5. P5.5 ∠0◦ mA and IC = 0. P5. Two circuit elements are connected in series as shown in Fig.2◦ A. . Repeat problem P5-7 if IR = 0. and XL2 = 15 Ω. (c) Write the real and imaginary parts of I.10. (a) Determine the total impedance.5 ∠56. Repeat problem P5-10 if IR = 0. C = 25 𝜇 F.8◦ A and IC = 0. 5-11. P5. 5-13. The impedance of the second circuit element is Z2 = R2 + j XL2 . A resistor. where ZR = R Ω. (a) Determine the total impedance Z in rectangular form.7◦ mA and IC = 55.148 Chapter 5 Complex Numbers in Engineering (b) Determine I = IR + IL in both its rectangular and polar forms.37 ∠68. For a particular design j𝜔C R = 100 Ω. capacitor. (c) Write the real and imaginary parts of I. The impedance of the first circuit element is Z1 = R1 + j XL1 . 5-10. Assume IR = 83. 5-9. In the parallel RC circuit shown in Fig. (b) Determine I = IR + IL in both its rectangular and polar forms.707 ∠45◦ A and IL = 0.16. The output voltage across the capacitor in a series RLC circuit is measured by an oscilloscope as vo (t) = 15 cos(120 𝜋 t − 60◦ ) V. The total impedance of the circuit is Z = ZR + ZL + ZC . R Figure P5.15 RLC circuit for problem P5-15. Z = Z1 + Z2 . 5-16. the total current I is the sum of the currents flowing through the resistor (IR ) and the capacitor (IC ). R2 = 5 Ω.15. XL1 = 25 Ω.6 ∠30◦ 𝜇A and IL = 50 ∠−60◦ 𝜇A. 5-8. IL I ↑ IR L R Figure P5.2 ∠90◦ mA. Repeat problem P5-7 if IR = 86. where R1 = 10 Ω. 5-12. Repeat problem P5-10 if IR = 0.707 ∠−45◦ A. 5-15. (a) Write IR and IL in rectangular form. and 𝜔 = 120 𝜋 rad/s. Write the voltage vo (t) in the exponential form.2 ∠−33. The current flowing through the resistor in a parallel RL circuit is given as iR (t) = 5 sin(5 𝜋 t + 30◦ ) mA. 5-14.7 A parallel RL circuit for problem P5-7. L = 500 mH.3◦ mA.10 A parallel RC circuit for problem P5-10. (c) Determine the complex conjugate Z∗ and compute the product Z Z∗ . Write the current iR (t) in the exponential form. ZL = j 𝜔 L Ω.929 ∠−21. (b) Determine the total impedance Z in polar form. and 1 ZC = Ω. and an inductor are connected in series as shown in Fig. an inductor L. (b) Find V if I = 7. (b) Compute the quantity Z1 + Z2 + Z3 . R C L Figure P5. L = 15 mH. where C = 5 𝜇F. and j = −1. R = 4 Ω and XC = 2 Ω. and a capacitor C. and the impedance of the capacitor is Z3 = √ 1 Ω. 5-17. 5-19. I V R C Figure P5. and express the result in both rectangular and polar forms.19 consists of a resistor R. The impedance of the resistor is Z1 = R Ω. 5-18.18 Impedance of series-parallel combination of circuit elements for problem P5-18. Write the Z1 + Z2 total impedance in both its rectangular and polar forms. L √ = 0.16 Two circuit elements in series for 149 (b) Write down the complex conjugate of Z2 and calculate the product Z2 Z2∗ . in other words find Z = |Z| ∠𝜃. 𝜔 = 120 𝜋 rad/s. and express the result in both rectangular and polar forms. The circuit shown in Fig. For a particular design.17 RC circuit subjected to an alternating voltage source for problem P5-17. C = 25 𝜇F. P5. For a particuj𝜔C lar design. and 𝜔 = 120 𝜋 rad/s. Z2 Z1 problem P5-16. R = 100 Ω. the impedance of the inductor is Z2 = j 𝜔 L Ω.17. An RC circuit is subjected to an alternating voltage source V as shown in Fig. (c) Compute the transfer function H = Z2 + Z3 and express the result Z1 + Z2 + Z3 in both rectangular and polar forms. (a) Write Z1 and Z2 as complex numbers in both their rectangular and polar forms. (a) Compute the quantity Z2 + Z3 . The relationship between the voltage and current is V = I Z. P5. (a) Find I if V = 120 ∠30◦ V. where j = −1.18. . P5. A series-parallel electric circuit consists of the components shown in Fig. (d) Determine the complex conjugate of H and compute the product H H∗. R = 100 Ω. Z1 R1 Z2 XL1 R2 XL2 Figure P5.15 H.Problems (b) Determine the magnitude and phase of the total impedance. (c) Calculate the total impedance Z = Z1 Z2 of the circuit.0 ∠45◦ A. where Z = R − j XC . The values of the impedance of the two −j and Z2 = components are Z1 = 𝜔C R + j 𝜔 L. 20. An electric circuit consists of two components as shown in Fig. P5. (c) Compute the total impedance of the Z1 Z2 two components Z = and Z1 + Z2 express the result in both rectangular and polar forms. (c) Determine the complex conjugate of Z and compute the product Z Z∗ .21. write Z in both rectangular and polar forms.22.20 Impedance of elements connected in parallel for problem P5-20. and ZC = Ω. find the (d) If VC = ZR + ZL + ZC voltage in both rectangular and polar forms. R V L + C VC − Figure P5. (a) Write Z1 and Z2 as complex numbers in both their rectangular and polar forms. and ZC as complex numbers in both their rectangular and polar forms. (a) Write Z1 and Z2 as complex numbers in polar form. L = 𝜋 500 𝜇F.19 Transfer function of series-parallel circuit for problem P5-19. XL = 250 Ω. ZL = 1 j 𝜔 L Ω. ZL . The voltage V1 is given by V1 = Figure P5. ZC V. (b) If the total impedance is Z = ZR + ZL + ZC .150 Chapter 5 Complex Numbers in Engineering R L C Figure P5. C = ing that R = 100 Ω. Z1 + Z2 Assuming that R1 = 50 Ω.2◦ V is applied to an circuit shown in Fig. 3𝜋 (a) Write ZR .21 Voltage division for problem P5-21. R2 = 50 Ω. XL = 100 Ω . where Z1 = R1 − j XC Ω and Z2 = R2 + j XL Ω. P5. and XC = 125 Ω. and XC = 100 Ω. where R1 = 75 Ω. and j2 = −1. A √ sinusoidal voltage source V = 110 2 ∠−23. P5. (b) Determine the complex conjugate of Z2 and compute the product Z2 Z2∗ . A sinusoidal voltage source V = 110 V of frequency 60 Hz (𝜔 = 120 𝜋 rad/s) Z2 V. 5-20. . (b) Determine V1 in both rectangular and polar forms. R2 = 100 Ω. The values of the impedance of the two components are Z1 = R1 + j XL and Z2 = R2 − j XC . where ZR = R Ω. Z1 Z2 R1 R2 XL XC is applied to an RLC circuit as shown in Fig. 5-21. Assumj𝜔C 500 mH. 5-22. P5. P5.22 Voltage division for problem P5-22. ZL . √ 1 and ZC = XC .25. an inductor L.24 A capacitor connected in series with a parallel combination of resistor and inductor for problem P5-24. which is given as ZLC = Figure P5. Z1 R1 C + R2 V Z2 V1 L − Figure P5. and C for problem P5-23. the impedances of R and C are given . 50 3 (a) Express the impedance ZR . L = 3 mH. ZL = j XL Ω. 5-23. and a capacitor C. where j = j 𝜔 C √ −1. ZL . 5-25. connected as shown in Fig. suppose that the total impedance of the circuit is Z = Z Z ZC + R L . Supj √ pose R = 120 Ω. Express the result in both ZL + ZC rectangular and polar forms.Problems (c) Determine the complex conjugate of Z1 and compute the product Z1 Z1∗ . The impedance of the resistor is ZR = R Ω. P5. (a) Write ZR . XL = 120 3 Ω. (c) Determine the complex conjugate of Z and compute the product Z Z∗ . and 1 XC = √ Ω. and 𝜔 = 1000 rad/s. Determine the ZR + ZL total impedance and express the result in both rectangular and polar forms. Suppose R = 9 Ω. and ZC as complex numbers in both rectangular and polar forms. L 151 ZL ZC . (d) Determine the complex conjugate of Z and compute the product Z Z∗ . 5-24. C = 250 𝜇F. (b) Now. and the impedance of the 1 capacitor is ZC = Ω. An electric circuit consists of a resistor R. the impedances of the various components are ZR = R Ω. C R C Z R L Figure P5. In the circuit shown in Fig. (b) Calculate the total impedance of the inductor-capacitor parallel combination. (c) Now suppose that the total impedance of the circuit is Z = ZR + ZLC . Determine the total impedance in both rectangular and polar forms. L.24.23. In the RC circuit shown in Fig.23 Series-parallel combination of R. and ZC as complex numbers in both rectangular and polar form. the impedance of the inductor is ZL = j𝜔L Ω. where j = −1. (d) Knowing the current in part (c). find the current V I= as a complex number in both Z rectangular and polar forms. show V = VR + VL . P5. find the current V I= as a complex number in both Z rectangular and polar forms. (e) Show that the KVL is satisfied for the circuit shown in Fig. I R V C (a) Express the impedances ZR and ZL as complex numbers in both rectangular and polar forms.25 An RC circuit for problem P5-25. Suppose R = 100 Ω and XL = 50 Ω. find the voltage phasors VR = I ZR and VC = I ZC in both rectangular and polar forms. R = j𝜔C 100 Ω. The total impedance of the circuit is Z = ZR + ZL + ZC . I Figure P5. and 𝜔 = 120 𝜋 rad/s. find the voltage phasors VR = I ZR and VL = I ZL in both rectangular and polar forms. (a) Express the impedances ZR . (b) Find the total impedance Z = ZR + ZL as a complex number in both rectangular and polar forms. in other words. V L C Figure P5.27 RLC circuit for problem P5-27. (e) Show that the KVL is satisfied for the circuit shown in Fig. C = 25 𝜇F. I R V L Figure P5.152 Chapter 5 Complex Numbers in Engineering as Z√ R = R Ω and ZC = −j XC Ω. (b) Find the total impedance Z = ZR + ZC as a complex number in both rectangular and polar forms. . (a) Express the impedances ZR and ZC as complex numbers in both rectangular and polar forms. capacitor. 5-27. L = 500 mH. P5. where j = −1. Suppose R = 100 Ω and XC = 50 Ω. P5. (d) Knowing the current in part (c). in other words.25. and ZC as complex numbers in both rectangular and polar forms. where j = −1. A resistor. the impedances of R and L are given as Z √R = R Ω and ZC = j XL Ω.26. ZL . and ZC = 1 Ω.27. For a particular design. In the RL circuit shown in Fig. R 5-26.26. P5.26 An RL circuit for problem P5-26. (c) If V = 100 ∠0◦ V. (c) If V = 100 ∠0◦ V. and inductor are connected in series as shown in Fig. (b) Determine the total impedance Z in both its rectangular and polar forms. where ZR = R Ω. ZL = j 𝜔 L Ω. show V = VR + VC . find the current V I= as a complex number in both Z rectangular and polar forms. In the current divider circuit shown in Fig.. Write VR and VC as comj plex numbers in both rectangular and polar forms if VR = I1 × ZR and VC = I2 × ZC . 5-30.Problems (c) If V = 100 ∠0◦ V. P5. (a) Write I1 and I2 in rectangular form.28. Suppose I1 = 1 ∠0◦ and I2 = 1 ∠90◦ . 5-29. √ √ Suppose I1 = 2 ∠45◦ and I2 = 2 ∠−45◦ . I = I1 + I2 ). P5. In the current divider circuit shown in Fig. Write VR and VL as complex numbers in both rectangular and polar forms if VR = I1 × ZR and VL = I2 × ZL . (c) Show I = I1 + I2 . show V = VR + VL + VC ).30. P5. I = I1 + I2 ). 5-31. both measured in mA. (c) Suppose ZR = 1000 Ω and ZC = 103 Ω.28 A current divider circuit for problems P5-28 and P5-29.27 (i.e. the currents flowing through the resistor I1 and capacitor I2 are given by ZC I I1 = ZR + ZC ZR I I2 = ZR + ZC − Figure P5. both measured in mA. P5. I + V L I2 I1 + VL − + R VR − Figure P5. the currents flowing through the resistor I1 and inductor I2 are .e.30 A current divider circuit for problems P5-30 and P5-31. determine I1 and I2 in both rectangular and polar forms. (d) Show that the KVL is satisfied for the circuit shown in Fig. 5-28.e. express ZR and ZC as complex numbers in both rectangular and polar forms.30. In the current divider circuit shown in Fig. In the current divider circuit shown in Fig. the sum of the current phasors I1 and I2 is equal to the total current phasor I (i. (a) Write I1 and I2 in rectangular form. P5. (b) Determine the total current phasor I in both rectangular and polar forms. I + V I2 C I1 + + VC R VR − − − 153 where ZR = R Ω is the impedance of X the resistor and ZC = C Ω is the j impedance of the capacitor.28. (b) Determine the total current phasor I in both rectangular and polar forms.. the sum of the current phasors I1 and I2 is equal to the total current phasor I (i. (c) Suppose ZR = 1000 Ω and ZL = j 1000 Ω. (a) If R = 1 k Ω and XC = 103 Ω.. (b) If I = 1 mA. determine Vo in both rectangular and polar forms. (c) Show I = I1 + I2 .32. the output voltage Vo is given by Z Vo = − C Vin ZR C Vin where ZR = R Ω is the impedance of the resistor and ZC = −j XC Ω is the impedance of the capacitor.33. (a) If R = 2 kΩ and XC = 1 k Ω. +Vcc − + + −Vcc Vo − Figure P5. In the Op–Amp circuit shown in Fig. determine Vo in both rectangular and polar forms. (a) If R1 = 1 kΩ. R2 = 2 kΩ. express ZR and ZL as complex numbers in both rectangular and polar forms. P5.154 Chapter 5 Complex Numbers in Engineering given by I1 = ZL I ZR + ZL I2 = ZR I ZR + ZL 5-33. P5. ZR2 = R2 Ω is the impedance of the resistor R2.33 An Op–Amp circuit for problem P5-33. 5-34. and XC = 2 k Ω. (a) If R = 1 kΩ and XC = 2 k Ω. (b) If Vin = 10 ∠0◦ V. + − R + Vin − +Vcc − + + −Vcc Vo − Figure P5. the output voltage Vo is given by Vo = − where ZR = R Ω is the impedance of the resistor and ZC = −j XC Ω is the impedance of the capacitor. P5.32 An Op–Amp circuit for problem P5-32. and ZC = −j XC Ω is the impedance of the capacitor. In the Op–Amp circuit shown in Fig.34. determine Vo in both rectangular and polar forms. determine I1 and I2 in both rectangular and polar forms. express ZR and ZC as complex numbers in both rectangular and polar forms. and ZC as complex numbers in both rectangular and polar forms. (b) If Vin = 5 ∠90◦ V. . (a) If R = 1 kΩ and XL = 103 Ω. In the Op–Amp circuit shown in Fig. (b) If Vin = 2 ∠45◦ V. the output voltage Vo is given by Vo = − C ZR V ZC in ZR2 + ZC Vin ZR1 where ZR1 = R1 Ω is the impedance of the resistor R1. ZR2 . R 5-32. express ZR and ZC as complex numbers in both rectangular and polar forms. express ZR1 . (b) If I = 1 mA. where ZR = R Ω is the impedance of the resistor and ZL = j XL Ω is the impedance of the inductor. The Delta-to-Yee (Δ-Y) and Yee-toDelta (Y-Δ) conversions are used in electrical circuits to find the equivalent impedances of complex circuits.Problems C R2 R1 Vin given by +Vcc − + Vo = − + + − −Vcc Vo Figure P5.5 k Ω. the impedances Za . Z2 . 5-36. and Z3 can be written in term of . +Vcc − + + −Vcc Vo − Figure P5. and ZC2 = −j XC2 Ω is the impedance of the capacitor C2. ZC1 = −j XC1 Ω is the impedance of the capacitor C1. express ZR1 . In the Op–Amp circuit circuit shown in Fig.37.36 An Op–Amp circuit for problem R2 R1 ZR2 + ZC2 V ZR1 + ZC1 in where ZR1 = R1 Ω is the impedance of the resistor R1. R2 = 1 kΩ and XC = 0.5 k Ω. − Vo = − 155 P5-35. and ZC = −j XC Ω is the impedance of the capacitor.5 kΩ. ZC1 . In the Op–Amp circuit shown in Fig. ZR2 = R2 Ω is the impedance of the resistor R2. (b) If Vin = 1. In the circuit shown in Fig. ZR2 = R2 Ω is the impedance of the resistor R2. the output voltage Vo is given by ZR2 V ZR1 + ZC in where ZR1 = R1 Ω is the impedance of the resistor R1. determine Vo in both rectangular and polar forms. The impedances Z1 .35.35 An Op–Amp circuit for problem P5-35. the output voltage Vo is 5-37. and XC2 = 5 k Ω. and ZC2 as complex numbers in both rectangular and polar forms. and ZC as complex numbers in both rectangular and polar forms. Z2 . determine Vo in both rectangular and polar forms. ZR2 . (a) If R1 = 10 kΩ. (a) If R1 = 1. Zb . 5-35. P5. ZR2 . R2 = 5 kΩ. express ZR1 .36. XC1 = 2. and Zc connected in the Delta interconnection can be converted into their equivalent impedances Z1 . P5. 5.5 ∠0◦ V. Vin + − C R2 R1 C2 +Vcc C1 − + Vin + − + −Vcc Vo − Figure P5. (b) If Vin = 1 ∠−45◦ V. and Z3 connected in the Yee interconnection.34 An Op–Amp circuit for problem P5-34. 7) (a) If Za = 20 Ω. and Zc connected in the Delta interconnection as Za = Zb = Zc = Z1 Z2 + Z2 Z3 + Z3 Z1 Z3 Z1 Z2 + Z2 Z3 + Z3 Z1 Z2 Z1 Z2 + Z2 Z3 + Z3 Z1 Z1 . and Z3 in both rectangular and polar forms.33 Ω. Z2 = 17. Zb . and Zc = 20 + j 50 Ω. express Z1 . . and Zc as Z1 = Z2 = Z3 = 5-38. and Z3 = 3. Za Zb Za + Zb + Zc Za Zc Za + Zb + Zc Zc Zb Za + Zb + Zc .5 Ω. and Zc = 20 + j 10 Ω. (5.5 + j 10. Zb . and Zc in both rectangular and polar forms. Z2 . 5. (5.5 + j 7. the impedances Z1 .8) (a) If Z1 = 2.33 + j 3. 5-39.0 − j 1.66 Ω. Zb . (b) Determine Za .5 Ω. Repeat problem P5-37 if Za = 10 Ω. and Z3 connected in the Yee interconnection can be converted into their equivalent impedances Za . Z2 = 5.37 Impedances connected in Delta and Yee interconnections.75 Ω. Zb . Zb = −j 10 Ω. Z3 Zb Zc Z2 Z1 Za Figure P5. Zb = j 20 Ω.37. express Za . In the circuit shown in Fig. Z2 . (b) Determine Z1 . and Zc as complex numbers in both rectangular and polar forms. and Z3 = 3.6 Ω. 5-40.75 − j 8. and Z3 as complex numbers in polar form. Z2 .156 Chapter 5 Complex Numbers in Engineering impedances Za . Repeat problem P5-39 if Z1 = 3.5 − j 2. A few examples of a sinusoid are the motion of a one-link planar robot rotating at a constant rate. phase angle.1 ONE-LINK PLANAR ROBOT AS A SINUSOID A one-link planar robot of length l and angle 𝜃 is shown in Fig. In this chapter. frequency (both linear and angular). whereas in many other parts of the world this frequency is 50 Hz.1.3. and the voltage waveform of an electric power source. and phase shift. These functions are the most important signals because all other signals can be constructed from sine and cosine signals. The sinusoid has the form of a sine (sin) or a cosine (cos) function (discussed in Chapter 3) and has applications in all engineering disciplines. the plots of y = l sin 𝜃 and x = l cos 𝜃 are shown in Figs. 6. 6. 157 . the frequency of the voltage waveform associated with electrical power in North America is 60 cycles per second (Hz). the oscillation of an undamped spring-mass system. For example.1 A one-link planar robot. 6. the example of a one-link robot rotating at a constant rate will be used to develop the general form of a sinusoid and explain its amplitude. Varying 𝜃 from 0 to 2𝜋 radians and assuming l = 1 (l has units of x and y). It was shown in Chapter 3 that the tip of the robot has coordinates x = l cos 𝜃 and y = l sin 𝜃. The sum of sinusoids of the same frequency will also be explained in the context of both electrical and mechanical systems. y l θ x y x Figure 6.Sinusoids in Engineering CHAPTER 6 A sinusoid is a signal that describes a smooth repetitive motion of an object that oscillates at a constant rate (frequency) about an equilibrium point. respectively.2 and 6. Figures 6. 2π θ . 6. for 0 ≤ 𝜃 ≤ 2 𝜋. It can be seen from Fig.2 y-coordinate of one-link robot with l = 1 (sine function). thus completing one full cycle.3 x-coordinate of one-link robot with l = 1 (cosine function). rad . 6. the plot of sine and cosine functions will complete 1 cos θ 0 0 π 2 π 3π 2 −1 Figure 6. respectively. it can be seen that cos 𝜃 goes from 1 (for 𝜃 = 0) to 0 (for 𝜃 = 𝜋∕2) to −1 (for 𝜃 = 𝜋) to 0 (for 𝜃 = 3 𝜋∕2) and back to 1 (for 𝜃 = 2 𝜋).158 Chapter 6 Sinusoids in Engineering 1 sin θ 0 π 2 0 π 3π 2 2π θ . thus completing one full cycle.2 that sin 𝜃 goes from 0 (for 𝜃 = 0) to 1 (for 𝜃 = 𝜋∕2) and back to 0 (for 𝜃 = 𝜋) to −1 (for 𝜃 = 3 𝜋∕2) back to 0 (for 𝜃 = 2 𝜋). From Fig. rad −1 Figure 6. Note that the minimum value of the sin and cos functions is −1 and the maximum value of the sin and cos functions is 1.4 and 6.5 show two cycles of the sine and cosine functions.3. and it can be seen from these figures that sin(𝜃 + 2 𝜋) = sin(𝜃) and cos(𝜃 + 2 𝜋) = cos(𝜃). Similarly. 5 Two cycles of the cosine function. Thus. 6. 3π 2 3π 7π 2 4π θ .6.2 ANGULAR MOTION OF THE ONE-LINK PLANAR ROBOT Suppose now that the one-link planar robot shown in Fig. 1 cos θ 0 0 π 2 π 3π 2 2π −1 Figure 6. rad . as shown in Fig. another cycle from 4 𝜋 to 6 𝜋 and so on for every 2 𝜋.1 is rotating with an angular frequency 𝜔.4 Two cycles of the sine function. 6. the period of the sine and cosine functions can also be written as 360◦ . Since 𝜋 = 180◦ . rad −1 Figure 6. 6.6.2 Angular Motion of the One-Link Planar Robot 159 1 sin θ 0 0 π 2 π 2π 3π 2 3π 2 3π 7π 2 4π θ . both the sine and cosine functions are called periodic functions with period T = 2 𝜋 rad. and y-components oscillate between l and −l. l y = l sin t 0 0 π 2 π 3π 2 2π t. and it was found that 𝜔 = 1 rad/s (i. it took 2 𝜋s to complete one cycle of the sinusoidal signals.6 A one-link planar robot rotating at a constant angular frequency 𝜔. The resulting plots of y = l sin t and x = l cos t are shown in Figs. ..7 and 6. 6. Suppose the robot starts from 𝜃 = 0 at time t = 0 s and takes t = 2 𝜋s to complete one revolution. The angle traveled in time t is given by 𝜃 = 𝜔 t. The x. y(t) = l sin(𝜃) = l sin(𝜔 t) and x(t) = l cos(𝜃) = l cos(𝜔 t). the robot went through 1 radian of rotation).8. 6.7 and 6. s −l Figure 6. where the length l is the amplitude of the sinusoids. Therefore. respectively. in 1 s.2.8. 6.e.160 Chapter 6 Sinusoids in Engineering y l θ =ωt y x x Figure 6.7 The y-component of the one-link planar robot completing one cycle in 2𝜋 s. the angular fre2 𝜋rad 𝜃 = 1 rad/s. Since 𝜃 = 𝜔 t.1 Relations between Frequency and Period In Figs. and the time period to complete one cycle is quency is 𝜔 = = t 2 𝜋s T = 2 𝜋s. . 𝝎 Since 𝜔 = 2 𝜋f . Therefore.e. The angular frequency is 𝜔 = 2 𝜋s period is T = 2 𝜋∕2 = 𝜋s.6.10. 𝜔. the revolutions (i. 6.2 Angular Motion of the One-Link Planar Robot 161 l x = l cos t 0 0 π 2 π 3π 2 2π t. assume that a one-link robot of length l goes through two 4 𝜋rad = 2 rad/s. the relationship between the angular frequency 𝜔 and linear frequency f is given by 𝝎 = 2 𝝅f . For example. T T= 2𝝅 . which means f is the reciprocal of T. f = 1 . Solving for T gives The above relations allow the computation of f . and their plots are as shown in Figs. s −l Figure 6.9 and 6. This is called the linear frequency or simply frequency f . y(t) = l sin(𝜔 t) = l sin(2 t) and x(t) = l cos(𝜔 t) = l cos(2 t). or T when only one of the three is given. a robot rotating at 1 rad/s would go through 1/2𝜋 cycles in 1s. and the frequency is f = 2∕2 𝜋 = 1∕𝜋Hz. Therefore. . with units of cycle/s (s−1 ).8 The x-component of the one-link planar robot completing one cycle in 2𝜋s. In other words. T 𝝎= 2𝝅 . respectively. the period T is defined as the number of seconds per cycle. By definition. Since one revolution (cycle) = 2 𝜋rad. 4 𝜋rad) in 2 𝜋s. 10 The x-component of one-link planar robot completing one cycle in 2𝜋s or 𝜔 = 2 rad/s.9 The y-component of one-link planar robot completing two cycles in 2𝜋s or 𝜔 = 2 rad/s. starts rotating from an initial position 𝜃 = 𝜋/8 rad and takes T = 1 s to complete one revolution. PHASE SHIFT. 6. s −l Figure 6.and y-components are given by x(t) = l cos 𝜃 = l cos (𝜔 t + 𝜙) ) ( 𝜋 = l cos 𝜔 t + 8 . as shown in Fig.11. AND TIME SHIFT Suppose now that a robot of length l = 10 in. 6. the x.3 PHASE ANGLE. l x = l cos (2t) 0 0 π 2 π 3π 2 2π t. At any time t. s −l Figure 6.162 Chapter 6 Sinusoids in Engineering l y = l sin (2t) 0 π 2 0 π 3π 2 2π t. 12. the sinusoid shifts to the left as shown in Fig. and 𝜙 = 𝜋/8 rad. Therefore. = 16 time shift = Note that the phase angle used to calculate the time shift must be in rad. the phase shift is a shift in radians or degrees.11 One-link planar robot starting rotation from an angle of 𝜋/8 rad. the x. The time shift is the time it takes the robot moving at a speed 𝜔 to pass through the phase shift 𝜙. the value of the sinusoid given by equation (6.1) where 𝜙 = 𝜋/8 is called the phase angle.and y-components of the one-link robot are given by ) ( 𝜋 x(t) = 10 cos 2 𝜋t + 8 and ) ( 𝜋 . the sinusoid will shift to the right. Setting 𝜃 = 𝜔 t gives Phase angle 𝜙 = Angular frequency 𝜔 𝜋 = 8 2𝜋 1 s. y(t) = 10 sin 2𝜋t + 8 (6.6. = l sin 𝜔 t + 8 Since l = 10. 6. Since the phase angle is positive. the sinusoid given by equation (6. Phase Shift. but if the phase angle is negative. y) ω θ t ngle se a π ϕ= /8 ha =P x Figure 6. If the phase angle is positive.1) is not zero at time t = 0.1) is shifted to the left. 𝜔 = 2 𝜋∕T = 2 𝜋∕1 = 2 𝜋rad/s.3 Phase Angle. Since 𝜙 represents a shift from the zero phase to a phase of 𝜋/8 rad. For example. and Time Shift 163 y t=0 10 (x. and y(t) = l sin 𝜃 = l sin (𝜔 t + 𝜙) ) ( 𝜋 . . it is sometimes called a phase shift. Example 6-1 Consider a cart of mass m moving on frictionless rollers as shown in Fig.2) where A is the amplitude. and 𝜙 is the phase angle.164 Chapter 6 Sinusoids in Engineering y (t) 10 10 sin − π (8) 1 16 15 16 0 1 t. 6. s 1 16 −10 Figure 6. The mass is attached to the end of a spring of stiffness k.12 Plot of the sine function shifted to the left a phase angle of 𝜋∕8 (positive phase angle). 𝜔 is the angular frequency.4 GENERAL FORM OF A SINUSOID The general expression of a sinusoid is x(t) = A sin(𝜔 t + 𝜙) (6.13.13 Harmonic motion of a spring-mass system. x(t) = 2 sin 6 𝜋t + 2 (6. x(t) k m Frictionless rollers Figure 6. 6.3) . Suppose that the position of the mass x(t) is given by ) ( 𝜋 m. substitute t = 2.6.0 s. 12 (b) To find x(t) at t = 2. 2𝜋 2𝜋 and the period of the harmonic motion is given by T= 1 1 = s.3).0 in equation (6. linear and angular frequencies. . the linear frequency f is given by f = 6𝜋 𝜔 = = 3 Hz. period.0 s. (b) Find x(t) at t = 2. which gives ) ( 𝜋 x(2) = 2 sin 6 𝜋(2) + 2 ) ( 𝜋 = 2 sin 12 𝜋 + 2 = 2 sin(12.5 𝜋 − 12 𝜋) ( ) 𝜋 = 2 sin 2 = 2. (d) Plot the displacement x(t) for 0 ≤ t ≤ 3 s. and time shift. Solution ) ( 𝜋 to the general (a) Comparing the position of the mass x(t) = 2 sin 6 𝜋t + 2 expression of equation (6. (c) Find the time required for the system to reach its maximum negative displacement.4 General Form of a Sinusoid 165 (a) Determine the amplitude.2) gives Amplitude A = 2 m Angular frequency 𝜔 = 6𝜋 rad/s 𝜋 Phase angle 𝜙 = rad.0 m. 2 Since the angular frequency is 𝜔 = 2 𝜋f . f 3 The time shift can be determined from the phase angle and the angular frequency as 𝜙 𝜔 ) ( )( 𝜋 1 = 2 6𝜋 Time shift = = 1 s. phase angle.5 𝜋) = 2 sin(12. (c) The displacement reaches the first maximum negative displacement of −2 m ) ( 𝜋 3𝜋 𝜋 = −1 or 𝜃 = 6 𝜋t + = . It should be noted that an integer multiple of 2 𝜋 can always be added or subtracted from the argument of sine and cosine functions. m 2 0 0 1 2 3 t. The resulting amplitude and phase are different from the amplitude and phase of the two original sinusoids. 6. 6 (d) The plot of the displacement x(t) for 0 ≤ t ≤ 3 s is shown in Fig.5 𝜋). .14 Harmonic motion of the mass-spring system for 3 seconds. s −2 Figure 6.14. This is done because the sine and cosine functions are periodic with a period of 2 𝜋. t= x(t). 6. as illustrated with the following example.5 𝜋 to find the value of sin(12. 12 𝜋 was subtracted from the angle 12. Solving for t gives when sin (𝜃) = sin 6 𝜋t + 2 2 2 3𝜋 𝜋 6 𝜋t + = 2 2 3𝜋 𝜋 ⇒ 6 𝜋t = − 2 2 =𝜋 𝜋 ⇒ t= 6𝜋 or 1 s.5 ADDITION OF SINUSOIDS OF THE SAME FREQUENCY Adding two sinusoids of the same frequency but different amplitudes and phases results in another sinusoid (sin or cos) of the same frequency.166 Chapter 6 Sinusoids in Engineering In obtaining x(2). 5) where the objective is to find M and 𝜙.7) cosines : M sin(𝜙) = 8. In terms of a sine function.6) Equating the coefficients of sin(2 t) and cos(2 t) on both sides of equation (6.6.5 Addition of Sinusoids of the Same Frequency Example 6-2 167 Consider an electrical circuit with two elements R and L connected in series as shown in Fig. (6. or v(t) = 6 sin(2 t) + 8 cos(2 t) V. (6. equation (6. 6.16. (6. equations (6.4) The total voltage given by equation (6. the current i(t) = 6 sin(2 t) A flowing through the circuit produces two voltages: vR = 6 sin(2 t) V across the resistor and vL = 8 cos(2 t) V across the inductor. 6. 6.7) and (6.15. + 1Ω i(t) ↑ = 6 sin (2t) A 2 H 3 + vR = 6 sin (2t) V − v = vR + vL + vL = 8 cos (2t) V − − Figure 6.8) are converted to polar form as shown in Fig.5) can be written as 6 sin(2 t) + 8 cos(2 t) = (M cos𝜙) sin(2 t) + (M sin𝜙) cos(2 t).15.15 Addition of sinusoids in an RL circuit.8) To determine the magnitude M and phase 𝜙. The total voltage v(t) across the current source can be obtained using KVL as v(t) = vR (t) + vL (t). 6) = 53. In Fig. . Therefore.6) gives sines : M cos(𝜙) = 6 (6. The objective is to find the amplitude and the phase angle of the resulting sinusoid.13◦ . Using the trigonometric identity sin(A + B) = sin (A)cos (B) + cos (A) sin (B) on the right side.4) can be written as one sinusoid (sine or cosine) of frequency 2 rad/s. √ M = 62 + 82 = 10 𝜙 = atan2(8. v(t) = 6 sin(2 t) + 8 cos(2 t) = M sin(2 t + 𝜙). (6. 927 = 2 = 0.17 Voltage and current relationship for an RL circuit. t= The plot of the voltage and current waveforms is shown in Fig. 2.17. 6. the linear frequency is f = 𝜔∕2 𝜋 = 2∕2 𝜋 = 1∕𝜋Hz.927 rad.13◦ = (53. 10 Voltage 6 Current 0 −0.142 s.13◦ ) V.142 t.464 s (time shift). the angular frequency is 𝜔 = 2 rad/s. v(t) = 6 sin(2 t) + 8 cos(2 t) = 10 sin(2 t + 53.464 s. 6.13◦ ) (𝜋rad∕180◦ ) = 0. The amplitude of the voltage sinusoid is 10 V. the phase angle = 53.16 x Determination of magnitude and phase of the resulting sinusoid in an RL circuit. It can be seen from Fig. the period is T = 𝜋 = 3. and the time shift can be calculated as 𝜙 𝜔 0.464 0 −6 −10 Figure 6. s .17 that the voltage waveform is shifted to the left by 0. Therefore.678 3.168 Chapter 6 Sinusoids in Engineering y M M sin ϕ = 8 ϕ M cos ϕ = 6 Figure 6. Using the trigonometric identity cos(A + B) = cos (A) cos (B) − sin (A) sin (B) on the right side of equation (6. where the voltage waveform lags the current. 6. Therefore. v(t) = 6 sin(2 t) + 8 cos(2 t) = 10 cos(2 t − 36.12) are converted to polar form as shown in Fig.87◦ ) V.18 Determination of magnitude and phase for a cosine function. This expression can also be obtained directly from the sine function using the trig identity sin 𝜃 = cos(𝜃 − 90◦ ). 8) = −36.11) and (6. (6.87◦ ). equations (6. y M cos ϕ 1 = 8 x ϕ1 M M sin ϕ 1 = −6 Figure 6. It will be shown later that it is opposite in an RC circuit.13◦ ) − 90◦ ) = 10 cos(2 t − 36.5 Addition of Sinusoids of the Same Frequency 169 In other words.10) gives sines : M sin(𝜙1 ) = −6 M cos(𝜙1 ) = 8. √ M = 62 + 82 = 10 𝜙1 = atan2(−6. cosines : (6. .6.18. Therefore. (6. 10 sin(2 t + 53. Note: The voltage v(t) can also be represented as one sinusoid using the cosine function as v(t) = 6 sin(2 t) + 8 cos(2 t) = M cos(2 t + 𝜙1 ).10) Equating the coefficients of sin(2 t) and cos(2 t) on both sides of equation (6.12) To determine the magnitude M and phase 𝜙1 .13◦ ) = 10 cos ( (2 t + 53.3◦ .9) The amplitude M and the phase angle 𝜙1 can be determined using a procedure similar to that outlined above. the voltage in the RL circuit leads the current by 53.11) (6.87◦ Therefore.9) gives 6 sin(2 t) + 8 cos(2 t) = (−M sin𝜙1 ) sin(2 t) + (M cos𝜙1 ) cos(2 t). In Fig. Example 6-3 Consider the RC circuit shown in Fig.19. the results of this example can be expressed as follows: √ A cos 𝜔 t + B sin 𝜔 t = A2 + B2 cos ( 𝜔 t − atan2(B.16) are converted to the polar form as shown in Fig. (6. equations (6. 6.18 cos(120 𝜋t − 26.20.13) can be written as a single sinusoid of frequency 120 𝜋rad/s. which gives v(t) = 11.14) gives cosines : sines : M cos(𝜙2 ) = 10 M sin(𝜙2 ) = −5. 6.15) and (6. (6. The objective is to find the amplitude and the phase angle of the sinusoid.14) where the trigonometric identity cos (A + B) = cos A cos B − sin A sin B is employed on the right-hand side of the equation.19 200 μ F 120π Addition of sinusoids in an RC circuit.57◦ ) V.57◦ . (6. 6. Equating the coefficients of sin(120 𝜋t) and cos(120 𝜋t) on both sides of equation (6. A) ) √ A cos 𝜔 t + B sin 𝜔 t = A2 + B2 sin ( 𝜔 t + atan2(A. The total voltage v(t) can be obtained using KVL as v(t) = vR (t) + vC (t) = 10 cos(120 𝜋t) + 5 sin(120 𝜋t).16) To determine the magnitude M and phase angle 𝜙2 . The total voltage can be written as a cosine function as 10 cos(120𝜋t) + 5 sin(120𝜋t) = M cos(120𝜋t + 𝜙2 ) = (Mcos𝜙2 ) cos(120𝜋t) + (−Msin𝜙2 )sin(120𝜋t). √ M = 102 + 52 = 11. + + vR = 10 cos 120π t V − v = vR + vC + vC = 5 sin 120π t V − − 1 kΩ i(t) ↑ = cos(120π t)mA C C= Figure 6.13) The total voltage given by equation (6. the current i(t) = cos(120 𝜋t) mA flowing through the circuit produces two voltages: vR = 10 cos(120 𝜋t) across the resistor and vC = 5 sin(120 𝜋t) across the capacitor. 10) = −26.19.15) (6. Therefore. B) ) . .18 𝜙2 = atan2(−5.170 Chapter 6 Sinusoids in Engineering In general. 11.18 Voltage in V 1. the voltage in this RC circuit lags the current by 26. 6.22. the phase angle is −26. It can be seen from the figure that the voltage waveform is shifted to the right by 1.21.20 M sin ϕ 2 = −5 Determination of magnitude and phase of the resulting sinusoid in an RC circuit. The load applied to the hip implant is given by F(t) = 250 sin(6 𝜋t) + 1250 N .7 ms. 6. In other words.57◦ . as shown in Fig.23 ms Current in mA 1. The plot of the voltage and current waveforms is shown in Fig.0 0 1.0 0 1.57◦ . ms 16.7 −11.6. Example 6-4 A hip implant is subjected to a cyclic load during a fatigue test.23 ms (time shift).8 V.23 −1.18 Figure 6.5 Addition of Sinusoids of the Same Frequency 171 y M cos ϕ 2 = 10 x ϕ2 M Figure 6. the angular frequency is 𝜔 = 120 𝜋rad/s.23 ms. the period is T = 16. and the time shift is 1. the linear frequency is f = 𝜔/2𝜋 = 60 Hz.21 Voltage and current relationship for an RC circuit.23 ms t. The amplitude of the voltage sinusoid is 11. Frequency : 𝜔 = 2 𝜋f = 6𝜋 Period : T= ⇒ f = 3 Hz.) Vertical shift = 1250 N. (b) Plot one cycle of F(t) and indicate the earliest time when the force reaches its maximum value. and vertical shift (in Newtons) of the load profile.2) gives amplitude = 250 N.22 Hip implant subjected to a cyclic load. (b) A plot of one cycle of the cyclic force F(t) is shown in Fig.23. .172 Chapter 6 Sinusoids in Engineering (a) Write down the amplitude. Solution (a) Comparing the force F(t) = 250 sin(𝜋t) + 1250 to the general form of equation (6. 1 f 1 s. 6. = Phase angle : Time shift : 6 𝜋t = 0 ⇒ t = 0 s (There is no time shift. F(t) Figure 6. It can be seen from this figure that the earliest time the force has a maximum value of 1500 N is at time tmax = 1∕12 s. time shift (in seconds). The times when the force F(t) is maximum can also be found analytically as 250 sin(6 𝜋tmax ) + 1250 = 1500 = 250 ⇒ ⇒ 250 sin(6 𝜋tmax ) sin(6 𝜋tmax ) = 1. period (in seconds). phase angle (in degrees). 3 𝜙 = 0◦ . frequency (in Hertz). N 0 0 1/12 Figure 6.and y-components as a function of time. and time shift. Also find the amplitude. phase angle.Problems 173 Therefore. period. s 1/4 One cycle of the force F(t).500 1.1 Rotating one-link robot starting at Figure P6. ⋅ ⋅ ⋅ s.1. plot the x. and time shift. . tmax = 1. If l = 5 in. y y t= t=0 l 0 l x π /6 x Figure P6. 12 4 Therefore.23 1/6 1/3 t. the earliest time the maximum force occurs is t = 1∕12s.250 1. ⋅ ⋅ ⋅. 2 2 which gives 1 1 .000 Force. frequency. . 𝜃 = 30◦ . Also find the amplitude.2. plot the x.and y-components as a function of time. phase angle. The tip of a one-link robot is located at 𝜃 = 0 at time t = 0 s as shown in Fig. P6.2 Rotating one-link robot starting at 𝜃 = 0◦ . 6 𝜋tmax = 𝜋 3𝜋 . The tip of a one-link robot is located at 𝜃 = 𝜋∕6 rad at time t = 0 s. It takes 1 s for the robot to move from 𝜃 = 0 to 𝜃 = 2 𝜋rad.. 6-2. PROBLEMS 6-1. as shown in Fig. If l = 10 in. P6.. It takes 2 s for the robot to move from 𝜃 = 𝜋∕6 rad to 𝜃 = 𝜋∕6 + 2 𝜋rad. period. frequency. P6. P6. y t= 0 6-3. and phase angle of the motion. It takes 2 s for the robot to move from 𝜃 = 3 𝜋∕4 rad to 𝜃 = 3 𝜋∕4 + 2 𝜋rad. 6-4. plot the x.6 Rotating one-link robot starting at y l Figure P6.and y-components as a function of time. write the expression for x(t).and y-components as a function of time. It takes 3 s for the robot to move from 𝜃 = 𝜋rad to 𝜃 = 3 𝜋rad. t=0 π /2 x Figure P6. P6.5. Also find the amplitude. period. 6-8.and y-components as a function of time. and time shift. 6-5.7. 6-6. The tip of a one-link robot is located at 𝜃 = 3 𝜋∕4 rad at time t = 0 s as shown in Fig.174 Chapter 6 Sinusoids in Engineering Also find the amplitude. Also. If l = 5 cm. and time shift.3 Rotating one-link robot starting at 𝜃 = −45◦ . period. phase angle. 6-7. frequency. period. The tip of a one-link robot is located at 𝜃 = 𝜋∕2 rad at t = 0 s as shown in Fig.8. Repeat problem P6-8 for the sinusoidal motion shown in Fig. period. l 3π 4 x y − x π 4 l t= 0 Figure P6. Also find the amplitude. and phase angle of the motion. frequency. and time shift.4.4 Rotating one-link robot starting at 𝜃 = 90◦ . period.5 Rotating one-link robot starting at 𝜃 = 180◦ . and time shift. It takes 4 s for the robot to move from 𝜃 = 𝜋∕2 rad to 𝜃 = 𝜋∕2 + 2 𝜋rad. phase angle. 𝜃 = 135◦ . A spring-mass system moving in the x-direction has a sinusoidal motion as shown in Fig.3. 6-9. A spring-mass system moving in the y-direction has a sinusoidal motion as shown in Fig. P6. The tip of a one-link robot is located at 𝜃 = 𝜋rad at time t = 0 s as shown in Fig. plot the x. The tip of a one-link robot is located at 𝜃 = −𝜋∕4 rad at time t = 0 s as shown in Fig. frequency. period.9. y t=0 l π x Figure P6. If l = 10 cm.6. . P6. frequency. phase angle. The robot is rotating at an angular frequency of 2 𝜋rad/s. P6. frequency. P6. Determine the amplitude. Also. If l = 20 cm.and y-components as a function of time. Also find the amplitude. If l = 15 cm. phase angle. Determine the amplitude. plot the x. write the expression for y(t). plot the x. frequency. cm 0 k 1 30 0 1 t. s 0.5 t. s 15 x(t) m −5 Frictionless rollers Figure P6.4 x(t) m Frictionless rollers −10 Figure P6.9 Motion of a spring-mass system in the x-direction for problem P6-9. in. cm y(t) 0 0 0.9 t.7 Sinusoidal motion of a spring-mass system in the y-direction for problem P6-7.1 k 0 0 0.Problems 175 2 y(t). 5 x(t). 0.8 Sinusoidal motion of a spring-mass system in the x-direction for problem P6-8. −0.25 m k −2 Figure P6. 10 x(t). s . k x(t) m Frictionless rollers Figure P6. 0 0 k 0. The position of a spring-mass system shown in Fig. P6. Suppose the spring-mass system of problem P6-11 is displaced −10 cm and takes 𝜋∕4 s to complete a cycle. frequency. (c) Plot one complete cycle of x(t). The system then vibrates under a simple harmonic motion in the horizontal direction.13 is given by x(t) = 10 sin(4 𝜋t − 𝜋2 ) cm.)P6.176 Chapter 6 Sinusoids in Engineering 10 x(t).6 1. (c) Plot one complete cycle of x(t). and time shift on the graph. 6-11. and time shift of the position x(t). 6-12. period. . 6-13.13 A spring-mass system for problem P6-13. frequency. A spring-mass system is displaced x = 10 cm and let go. and indicate the amplitude and period on the graph. frequency. (b) Find the time required for the system to reach the first maximum displacement. (a) Find the amplitude. P6. period. and indicate the amplitude.10. in. (b) Find the time required for the system to reach the equilibrium point (i. period. (b) The time required for the system to reach −10 cm. s x(t) m Frictionless rollers −10 Figure P6. 6-14. and period of the motion. it travels back and forth from 10 cm to −10 cm. 6-10. (b) Find the time required for the system to reach x(t) = 0 cm and x(t) = 10 cm for the first time (after t = 0). (c) Plot one complete cycle of x(t). The position of a spring-mass system shown ( in Fig.13 is given by x(t) = 8 sin 2 𝜋t + 𝜋 4 cm.1 0. and time shift of the position of the mass. and period of the motion.10 Motion of a spring-mass system in the x-direction for problem P6-10. x(t) = 0). If it takes the system 𝜋∕2 s to complete one cycle of the harmonic motion. Repeat problem P6-8 for the sinusoidal motion shown in Fig. frequency. (a) Find the amplitude.e. in other words.1 t. and indicate the amplitude and the time shift on the graph. determine (a) The amplitude. (a) Find the amplitude. P6.86◦ ) mA is applied to the RC circuit shown in Fig. Note g = and 𝜃(t) = 5 sin L 32.8 m/s2 Figure P6. and time shift of the position of the mass. Write v(t) in the form v(t) = M sin(100 t + 𝜃).18. (c) Plot one complete cycle of x(t).61 cos(120 𝜋t − 75.e.28 V. P6. and indicate the amplitude and period on the graph.28 sin 100 t − 𝜋 4 . and time shift (in seconds) of the voltage v(t). 𝜃(t) = 0). and indicate the earliest time (after t = 0) when voltage is 28.86◦ ) V vC (t) = 10. and indicate the amplitude and the time shift on the graph. and period of oscillation of 𝜃(t). A sinusoidal current i(t) = 0. (c) Plot one complete cycle of 𝜃(t).18 RC circuit for problem P6-18. (a) Find the amplitude.13 is given by x(t) = 5 cos(𝜋t) cm.1 sin(100 t) amps is applied to the RC circuit shown in Fig. (b) Suppose now ) that v(t) = 28. find M and 𝜃. A simple pendulum of length L = 100 cm is shown in Fig. The voltage across the resistor and capacitor are given by vR (t) = 40 cos(120 𝜋t + 14.. g t + 𝜋2 . in other words. (c) Plot one (cycle of the ) voltage v(t) = 28.16. (a) The voltage applied to the circuit is given by v(t) = vR (t) + vC (t). period (in seconds). (a) Find the amplitude. x(t) = −5 cm). (b) Find the time required for the simple pendulum to reach its first zero angular displacement (i.e. 6-19. A sinusoidal current i(t) = 200 cos (120 𝜋t + 14. P6.28 ( sin 100 t − L 177 𝜋 4 volts.5 cos L 6-17. 6-16. The angular displacement 𝜃(t) in radians is given by (√ ) g t .16 A simple pendulum. phase angle (in degrees). period.18. frequency. frequency (in Hz). The position of a spring-mass system shown in Fig.Problems (c) Plot one complete cycle of x(t). 𝜃(t) = 0. i(t) + θ g = 9. 6-18.2 ft/s2 .14◦ ) V where t is in seconds. The voltage across the resistor and capacitor are given by vR (t) = 20 sin(100 t) V vC (t) = −20 cos(100 t) V where t is in seconds. Repeat problem (√P6-16 if) L = 10 in.. frequency. P6. 6-15. + vR(t) − + vc(t) − 200 Ω v(t) − 50 μF Figure P6. . Write down the amplitude. and indicate the amplitude and the time shift on the graph. (b) Find the time required for the system to reach its first maximum negative displacement (i. and indicate the earliest time (after t = 0) when voltage is −41. in other words. (a) Write down the amplitude. (b) Plot one cycle of the voltage vL (t). phase shift (in radians). (c) Plot one cycle of the voltage v(t) = 41. . and time shift (in ms) of the current i(t) = 0. Write down the amplitude. and indicate the earliest time after t = 0 when the voltage is minimum. (c) The total voltage across the circuit is given by v(t) = vR (t) + vL (t). and time shift (in seconds) of the voltage vL (t). phase angle (in degrees). The voltage across the resistor and inductor are 𝜋t) and given by vR((t) = 0. as shown in Fig. The voltage across the resistor and inductor are 𝜋t) and given by vR((t) = 10 cos(120 ) vL (t) = 12 cos 120 𝜋t + t is in seconds. frequency (in Hz). period (in seconds). Write v(t) in the form v(t) = M cos(20𝜋t + 𝜃). 𝜋 2 volts. and C of vR (t) = 7. find M and 𝜃. 6-20. phase shift (in degrees).07 sin(1000 t + 45◦ ) V vL (t) = 7. A sinusoidal voltage v(t) = 10 sin (1000 t) V is applied to the RLC circuit shown in Fig. phase shift (in degrees). as shown in Fig. frequency (in Hz). (b) Plot one cycle of the voltage vL (t). (c) The total voltage across the circuit is given by v(t) = vR (t) + vL (t). frequency (in Hz). P6. A series RL circuit is subjected to a sinusoidal voltage of frequency 20 𝜋rad/s. where (a) Write down the amplitude. 6-22. find M and 𝜃. and time shift (in seconds) of the voltage v(t). 6-21.07 sin(1000 t + 135◦ ) V vC (t) = 14. The current i(t) = 10 cos(120 𝜋t) A is flowing through the circuit. period (in seconds). P6.1 sin 20 𝜋t + t is in seconds. A series RL circuit is subjected to a sinusoidal voltage of frequency 120 𝜋rad/s.14 sin(1000 t − 45◦ ) V. period (in seconds). (b) Suppose now that v(t) = 41. find M and 𝜃. P6. L.38 V. The current i(t) = 0. frequency (in Hz). Write v(t) in the form v(t) = M cos(120𝜋t + 𝜃). The current i(t) = 500 sin(20 𝜋t) mA is flowing through the circuit.707 sin(1000 t + 45◦ ) A. period (in seconds). where i(t) v(t) + − + R vR (t) − + L vL (t) − 10 R = 1 Ω. and time shift (in seconds) of the voltage vL (t). Write v(t) in the form v(t) = M cos(120 𝜋t + 𝜃).38 cos (120 𝜋t) volts. L = π mH Figure P6.178 Chapter 6 Sinusoids in Engineering (a) The voltage applied to the circuit is given by v(t) = vR (t) + vC (t). in other words.707sin(1000 t + 45◦ ) flowing through the circuit produces voltages across R.22. in other words.20. and indicate the earliest time after t = 0 when the voltage is maximum.5 sin(20 ) vL (t) = 0.20 A series RL circuit for problem P6-20.20. (a) Write down the amplitude. 𝜋 2 volts.38 cos (120 𝜋t). A parallel RL circuit is subjected to a sinusoidal voltage of frequency 10 𝜋rad/s. in other words.1414 sin 10𝜋t + 4 mA. The currents i1 (t) and i2 (t) are given by i1 (t) = 100 cos(10 𝜋t) mA i2 (t) = 100 sin(10 𝜋t) mA.707 sin(1000t + 45°) A v(t) = 10 sin(1000t) V + vR − + vL − + vC − + − R = 10 Ω L = 10 mH C = 50 μF Figure P6. find M and 𝜃.24. find M and 𝜃.23. plot one cycle of the current i(t). and clearly indicate the earliest time after t = 0 at which it reaches its maximum value. plot one cycle of the current i(t). (c) Given your results of part (b).14 sin 120𝜋t − 4 A. ) ( 𝜋 (b) Suppose i(t) = 0.707 A. find M and 𝜃. The currents i1 (t) and i2 (t) are given by i1 (t) = 10 sin(120 𝜋t) A ) ( 𝜋 A. i(t) i1 (t) i2 (t) v(t) = 120 sin(120 π t) V + − L R = 12 Ω 100 L = π mH Figure P6.Problems ) ( 𝜋 (b) Suppose i(t) = 14. phase shift (in degrees).707 sin(1000 t + 45◦ ) A and indicate the earliest time (after t = 0) when the current is 0. and time shift (in seconds). P6. (c) Using trigonometric identities. (a) Given that i(t) = i1 (t) + i2 (t). phase shift (in degrees). A parallel RL circuit is subjected to a sinusoidal voltage of frequency 120 𝜋rad/s. and clearly indicate the earliest time . i(t) = 0. period (in seconds). as shown in Fig. show that v1 (t) = vR (t) + vC (t) = 15 sin 1000 t − 5 cos 1000 t.23 A parallel RL circuit for problem P6-23. (c) Given your results of part (b). P6. in other words. write i(t) in the form i(t) = M sin (120 𝜋t + 𝜃). write i(t) in the form i(t) = M sin(10 𝜋t + 𝜃). frequency (in Hz). Determine the amplitude.24 A parallel RL circuit for problem P6-24. 179 6-24. (d) Write v1 (t) obtained in part (c) in the form v1 (t) = M cos(1000 t + 𝜃). frequency (in Hz). i(t) i1(t) i2(t) v(t) = 10 sin(10 π t) V + − L R = 100 Ω 1 L = π mH Figure P6. (b) Plot one cycle of the current i(t) = 0. i2 (t) = 10 sin 120 𝜋t − 2 (a) Given that i(t) = i1 (t) + i2 (t). Determine the amplitude. period (in seconds). in other words. 6-23. and time shift (in seconds).22 RLC circuit for problem P6-22. as shown in Fig. P6. (b) Plot one cycle of the voltage v1 (t). Repeat problem P6-27 if v1 (t) = 10 cos(100 𝜋t + 90◦ ) V and v2 (t) = 3 sin (100 𝜋t + 𝜋4 ) V. given by (c) The output voltage vo (t) is ) (√ vo (t) = − 2 v1 (t) + v2 (t) .27 An Op–Amp circuit for problem P6-27. (b) Suppose now that i(t) = 4 sin ) ( 𝜋 A. in other words. 6-28.180 Chapter 6 Sinusoids in Engineering after t = 0 at which it reaches its maximum value. period (in seconds). Consider the RC circuit shown in Fig. P6. ( √ 6-29. (a) Write down the amplitude.29. find M and 𝜙. Determine the 120 𝜋t + 3 amplitude. write i(t) in the form i(t) = M sin (120 𝜋t + 𝜙) A. in other words. and time shift (in seconds). phase shift (in degrees). i(t) i1(t) i2(t) v(t) + − C R 6-27. 4 (a) Given that i(t) = i1 (t) + i2 (t). write i(t) in the form i(t) = M sin(100 t + 𝜙) A. plot one cycle of the current i(t).28 kΩ 20 kΩ −+ +VCC v2 (t) −+ Figure P6. (b) Suppose now that i(t) = 0. P6.27. Two voltages v1 (t) = 10 sin(100 𝜋t − 45◦ ) V and v2 (t) = 10 sin(100 𝜋t) V are applied to the Op–Amp circuit shown in Fig.51 sin (100 t + 146. where the currents are i1 (t) = 100 cos(100 𝜋t + 𝜋4 ) mA and i2 (t) = 500 cos(100 t + 3𝜋 ) mA. where the currents are (t) = 2 sin(120 𝜋t) A and i2 (t) = i1 √ 2 3 sin(120 𝜋t + 𝜋2 ) A.28 kΩ P6-25. Write vo (t) in the form vo (t) = M cos (100𝜋t + 𝜃 ◦ ). in other words. and clearly indicate the earliest time after t = 0 at which it reaches its maximum value. Two voltages v (t) = 10 2 sin 500 𝜋t − ) 1 3𝜋 V and v2 (t) = 5 sin(500𝜋t) V are 4 applied to the Op–Amp circuit shown in Fig. Consider the RC circuit shown in Fig. Determine the amplitude. frequency (in Hz). . (c) Given your results of part (b). find M and 𝜃). period (in seconds). (a) Given that i(t) = i1 (t) + i2 (t). 6-25. v1 (t) 28. P6. plot one cycle of the current i(t). and time shift (in seconds) of the voltage v1 (t). phase shift + −VCC vo (t) − Figure P6.25. period (in seconds).31◦ )A.25 A parallel RC circuit for problem − + 28. 6-26. frequency (in Hz). find M and 𝜙. (c) Given your results of part (b). frequency (in Hz). (in degrees). and time shift (in seconds).25. and clearly indicate the earliest time after t = 0 at which it reaches its maximum value. phase shift (in radians). and indicate the earliest time after t = 0 when the voltage is 10 V. period (in seconds). frequency (in Hz). phase angle or phase shift (in radians). in other words. phase shift (in degrees). Write vo (t) in the form vo (t) = M sin(500𝜋t + 𝜃) (i. and time shift (in seconds) of the position of the first mass y1 (t). frequency (in Hz).. Two oscillating masses are connected by a spring as shown in Fig.28 kΩ + +Vcc − v2(t) 181 28. where t is in seconds. where t is in seconds. frequency (in Hz). in other words. 6-30. 6-32. (c) The vertical distance between the two masses is given by 𝛿(t) = y1 (t) − y2 (t).29 An Op–Amp circuit for problem P6-29. frequency (in Hz). (a) Write down the amplitude. find M and 𝜃). and indicate the earliest time after t = 0 when the position is zero.30 A pair of springs and masses for problem P6-30.28 kΩ − + − + +Vcc − + −Vcc + −Vcc vo(t) − Figure P6. A pair of springs and masses vibrate under simple harmonic motion. (b) Plot one cycle of the position y1 (t). and indicate the earliest time after t = 0 when the position is zero. and indicate the earliest time √ after t = 0 when the voltage is 10 2 V. (b) Plot one cycle of the voltage v1 (t). The positions of the masses ) are given by ( in inches √ 𝜋 and x2 (t) = x1 (t) = 5 2 cos 2 𝜋t + 4 10 cos(2 𝜋t).e. P6. as shown in Fig. Suppose the positions of the masses in problem P6-30 are given by y1(t) y2(t) Figure P6. . (c) The output voltage vo (t) is given by vo (t) = v1 (t) + v2 (t). and time shift (in seconds) of the voltage v1 (t).Problems v1(t) 28. period (in seconds).30. The positions of the masses in inches are given ) ( √ 𝜋 and by y1 (t) = 5 2 cos 2 𝜋t + 4 y2 (t) = 10 cos(2 𝜋t). where t is in seconds. find M and 𝜃. (a) Write down the amplitude. and time shift (in seconds) of the position of the first mass y1 (t). period (in seconds).32. (a) Write down the amplitude. (a) Write down the amplitude. (b) Plot one cycle of the position y1 (t). (c) Write 𝛿(t) = y1 (t) − y2 (t) in the form 𝛿(t) = M cos(4 𝜋t + 𝜃 ◦ ). period (in seconds). P6. find M and 𝜃. ) ( 𝜋 and y2 (t) = y1 (t) = 8 sin 4 𝜋t + 3 6 cos(4 𝜋t). Write 𝛿(t) in the form 𝛿(t) = M sin(2 𝜋t + 𝜃 ◦ ). 6-31.28 kΩ 28. phase shift (in degrees). (c) Write 𝛿(t) = x2 (t) − x1 (t) in the form 𝛿(t) = M cos(4 𝜋t + 𝜙). vab (t) = van (t) − vbn (t).e. and phase shift (in seconds) of the voltage across the motor. and indicate the earliest time after t = 0 when the position is zero. frequency (in Hz). find V and 𝜃).e. find A and B). (c) Write the line-to-line voltage vab (t) in the form vab (t) = A sin(120 𝜋t) + B cos(120 𝜋t) (i. 6-33. (b) Plot one cycle of the line-to-neutral voltage vbn (t). frequency (in Hz). (b) Plot one cycle of the position x1 (t). period. In the three-phase circuit shown in Fig. (c) The elongation of the spring is given by 𝛿(t) = x2 (t) − x1 (t). phase angle (in degrees). in other words. (a) Write down the amplitude..182 Chapter 6 Sinusoids in Engineering x1(t) x2(t) Figure P6. find V and 𝜃).36. where t is in seconds. and indicate the earliest time when the voltage is maximum. Write VT (t) in the form VT (t) = V cos (120 𝜋t + 𝜃) (i. and time shift (in seconds) of the position of the first mass x1 (t). and time shift (in seconds) of the position of the first mass x1 (t). 6-36. VM (t). Repeat problem P6-34 if VH (t) = 100 sin(120 𝜋t) V and VM (t) = 164 sin (120 𝜋t + 70◦ ) V.32 Two oscillating masses for problem P6-32. phase shift (in degrees). find A and B). and phase shift (in seconds) of the voltage. (d) Write vab (t) in the form vab (t) = V cos(120 𝜋t + 𝜃) (i. . vbn (t). (b) Plot one cycle of the position x1 (t). find M and 𝜙. the heater and motor are given by VH (t) = 66 cos(120 𝜋t) V and VM (t) = 180 cos(120 𝜋t + 𝜋3 ) V.e. (d) The total voltage is VH (t) + VM (t). P6. phase angle (in degrees).34.. Write 𝛿(t) in the form 𝛿(t) = M sin(2 𝜋t + 𝜙). (a) Write down the amplitude. 6-35. A manufacturing plant employs a heater and a conveyer belt motor on the same 220 V service line as shown in Fig.34 Conveyer motor and heater connected across the 220 V service line. (b) Plot one cycle of the motor voltage VM (t). phase shift (in degrees). where t is in seconds. period. (c) Write the voltage VM (t) in the form VM (t) = A sin(120 𝜋t) + B cos (120 𝜋t) (i. The voltages across + VT(t) + VH(t) − + VM(t) − Heater Motor − Figure P6. find M and 𝜙. frequency (in Hz).. P6. and indicate the earliest time after t = 0 when the position is zero. and indicate the earliest time when the voltage is maximum. in other words. period (in seconds). 6-34. (a) Write down the amplitude. where van (t) = 120 sin(120 𝜋t) V and vbn (t) = 120 sin(120 𝜋t − 120◦ ) V are the line-toneutral voltages and vab is the line-toline voltage of the three-phase system.. Now assume that the positions of two masses in problem P6-31 ) are given ( 𝜋 and x2 (t) = by x1 (t) = 8 sin 4 𝜋t + 4 16 cos(4 𝜋t).e. where vbn (t) = 120 sin(120 𝜋t − 120◦ ) V and vcn (t) = 120 sin(120 𝜋t + 120◦ ) V are the line-to-neutral voltages and vbc is the line-to-line voltage of the threephase system. period (in seconds). find V and 𝜃). (d) Write vbc (t) in the form vbc (t) = V cos (120 𝜋t + 𝜃) (i. 6-39. P6. and indicate the earliest time when the voltage reaches 60 V. (d) Write vca (t) in the form vca (t) = V cos(120 𝜋t + 𝜃) (i. The hip implant shown in Fig. P6. phase angle (in degrees). period.Problems c van vcn a n vbn b Figure P6.36 A balanced three-phase circuit.e. In the three-phase circuit shown in Fig. vbc (t) = vbn (t) − vcn (t). 6-38. (b) Plot one cycle of the line-to-neutral voltage vcn (t).. time shift (in seconds). frequency (in Hz).. if F(t) = .36. where vcn (t) = 120 sin(120 𝜋t + 120◦ ) V and van (t) = 120 cos(120 𝜋t − 90◦ ) V are the line-to-neutral voltages and vca is the line-to-line voltage of the threephase system. find V and 𝜃).39 Hip implant subjected to a cyclic load. (b) Plot one cycle of F(t) and indicate the earliest time when the force reaches its maximum value. find A and B). 6-37. (a) Write down the amplitude. phase angle (in degrees). vca (t) = vcn (t) − van (t). find A and B). (a) Write down the amplitude. P6.e. van (t). and phase shift (in seconds) of the voltage. and indicate the earliest time when the voltage is minimum. (c) Write the line-to-line voltage vca (t) in the form vca (t) = A sin(120 𝜋t) + B cos(120 𝜋t) (i. In the three-phase circuit shown in Fig.. Repeat problem P6-39 15 sin(10 𝜋t) + 75 N.36. and phase shift (in seconds) of the voltage. The load applied to the hip implant is given by F(t) = 200 sin(5 𝜋t) + 1000 N F(t) Figure P6. frequency (in Hz).. 6-40. period.e. and vertical shift (in N) of the load profile. vcn (t). phase angle (in degrees).39 is subjected to a cyclic load during a fatigue test. frequency (in Hz). (a) Write down the amplitude.e. 183 (b) Plot one cycle of the line-to-neutral voltage van (t). (c) Write the line-to-line voltage vbc (t) in the form vbc (t) = A sin(120 𝜋t) + B cos(120 𝜋t) (i. 1) (7. while the matrix algebra method and Cramer’s rule are explained in detail.1 A two-loop resistive circuit. graphical method. dynamics. 7. While the examples given are limited to 2 × 2 systems of equations.1. the matrix algebra approach is applicable to linear systems having any number of unknowns and is suitable for immediate implementation in MATLAB.CHAPTER 7 7. In this chapter. Using a combination of Kirchhoff’s voltage law (KVL) and Ohm’s law. and DC circuit analysis. It is assumed that the students are already familiar with the substitution and graphical methods from their high school algebra course. 184 (7. a system of two equations with two unknowns I1 and I2 can be obtained as 10 I1 + 4 I2 = 6 12 I2 + 4 I1 = 9.1 Systems of Equations in Engineering INTRODUCTION The solution of a system of linear equations is an important topic for all engineering disciplines. 6Ω 6V + − Figure 7. and Cramer’s rule.2 SOLUTION OF A TWO-LOOP CIRCUIT Consider a two-loop resistive circuit with unknown currents I1 and I2 as shown in Fig. statics. the solution of 2 × 2 systems of equations will be carried out using four different methods: substitution method.2) 8Ω I1 4Ω I2 + − 9V . The objective of this chapter is to be able to solve the systems of equations encountered in beginning engineering courses such as physics. matrix algebra method. 7. 6) are linear equations of the form y = m x + b. 10 (7. 0. the solution of the system of equations (7.2) represent a system of equations for I1 and I2 that can be solved using the four different methods outlined as follows: 1.6346) 10 I1 = 0.3462 A.6) is the intersection point of the two lines.4) The current I1 can now be obtained by substituting the value of the second variable I2 from equation (7. 2.2).2 Solution of a Two-Loop Circuit 185 Equations (7.5) and (7. (7. I2 ) = (0.4 I2 = 6.5) and (7.3) into equation (7.3) The current I2 can now be solved by substituting I1 from equation (7. which gives ( ) 6 − 4 I2 =9 12 I2 + 4 10 12 I2 + 2. solving equation (7. Substitution Method: Solving equation (7.6 I2 = 10. I1 = Therefore. 3 4 (7. The plot of the two straight lines along with their intersection point is .3462 A.2) for I2 gives 4 I1 + 12 I2 = 9 12 I2 = −4 I1 + 9 1 3 I 2 = − I1 + . Graphical Method: Begin by assuming I1 as the independent variable and I2 as the dependent variable. The simultaneous solution of equations (7.7.6346 A.1) for the first variable I1 gives 10 I1 = 6 − 4 I2 6 − 4 I2 I1 = .5) Similarly.6 6.4) into equation (7. Solving equation (7.6346 A).4 − 1.3) as 6 − 4(0.2) is given by (I1 . 2 2 (7.1) and (7.6) Equations (7.4 I2 = 0.6 I2 = 9 10.1) and (7.1) for the dependent variable I2 gives 10 I1 + 4 I2 = 6 4 I2 = −10 I1 + 6 5 3 I2 = − I1 + . 186 Chapter 7 Systems of Equations in Engineering shown in Fig.0 I1.2).7) and (7.2. therefore. one equation can be obtained by performing linear operations on the other equation).8) (7. 7.9) . Rewriting the system of equations (7. Matrix Algebra Method: The matrix algebra method can also be used to solve the system of equations given by equations (7.35. the two equations are dependent (i.7) (7.35 A. they are the same line) and the system of equations has infinitely many solutions. is the solution of the 2 × 2 system of equations (7.63) 1. if the two lines do not intersect. 0.8) in matrix form yields [ ][ ] [ ] 6 10 4 I1 = .5 1. I2.1) and (7. In this case.2 Plot of 2 × 2 system of equations (7. 0. 9 4 12 I2 (7.5 Solution: (I1. A 1 3 I2 = −− I1 +− 3 4 Figure 7. Also. A 2.1) and (7. I2 ) ≈ (0.0 0.0 0.1) and (7.0 1.2). The intersection point. (I1 . then one of the following two possibilities exist: (i) The two lines are parallel lines (same slope but different y-intercepts) and the system of equations has no solution. this method is generally not used when an accurate result is needed.0 0. I2) = (0. (ii) The two lines are parallel lines with same slope and y-intercept (the two lines lie on top of each other.. 3.0 5 3 I2 = −− I1 +− 2 2 1. Now. Note that the graphical method gives only approximate results.1) and (7.63 A).5 2. writing equations (7.2).e.2) in the form so that the two variables line up gives 10 I1 + 4 I2 = 6 4 I1 + 12 I2 = 9.5 0. For any system of the form Ax = b. Now.9). For a 2 × 2 system of equations where [ ] a b A= . [ ] 10 4 A= 4 12 [ ] a b = . In other words.12) is a 2 × 1 matrix on the right-hand side (RHS) of equation (7.11) is a 2 × 1 matrix (column vector) of unknowns.2 Solution of a Two-Loop Circuit Equation (7. the system of equations Ax = b has no solution.9) is of the form Ax = b.7. Note that if Δ = |A| = 0.10) [ ] I x= 1 I2 (7. for the two-loop circuit problem. c d the inverse of the matrix A is given by A−1 = 1 Δ [ d −b −c a ] where Δ = |A| is the determinant of matrix A and is given by |a b| | Δ = || | |c d| = a d − b c. c d The inverse of matrix A is given by −1 A 1 = Δ = 1 Δ [ ] [ d −b −c a ] 12 −4 −4 10 . and [ ] 6 b= 9 (7. 187 (7. the solution is given by x = A−1 b where A−1 is the inverse of the matrix A. A−1 does not exist. where [ ] 10 4 A= 4 12 is a 2 × 2 coefficient matrix. 4. … xi = i |A| |A| |A| where |Ai | is obtained by replacing the ith column of the matrix A with the column vector b.13) =⎢ 5 ⎥⎥ ⎢− 1 ⎣ 26 52 ⎦ [ ] I The solution of the system of equations x = 1 can now be found by multiplying I2 A−1 (given by equation (7.3462A.6346 The solution of the system of equations (7. (7. 0. I2 ) = (0. 0.12)) as x = A−1 b 1⎤ [ ] [ ] ⎡ 3 ⎢ 26 − 26 ⎥ 6 I1 =⎢ I2 5 ⎥⎥ 9 ⎢− 1 ⎣ 26 52 ⎦ ( ) ⎡ 3 (6) + − 1 (9) ⎤ ⎥ ⎢ 26 26 ) ( ) = ⎢( ⎥ 5 1 ⎥ ⎢ − (6) + (9) ⎦ ⎣ 26 52 ⎡ 18 − 9 ⎤ ⎢ ⎥ 26 =⎢ ⎥ −12 + 45 ⎢ ⎥ ⎣ ⎦ 52 [ ] 0.188 Chapter 7 Systems of Equations in Engineering where Δ = |A| = a d − c b = (10) (12) − (4) (4) = 104. Therefore. the inverse of matrix A can be calculated as [ ] 12 −4 1 −1 A = 104 −4 10 1⎤ ⎡ 3 ⎢ 26 − 26 ⎥ .3462 = . Writing the 2 × 2 system of equations a11 x1 + a12 x2 = b1 a21 x1 + a22 x2 = b2 in matrix form as [ a11 a12 a21 a22 ][ ] [ ] x1 b = 1 .1) and (7. x2 = 2 . Cramer’s Rule: For any system Ax = b.6346A).13)) with the column matrix b (given by equation (7. the solution of the system of equations is given by |A | |A | |A | x1 = 1 .2) is therefore given by (I1 . x2 b2 . the currents I1 and I2 can be determined as |6 4 | | | |9 12| | | I1 = |10 4 | | | | 4 12| | | 6(12) − 9(4) = 10(12) − 4(4) 36 104 = 0. the 2 × 2 system of equations is [ ][ ] [ ] 10 4 I1 6 = . 4 12 I2 9 Using Cramer’s rule.6346 A.7. |10 6| | | | 4 9| | | I2 = |10 4 | | | | 4 12| | | = = 10(9) − 4(6) 10(12) − 4(4) 66 104 = 0.3462 A. a11 a22 − a12 a21 For the two-loop circuit. I1 = 0. . Note that Cramer’s rule is probably fastest for solving 2 × 2 systems but not faster than MATLAB.6346 A.3462 A and I2 = 0.2 Solution of a Two-Loop Circuit 189 Cramer’s rule gives the solution of the system of equations as x1 = = x2 = = | b1 a12 | | | |b a | | 2 22 | | a11 a12 | | | |a | | 21 a22 | a22 b1 − a12 b2 . a11 a22 − a12 a21 | a11 b1 | | | | |a | 21 b2 | | a11 a12 | | | |a | | 21 a22 | a11 b2 − a21 b1 . = Therefore. Similarly.7071 T1 + 0.8165 T1 . matrix algebra. This implies that all the forces in the x. Since the system shown in Fig.14) and (7.5 T2 = 95.8660 T1 0 = .3 Chapter 7 Systems of Equations in Engineering TENSION IN CABLES An object weighing 95 N is hanging from a roof with two cables as shown in Fig. and Cramer’s rule. the second variable T2 is found in terms of the first variable T1 as 0.7071 0.3.8660 T2 = 0. respectively.and y-directions are given by T2 cos(30◦ ) N and T2 sin(30◦ ) N.7071 T1 + 0. the sum of all the forces shown in the free-body diagram must be equal to zero.17) . (7. (7.3 is in equilibrium. the matrix algebra method.190 7. (7. the components of the tension T⃗ 2 in the x. Determine the tension in each cable using the substitution.and y-directions are given by −T1 cos(45◦ ) N and T1 sin(45◦ ) N. respectively.15) Equations (7. 7. and Cramer’s rule methods.8660 T2 = 0. Summing all the forces in the x-direction gives −T1 cos(45◦ ) + T2 cos(30◦ ) = 0 −0.3 A 95 N object hanging from two cables. 7.5 95 The solution of the system of equations (T1 and T2 ) will now be obtained using three methods: the substitution method.14) T1 T2 45° 30° x 95 N 95 N Figure 7. summing the forces in the y-direction yields T1 sin(45◦ ) + T2 sin(30◦ ) = 95 0.15) make a 2 × 2 system of equations with two unknowns T1 and T2 that can be written in matrix form as [ ][ ] [ ] −0. The components of the tension T⃗ 1 in the x.and y-directions are equal to zero (see Chapter 4). The components of the object weight is 0 N in the x-direction and −95 N in the y-direction.16) T2 0. 45° 30° y (7. Substitution Method: Using equation (7.7071 T1 T2 = 0. 1. Similarly.7071 0.14). 14) and (7.5 −0.8660 1 −1 A = −0.9659 [ ] 0.5 [ ] T x= 1 T2 [ ] 0 b= .21) where Δ = a d − b c. therefore. 2. a = −0.8660) = −0.17) yields T2 = 0.8165 T1 ) = 95 1.7071 [ ] −0.7071 0.7.20) [ ] T1 can be found T2 by solving equation (7.7321 (7.6 N.7321 0.7071. x. b = 0.9659 −0.17) into equation (7.22) .15) gives 0. and b are given by [ ] −0. T1 = 85.7071 0. the solution of the 2 × 2 system of equations x = (7. for this example. substituting T1 obtained in equation (7. (7. Since. c = 0.5177 0.5) − (0.17 N. Therefore.8165 (85.8660.2 N and T2 = 69.19) Ax = b where matrices A.7071 −0.15) are now determined using the matrix algebra method. 95 Therefore.17) = 69.15) in the matrix form as (7.7071) (0.115 T1 = 95 T1 = 85. Δ = (−0.7071 T1 + 0.8966 = .8660 A= 0.7071.18) into equation (7. Matrix Algebra Method: The two unknowns (T1 and T2 ) in the 2 × 2 system of equations (7.7071) (0.3 Tension in Cables 191 Substituting T2 from equation (7. If A = .55 N.14) and (7. 0. then c d [ ] 1 d −b −1 A = Δ −c a (7. Write equations (7. and d = 0.18) Now.5 (0.5.19) as x = A−1 b where A−1 [ ] a b is the inverse of matrix A. 7071 0 | | | | | | 0.5 || | T1 = | −0.2 69.6 N.7321(95) [ = 85.2 N T2 = | −0.16 −0.7071(0.19).6 N.9659 = 69.15) are now determined using Cramer’s rule. 3.8966(95) T1 = T2 0 + 0. Using matrix equation (7.8660 | | | | | | 95 0.27 −0.192 Chapter 7 Systems of Equations in Engineering Substituting matrices A−1 from equation (7.7321 [ [ ] 0 + 0.8966 0. Cramer’s Rule: The two unknowns (T1 and T2 ) in the 2 × 2 system of equations (7. Therefore.8660) −0. the tensions T1 and T2 can be found as | 0 0.22) and b from equation (7.9659 = −0.2 N and T2 = 69.6 N.20) into equation (7.14) and (7.866 | | | | | | 0.8660) = −82.7071(95) − 0 −0. T1 = 85.7071(0.7071 95 | | | −0.5 || | = 0 − 95(0. Therefore.5) − 0.21) gives [ x= −0.9659 = 85.2 N and T2 = 69.7321 0. T1 = 85.6 ][ 0 ] 95 ] ] .7071 0.5177 0. .7071 0.9659 = −67. R1 where x = .24) gives 6 (4800 − R2 ) − 4 R2 = 0 28.4. 7.25) Substituting R1 from equation (7. Solution (a) Substitution Method: Using equation (7. Therefore.25) into equation (7. 800 R2 = 2880 lb. substituting R2 from equation (7. Perform all computations by hand and show all steps.24) in the matrix form A x = b. the reaction forces R1 and R2 satisfy the equation: R1 + R2 − 4800 = 0 (7.23) and (7.26) into equation (7.7. R1 = 1920 lb and R2 = 2880 lb.26) . (b) Write the system [ ] of equations (7.23) 6 R1 − 4 R2 = 0. (d) Find R1 and R2 using Cramer’s rule. If the weight is W = 4800 lb.4 Reaction forces acting on a vehicle. Now. R2 (c) Find R1 and R2 using the matrix algebra method. (7.4 Further Examples of Systems of Equations in Engineering 7. (7. find R1 in terms of R2 as R1 = 4800 − R2 .24) Also. (7. suppose that (a) Find R1 and R2 using the substitution method.23).4 Example 7-1 193 FURTHER EXAMPLES OF SYSTEMS OF EQUATIONS IN ENGINEERING Reaction Forces on a Vehicle: The weight of a vehicle is supported by reaction forces at its front and rear wheels as shown in Fig. 800 − 6 R2 − 4 R2 = 0 10 R2 = 28. W R1 R2 Figure 7.25) yields R1 = 4800 − 2880 = 1920 lb. 31) (7.27).6)(4800) + 0 [ ] 1920 = .32) Substituting equation (7.31).29) as [ ][ ] 0.1 4800 x= 0. R1 = 1920 lb and R2 = 2880 lb.32) in equation (7.1 0 ] [ ] [ (0.1 The reaction forces can now be found by multiplying A−1 in equation (7.24) in the matrix form gives [ ][ ] [ ] 1 1 R1 4800 = .4)(4800) + 0 R1 = R2 (0.4 0. the matrices A.4 0.33) with matrix b given in equation (7. 6 −4 R2 0 (7. (7.1 = . the inverse of matrix A is given by [ ] −4 −1 1 A−1 = −10 −6 1 [ ] 0.27) (c) Matrix Algebra Method: From the matrix equation (7.28) 6 −4 [ ] 4800 b= (7. 2880 Therefore. . (7.23) and (7.29) 0 [ ] R1 .30) x= R2 The reaction forces can be found by finding the inverse of matrix A and then multiplying this with column matrix b as x = A−1 b where A and −1 [ ] 1 −4 −1 = |A| −6 1 |1 1 | | |A| = || | | 6 −4 | = (1)(−4) − (6)(1) = −10. (7. x.6 −0. and b are given by [ ] 1 1 A= (7.194 Chapter 7 Systems of Equations in Engineering (b) Writing equations (7.33) 0.6 −0. 34) (7.4 Further Examples of Systems of Equations in Engineering 195 (d) Cramer’s Rule: The reaction forces R1 and R2 can be found by solving the system of equations (7. F2 (c) Find F1 and F2 using the matrix algebra method.35) in the matrix form A x = b. (d) Find F1 and F2 using Cramer’s rule. (b) Write the system [ ] of equations (7. Perform all computations by hand and show all steps.5 A truss subjected to external forces. Therefore. F where x = 1 . R1 = 1920 lb and R2 = 2880 lb.35) (a) Find F1 and F2 using the substitution method. | 1 4800 | | | |6 0 || | −10 (1)(0) − (6)(4800) −10 2880. F1 6 ft 200 lb 200 lb 6 ft 100 lb F2 8 ft Figure 7. 7.8 F2 − 200 = 0 0.23) and (7.8 F1 + 0.6 F2 − 100 = 0.24) using Cramer’s rule as R1 = = = R2 = = = | 4800 1 | | | | 0 −4 || | −10 (4800)(−4) − (0)(1) −10 1920. (7.5.7. The forces F1 and F2 satisfy the following system of equations: 0.34) and (7.6 F1 − 0. Example 7-2 External Forces Acting on a Truss: A two-bar truss is subjected to external forces in both the horizontal and vertical directions as shown in Fig. 100 lb . (7.35) in the matrix form yields ] [ ][ ] [ 200 0.8 A= (7.35) gives 0.67 lb.96. 100 0.8)(−0.6 0.41) F2 The forces F1 and F2 can be found by finding the inverse of matrix A and then multiplying this with matrix b as x = A−1 b where A−1 = [ ] 1 −0.36) Substituting F1 from equation (7.67) F2 = 2 = 41.8 F1 = 200 − 0. (7.6 | = (0.42) and | 0.67 = 208.33 lb.8 |A| −0.33 lb and F2 = 41.8 0.39) 0.37) into equation (7.6 (250 − F2 ) − 0.6)(0.34).36) yields F1 = 250 − 41.196 Chapter 7 Systems of Equations in Engineering Solution (a) Substitution Method: Using equation (7.8 0.38) in the form A x = b gives [ ] 0.67 lb.8 F1 = .6 −0.67 (250 − 166.34) and (7. (7.6 F2 = 100 250 − 2 F2 = 166.6 F2 (7. (b) Writing equations (7.37) Now.6 −0.6 −0.8 | | |A| = || | | 0. substituting F2 from equation (7. Therefore.38) (c) Matrix Algebra Method: Writing matrix equation in (7.36) into equation (7.8 F2 F1 = 250 − F2 . (7.8) = −0.8 (7. F1 = 208.6) − (0.40) 100 [ ] F x= 1 .6 −0.6 [ ] 200 b= (7.43) . find force F1 in terms of F2 as 0.8 0. 7.96 (0.6 0.96 = 41.8 [ ] 0.8)(100) − (0.67 Therefore.8 | | | | 100 −0.35) using Cramer’s rule as | 200 0. F1 = 208.625)(200) + (0.625)(200) + (−0.67 lb Therefore.833 The forces F1 and F2 can now be found by multiplying A−1 in equation (7.46) .625 −0.7.96 (200)(−0.45) (7.6 100 | | | F2 = −0.833 = .833)(100) [ ] 208.96 = 208.8) = −0.67 lb.43) in equation (7.6 −0.42).44) 0. An analysis of the Op-Amp circuit shows that the conductances G1 and G2 in mho () satisfy the following system of equations: 10 G1 + 5 G2 = 125 9 G1 − 19 = 4 G2 .8 200 | | | | 0. the inverse of matrix A is given by [ ] −0.625 −0.96 −0.6.4 Further Examples of Systems of Equations in Engineering 197 Substituting equation (7.33 = lb 41.833 100 [ ] [ ] F1 (0.40) as [ ][ ] 0.44) by column matrix b given in equation (7. (7.6)(200) = −0.8 1 A−1 = −0.625 0.33 lb | 0.6) − (100)(0.625 0.34) and (7. F1 = 208.6 | | | F1 = −0.67 lb. Example 7-3 Summing Op-Amp Circuit: A summing Op-Amp circuit is shown in Fig. (7.833 200 x= 0. (d) Cramer’s Rule: The forces F1 and F2 can be found by solving the system of equations (7.33 lb and R2 = 41.33 lb and F2 = 41.833)(100) = F2 (0. write equations (7.46) in the matrix form A x = b. (b) Write the system [ ] of equations (7.5 G2 ) − 19 = 4 G2 93. (d) Find G1 and G2 using Cramer’s rule.50) in matrix form as [ ][ ] [ ] 10 5 G1 125 = .5 G2 .5 G2 G2 = 11 .5 G2 = 4 G2 93.49) and (7. V1 GF G1 −+ V2 +Vcc − + G2 −+ + −Vcc Vo − Figure 7. Perform all computations by hand and show all steps.47) into equation (7.0 . find the admittance G1 in terms of G2 as 10 G1 = 125 − 5 G2 G1 = 12.46) gives 9 (12. (7. Therefore.46) in the form 10 G1 + 5 G2 = 125 (7.5(11) = 7.5 − 0.49) 9 G1 − 4 G2 = 19.48) into equation (7.48) Now.47) yields G1 = 12.5 − 0.5 − 0. (7.47) Substituting G1 from equation (7.45) and (7. (b) Rewrite equations (7. G1 where x = . (7.198 Chapter 7 Systems of Equations in Engineering (a) Find G1 and G2 using the substitution method. 9 −4 G2 19 (7.51) .50) Now.45).45) and (7. substituting G2 from equation (7.5 − 4.5 = 8. G2 (c) Find G1 and G2 using the matrix algebra method. Solution (a) Substitution Method: Using equation (7. G1 = 7  and G2 = 11 .6 A summing Op-Amp circuit. 55) |A| −9 10 and | 10 5 | | |A| = || | | 9 −4 | = (10)(−4) − (5)(9) = −85. the inverse of matrix A is given as [ ] −4 −5 1 −1 A = −85 −9 10 5 ⎤ ⎡ 4 ⎢ 85 85 ⎥ (7. (7.54) G2 The admittance G1 and G2 can be found by finding the inverse of matrix A and then multiplying this with matrix b as x = A−1 b where [ ] 1 −4 −5 −1 A = (7.57) =⎢ ⎥. ⎢ 9 − 10 ⎥ ⎣ 85 85 ⎦ The admittance G1 and G2 can now be found by multiplying A−1 in equation (7.55).64) in the form A x = b gives [ ] 10 5 A= (7. G1 = 7  and G2 = 11 . (7.52) 9 −4 [ ] 125 b= (7.57) and the column matrix b given in equation (7.7.53) 19 [ ] G1 x= .56) in equation (7.53) as 5 ⎤[ ] ⎡ 4 ⎢ 85 85 ⎥ 125 x=⎢ ⎥ 19 ⎢ 9 − 10 ⎥ ⎣ 85 85 ⎦ ( ) ( ) ⎡ 4 ⎤ 5 (125) + (19) ⎥ [ ] ⎢ 85 85 G1 ⎥ = ⎢( ) ( ) ⎢ 9 ⎥ G2 10 (125) + − (19)⎥ ⎢ 85 85 ⎣ ⎦ (100 + 19) ⎡ ⎤ [ ] ⎢ ⎥ 7 17 =⎢ ⎥ = 11 .4 Further Examples of Systems of Equations in Engineering 199 (c) Matrix Algebra Method: Writing the matrix equation in (7.56) Substituting equation (7. . (225 − 38) ⎢ ⎥ ⎣ ⎦ 17 Therefore. 59) where Rx and Ry are the reactions at the ankle. (b) Find Fm and WF using the substitution method.59) in the matrix form A x = b.45) and (7. (c) Write the system [ ] of equations (7. 70 (a) Suppose Rx = √ N and Ry = 120 N.58) (7.200 Chapter 7 Systems of Equations in Engineering (d) Cramer’s Rule: The admittance G1 and G2 can be found by solving the system of equations (7.46) using Cramer’s rule as | 125 5 | | | | | | 19 −4 | | | G1 = −85 = (125)(−4) − (19)(5) −85 =7 | 10 125 | | | | | | 9 19 | | | G2 = −85 = (10)(19) − (9)(125) −85 = 11 . WF (d) Find Fm and WF using the matrix algebra method. The free-body diagram of the driver’s foot is also shown. 7. as shown in Fig. . Write the system of equations in terms 3 of Fm and WF . Based on the x-y coordinate system shown. Therefore. Example 7-4 Force on the Gastrocnemius Muscle: A driver applies a steady force of FP = 30 N against a gas pedal. Perform all computations by hand and show all steps. (e) Find Fm and WF using Cramer’s rule. Fm where x = . the force of the gastrocnemius muscle Fm and the weight of the foot WF satisfy the following system of equations: Fm cos 60◦ − WF cos 30◦ = Rx Fm sin 60◦ − WF sin 30◦ = Ry − FP . G1 = 7  and G2 = 11 .7.58) and (7. (7. 62) yields Fm = 80.732(20) = 115.4 Further Examples of Systems of Equations in Engineering 201 Fp Driver Fm WF Rx 60° Ry Gas pedal y x 60° Figure 7.8 + 1.60) 0.61) in the matrix form A x = b gives ] [ ] [ ][ 40.5 WF = 90 WF = 20 N.5 WF = 90 70 + 1.8 + 1.732 WF . (7.866 −0.4 + 0. Therefore.8 + 1.63) Now.4 N and WF = 20 N.60).61) gives 0.5 .62) into equation (7.866 Fm = (7.5 Fm − 0.5 WF − 0.64) WF 90 0.63) into equation (7. (7. Fm = 115.5 −0.60) and (7.4 (7.62) Substituting Fm from equation (7.866 (80. substituting WF from equation (7.59) in terms of given information yields 0.7.4 0. (c) Writing equations (7. (7.61) (b) Substitution Method: Using equation (7.866 Fm − 0.5 Fm = 40.732 WF ) − 0. Solution (a) Rewriting equations (7. find the force Fm in terms of WF as 0.5 WF = 90.7 A driver applying a steady force against the gas pedal.58) and (7.866 WF Fm = 80.866 WF = 40.4 N. 66) (7.0 1.866 −0.0 (90) [ ] 115. |A| −0.5 0.866 | | |A| = || | | 0. .866)(−0.66) as [ x= −1.732 (40. WF (7.5 ] [ −1.67) (d) Matrix Algebra Method: The forces Fm and WF can be found by finding the inverse of the matrix A and then multiplying this by vector b as x = A−1 b.5) − (0.866 1 −1 A = 0.4) + 1. (7. 90 [ ] Fm and x = .0 Therefore.4 N and WF = 20 N.5 (7.732 (90) Fm = WF −1.732 −1. 0. 20.0 1.732 1.5 0.4 90 ] ] [ ] −1. the inverse of matrix A is given as [ ] −0.732 .732 1.69) in equation (7.68). (7.5 −0.4) + 1.4 = N. Fm = 115.866 1 .202 Chapter 7 Systems of Equations in Engineering where [ ] 0.5)(−0.70) = −1.69) Substituting equation (7.866 0.866 .5 [ ] 40.65) A= (7.5. where A−1 = [ ] −0.0 (40.5 | and = (0.4 b= .866 0.68) | 0.5 −0.866 −0.866) = 0.0 [ ][ 40.5 −0.70) and the column matrix b given in equation (7.0 The forces Fm and WF can now be found by multiplying A−1 in equation (7. The required volumes of the two spray solutions v1 and v2 can be determined from a system of equations describing conditions for volume and concentration. as v1 + v2 = V c1 v1 + c2 v2 = C V (7.12 v1 + 0.61) using Cramer’s rule as Fm = = = WF = = = | 40. Example 7-5 Two-Component Blending of Liquids: An environmental engineer wishes to blend a single mixture of insecticide spray solution of volume V = 1000 L and concentration C = 0.74) . and C = 0.12 and c2 = 0.5) − (90)(−0.5 (0.866) 0.4)(−0.72) in the matrix form A x = b. (b) Find v1 and v2 using the substitution method.17. respectively.71) and (7.0 N.71) and (7.72) in terms of the given information yields v1 + v2 = 1000 0. v2 (d) Find v1 and v2 using the matrix algebra method.4 | | | | 0.866 90 | | | 0.4 N | 0. (7.4 N and WF = 20 N.12.5 || | 0.866 | | | | 90 −0.17 v2 = 150. (c) Write the system [ ] of equations (7.17.5 20.73) (7.4 Further Examples of Systems of Equations in Engineering 203 (e) Cramer’s Rule: The forces Fm and WF can be found by solving the system of equations (7.72) (a) Knowing that V = 1000 L. Solution (a) Rewriting equations (7.5)(90) − (0.5 (40. c2 = 0. c1 = 0. (e) Find v1 and v2 using Cramer’s rule. v where x = 1 .71) (7.5 115.5 40. rewrite the system of equations in terms of v1 and v2 .866)(40.4) 0.15. Therefore.60) and (7.7.4 −0.15 from two spray solutions of concentrations c1 = 0. Perform all computations by hand and show all steps. Fm = 115. 17 [ ] 1000 b= .76) into equation (7. 150 [ ] v and x = 1 .82) . 0.17 (1000 − v1 ) = 150 0.12 v1 + 0.17) − (0.80) (d) Matrix Algebra Method: The volumes v1 and v2 can be found by finding the inverse of the matrix A and then multiplying this by column vector b as x = A−1 b. v2 (7.79) (7.05 v1 = −20 v1 = 400.05. (c) Writing equations (7.17 | = (1)(0.76) Now.17 v2 150 where ] 1 1 A= .75) into equation (7.204 Chapter 7 Systems of Equations in Engineering (b) Substitution Method: Using equation (7. where A−1 = and [ ] 0. (7. find the volume v2 in terms of v1 as v2 = 1000 − v1 . Therefore.74) gives 0.75) Substituting v2 from equation (7.17 −1 1 |A| −0.81) | 1 1 || |A| = || | | 0.73) and (7. 0.12 v1 + 170 − 0. v1 = 400 L and v2 = 600 L.73). (7.78) (7.74) in matrix form A x = b gives [ ][ ] [ ] 1 1 v1 1000 = .75) yields v2 = 1000 − 400 = 600.12 0. substituting v1 from equation (7.77) [ (7.17 v1 = 150 −0.12 1 (7. (7.12)(1) = 0.12 0.12 0. 4 −20 −2. (e) Cramer’s Rule: The volumes v1 and v2 can be found by solving the system of equations (7.4 Further Examples of Systems of Equations in Engineering 205 Substituting equation (7.17 | | | 0.81). v1 = 400 L and v2 = 600 L.12 150 | | | v2 = 0.05 (1000)(0.05 = (1)(150) − (0.7. Therefore. v1 = 400 L and v2 = 600 L.12)(1000) 0.74) using Cramer’s rule as v1 = = | 1000 1 | | | | | | 150 0.4 20 ] .4 (1000) + 20 (150) [ ] 400 = .17 −1 1 −1 A = 0.82) in equation (7. (7. the inverse of matrix A is given as [ ] 0.17) − (150)(1) 0. 600 Therefore.79) as [ ][ ] 3.05 = 600.4 −20 1000 x= 150 −2.4 20 ] [ ] [ 3.83) and the column vector b given in equation (7.05 = 400 | 1 1000 | | | | | | 0.12 1 [ = 3.4 (1000) − 20 (150) v1 = v2 −2. .83) The volumes v1 and v2 can now be found by multiplying A−1 in equation (7.73) and (7.05 −0. 334 V 1100 I1 + 1000 I2 − 9 = 0 1100 I2 + 1000 I1 = 0. The voltages V1 and V2 (in V) satisfy the following system of . I2 (c) Find I1 and I2 using the matrix algebra method. (d) Find I1 and I2 using Cramer’s rule. 8Ω 246 V + − (a) Find I1 and I2 using the substitution method.89) in[the] matrix form I AI = b. (7. where I = 1 . (b) Write the system of equations (7. Consider the two-loop circuit shown in Fig.85) in[the] matrix form I A I = b.1 Two-loop circuit for problem P7-1. I2 (c) Find I1 and I2 using the matrix algebra method.1. Perform all computations by hand and show all steps. Consider the two-loop circuit shown in Fig. P7.89) 100 Ω 1 kΩ I2 Figure P7. Consider the two-node circuit shown in Fig. where I = 1 .84) (7. (a) Find I1 and I2 using the substitution method.2. The currents I1 and I2 (in A) satisfy the following system of equations: 16 I1 − 9 I2 = 110 20 I2 − 9 I1 + 110 = 0. 7-4.88) (7.84) and (7.86) and (7. (d) Find I1 and I2 using Cramer’s rule. 100 Ω 9V + − I1 (7. (b) Write the system of equations (7. Perform all computations by hand and show all steps.3.206 Chapter 7 Systems of Equations in Engineering PROBLEMS 7-1. I2 (c) Find I1 and I2 using the matrix algebra method. 7Ω 110 V + − (7. Consider the two-loop circuit shown in Fig.85) 11 Ω I1 9Ω I2 + − 110 V 7-3. (d) Find I1 and I2 using Cramer’s rule. (a) Find I1 and I2 using the substitution method. where I = 1 .87) in[the] matrix form I A I = b.4. 7-2. P7.88) and (7. P7. The currents I1 and I2 (in A) satisfy the following system of equations: 18 I1 − 10 I2 − 246 = 0 22 I2 − 10 I1 = −334.86) (7.2 Two-loop circuit for problem P7-2. P7. The currents I1 and I2 (in A) satisfy the following system of equations: Figure P7.87) 12 Ω I1 10 Ω I2 + − Figure P7. (b) Write the system of equations (7. Perform all computations by hand and show all steps.3 Two-loop circuit for problem P7-3. 7-5. 20 Ω 10 V + − V1 10 Ω 50 Ω V2 (7.1 V2 = 4 0. Perform all computations by hand and show all steps. (b) Write the system of equations (7. where V = . Consider the two-node circuit shown in Fig.90) (7. (a) Find V1 and V2 using the substitution method.1 V1 + 2 = 0.93) in the form [ matrix ] V1 A V = b. 21 V 0. Perform all computations by hand and show all steps.7 yields the following system of . P7.5 Two-node circuit for problem P7-5. where V = . (b) Write the system of equations (7. V2 (c) Find V1 and V2 using the matrix algebra method. P7.Problems equations: 10 Ω 10 V V1 + − 20 Ω V2 20 Ω (7.94) (7.91) 30 Ω 20 Ω + − 20 V 7-6. (b) Write the system of equations (7. Perform all computations by hand and show all steps.95) in the form A V = [ matrix ] V1 b.4 Two-node circuit for problem P7-4.90) and (7.95) V2 5Ω ↑ 4 V1 − V2 = 20 −3 V1 + 8 V2 = 40. V1 10 Ω 4A ↑ 10 Ω (7. The voltages V1 and V2 (in V) satisfy the following system of equations: 17 V1 = 10 V2 + 50 (7. V2 (c) Find V1 and V2 using the matrix algebra method. The voltages V1 and V2 (in V) satisfy the following system of equations: Figure P7. (d) Find V1 and V2 using Cramer’s rule. 207 2A Figure P7.6 Two-node circuit for problem P7-6. V2 (c) Find V1 and V2 using the matrix algebra method. Consider the two-node circuit shown in Fig.2 V1 − 0.93) 30 Ω 20 Ω + − Figure P7.3 V2 − 0.5. (a) Find V1 and V2 using the substitution method.91) in the form [ matrix ] V1 A V = b.92) and (7.6.94) and (7.92) 11 V2 − 6V1 − 42 = 0. where V = . (d) Find V1 and V2 using Cramer’s rule. 7-7. (a) Find V1 and V2 using the substitution method. P7. (d) Find V1 and V2 using Cramer’s rule. An analysis of the circuit shown in Fig. (a) Write the system of equations (7. Perform all computations by hand and show all steps.5 T1 = 0.125. where G = .5 T2 + 0.97) R1 R2 R5 20 Ω 10 Ω 20 Ω R3 = 40 Ω + R4 V2 =5Ω − + V1 − + − 7-9.99) in the form [ matrix ] G1 A G = b.98) (7. V2 (c) Find V1 and V2 using the matrix algebra method. The tensions T1 and T2 satisfy the following system of equations: 0. A 20 kg object is suspended by two cables as shown in Fig.100) 10 G1 + 10 G2 = 0.100) and (7.101) in the matrix form A T = b.7 Two-node circuit for problem P7-7. (7.2 (7. where G = . (7. G2 (c) Find G1 and G2 using the matrix algebra method. where V = .96) 2 V1 − 7 V2 + 10 = 0.99) (a) Find G1 and G2 using the substitution method.101) in the [ matrix ] G1 form A G = b. P7. (d) Find G1 and G2 using Cramer’s rule.10. (b) Write the system of equations (7. Write the system of equations (7.866 T1 = 196. An analysis of the Op-Amp circuit shows that the admittances G1 and G2 in mho () satisfy the following system of equations: 5 G1 − 145 = −10 G2 −9 G2 + 71 = −4 G1 .6. G2 Find G1 and G2 using the matrix algebra method. Find G1 and G2 using Cramer’s rule. where [ ] T1 .98) and (7. Find G1 and G2 using the substitution method. T= T2 . 7-8. 7.10 A 20 kg object suspended by two cables for problem P7-10.97) in the form [ matrix ] V1 AV = b. (c) (d) 7-10. A summing Op-Amp circuit is shown in Fig.208 Chapter 7 Systems of Equations in Engineering equations: −4 V2 + 7 V1 = 0 (7. (7.6. 7.103) 0.96) and (7. A summing Op-Amp circuit is shown in Fig.101) 5 G1 + 15 G2 = 0. 60° 30° T1 y T2 60° x 20 kg g = 9.866 T2 (7. (b) Write the system of equations (7. Perform all computations by hand and show all steps. (a) Find V1 and V2 using the substitution method. An analysis of the Op-Amp circuit shows that the admittances G1 and G2 in mho () satisfy the following system of equations: 10 V (a) (b) Figure P7. (d) Find V1 and V2 using Cramer’s rule.102) (7. Perform all computations by hand and show all steps.8 m/sec2 30° W Figure P7.100) and (7. find matrices A and b. (a) Write the system of equations (7. A 100 lb weight is suspended by two cables as shown in Fig. F= F2 In other words.13. A two-bar truss supports a weight of W = 750 lb as shown in Fig. 7-13. What are the dimensions of A and b? (b) Find T1 and T2 using the matrix algebra method.1.707 F1 = F2 0.107) in the matrix form A F = b.108) (7.105) (a) Write the system of equations (7. A 10 kg object is suspended from a twobar truss shown in Fig. Perform all matrix computation by hand and show all steps. Perform all matrix computation by hand. . (b) Find F1 and F2 using the matrix algebra method. where [ ] F1 .106) (7.11.8 T2 = 100 0. P7. Perform all matrix computation by hand and show all steps. where [ ] T1 .Problems In other words. (c) Find T1 and T2 using Cramer’s rule. x= T2 In other words.104) (7.87° 53.6 T2 = 0. The forces F1 and F2 satisfy the following system of equations: 0. y x 100 lb 209 (7. (7. P7.104) and (7. The forces F1 and F2 satisfy the following system of equations: 0. 36.107) 45° x 45° F2 W W Figure P7.5 F1 = 750. (7. find matrices A and b. P7.12.13° T2 T1 F1 y 30° x 30° F2 W W Figure P7. (b) Find T1 and T2 using the matrix algebra method.707 F1 = 98. 7-11. (c) Find T1 and T2 using Cramer’s rule.105) in the matrix form A x = b. The tensions T1 and T2 satisfy the following system of equations: 0. y 7-12.13 An object suspended from a two-bar truss for problem P7-13.11 A 100 lb weight suspended by two cables for problem P7-11.109) 100 lb F1 Figure P7.866 F1 = F2 0.12 A two-bar truss supporting a weight for problem P7-12.6 T1 + 0. (c) Find F1 and F2 using Cramer’s rule.8 T1 − 0.106) and (7. find matrices A and b. Perform all matrix computation by hand and show all steps.114) (7. 59.111) in the matrix form A F = b.707 F2 = 0. P7. (7. (c) Find F1 and F2 using Cramer’s rule. find matrices A and b.515F2 . In other words. where x In other words.115) in the matrix . The forces F1 and F2 satisfy the following system of equations: In other words.3° 3 F1 Figure P7. (b) Find F1 and F2 using the matrix algebra method. P7.112) (7. (b) Write the system of equations (7. F= F2 [ (7.114) and (7. find matrices A and b. The weight of a vehicle is supported by reaction forces at its front and rear wheels as shown in Fig.113) (7.108) and (7. find matrices A and b. 1 60° FBD F 30° 3 F2 F2 F1 Figure P7.14 A 100 N force applied to a two-bar truss for problem P7-14. P7.115) (a) Find R1 and R2 using the substitution method. (a) Write the system of equations (7. (b) Find F1 and F2 using the matrix algebra method.5548 F1 − 0.15 A 200 lb force applied to a two-bar truss for problem P7-15.8572 F2 = −100 0. If the weight of the vehicle is W = 4800 lb. the reaction forces R1 and R2 satisfy the following system of equations: R1 + R2 − 4800 = 0 6 R1 − 4 R2 = 0.112) and (7. where [ ] F1 .210 Chapter 7 Systems of Equations in Engineering (a) Write the system of equations (7. A force F = 200 lb is applied to a twobar truss as shown in Fig. (c) Find F1 and F2 using Cramer’s rule. (c) Find F1 and F2 using Cramer’s rule. (7. (b) Find F1 and F2 using the matrix algebra method. Perform all matrix computation by hand and show all steps.16. where ] F1 .0° y 2 F y ] F1 .832 F1 = 0. F= F2 [ 7-14. 7-16. A F = 100 N force is applied to a twobar truss as shown in Fig. F= F2 7-15.109) in the matrix form A F = b. The forces F1 and F2 satisfy the following system of equations: 0.15.14. 2 F FBD F x 1 56.110) −0.866 F1 + 0. Perform all matrix computation by hand and show all steps.5 F1 − 0.707 F2 = 200 0.110) and (7.111) (a) Write the system of equations (7.113) in the matrix form A F = b. R2 (a) Write the system of equations (7.16 A vehicle supported by reaction g forces.124) (7. (c) Find F and N using Cramer’s rule. 7-20. The forces Ff and N satisfy the following system of equations: 0.117) 7-18. N In other words. P7.118) and (7.8◦ ) as shown in Fig. (7. (7.120) and (7. (7. Perform all matrix computation by hand and show all steps. Solve parts (a)–(c) of problem P7-19 for system of equations given by (7. where [ ] R1 x= .9285 F = 0. A 100 lb crate weighing W = 100 lb is pushed against an incline (𝜃 = 30◦ ) with a force of Fa = 30 lb as shown in Fig.122) (7. (d) Find R1 and R2 using Cramer’s rule. FBD F y R1 211 x N N θ W θ Figure P7. A vehicle weighing W = 10 kN is parked on an inclined driveway (𝜃 = 21.5736 F = 2000.5 Ff − 100 = 0. P7. The forces F and N satisfy the following system of equations: −0.19. find matrices A and b.5N + 30 = 0 0.Problems form A x = b.119). where [ ] F x= . driveway.8192 F + 0.8192 N + 0.116) (7. Solve problem P7-16 if the system of equations are given by (7. 7-21.116) and (7.866 Ff − 0. Perform all matrix computations by hand and show all steps. A vehicle weighing W = 2000 lb is parked on an inclined driveway (𝜃 = 35◦ ) as shown in Fig. In other words. (c) Find R1 and R2 using the matrix algebra method.118) (7.3714 N (7.121). The forces F and N satisfy the following system of equations: 0. R1 + R2 = 5000 4 R1 = 6 R2 .121) where forces F and N are in kN.123) where forces F and N are in lb. R1 + R2 − 3000 = 0 7 R1 − 3 R2 = 0.117).866 N + 0.3714 F − 10 = 0 (7.5736 N = 0 0.119) 7-19. Solve problem P7-16 if the system of equations are given by equations (7.123).19. P7.120) 0.122) and (7.9285 N + 0. in the matrix form A x = b. (b) Find F and N using the matrix algebra method.21. W W R2 F Figure P7. find matrices A and b.125) .19 A truck parked on an inclined 7-17. (7. 131) Solve parts (a)–(d) of problem P7-23 for system of equations (7. where [ ] Nf .131).3. find matrices A and b. A crate weighing W = 500 N is pushed against an incline (𝜃 = 20◦ ) with a force of Fa = 100 N as shown in Fig.3 Nf = 200 a Nf + 342 = 9.125) in the matrix form A x = b.127) Solve parts (a)–(c) of problem P7-21 for system of equations given by (7.5 = 9. The normal force Nf and acceleration a in m/s2 satisfy the following system of equations: (7.25.212 Chapter 7 Systems of Equations in Engineering (a) Write the system of equations (7. FBD Ff θ y x (a) Find Nf and a using the substitution method. (7. x= N In other words.23. (c) Find Nf and a using the matrix algebra method.81 × 200.342 N (7. (b) Write the system of equations (7. where [ ] Ff . (7. P7.342 Ff − 500 = 0. The weight of a vehicle is supported by reaction forces at its front and rear wheels as shown in Fig. (b) Find Ff and N using the matrix algebra method. P7. (7. 7-22.128) and (7.23 Desk being pulled by a force F. The normal force Nf and acceleration a in m/s2 satisfy the following system of equations: (7.129) in the matrix form A x = b.4 Nf = 100 a Nf + 565.21 A crate being pushed by a force.4.129) F = 800 N 45° F = 0. It is pulled with a force of F = 800 N at an angle of 𝜃 = 45◦ as shown in Fig. (d) Find Nf and a using Cramer’s rule. (c) Find Ff and N using Cramer’s rule.21. P7.4 NF 45° NF Figure P7.128) 565. 7-23. 7-24. 7-25. A 100 kg desk rests on a horizontal plane with a coefficient of friction of 𝜇 = 0. Fa Fa W In other words. Perform all matrix computation by hand and show all steps. The forces Ff and N satisfy the following system of equations: 0.133) .130) 939. The reaction forces R1 and R2 satisfy the following system of equations: R1 + R2 − m g = 0 l1 R1 − l2 R2 − m a k = 0.130) and (7.7 − 0.127). find matrices A and b. x= a θ 981 N 800 N N W Figure P7. It is pulled with a force of F = 1000 N at an angle of 𝜃 = 20◦ as shown in Fig.9397 Ff + 100 = 0.9397 N + 0. (7. P7.132) (7. A 200 kg desk rests on a horizontal plane with a coefficient of friction of 𝜇 = 0.81 × 100.124) and (7.23.126) and (7.6 − 0.126) 0. (b) Find Fm and WF using the substitution method. if l1 = 2 m. find R1 and R2 using the substitution method.29 A driver applying a steady force against the gas pedal. if l1 = 2 m. k = 1.Problems (a) Write the system of equations (7.135) in terms of Fm and WF . m = 1000 kg. (e) Find R1 and R2 using Cramer’s rule.25 A vehicle supported by reaction WF Rx Figure P7. The free-body diagram of the driver’s foot is also shown. x= R2 213 x-y coordinate system shown. k = 1. rewrite the system of equations (7. g = 9.81 m/s2 . k = 1. l2 = 2 m. where x = . l2 = 1.5 m. Repeat problem P7-25.5 m. g = 9. k = 1.5 m.134) Fm cos 45◦ − WF cos 30◦ = Rx Fm sin 45◦ − WF sin 30◦ = Ry − FP .29. if l1 = 2 m. and a = −5 m/s2 . 7-27. the force of the gastrocnemius muscle Fm and the weight of the foot WF satisfy the following system of equations: (7.81 m/s2 .5 m/s2 . m = 1000 kg. (c) Write the system of equations obtained in part (a) in the [ matrix ] Fm form A x = b.135) Driver In other words. Repeat problem P7-25. m = 1200 kg.81 m/s2 . (b) For the system of equations found in part (a). find matrices A and b. (7. A driver applies a steady force of FP = 25 N against a gas pedal. and a = 4.81 m/s2 . g = 9.5 m/s2 7-29.134) and (7. m = 1200 kg. where [ ] R1 . l2 = 2 m. and a = −4. Repeat problem P7-25. WF . (d) Find R1 and R2 using the matrix algebra method. Based on the Ry y forces.5 m. l2 = 1.5 m. and a = 5 m/s2 . g = 9. 7-28. Gas pedal W = mg Fp F ma G k l1 l2 Fm R1 R2 45° 7-26.133) if l1 = 2 m.5 m. (c) Write the system of equations in the matrix form A x = b. P7. √ (a) Knowing that Rx = 70∕ 2 N and Ry = 105 N. as shown in Fig. 60° x Figure P7.132) and (7. 1 s Ω 100 —– s V (7. (e) Find v1 and v2 using Cramer’s rule. The required volumes of the two spray solutions v1 and v2 can be determined from a system of equations describing conditions for volume and concentration. c1 = 0.137) (b) Find v1 and v2 using the substitution method.4. c1 = 0. I1 Figure P7. v2 (d) Find v1 and v2 using the matrix algebra method. Perform all computations by hand and show all steps.33 Two-loop circuit for problem P7-33. Consider the two-loop circuit shown in Fig.138) and (7. . c2 = 0. −0. 7-30. Perform all computations by hand and show all steps.138) 7-33.136) (a) Knowing that V = 500 L.2 s + s 10 — s Ω 10 Ω 100 —– s V + − I1 0.2. + − (7.2 from two spray solutions of concentration c1 = 0.15.25.2 s + 10) I1 − 0. I2 (c) Find I1 and I2 using the matrix algebra method.25.214 Chapter 7 Systems of Equations in Engineering (d) Find Fm and WF using the matrix algebra method. The currents I1 and I2 (in A) satisfy the following system of 1Ω I2 (a) Find I1 and I2 using the substitution method. −1 I1 + 1 + s (e) Find Fm and WF using Cramer’s rule.32.140) (0.32 Two-loop circuit for problem P7-32. 7-32. 7-31. rewrite the system of equations (7. respectively as v1 + v2 = V c1 v1 + c2 v2 = C V (c) Write the system of equations obtained in part (a) in the [ ]matrix v form A x = b.139) in the [ ]matrix I form A I = b.2 s I1 + 0.2 s I2 = s ( ) 10 (7. (a) Find I1 and I2 using the substitution method. The currents I1 and I2 (in A) satisfy the following system of equations: 100 (7.139) 2 − s Ω 0. (d) Find I1 and I2 using Cramer’s rule.137) in terms of v1 and v2 .141) I2 = 0. (7. An environmental engineer wishes to blend a single mixture of insecticide spray solution of volume V = 500 L with a specified concentration C = 0. Consider the two-loop circuit shown in Fig.2 s Ω I2 Figure P7. and C = 0. P7.15 and c2 = 0.33.136) and (7. where I = 1 . (7.1 s + 1) I1 − I2 = s ( ) 2 I2 = 0.2. equations: 100 (0. Repeat problem P7-30 if V = 400 L. where x = 1 . (b) Write the system of equations (7. C = 0. and c2 = 0.25. P7. 7-34.1 — s A ↑ 10 Ω (a) Write the system of equations (7.1 + V2 − 0.145) in the matrix form ]A X(s) = b. Consider the two-node circuit shown in Fig.1 V2 = (7. The voltages V1 and V2 (in V) satisfy the following system of equations: ) ( 0.2 ↑ — s A 5 — s Ω Figure P7.2 5 0. (c) Find the expressions of X1 (s) and X2 (s) using Cramer’s rule.1 V1 = .35.140) and (7. I2 (c) Find I1 and I2 using the matrix algebra method. X1 X2 m2 = 2 kg 50 (X1 − X2) m1 = 2 kg f = 100 N . A mechanical system along with its free body diagram is shown in Fig.Problems (b) Write the system of equations (7.144) and (7. (7. The linear displacements X1 (s) and X2 (s) of the two springs in the s-domain satisfy the following system of equations: 50 40 X1 (s) − 40 X2 (s) = s (2 ) −40 X1 (s) + s + 65 X2 (s) = 0 X1 X2 f = 50 N (7.142) and (7. and the displacements X1 (s) and X2 (s) of the two masses in the s-domain satisfy the X1 k1 = 40 N (7. where f = 50 N is the applied force. (d) Find I1 and I2 using Cramer’s rule. V2 0. V2 X2 m = 1 kg (c) Find V1 and V2 using Cramer’s rule. P7.2 s Ω 215 25 X2 m 40 (X1 − X2) f = 50 N Figure P7. where I = 1 .34.145) 7-36. The mass m1 is pulled with a force f = 100 N. Fig.142) s s ( ) 0.144) (a) Write the system of equations (7. P7.1 5 + 0.34 Two-loop circuit for problem P7-34. X2 k2 = 50 N m2 = 2 kg k1 = 50 N X1 m1 = 2 kg f = 100 N 50 X2 Figure P7.36 shows a system with two-mass elements and two springs. 7-35. X2 (s) (b) Find the expressions of X1 (s) and X2 (s) using the matrix algebra method.141) in the [ ]matrix I form A I = b. k2 = 25 N (b) Find V1 and V2 using the matrix algebra method. 0.143) in the [ matrix ] V1 form A V = b.35 Mechanical system for problem P7-35.143) s s V1 0.36 Mechanical system for problem P7-36. P7. where X(s) = [ X1 (s) .1 V1 − 0. where V = . Repeat parts (a)–(d) of problem P7-37 if Composite A requires 1.0 ounces of polymer resin. 7-37. By the finite F = 5000 lb 24 in. 7-38. respectively.39 Finite element idealization of truss structure subject to loading.146) ( ) −50 X1 (s) + 2 s2 + 100 X2 (s) = 0 (7. The student knows that Composite A requires 0. respectively.15 pounds of carbon fiber and 1. A structural engineer is performing a finite element analysis on an aluminum truss support structure subjected to a load as shown in Fig.75 xA + 0.2 pounds of carbon fiber and 2.15 xA + 3.146) and (7. the amount of each composite to manufacture satisfies the system of equations (7.6 xA + 2. Perform computation by hand and show all steps. (7. where x = A .5 xB = 17 (7. X2 (s) (b) Find the expressions of X1 (s) and X2 (s) using the matrix algebra method.151) 7-39.16 xA + 2.6 ounces of polymer resin while Composite B requires 3. A co-op student at a composite materials manufacturing company is attempting to determine how much of two different composite materials to make from the available quantities of carbon fiber and resin. P7.75 pounds of carbon fiber and 0. xB (c) Find xA and xB using the matrix algebra method.149) where xA and xB are the amounts of Composite A and B (in pounds).150) 1. 0. where X(s) = [ X1 (s) .146) in the matrix form ]A X(s) = b.216 Chapter 7 Systems of Equations in Engineering following system of equations: (b) Write the system of equations (7.150) and (7.39. (7. ( ) 100 2 s2 + 50 X1 (s) − 50 X2 (s) = s (7.147) (a) Write the system of equations (7. If the available quantities of carbon fiber and resin are 17 pounds and 12 ounces.5 ounces of polymer resin while Composite B requires 1. Figure P7.149) in the [ ]matrix x form A x = b.0 xB = 10. respectively.148) 1. (c) Find the expressions of X1 (s) and X2 (s) using Cramer’s rule. .78 ounces of polymer resin.78 xB = 12. If the available quantities of carbon fiber and resin are 15 pounds and 10 ounces. (a) Find xA and xB using the substitution method. (d) Find xA and xB using Cramer’s rule. the amount of each composite to manufacture satisfies the system of equations: 0.6 pounds of carbon fiber and 0.151). 12 in.148) and (7. 36 in.2 xB = 15 (7. where x = .153) in the [ ] matrix u form A x = b.02.Problems element method.152) and (7. v (c) Find u and v using the matrix algebra method.155) in the [ ] matrix u form A x = b.99 v = 0.40 Finite element idealization of crane structure subject to loading. 7-40. (d) Find u and v using Cramer’s rule.49 v = −0. The finite element idealization of the crane structure picking up a 2000 lb load is shown in Fig. .40.33 v = −0. v (c) Find u and v using the matrix algebra method. A structural engineer is designing a crane that is to be used to unload barges at a dock.08 v = 0 1. can be determined from the system of equations: 2. where x = .05 3. (b) Write the system of equations (7.47 u + 1.152) (7. (a) Find u and v using the substitution method. respectively. (b) Write the system of equations (7.57 u + 3. (7.154) (7. Perform computation by hand and show all steps. (d) Find u and v using Cramer’s rule. the displacements in the horizontal and vertical directions at the node where the load is applied can be determined from the system of equations: 2.33 u + 6. The displacement in the horizontal and vertical directions at the end of the crane 217 20 ft F = 2000 lb 15 ft 40 ft Figure P7. She is using a technique called finite element analysis to determine the displacement at the end of the crane. P7.08 u + 0. (a) Find u and v using the substitution method.155) where u and v are the the displacements in the horizontal and vertical directions. Perform computation by hand and show all steps.153) where u and v are the the displacements in the horizontal and vertical directions. (7.154) and (7. respectively. 452 s Figure 8.CHAPTER 8 8.1 A ball dropped from a height of 1 meter. The professor then poses the following questions: (a) What is the average velocity of the ball? (b) What is the speed of the ball at impact? (c) How fast is the ball accelerating? t=0s 1m t = 0. an engineering professor asks a student to drop a ball (shown in Fig.1. and mechanics of materials are emphasized. 8.1 What Is a Derivative? To explain what a derivative is.1) from a height of y = 1. 218 y(t) y=0 . Using a high-resolution stopwatch. electric circuits.452 s.0 m to find the time when it impacts the ground. the student measures the time at impact as t = 0. The concepts of maxima and minima along with the applications of derivatives to solve engineering problems in dynamics. 8.1 Derivatives in Engineering INTRODUCTION This chapter will discuss what a derivative is and why it is important in engineering. 452) = lim t → 0.e. Note that the negative sign means the ball is moving in the negative y-direction.3] [0.1 Additional data collected from the dropped ball.0 0.452) − y(t) ..2 Average velocity of the ball in different intervals. v(t) = lim Δ t → 0 where Δ t = 0.e.1 Introduction 219 Using the given information. the student repeats the experiment and collects the data given in Table 8.44 −4.215 0 The student then calculates the average velocity v̄ = Δy∕Δt in each interval.3.2] [0.. i.559 0. m/s −0. but claims that he/she would need an infinite (∞) number of data points to get it exactly right. 0.2 0.13 m/s as the speed of impact with ground.1 − 0 age velocity in the remaining intervals is given in Table 8.452 − t The professor suggests that this looks like the definition of a derivative.452 − t. 0. 0. i.4.1 0.3 0.2. TABLE 8. 0.452 − 0 =− 1.45 −3. 0. i.4] [0. Interval [0.4 0.490 m/s.e. 0. TABLE 8.47 −2. t.1].21 m/s. (b) Speed at Impact: The student finds that there is not enough information to find the speed of ball when it impacts the ground. ῡ = y2 − y1 𝚫y Total distance = = t2 − t1 𝚫t Total time =− 0 − 1. v̄ = 0. v(t = 0.1] [0. Using an ultrasonic motion detector in the laboratory.0 0.490 −1.8.2.452 = −2. The averFor example. in the interval t = [0.452 y(0. y(t + Δ t) − y(t) dy = Δt dt .0 = −0.1.1. m 0 0..0 0.951 − 1.804 0. s y(t). the student provides the following answers: ̄ The average velocity is the total distance traveled per unit (a) Average Velocity. 0. 0. υ: time.452] v̄ .951 0.13 The student proposes an approximate answer of −4.452 1. 905 t2 . The velocity at any time is thus calculated by taking the derivative as v(t) = dy dt = ) d ( 1. TABLE 8. df (t) dt sin(𝜔 t) 𝜔 cos(𝜔 t) cos(𝜔 t) −𝜔 sin(𝜔 t) eat a eat tn n tn−1 c f (t) c c df (t) dt 0 df1 (t) df (t) + c2 2 dt dt dg(t) df (t) f (t) + g(t) dt dt dg(t) df × dg dt c1 f1 (t) + c2 f2 (t) c1 f (t) .0 − 4. n. Note that 𝜔. f (t) Derivative.0 − 4. and c2 are constants and not functions of t.905 dt dt = 0 − 4. c.3 Some common derivatives used in engineering.905 t2 dt = d d ( 2) t (1.81 t m/s. a.905(2 t) = −9. which gives y(t) = 1.0) − 4.220 Chapter 8 Derivatives in Engineering The derivatives of some common functions in engineering are given below. c1 . Function. g(t) f (g(t)) The professor then suggests a quadratic curve fit of the measured data. . 43 m/sec y(t) = 4. (a) How long does it take for the ball to reach its maximum height? (b) What is the velocity at y = ymax ? (c) What is the maximum height ymax achieved by the ball? v(0) = 4.43 t − 4.43 m/s as shown in Fig.3. 8.e.2.. The professor suggests that the height of the ball is governed by the quadratic equation y(t) = 4. the acceleration due to gravity is constant and is equal to −9. if v(t) = −9. (8.905 t2 m. a(t) = lim Δ t → 0 = dv(t) dt = d dy(t) dt dt = d2 y(t) dt2 Δ v(t) Δt .2 Maxima and Minima 221 (c) The student is now asked to find the acceleration without taking any more data.81 m/s2 . and plotted as shown in Fig.1) .905 t2 m Figure 8. Hence.2 MAXIMA AND MINIMA Suppose now that the ball is thrown upward with an initial velocity vo = 4.2 A ball thrown upward. i. Thus. 8. then a(t) = d (−9. The acceleration is the rate of change of velocity.81 t.43 t − 4.81 t) dt = −9. 8.81 m/s2 .8. 8.4 The derivative as the slope of the tangent line. Based on the definition of the derivative.4. m ymax 0 tmax 0 t. the slope of the tangent line is zero (Fig.43 − 9.3 The height of the ball thrown upward. For the problem at hand.43 t − 4.222 Chapter 8 Derivatives in Engineering y(t). as shown in Fig. v(t) = (8. m Δy Δt 0 0 t t + Δt t. at the time when y = ymax . 8.81 t m/s.2) .905 t2 ) dt = 4. dy Δy — = lim — dt Δt Δt → 0 y(t). the velocity is given by dy(t) dt d = (4.5). the velocity v(t) at any time t is the slope of the line tangent to y(t) at that instant. s Figure 8. s Figure 8. Therefore. 81 or tmax = 0. reduces to 0 at t = 0. Hence.4515)2 = 1.2 Maxima and Minima dy — = v(t) = 0 dt y(t). v(t) = = 0. m ymax 0 0 223 t. the velocity at y = ymax is zero. and reaches a minimum value (−4. setting v(t) in equation (8. v(t) = 4. It can be seen that the velocity is maximum at t = 0 s (initial velocity = 4.8. .5 The slope of the tangent line at maximum height.43 9.6. s tmax Figure 8. (b) What is the velocity at y = ymax ? Since the slope of the height at t = tmax is zero.903 s.2). is shown in Fig.43 m/s) at t = 0. 8.2) to zero gives dt 4. the student answers the professor’s questions as follows: (a) How long does it take to reach maximum height? At the time of maximum height dy(t) t = tmax . (c) The maximum height: The maximum height can now be obtained by substituting t = tmax = 0.43 m/s). Given equation (8.1) for y(t) ymax = y(tmax ) = 4.4515) − 4.4515 s.43 − 9.43 (0.4515 s for the ball to reach the maximum height.4515 s (t = tmax ). The plot of the velocity.0 m.905 (0. Therefore.81 t for times t = 0 to t = 0.4515 s in equation (8.43 − 9.81tmax = 0 tmax = 4. it takes 0.903 sec. t . So if the derivative is zero at both maxima and minima. 8.7. As discussed earlier.6 The velocity profile of the ball thrown upward. m/s 4. how can one tell whether the value of the function found earlier is a maximum or a minimum? Consider the function shown in Fig. ——– < 0 (maxima) dt dt2 dy(t) —— < 0 dt dy(t) —— > 0 dt dy(t) d2y(t) —— = 0.7 Plot of a function with local maximum and minimum values.43 0 tmax = 0. the derivative of a function at a y(t) dy(t) —— > 0 dt dy(t) d2y(t) —— = 0.903 Figure 8. ——– > 0 (minima) dt dt2 dy(t) —— < 0 dt 0 0 Figure 8. which has local maximum and minimum values.4515 t.224 Chapter 8 Derivatives in Engineering v(t). It should be noted that the derivative of a function is zero both at the points where the value of the function is maximum (maxima) and where the value of the function is minimum (minima). s 0.43 0 −4. At its maximum value. Therefore. find f ′ (t) = df (t) .7 changes from positive to negative.0 m. however. dt dt2 To test whether the point where the slope (first derivative) of the trajectory of the ball thrown upward is zero is a maximum or a minimum. and acceleration. dt dt2 At a Local Minimum: d2 y(t) dy(t) > 0 = 0. the function has a local minimum for these values. the point where the slope of the trajectory of the ball is zero is a maximum and thus the maximum height is 1. if the second derivative is positive. f (t). in other words. determine the second derivative f ′′ (t) = d2 f (t) dt2 of the function. the student obtains the second derivative of the height as d2 y(t) d (4.3 Applications of Derivatives in Dynamics 225 point is the slope of the tangent line at that point. At its minimum value. If the second ( d2 f (t) derivative is negative dt2 < 0 .81 t) = −9.81 < 0. the procedure of finding the local maxima and minima of any function f (t) is as follows: dt2 = (a) Find the derivative of the function with respect to t. velocity. dt Therefore. This section also demonstrates the application of derivatives in sketching plots of position.43 − 9.3 APPLICATIONS OF DERIVATIVES IN DYNAMICS This section demonstrates the application of derivatives in determining the velocity and acceleration of an object if the position of the object is given. . the derivative (slope) of the function changes from negative to positive. to test for maxima and minima. dt (b) Find the solution of the equation f ′ (t) = 0. the function has a local maximum for those values of t. find the values of t where the function has a local maximum or a local minimum. at the values of t found in step (b) to find the maximum and minimum values. (e) Evaluate the function. In other words.8. the derivative (slope) of the function shown in Fig. 8. (d) Evaluate the second derivative at )the values of t found in step (b). In general. 8. in other words. the rate of change of the derivative (or the second derivative of the function) is negative at maxima and positive at minima. (c) To find which values of t gives the local maximum(and which values ) of t gives the local minimum. the following rules apply: At a Local Maximum: d2 y(t) dy(t) < 0 = 0. dt .. velocity v(t) is the instantaneous rate of change of the position (i.3. the velocity v(t) is the slope of the position x(t) as shown in Fig. s Figure 8. m Parabolic blend Linear function t. Velocity. parabolic position). and Acceleration Suppose the position x(t) of an object is defined by a linear function with parabolic blends. x(t).8 Position of an object as a linear function with parabolic blends. the derivative of the position) and is given by v(t) = limΔt → 0 Δx(t) Δt or v(t) = dx(t) . 8. dt Therefore. cruising at that constant speed (linear position).226 Chapter 8 Derivatives in Engineering 8.9 that the object is starting from rest. It can be seen from Fig. 8.e. and then coming to stop with maximum braking (maximum negative acceleration.8. and comes to rest again after moving at a linear velocity with a negative slope. The acceleration a(t) is the instantaneous rate of change of the velocity (i. moves at a linear velocity with positive slope until it reaches a constant velocity.1 Position. As discussed previously.9. cruises at that constant velocity. This motion is similar to a vehicle starting from rest and accelerating with a maximum positive acceleration (parabolic position) to reach a constant speed. 8.e.. the derivative of the velocity): a(t) = dv(t) . as shown in Fig. and then comes to rest with maximum braking (constant negative acceleration).3 Applications of Derivatives in Dynamics 227 v(t). 8. Therefore.10 that the object starts with maximum positive acceleration until it reaches a constant velocity. a(t). 8. or a(t) = d2 x(t) dt2 .10. 8. and acceleration at t = 0.9 Velocity of the object moving as a linear function with parabolic blends. velocity. the acceleration a(t) is the slope of the velocity v(t). m/s2 0 t.8.5 seconds if (a) x(t) = sin (2 𝜋t) m (b) x(t) = 3 t3 − 4 t2 + 2 t + 6 m (c) x(t) = 20 cos(3 𝜋t) − 5 t2 m . The following examples will provide some practice in taking basic derivatives using the formulas in Table 8. s 0 Figure 8.3. Example 8-1 The motion of the particle shown in Fig. m/s Constant velocity Li y ne cit ar ine lo ve ar ve loc ity L t.11 is defined by its position x(t).10 Acceleration of the object moving as a linear function with parabolic blends. which is shown in Fig. Determine the position. cruises with zero acceleration. s Figure 8. It can be seen from Fig. 5)) = sin 𝜋 = 0 m v(0.228 Chapter 8 Derivatives in Engineering v(t) x(t) Figure 8. Solution (a) The velocity and acceleration of the particle can be obtained by finding the first and second derivatives of x(t).3) the velocity is v(t) = = dx(t) dt d (sin 2 𝜋t) dt or v(t) = 2 𝜋cos 2 𝜋t m/s. Since x(t) = sin 2 𝜋t m.6) . (8. (8.4) The acceleration of the particle can now be found by differentiating the velocity as a(t) = = dv(t) dt d (2 𝜋cos 2 𝜋t) dt = (2 𝜋) d (cos 2 𝜋t) dt = (2 𝜋) (−2 𝜋 sin 2 𝜋t) or a(t) = −4 𝜋 2 sin 2 𝜋t m/s2 . velocity. (8.5) = sin (2 𝜋(0. respectively. (b) The position of the particle is given by x(t) = 3 t3 − 4 t2 + 2 t + 6 m.3).5)) = −4 𝜋 2 sin 𝜋 = 0 m/s2 . (8. and (8. (8.4).5) = 2 𝜋cos (2 𝜋(0. and acceleration of the particle at t = 0.5 s can now be calculated by substituting t = 0.5)) = 2 𝜋cos 𝜋 = −2 𝜋 m/s a(0.5) = −4 𝜋 2 sin (2 𝜋(0.5) as x(0.5 in equations (8.5) The position.11 A particle moving in the horizontal direction. 5) + 6 = 6.25 m/s a(0.7).7) The acceleration of the particle can now be obtained by differentiating equation (8.5) = 9 (0. and acceleration of the particle at t = 0. (8. (8.8.5)2 + 2 (0.8) The position.5) + 2 = 0.6) as dx(t) v(t) = dt d = (3 t3 − 4 t2 + 2 t + 6) dt d d d d = 3 (t3 ) − 4 (t2 ) + 2 (t) + 6 (1) dt dt dt dt = 3 (3 t2 ) − 4 (2t) + 2 (1) + 6 (0) or v(t) = 9 t2 − 8 t + 2 m/s.9) The velocity of the particle can be calculated by differentiating equation (8.3 Applications of Derivatives in Dynamics 229 The velocity of the particle can be calculated by differentiating equation (8. (8.5)2 − 8 (0.5) = 3 (0.5)3 − 4 (0. (c) The position of the particle is given by x(t) = 20 cos(3 𝜋t) − 5 t2 m. and (8.5 s can now be calculated by substituting t = 0.7) as dv(t) dt d = (9 t2 − 8 t + 2) dt d d d = 9 (t2 ) − 8 (t) + 2 (1) dt dt dt = 9 (2 t) − 8 (1) + 2 (0) a(t) = or a(t) = 18 t − 8 m/s2 .9) as dx(t) v(t) = dt d (20 cos(3 𝜋t) − 5 t2 ) = dt d d = 20 (cos (3 𝜋t) ) − 5 (t2 ) dt dt = 20 (−3 𝜋 sin (3 𝜋t) ) − 5 (2 t) . (8.5 in equations (8.5) = 18 (0.5) − 8 = 1.6).0 m/s2 .375 m v(0. velocity.8) as x(0. 5 m/s a(0.5)) − 10 (0.10) The acceleration of the particle can now be obtained by differentiating equation (8. velocity.5)) − 5 (0.25) 2 = 0 − 1.5) = 20 cos (3 𝜋(0.5)2 ( ) 3𝜋 = 20 cos − 5 (0. (8.25 m v(0. .5 in equations (8.10) as a(t) = = dv(t) dt d (−60 𝜋 sin(3 𝜋t) − 10 t) dt = −60 𝜋 d d (sin(3 𝜋t)) − 10 (t) dt dt = −60 𝜋 (3 𝜋 cos(3 𝜋t)) − 10 (1).11) The position.5 s can now be calculated by substituting t = 0.5)) − 10 ( ) 3𝜋 2 = −180 𝜋 cos − 10 2 = −180 𝜋 2 (0) − 10 = −10 m/s2 .9). and (8.5) ) ( 3𝜋 −5 = −60 𝜋 sin 2 = 60 𝜋 − 5 = 183. (8.230 Chapter 8 Derivatives in Engineering or v(t) = −60 𝜋 sin(3 𝜋t) − 10 t m/s.5) = −60 𝜋 sin (3 𝜋(0. or a(t) = −180 𝜋 2 cos(3 𝜋t) − 10 m/s2 (8.25 = −1.11) as x(0. and acceleration of the particle at t = 0.5) = −180 𝜋 2 cos (3 𝜋(0.10). 3 Applications of Derivatives in Dynamics 231 The following example will illustrate how derivatives can be used to help sketch functions. 3 (8. For example. Solution (a) The velocity of the particle can be calculated by differentiating equation (8.12 The position of a particle in the vertical plane. factoring equation (8.14) The quadratic equation (8.14) can be solved using one of the methods discussed in Chapter 2.12 is defined by its position y(t) as 8-2 y(t) = 1 3 t − 5 t2 + 21 t + 10 m.12) (a) Determine the value of the position and acceleration when the velocity is zero. v(t) 1 y(t) = − t3 − 5 t2 + 21 t + 10 m 3 Figure 8. (8.15) .13) equal to zero as t2 − 10 t + 21 = 0.12) as dy(t) v(t) = dt ) ( d 1 3 2 t − 5 t + 21 t + 10 = dt 3 1 d 3 d d d (t ) − 5 (t2 ) + 21 (t) + 10 (1) 3 dt dt dt dt 1 = (3 t2 ) − 5 (2t) + 21 (1) + 10 (0) 3 = or v(t) = t2 − 10 t + 21 m/s. 8. Example The motion of a particle shown in Fig. (8.13) The time when the velocity is zero can be obtained by setting equation (8.8. (8.14) gives (t − 3) (t − 7) = 0. (b) Use the results of part (a) to sketch the graph of the position y(t) for 0 ≤ t ≤ 9 s. Therefore. = 2 2 or t = 3. 7 s.14) can also be solved using the quadratic formula.15) are given as: t−3=0 ⇒ t=3s t−7=0 ⇒ t=7s Note that quadratic equation (8. the velocity is zero at both t = 3 s and t = 7 s.16) as y(7) = 1 (7)3 − 5 (7)2 + 21 (7) + 10 = 26. The acceleration of the particle can be obtained by differentiating the velocity of the particle (equation (8.3 m 3 a(7) = 2(7) − 10 = 4 m/s2 . Similarly.232 Chapter 8 Derivatives in Engineering The two solutions of equation (8. an expression for the acceleration is needed.12) and (8. (8.16) The position and acceleration at time t = 3 s can be found by substituting t = 3 in equations (8. To evaluate the acceleration at these times.13)) as dv(t) dt d 2 = (t − 10 t + 21) dt d d d 2 (t ) − 10 (t) + 21 (1) = dt dt dt a(t) = or a(t) = 2 t − 10 m/s2 . which gives √ 10 ± 102 − 4(1)(21) t= 2(1) √ 10 ± 16 = 2 10 ± 4 = 2 10 − 4 10 + 4 .16) as y(3) = 1 (3)3 − 5 (3)2 + 21 (3) + 10 = 37 m 3 a(3) = 2(3) − 10 = −4 m/s2 .12) and (8. the position and acceleration at t = 7 s can be found by substituting t = 7 in equations (8. . Since the acceleration at t = 3 s is negative (a(3) = −4 m/s2 ). To check whether y(t) has a local minimum or maximum. s The approximate sketch of the position y(t) of Example 8-2.13. which begins with a plot of the acceleration a(t). 8. . 8.14. the second derivative (acceleration) test is applied. Knowing that the particle starts from rest at position x = 0 and travels a total of 16 m. the position y(7) = 26. the position y(3) = 37 m is a local maximum. m Local maxima 37 26.13 3 7 t. along with the positions of the particle at t = 0 (y(0) = 10 m) and t = 9 (y(9) = 37 m).3 Local minima 0 0 Figure 8.3 Applications of Derivatives in Dynamics 233 (b) The results of part (a) can be used to sketch the graph of the position y(t). y(t).8.. Such is the case in the following example. Since the acceleration at t = 7 s is positive (a(7) = 4 m/s2 ). What this means is that the position y(t) has a local minimum or maximum at t = 3 s and t = 7 s.e.3 m is a local minimum. This information. Example 8-3 The acceleration of a vehicle is measured as shown in Fig. Derivatives are frequently used in engineering to help sketch functions for which no equation is given. Since the velocity is the derivative of the position. the slope is zero). It was shown in part (a) that the velocity of the particle is zero at t = 3 s and t = 7 s. sketch plots of the position x(t) and velocity v(t). the derivative of the position at t = 3 s and t = 7 s is zero (i. as shown in Fig. can be used to sketch the position y(t). s Figure 8. the position of the vehicle must be a quadratic equation of the form x(t) = t2 + C.e. therefore.14. v(t).14 Acceleration of a vehicle for Example 8-3. a(t) is the slope of v(t)).15. s −2 x(t) Figure 8. v(t) is the slope of x(t)). to construct the position x(t). dx(t) Knowing that x(0) = 0 m and v(t) = dt (i. v(t) = dt = 2 t m/s. use the velocity.15 The velocity profile for Example 8-3. dt The graph of the velocity profile is shown in Fig.3. each interval can be analyzed as follows: dv(t) 0 ≤ t ≤ 2s: = a(t) = 2 ⇒ v(t) is a line with slope = 2. 8.. From Table 8. dt dv(t) 2 < t ≤ 4s: = a(t) = 0 ⇒ v(t) is constant. v(t). m/s2 2 0 2 4 6 t. Solution (a) Plot of Velocity: The velocity of the vehicle can be obtained from the acceldv(t) eration profile given in Fig.17) . 8.. (8. m/s 4 −2 2 1 1 0 2 4 6 t. Knowing that v(0) = 0 m/s and a(t) = dt (i. each interval can be analyzed as follows: (i) 0 ≤ t ≤ 2 s: v(t) is a straight line with a slope of 2 starting from origin dx(t) (v(0) = 0). dt dv(t) 4 < t ≤ 6s: = a(t) = −2 ⇒ v(t) is a line with slope = −2.e.234 Chapter 8 Derivatives in Engineering a(t). (b) Plot of Position: Now. x(t) is a straight line with a slope of 4 m/s starting with a value of 4 m at t = 2 s as shown in Fig. Therefore. for example. 8. dx(t) (ii) 2 < t ≤ 4 s: v(t) has a constant value of 4 m/s. for example. x(t) = t2 is a quadratic function with a positive slope (concave up) and x(2) = 4 m. v(t) = d 2 (t dt dx(t) dt 235 = + C) = 2 t m/s. Therefore.16 2 4 The position of the particle for Example 8-3. for 0 ≤ t ≤ 2 s. m 16 12 4 0 0 Figure 8. therefore. 8. So during the two seconds between t = 2 and t = 4. 8. s . The equation of position for 2 < t ≤ 4 s can be written as x(t) = 4 + 4 (t − 2) = 4 t − 4 m. Therefore. the position increases by 4 m every second. its position at time t = 4 s will be 4 m + 8 m = 12 m. x(t) is a quadratic function with decreasing slope (concave down) starting at x(4) = 12 m and ending at x(6) = 16 m with zero slope. And since the position at time t = 2 s was 4 m.16. 6 t.8. Since the slope is 4 m/s. The value of C is obtained by evaluating equation (8.16.3 Applications of Derivatives in Dynamics This can be checked by taking the derivative. (iii) 4 < t ≤ 6 s: v(t) is a straight line with a slope of −2 m/s. its position increases by 8 m. x(t).16. as shown in Fig. The resulting graph of the position is shown in Fig.17) at t = 0 and substituting the value of x(0) = 0 as x(0) = 0 + C 0 = 0+C C = 0. v(t) = dt = 4.17) is correct. the equation of position given by (8. (b) The acceleration a(t) of the cart is obtained by differentiating the position v(t) as dv(t) a(t) = dt = d (−2 𝜋 sin(2 𝜋t)) dt = −2 𝜋 d (sin(2 𝜋t)) dt = −(2 𝜋)2 sin(2 𝜋t) m/s2 . (a) Find the velocity of the cart.17 Solution A cart moving on frictionless rollers. for example d2 sin(𝜔t) = −𝜔2 sin(𝜔t) dt2 d2 cos(𝜔t) = −𝜔2 cos(𝜔t) dt2 . (a) The velocity v(t) of the cart is obtained by differentiating the position x(t) as v(t) = = dx(t) dt d (cos(2 𝜋t)) dt = −2 𝜋 sin(2 𝜋t) m/s.236 Chapter 8 Derivatives in Engineering Example 8-4 The position of the cart moving on frictionless rollers shown in Fig. Note that the second derivative of sin(𝜔t) or cos(𝜔t) is the same function scaled by −𝜔2 . where 𝜔 = 2 𝜋. 8. (b) Show that the acceleration of the cart is given by a(t) = −𝜔2 cos(𝜔t) m/s2 .17 is given by x(t) = cos(𝜔t) m. where 𝜔 = 2 𝜋. x(t) k m Frictionless rollers Figure 8. 18. (b) The values of the displacement and acceleration when the velocity is zero.e.. (i. 2 2 (8. v(t) = = 0). The solutions of equation (8. The resulting displacement of the beam is given by v (8. This is done by equating the derivative of the dy(t) displacement (or the velocity) to zero.3 Applications of Derivatives in Dynamics Example 8-5 237 An object of mass m moving at velocity v0 impacts a cantilever beam (of length l and flexural rigidity EI) as shown in Fig. 8.18) y(t) = 0 sin 𝜔t 𝜔 √ 3EI where 𝜔 = is the angular frequency of displacement.19) to zero gives v0 cos 𝜔tmax = 0 ⇒ cos 𝜔tmax = 0. Find the following: m l3 (a) The maximum displacement ymax .8. Since v0 and 𝜔 dt are constants. the derivative is given by dy(t) dt v0 d = (sin 𝜔t) 𝜔 dt v = 0 (𝜔 cos 𝜔t) 𝜔 v(t) = or v(t) = v0 cos 𝜔t m/s. (8. l v0. …. .20) are 𝜔tmax = 𝜋 3𝜋 . m/s m y(t) Figure 8. Solution (a) The maximum displacement can be found by first finding the time tmax when the displacement is maximum.20) .19) Equating equation (8.18 A mass impacting a cantilever beam. 19) as tmax = d2 y(t) dt2 d (cos 𝜔t) dt = v0 (−𝜔 sin 𝜔t) = vo or d2 y(t) dt2 = −v0 𝜔 sin 𝜔t m/s2 .238 Chapter 8 Derivatives in Engineering or 𝜋 3𝜋 . (8.21) 2𝜔 2𝜔 Therefore.21).22) 𝜋 The value of the second derivative of the displacement at is given by 2𝜔 ) ( 𝜋 d2 y ( ) 2 𝜔 = −v 𝜔 sin 𝜋 < 0. the displacement of the beam has local maxima or minima at the values of t given by equation (8. the value of the second derivative of the displacement at is given 2𝜔 by ( ) 3𝜋 d2 y ( ) 2𝜔 3𝜋 = −v0 𝜔 sin > 0. To find the time when the displacement is maximum. 2 dt2 Therefore. the displacement is maximum at time 𝜋 . . The second derivative of the displacement is obtained by differentiating equation (8. (8. tmax = 2𝜔 The maximum displacement can be found by substituting tmax = equation (8.18) as v0 sin(𝜔 tmax ) 𝜔 v 𝜋 = 0 sin 𝜔 2 v0 = 𝜔 v = √ 0 3EI m l3 ymax = or √ ymax = v0 m l3 . the second derivative rule is applied. ….23) 𝜋 into 2𝜔 . 3EI (8. 0 2 dt2 3𝜋 Similarly. It can be seen from Fig. y(t) v0 — ω 0 0 Figure 8. the acceleration when the velocity is zero is given by tmax = a(tmax ) = −v0 𝜔 sin (𝜔 tmax ) ( ) 𝜋 = −v0 𝜔 sin 2 = −v0 𝜔 √ 3EI . = −v0 m l3 .8.24) ymax = v0 3EI The acceleration found in part (a) is given by a(t) = −v0 𝜔 sin 𝜔t. Therefore. (8. The plot of the beam displacement given by equation (8.18) is shown in Fig.19 that the maximum value of the beam displacev v ment is simply the amplitude ymax = 0 = √ 0 and the time where the dis𝜔 3EI placement is maximum is given by 𝜔tmax = or m l3 𝜋 2 𝜋 . 8.19 π —– 2ω 3π —– 2ω π — ω 2π —– ω t.19. s Displacement plot to find maximum value. 8. 2𝜔 (b) The position when the velocity is zero is simply the maximum value √ m l3 .3 Applications of Derivatives in Dynamics 239 Note: The local maxima or minima of trigonometric functions can also be obtained without derivatives. The relationship between power and energy is also a derivative relationship.4 APPLICATIONS OF DERIVATIVES IN ELECTRIC CIRCUITS Derivatives play a very important role in electric circuits. the relationship between different variables in circuit elements is discussed briefly here. the voltage is the derivative of the electric potential energy with respect to charge.240 Chapter 8 Derivatives in Engineering Note: a(t) = −v0 𝜔 sin 𝜔t = −𝜔2 ( v0 𝜔 ) sin 𝜔t = −𝜔2 y(t). dq The current i(t) is the rate of change (i. Since the second derivative of a sinusoid is also a sinusoid of the same frequency (scaled by −𝜔2 ).. The voltage v(t) is the rate of change of electric potential energy w(t) (in joules (J)) per unit charge q(t) (in coulomb (C)). derivative) of electric charge per unit time (t in s). where v(t) is the voltage in volts (V) and i(t) is the current in amperes (A). dt Note that the power can be written as the product of voltage and current using the chain rule of derivatives: p(t) = p(t) = dw(t) dw dq = × . dt dq dt .20 Voltage and current in a circuit element. Note that the current always flows through the circuit element and the voltage is always across the element. this is a general result for harmonic motion of any system.20. written as v(t) = dw V. For example. Therefore.e. Before discussing the applications of derivatives in electric circuits. written as i(t) = dq(t) dt A. 8. written as dw(t) W. i(t) + v(t) Circuit element − Figure 8. the relationship between voltage and current for both the inductor and the capacitor is a derivative relationship. that is. the accelera- tion is maximum when the displacement y(t) is maximum.. derivative) of electric energy per unit time. Consider a circuit element as shown in Fig. The power p(t) (in watts (W)) is the rate of change (i. 8.e. By the chain rule of equation (8. then the derivative of composite function f (g(t)) with respect to t can be written as df dg df = × .4 Applications of Derivatives in Electric Circuits 241 or p(t) = v(t) × i(t). dt dg dt (8.26) For example. where g(t) = 2𝜋t. for example. if f is a function of g and g is a function of t. (8. df df dg = × dt dg dt d d = (sin(g)) × (2𝜋t) dt dt d = cos(g) × (2𝜋t) dt = cos(g) × (2𝜋) = 2𝜋 cos(2𝜋t).25) The chain rule of derivative is a rule for differentiating composition of functions.8. The chain rule is also useful in differentiating the power of sinusoidal function such as y1 (t) = sin2 (2𝜋t) or the power of the polynomial function such as y2 (t) = (2t + 10)2 . .26). The derivatives of these functions are obtained as ) dy1 d ( (sin(2𝜋t))2 = dt dt d = 2 (sin(2𝜋t))1 × (sin(2𝜋t)) dt = 2 sin(2𝜋t) × (2𝜋 cos(2𝜋t)) = 4𝜋 sin(2𝜋t) cos(2𝜋t) and ) dy2 d ( (2t + 10)2 = dt dt d = 2 (2t + 10)1 × (2t + 10) dt = 2 (2t + 10) × (2) = 4 (2t + 10) The following example will illustrate some of the derivative relationships discussed above. the function f (t) = sin(2𝜋t) can be written as f (t) = sin(g(t)). 242 Chapter 8 Derivatives in Engineering Example 8-6 For a particular circuit element.27) v(t) = 100 sin 250 𝜋t V.21 + − Element Voltage applied to a particular circuit element.28) i(t) v(t) Figure 8. (8. i(t).29) (b) Power: The power p(t) can be determined by multiplying the voltage given in equation (8. Solution (a) Current: The current i(t) can be determined by differentiating the charge q(t) as dq(t) i(t) = dt ( ) d 1 = sin 250 𝜋t dt 50 = 1 (250 𝜋 cos 250 𝜋t) 50 or i(t) = 5 𝜋 cos 250 𝜋t A.29) as p(t) = v(t) i(t) = (100 sin 250 𝜋t) (5 𝜋 cos 250 𝜋t) = 500 𝜋(sin 250 𝜋t) (cos 250 𝜋t) . Find the following quantities: (a) The current. (b) The power. 8. (8. p(t). the charge is 1 sin 250 𝜋t C 50 and the voltage supplied by the voltage source shown in Fig. (c) The maximum power pmax delivered to the circuit element by the voltage source.21 is q(t) = (8.28) and the current calculated in equation (8. 30) The power p(t) = 250 𝜋 sin 500 𝜋t W in equation (8.8.4. pmax = 250 𝜋W.22. Example 8-7 For the inductor shown in Fig.31).22 Inductor as a circuit element. 8.30) is obtained by using the double-angle trigonometric identity sin(2 𝜃) = 2 sin 𝜃 cos 𝜃 or sin(500 𝜋t) . Note that if the inductance is given in mH (1 mH (millihenry) = 10−3 H). Since −1 ≤ sin 500 𝜋t ≤ 1.22) is given by v(t) = L di(t) dt (8. it must be converted to H before using it in equation (8. where v(t) is the voltage across the inductor in V. if L = 100 mH and i(t) = t e−3t A. + i(t) v(t) L − Circuit element Figure 8. Therefore. i(t) is the current flowing through the inductor in A. (a) Find the voltage v(t) = L . (c) Use the results of (a) and (b) to sketch the current i(t). (8. dt (b) Find the value of the current when the voltage is zero. which is simply the amplitude of the power.3. and L is the inductance of the inductor in henry (H).31) .4 Applications of Derivatives in Electric Circuits 243 or p(t) = 250 𝜋sin 500 𝜋t W. the maximum value of a trigonometric function such as p(t) = 250 𝜋(sin 500 𝜋t) can be found without differentiating the function and equating the result to zero. which gives p(t) = 250 𝜋 sin 500 𝜋t. (sin 250 𝜋t) (cos 250 𝜋t) = 2 (c) Maximum Power Delivered to the Circuit: As discussed in Section 8.1 Current and Voltage in an Inductor The current-voltage relationship for an inductor element (Fig. di(t) . 8. the power delivered to the circuit element is maximum when sin 500 𝜋t = 1. 8. 00496 A). the approximate sketch of the current can be drawn . Using these values along with the values of the current at t = 1 s (i(1s) = 0.1 e−3t (−3 t + 1) = 0.1 di(t) . dt di(t) d where dt = dt (t e−3t ). first find the time t when the voltage is zero and then substitute this time in the expression for current. Since e−3t is never zero.244 Chapter 8 Derivatives in Engineering Solution (a) The voltage v(t)is determined as v(t) = L di(t) dt = (100 × 10−3 ) di(t) dt or v(t) = 0.32) gives ( ) ( ) ( −3t ) d −3t d d ( −3t ) te e = (t) (e ) + (t) dt dt dt ) ( ) ( = (t) −3 e−3t + (1) e−3t = e−3t (−3 t + 1) . i(0) = 0 A. (8. which gives ( ) ( ) ( ) −3 13 1 1 = e i 3 3 = 1 −1 e 3 or i = 0.1 e−3t (−3 t + 1) V. v(t) = 0. (c) Since the voltage is proportional to the derivative of the current. Therefore. at t = 0.0498 A) and at t = 2 s (i(2s) = 0. it follows that (−3 t + 1) = 0. the slope of the current is zero when the voltage is zero. To differentiate the product of two functions (t and −3t e ). Setting equation (8. dt dt dt (8.3) is used: d d d (f (t) g(t)) = f (t) (g(t)) + g(t) (f (t)). (8. Also. the product rule of differentiation (Table 8.123 A) at t = 1∕3 s. Therefore.34) (b) To find the current when the voltage is zero. which gives t = 1∕3 sec.123 A.34) equal to zero gives 0.32) Substituting f (t) = t and g(t) = e−3 t in equation (8. the current i(t) is maximum (imax = 0.33) Therefore. the value of the current when the voltage is zero is determined by substituting t = 1∕3 into the current i(t) = t e−3 t . Note that i = 0.1 0.23 1 1. as illustrated in the following example: Example 8-8 For the given input voltage (square wave) shown in Fig.06 0. Assume i(0) = 0 A and p(0) = 0 W.02 0 0 0. V i(t) v(t) + − 9 L 0 2 4 6 −9 Figure 8. no second derivative test is required! i(t).5 2 2.24. 8.04 0.24 A square wave voltage applied to an inductor.123 0. s Approximate sketch of the current waveform for Example 8-7.8.4 Applications of Derivatives in Electric Circuits 245 as shown in Fig.5 3 t. s . Hence. since it is the only location of zero slope and is greater than the values of i(t) at either t = 0 or t = 2 s. 8.5 Figure 8.08 0.123 A must be a maximum (as opposed to a minimum) value. derivatives are frequently used in circuits to sketch functions for which no equations are given. plot the current i(t) and the power p(t) if L = 500 mH. As seen in dynamics. v(t).12 0.23. 8 t.14 imax = 0. A 0. the value of the current at t = 2 s is 36 A. Therefore. the rate of change of the current is given by (500 × 10−3 ) di(t) = v(t) dt di(t) 1 = v(t) dt 0. Since p(t) = v(t) i(t) and the voltage is ± 9 V. the waveform for the current from 4 ≤ t ≤ 8 is the same as the waveform of the current from 0 ≤ t ≤ 4. the current-voltage relationship is given by v(t) = L . In the interval 0 ≤ t ≤ 2. 8. the slope of the current is twice the applied voltage. the current waveform is a straight line with a constant slope of ± 18 A/s. the power jumps down to (−9) (36) = −324 W. Therefore. Since dt the voltage is known. The waveform of the power is a straight line starting at 0 W with a slope of (9 × 18) = 162 W/s.25.25 Sketch of the current waveform for Example 8-8. A 36 0 0 2 4 6 8 t. Since the value of the current at t = 4 s is 0 (the same as at t = 0 s) and the applied voltage between interval 4 < t ≤ 8 is the same as the applied voltage between 0 ≤ t ≤ 4. Thus.35) or di(t) = 2 v(t). the power delivered to the inductor just before t = 2 sec is 324 W. the current waveform has a constant slope of ± 18 A/s. dt Therefore. the current waveform is a straight line starting at 36 A (at t = 2 s) with a slope of −18 A/s. the power is given by p(t) = (± 9) i(t) W. when a square wave voltage is applied to an inductor. Since the value of the power at t = 4 s is 0 (the same as at t = 0 s) and the applied voltage . s Figure 8. p(t) = 9 i(t) W. Just after t = 2 sec. therefore.246 Chapter 8 Derivatives in Engineering Solution di(t) For an inductor. This completes one cycle of the power. and the current has a negative slope of −18 A/s. Since v(t) = ± 9 V (constant in each interval). the voltage is −9 V and the current is 36 A. In the interval. in other words.5 (8. the waveform for the power is a straight line starting at −324 W with a slope of 162 W/s. This completes one cycle of the current waveform. Hence. Thus. Therefore. 2 < t ≤ 4. p(4) = −324 + 2(162) = 0 W. the resulting current is a i(t). the value of the current at t = 4 s is 0 A. The resulting plot of i(t) is shown in Fig. triangular wave. v(t) = −9 V. the current waveform is a straight line with a slope of 18 A/s that starts at 0 A (i(0) = 0 A). v(t) = 9 V. In the interval 0 ≤ t ≤ 2. Therefore. In the interval 2 < t ≤ 4. 36). + i(t) C v(t) − Circuit element Figure 8.36) where v(t) is the voltage across the capacitor in V. s −324 Figure 8. and C is the capacitance of the capacitor in farad (F).4 Applications of Derivatives in Electric Circuits 247 and current in the interval 4 ≤ t ≤ 8 are the same as the voltage and current in the 0 ≤ t ≤ 4. 8. and is typically referred to as a sawtooth curve. Example 8-9 Consider the capacitive element shown in Fig. the waveform for the power from 4 ≤ t ≤ 8 is the same as the waveform of the power from 0 ≤ t ≤ 4. i(t) is the current flowing through the capacitor in A. The resulting plot of p(t) is shown in Fig. W 324 0 0 2 4 6 8 t.27 with C = 25 𝜇F and v(t) = 20 e−500 t sin 5000 𝜋t V. 8. 8.4. 8. Find the current i(t).2 Current and Voltage in a Capacitor The current-voltage relationship for a capacitive element (Fig. p(t). it must be converted to F before using it in equation (8.26 Sketch of the power for Example 8-8. Note that if the capacitance is given in 𝜇F (10−6 F).27 Capacitor as a circuit element. dt (8.26.8. .27) is given by i(t) = C dv(t) . 5 𝜋e−500 t sin(5000 𝜋t + 92◦ ) A. Letting f (t) = e−500 t and g(t) = sin 5000 𝜋t. To differentiate the product of the two where dt dt functions e−500 t and sin 5000 𝜋t. dt dv(t) d = (20 e−500 t sin 5000 𝜋t). ms . ( ) i(t) = 25 × 10−6 10. Knowdq(t) dv(t) ing that i(t) = dt = C dt .28 Charging of a capacitor. Therefore. Example 8-10 The current shown in Fig. i(t).28 is used to charge a capacitor with C = 20 𝜇F. the product rule of differentiation is required. 1 2 4 5 t. plot the charge q(t) stored in the capacitor and the corresponding voltage v(t). A 4 2 + i(t) ↑ v(t) − C 0 −2 −4 Figure 8.000 e−500 t (10 𝜋cos 5000 𝜋t − sin 5000 𝜋t). i(t) = C dt Substituting the value of C gives dv(t) i(t) = (25 × 10−6 ) . this can also be written as i(t) = 2.248 Chapter 8 Derivatives in Engineering Solution The current i(t) can be found by using equation (8.000 e−500t (10 𝜋cos 5000 𝜋t − sin 5000 𝜋t) or i(t) = 0.36) as dv(t) . 8.25 e−500 t (10 𝜋cos 5000 𝜋t − sin 5000 𝜋t) A Using the results of Chapter 6. [ d d (20 e−500t sin 5000𝜋t) = 20 × e−500 t (sin 5000 𝜋t) dt dt ] d +(sin 5000 𝜋t) (e−500 t ) dt = 20 × [(e−500 t ) (5000𝜋 cos 5000 𝜋t) +(sin 5000 𝜋t)(−500 e−500 t )] = 10. Assume q(0) = v(0) = 0. dt dt dt C dt Therefore. . q(t). Substituting the value of C gives dq(t) dv(t) 1 = .29. as shown in Fig. the derivative (slope) of the voltage v(t) is equal to the derivative (slope) of the charge q(t) multiplied by the reciprocal of the capacitance.4 Applications of Derivatives in Electric Circuits Solution 249 dq(t) (a) Charge: Since i(t) = . dt 20 × 10−6 dt or dq(t) dv(t) = 50 × 103 . the relationship between the charge and the voltage is first derived as: i(t) = C dv(t) dv(t) dq(t) 1 dq(t) = ⇒ = . dt dt The plot of the voltage is thus the same as the plot of the charge with the ordinate scaled by 50 × 103 . 8.8. the plot of the charge q(t) stored in the capacitor can be drawn as shown in Fig.30. for example dq(t) dt dq(t) dt dq(t) dt dq(t) dt dq(t) dt 0 ≤ t ≤ 1 ms : 1 < t ≤ 2 ms : 2 < t ≤ 4 ms : 4 < t ≤ 5 ms : 5 < t ≤ ∞ ms : 4 C/s = = −2 C/s 2 C/s = = −4 C/s 0 C/s. the slope of the charge q(t) is given by each condt stant value of current in each interval. (b) Voltage: To find the voltage across the capacitor. 8. ms Charge on the capacitor in Example 8-10. mC qmax = 6 mC 6 1 1 4 4 2 4 −2 −4 2 1 0 Figure 8.29 1 2 4 5 t. = Therefore. the charge q(t) is in mC. Note that since time t is in ms. The product E I is called the flexural rigidity and is a measure of how stiff the beam is. Consider a beam with elastic modulus E (lb/in.5 APPLICATIONS OF DERIVATIVES IN STRENGTH OF MATERIALS In this section.37) . V vmax = 300 V 300 200 100 0 1 2 4 t. the derivative relationship for beams under transverse loading conditions will be discussed.250 Chapter 8 Derivatives in Engineering v(t). ms 5 Figure 8. 8. the beam deflects in the y-direction with dy(x) in radians. a deflection of y(x) (in. or N-m (8. 8. and results are used to sketch the deflection. or m) and a slope of 𝜃(x) = dx y q(x) Beam with flexural rigidity EI x θ (x) y(x) Figure 8. the voltage is measured in volts. The locations and values of maximum deflections are obtained using derivatives. The internal moments and forces in the beam shown in Fig.2 or N/m2 ) and second moment of area I (in.30 Voltage across the capacitor.31. This section also considers the application of derivatives to maximum stress under axial loading and torsion. Note that while the time is still measured in ms. or N/m).31 A beam loaded in the y-direction. If the beam is loaded with a distributed transverse load q(x) (lb/in.32 are given by the expressions Bending Moment: M(x) = EI d 𝜽(x) dx = EI d2 y(x) dx2 lb-in.4 or m4 ) as shown in Fig. 8. (8. More detailed background on the above relations can be found in any book on strength of materials.40).33. If the deflection is given by ) P ( 3 x − 3 l x2 m.5 Applications of Derivatives in Strength of Materials V(x) = d M(x) = EI d3 y(x) lb or N dx3 dx dV(x) d4 y(x) = −EI lb/in. Solution Deflection: The deflection of the beam at the free end can be determined by substituting x = l in equation (8. Example 8-11 Consider a cantilever beam of length l loaded by a force P at the free end. 8.33 Cantilever beam with an end load P. y P l x y(x) Figure 8. which gives ] P [ 3 l − 3 l (l2 ) y(l) = 6 EI ) P ( −2 l3 = 6 EI or P l3 m.40) y(x) = 6 EI find the deflection and slope at the free end x = l.38) (8. or N/m Distributed Load: q(x) = − dx4 dx Shear Force: q(x) 251 (8.8. y(l) = − . as shown in Fig.39) M(x) V(x) Figure 8.32 The internal forces in a beam loaded by a distributed load. 3 EI This classic result is used in a range of mechanical and civil engineering courses. the deflection is given by ) P ( 3 4 x − 3 l2 x m. For 0 ≤ x ≤ 1/2.34. and the maximum slope by a factor of 4! ymax = − Example 8-12 Consider a simply supported beam of length l subjected to a central load P. . for example: 𝜃(l) = − P l3 3 EI P l2 𝜃max = − . 8. and I are all treated as constants. 𝜃(x) = 2 EI Note that the parameters P. E. equation (8.42) y(x) = 48 EI Determine the maximum defection ymax . l.41) (x2 − 2 l x) rad. However. as well as the slope at the end x = 0.41) as P 2 [l − 2 l (l)] 2 EI P (− l2 ) = 2 EI 𝜃(l) = or P l2 rad. The slope 𝜃(x) at the free end can now be determined by substituting x = l in. 2 EI It can be seen that doubling the load P would increase the maximum deflection and slope by a factor of 2. (8. doubling the length l would increase the maximum deflection by a factor of 8.252 Chapter 8 Derivatives in Engineering Slope: The slope of the deflection 𝜃(x) can be found by differentiating the deflection y(x) as dy(x) 𝜃(x) = dx [ )] P ( 3 d x − 3 l x2 = dx 6[EI ] P d d 3 = (x ) − 3 l (x2 ) 6 EI dx dx [ ] P 3 (x2 ) − 3 l (2 x) = 6 EI ) P ( 2 3x − 6lx = 6 EI or P (8. as shown in Fig. 2 EI Note: It can be seen by inspection that both the deflection and the slope of deflection are maximum at the free end. 8. 𝜃(x) is set to zero and the resulting equation is solved for the values of x as l P (4 x2 − l2 ) = 0 ⇒ 4 x2 − l2 = 0 ⇒ x = ± .34 Solution A simply supported beam with a central load P. The value of the maximum deflection can now be found by substituting x = l∕2 in . the deflection is maximum at x = l∕2. 16 EI 2 Since the deflection is given for 0 ≤ x ≤ l∕2. The slope of the deflection can dx be found as dy(x) dx [ ] P d = (4 x3 − 3 l2 x) dx 48 EI ] [ d d P 3 2 (4 x ) − (3 l x) = 48 EI dx dx [ ] P d d = 4 (x3 ) − 3 l2 (x) 48 EI dx dx 𝜃(x) = ] P [ 4(3 x2 ) − 3 l2 (1) 48 EI ) P ( 12 x2 − 3 l2 = 48 EI = or 𝜃(x) = P ( 2 2) 4x − l . 16 EI (8.43) To find the location of the maximum deflection.5 Applications of Derivatives in Strength of Materials y 253 P l 2 x y(x) l Figure 8. dy(x) The deflection is maximum when = 𝜃(x) = 0. 35 A simply supported beam with a sinusoidal load. y(x) = − sin y πx q(x) = w0 sin —– l x l Figure 8. 48 EI Likewise.35.254 Chapter 8 Derivatives in Engineering equation (8. 8.42) as ymax ( ) l =y 2 [ ( ) ( )] 3 P l l = − 3 l2 4 48 EI 2 2 P = 48 EI ( l3 3 l3 − 2 2 ) or P l3 m. the slope 𝜃(x) at x = 0 can be determined by substituting x = 0 in equation (8.43) as ymax = − P (4 ∗ 0 − l2 ) 16 EI P ( 2) −l = 16 EI 𝜃(0) = or 𝜃(0) = − Example 8-13 P l2 rad. (8. the moment M(x) and the shear force V(x).44) . 16 EI A simply supported beam of length l is subjected to a distributed load q(x) = ( ) 𝜋x as shown in Fig. If the deflection is given by l w0 sin w0 l4 ( 𝜋x l ) m 𝜋 4 EI find the slope 𝜃(x). the shear force V(x) is the derivative of the moment M(x).46) for M(x) gives [ ] ( ) 2 d w0 l 𝜋x V(x) = sin dx l 𝜋2 = ( )] w0 l 2 [ 𝜋 𝜋x cos 2 l l 𝜋 . = EI dx dx2 Substituting equation (8.5 Applications of Derivatives in Strength of Materials Solution 255 Slope: The slope of the deflection 𝜃(x) is given by dy(x) dx [ ] ( ) w0 l 4 d 𝜋x = − sin dx l 𝜋 4 EI 𝜃(x) = =− =− [ w0 l4 𝜋 4 EI ( )] 𝜋x d (sin dx l ( )] w0 l4 [ 𝜋 𝜋x cos l 𝜋 4 EI l or 𝜃(x) = − ( w0 l3 𝜋 3 EI cos 𝜋x l ) rad. (8. = EI dx dx3 Substituting equation (8. or V(x) = dM(x) d3 y(x) .45) for 𝜃(x) gives ] [ ( ) w0 l 3 d 𝜋x − M(x) = EI cos dx l 𝜋 3 EI =− ( )] w0 l3 [ 𝜋 𝜋x − sin l l 𝜋3 or w0 l2 M(x) = 𝜋2 ( sin 𝜋x l ) N-m.45) Moment: By definition.46) Shear Force: By definition. (8. or M(x) = EI d𝜃(x) d2 y(x) .8. the moment M(x) is obtained by multiplying the derivative of the slope 𝜃(x) by EI. the normal stress 𝜎 acts perpendicular to the cross section and has units of force per unit area (psi or N/m2 ). 8.256 Chapter 8 Derivatives in Engineering or ( ) w0 l 𝜋x N.37. l which matches the applied load in Fig.36 A rectangular bar under axial loading. A P θ Aθ Figure 8. To find the stress on an oblique plane. 8.5. as shown in Fig. The normal stress is given by P .1 Maximum Stress under Axial Loading In this section. consider an inclined section of the bar as shown in Fig.37 Inclined section of the rectangular bar. A normal stress 𝜎 results when a bar is subjected to an axial load P (through the centroid of the cross section). 8. cos 𝜋 l dV(x) The above answer can be checked by showing that q(x) = − as dx [ ( )] 𝜋x d w0 l cos q(x) = − dx 𝜋 l (8. the application of derivatives in finding maximum stress under axial loading is discussed. ( (8.48) q(x) = w0 sin 8.35. P A θ Aθ .36. Therefore. (8.49) A where A is the cross-sectional area of the section perpendicular to longitudinal axis of the bar. 𝝈= P P A Figure 8.47) V(x) = =− ( )] w0 l [ 𝜋 𝜋x − sin 𝜋 l l or ) 𝜋x N/m. Note that the resultant force in the axial direction must be equal to P to satisfy equilibrium. The free-body diagram of the forces acting on the oblique plane is shown in Fig. 8.38.8. 8. and V. The force perpendicular to the inclined cross section F produces a normal stress 𝜎𝜃 (shown in Fig. F. cos 𝜃 The force P can be resolved into components perpendicular to the inclined plane F and parallel to the inclined plane V. F. and V can be found by using the right triangle shown in Fig.38 Free-body diagram.5 Applications of Derivatives in Strength of Materials 257 The relationship between the cross-sectional area perpendicular to the longitudinal axis and the area of the inclined plane is given by A𝜃 cos 𝜃 = A A A𝜃 = .40) given by 𝜎𝜃 = F A𝜃 = P cos 𝜃 A∕cos 𝜃 = P cos2 𝜃 . 8. Aθ A F θ P θ P V Figure 8. A (8.39 as F = P cos 𝜃 V = P sin 𝜃 P V = P sin θ θ F = P cos θ Figure 8.39 Triangle showing force P. The relationship among P.50) . the normal and A shear stresses on the inclined cross section are given by Substituting 𝜎 = 𝜎𝜃 = 𝜎 cos2 𝜃 (8. The tangential force V produces a shear stress 𝜏𝜃 given as 𝜏𝜃 = V A𝜃 = P sin 𝜃 A∕cos 𝜃 = P sin 𝜃 cos 𝜃 .50) and (8.53) In general. Solution (a) First.54) . and brass fail due to maximum values of 𝜏𝜃 (shear stress).40 Normal and shear stresses acting on the inclined cross section.51). brittle materials like glass.51) P from (8.52) 𝜏𝜃 = 𝜎 sin 𝜃 cos 𝜃. and cast iron fail due to maximum values of 𝜎𝜃 (normal stress). aluminum. concrete.52) is d𝜎𝜃 d =𝜎 (cos2 (𝜃)) d𝜃 d𝜃 =𝜎 d (cos 𝜃 cos 𝜃) d𝜃 = 𝜎 (cos 𝜃 (−sin 𝜃) + cos 𝜃 (−sin 𝜃)) = −2 𝜎 (cos 𝜃 sin 𝜃). (8. and find their maximum values. (8.258 Chapter 8 Derivatives in Engineering Aθ A σθ σ τθ Figure 8. Example 8-14 Use derivatives to find the values of 𝜃 where 𝜎𝜃 and 𝜏𝜃 are maximum.49) into equations (8. find the derivative of 𝝈𝜽 with respect to 𝜽: The derivative of 𝜎𝜃 given by equation (8. A (8. ductile materials like steel. However. 52) gives 𝜎max = 𝜎 cos2 (0◦ ) = 𝜎. d2 𝜎𝜃 d𝜃 2 = −2 𝜎 < 0.8. in other words. 𝜎𝜃 has a local maximum or minimum. So 𝜎𝜃 has a maxi- mum value at 𝜃 = d2 𝜎𝜃 For 𝜃 = 90◦ . Therefore. therefore. Maximum Value of 𝝈𝜽 : Substituting 𝜃 = 0◦ in equation (8. For 𝜃 = 0◦ . the second derivative test is performed.54) to zero and solve the resulting equation for the value of 𝜃 between 0 and 90◦ where 𝜎𝜃 is maximum.5 Applications of Derivatives in Strength of Materials 259 (b) Next. at 𝜃 = 0◦ and 90◦ . The second derivative of 𝜎𝜃 is given by d2 𝜎𝜃 d𝜃 2 = d (−2 𝜎 cos 𝜃 sin 𝜃) d𝜃 d (cos 𝜃 sin 𝜃) d𝜃 ] [ d d (cos 𝜃) (sin 𝜃) = −2 𝜎 cos 𝜃 (sin 𝜃) + d𝜃 d𝜃 = −2 𝜎 = −2 𝜎 [cos 𝜃(cos 𝜃) + (−sin 𝜃) sin 𝜃] = −2 𝜎 (cos2 𝜃 − sin2 𝜃) or d2 𝜎𝜃 d𝜃 2 = −2 𝜎 cos 2 𝜃 where cos 2 𝜃 = cos2 𝜃 − sin2 𝜃. which gives cos 𝜃 = 0 ⇒ 𝜃 = 90◦ sin 𝜃 = 0 ⇒ 𝜃 = 0◦ . 𝜎𝜃 has a minimum 2 d𝜃 value at 𝜃 = 90◦ . −2 cos 𝜃 sin 𝜃 = 0. This means that the largest normal stress during axial loading is simply the applied stress 𝜎! . 0◦ . = −2 𝜎 cos(180◦ ) = 2 𝜎 > 0. or Therefore. To find the value of 𝜃 where 𝜎𝜃 has a maximum. equate the derivative in equation (8. 𝜃 = 0◦ and 90◦ are the critical points. equate the derivative in equation (8. Thus.260 Chapter 8 Derivatives in Engineering Value of 𝜽 where 𝝉𝜽 is maximum: (a) First. therefore. . To find whether 𝜏𝜃 has a maximum or minimum at 𝜃 = 45◦ . 𝜏𝜃 has a local maximum or minimum at 𝜃 = 45◦ . Since the second derivative is negative. d2 𝜏𝜃 d𝜃 2 0. d𝜃 (8. but at an angle of 45◦ . cos 45◦ (√ ) (√ ) 2 2 =𝜎 2 2 or 𝜏max = 𝜎 2 at 𝜃 = 45◦ . Therefore.53) is given by d𝜏𝜃 d = (𝜎 sin 𝜃 cos 𝜃) d𝜃 d𝜃 d =𝜎 (sin 𝜃 cos 𝜃) d𝜃 ( ) sin 2 𝜃 d =𝜎 d𝜃 2 𝜎 = (2 cos 2 𝜃) 2 or d𝜏𝜃 = 𝜎 cos 2 𝜃. the maximum shear stress during axial loading is equal to half the applied normal stress.55) (b) Next.53) gives 𝜏max = 𝜎 sin 45◦ . 𝜏𝜃 has a maximum value at 𝜃 = 45◦ .55) to zero and solve the resulting equation 𝜎 cos 2 𝜃 = 0 for the value of 𝜃 (between 0 and 90◦ ) where 𝜏𝜃 is maximum: cos 2 𝜃 = 0 2 𝜃 = 90◦ ⇒ ⇒ 𝜃 = 45◦ . This is why a tensile test of a steel specimen results in failure at a 45◦ angle. the second derivative test is performed. < Maximum Value of 𝝉𝜽 : Substituting 𝜃 = 45◦ in equation (8. The second derivative of 𝜏𝜃 is given by d2 𝜏𝜃 d𝜃 2 d (𝜎 cos 2 𝜃) d𝜃 = 𝜎 (−sin 2 𝜃)(2) = = −2 𝜎 sin 2 𝜃 For 0 ≤ 𝜃 ≤ 90◦ ⇒ 0 ≤ 2 𝜃 ≤ 180◦ ⇒ sin 2 𝜃 > 0. find the derivative of 𝝉𝜽 with respect to 𝜽: The derivative of 𝜏𝜃 given by equation (8. 41 Velocity profile of the skydiver jumping from a height of 12. 8.41.41. 8. t. dv .8. 0) (a) (i) 0 ≤ t ≤ 5 s: v(t) is linear with slope m= −161 − 0 5−0 = − 32. s . (c) Use the results of part (b) to sketch the acceleration a(t) for 0 ≤ t ≤ 301 s. −30) (31.6 261 FURTHER EXAMPLES OF DERIVATIVES IN ENGINEERING Example 8-15 The velocity of a skydiver jumping from a height of 12. (iii) 28 < t ≤ 31 s: v(t) is linear with slope m= = −161 − (−30) 28 − 31 −131 −3 = 43. (a) Find the equation of the velocity v(t) for the five time intervals shown in Fig.2 t ft/s.67 ft/s. −161) y = 0 ft (a) A skydiver (b) Velocity profile Figure 8. v(t). v(t) = −32.6 Further Examples of Derivatives in Engineering 8.2 ft/s.042 ft. ft/s y = 12. (ii) 5 < t ≤ 28 s: v(t) is constant at v(t) = −161 ft/s. −30) (5. find the acceleration a(t) of the skydiver for 0 ≤ t ≤ (b) Knowing that a(t) = dt 301 s.042 ft 0 0 −30 −161 5 28 31 271 (271. −161) (28. Solution 301 (301. Therefore.042 ft is shown in Fig. a(t) = (iii) 28 < t ≤ 3 s: dv(t) dt d = (43. v(t) = 43. a(t) = .67) dt = 43. −30). The value of b (y-intercept) can be found by substituting the data point (t. which gives ⇒ −30 = 43.3 t) dt = −32. v(t)) = (301. (iv) 31 < t ≤ 271 s: v(t) is constant at v(t) = −30 ft/s.2 ft/s2 .67(31) + b b = −1383. v(t)) = (31. (v) 271 < t ≤ 301 s: v(t) is linear with slope −30 − 0 271 − 301 −30 = −30 = 1 ft/s. The value of b (y-intercept) can be found by substituting the data point (t.67 ft/s.67 t − 1383.67 Therefore. (b) (i) 0 ≤ t ≤ 5 s: dv(t) dt d = (−32. v(t) = t − 301 ft/s.7 ft/s2 . v(t) = 43. v(t) = t + b ft/s.262 Chapter 8 Derivatives in Engineering Therefore. m= Therefore. which gives 0 = 1(301) + b ⇒ b = −301 Therefore. a(t) = (ii) 5 < t ≤ 28 s: dv(t) dt d = (−161) dt = 0 ft/s2 .7 t − 1383. 0).67 t + b ft/s. y) set at benchmark A. a(t) = (c) The acceleration of the skydiver found in part (b) can be drawn as shown in Fig.43. the engineer has defined the hilltop segment of the highway by a parabolic arc: (8. ft/s2 43. These grades pass through benchmarks A and B located as shown in Fig.56) y(x) = a x2 + b x which is tangent to the uphill grade at the origin. respectively.42 Example 8-16 The acceleration of the skydiver. 8.6 Further Examples of Derivatives in Engineering 263 (iv) 31 < t ≤ 271 s: dv(t) dt d = (−30) dt = 0 ft/s2 . .42.8.67 1 0 5 28 31 271 301 t.2 Figure 8. With the origin of the coordinate axes (x. A proposed highway traverses a hilltop bounded by uphill and downhill grades of 10% and −8%. s −32. a(t) = (v) 271 < t ≤ 301 s: dv(t) dt d = (t − 301) dt = 1 ft/s2 . 8. a(t). (d) Find the equation of the parabolic arc. (c) Given that at the downhill point of tangency (̄x).264 Chapter 8 Derivatives in Engineering (a) Find the slope of the line for the uphill grade and the value of b for the parabolic arc.56) and has the value b = 0. m 6 200 B Downhill grade −8% Figure 8. the initial slope is equal to the derivative of equation (8.58) d̂y dy = (8. m Downhill point of tangency (x¯. the initial slope of the arc is given by the coefficient b of the equation (8.60) Therefore. Solution (a) The initial slope of the parabolic arc is equal to the uphill grade.57) for the downhill grade. ȳ ) of the parabolic arc with the downhill grade.60) as dy = b = 0.1. y.56) evaluated at x = 0. both the elevation and the slope of the parabolic arc are equal to their respective values of the downhill line: y(̄x) = ŷ (̄x) (8. . compute its elevation. Since the arc is tangent to the uphill grade at the origin. the slope of the line for the uphill grade can be found by setting x = 0 in equation (8.59) | dx x=̄x dx x=̄x Determine the point of tangency (̄x.1. (8.43 Parabolic arc traversing highway hilltop. The derivative of the parabolic arc is given by dy d = (a x2 + b x) dx dx = 2 a x + b. y¯) Uphill grade 10% 0 A 0 x. which is expressed as a decimal fraction of 0. Also.1. (8. (b) Find the equation of the line ŷ = c x + d (8.61) | dx x=0 Hence. Therefore.57)) for the downhill grade is given by c = −0.09 + 0.64) and (8.62) Since this line passes through benchmark B.09 = 111. −6) into equation (8.60) gives dy = 2 a x̄ + 0. (8.08 x̄ + 10 x̄ (a x̄ + 0. the value of x̄ can be found as x̄ (−0.65) into equation (8.1 m. Its elevation ȳ is obtained by substituting this value of x̄ into .8.18) = 10.64) Evaluating the derivative of equation (8. the equation of the line can be written as ŷ = −0.66) into equation (8.58) gives a x̄ 2 + 0.62) as −6 = −0. (8.66) Evaluating equations (8. substituting the value of a x̄ from equation (8.59) gives 2 a x̄ + 0.62) at x̄ and substituting the results in equation (8.08(200) + d d = −6 + 16 d = 10.67) Now.08 a x̄ = −0.68) (8.08 x + d.08.1. (8.09 x̄ = 10 x̄ = 10 0.63) at x = x̄ gives d̂y = −0.56) and (8.10 = −0.67). Therefore.1 m from benchmark A.08.1 x̄ = −0.18) = 10 0.63) (c) Substituting x = x̄ in equation (8.69) Thus. (8. the equation of the line for the downhill slope is given by ŷ = −0. the point of tangency lies a horizontal distance of 111.6 Further Examples of Derivatives in Engineering 265 (b) The slope of the line (equation (8.08x + 10.65) Substituting equations (8. (8.09. | dx x=̄x (8. the value of the y-intercept d can be found by substituting the data point (200. | dx x=̄x (8. y(0) = H = 50 ft is the initial height of the rocket. (b) The velocity v(t) = (c) The d2 y(t) . P8.09 a= 111. (8. A model rocket is fired from the roof of a 50 ft tall building as shown in Fig. The height of the rocket is given by y(t) = y(0) + v(0) t − 1 2 g t ft 2 y(t) where y(t) is the height of the rocket at time t. v(0) = 150 ft/s is the initial velocity of the rocket.70) and (8. 8-2.11 m.61) into equation (8. (d) With the value of x̄ known. and g = 32. Figure P8. The height in the vertical plane of a ball thrown from the ground with an initial velocity of v(0) = 250 m/s satisfies the . Use your results to sketch y(t). v(0) = 49 m/s and g = 9. Find the following: (a) Write the quadratic equation for the height y(t) of the rocket.08(111.266 Chapter 8 Derivatives in Engineering equation (8.1. Repeat problem P8-1 if H = 15 m.08 x̄ + 10 = −0.2 ft/s2 is the acceleration due to gravity.70) The equation for the parabolic arc may now be written by substituting the values of a and b from equations (8. Therefore.1) + 10 = 1. PROBLEMS 8-1.09 −0.1 a = −0.00081 x2 + 0.1 x.11) m.8 m/2 . 1.56) as y = −0.63) as ȳ = ŷ (̄x) = −0.00081 m−1 .66) as a x̄ = −0. the downhill point of tangency is given by (111. dt a(t) = dv(t) dt = (d) The time required to reach the maximum height as well as the corresponding height ymax . 8-3.1. the coefficient a for the parabolic arc can be obtained from equation (8.1 acceleration dy(t) . dt2 H g A rocket fired from top of a building in problem P8-1. Determine the position. 8-7.5 is described by its position x(t). (b) The velocity v(t) = (c) The 267 acceleration y(t) = t3 − 12 t2 + 36 t + 20 m. 8-8. and acceleration at t = 3. 2 where g = 9. The motion of a particle moving in the horizontal direction is described by its position x(t). Use your results to sketch y(t). dy(t) .5 s if (a) x(t) = 4 cos(5 𝜋t) m.2 ft/s2 . The height of the particle is given by Figure P8. y(t) = t3 − 12 t2 + 36 t + 20 m 8-4. and acceleration at t = 1. P8. velocity. y(t) g 8-6. (b) Use your results in part (a) to sketch y(t) for 0 ≤ t ≤ 4 s. The motion of a particle in the vertical plane is shown in Fig. Repeat problem P8-3 if v(0) = 161 ft/s and g= 32. .3 A projectile in the vertical plane. dt2 dv(t) dt = (d) The time required to reach the maximum height.7. 3 2 √ 7 (b) x(t) = 3 t5 − 5 t2 + + 2 t m. (a) Find the values of position and acceleration when the velocity is zero. Find the following: (a) Write the quadratic equation for the height y(t) of the ball. The motion of a particle in the vertical plane is shown in Fig. velocity. Determine the position. Figure P8.0 s if v(t) x(t) Figure P8. as well as the corresponding height ymax . P8. 8-5. (b) x(t) = 4 t3 − 6 t2 + 7 t + 2 m. t (c) x(t) = 2 e4 t + 3 e−5 t + 2 (et − 1) m.8. relationship (a) x(t) = 4 cos y(t) = v(0) t − 1 2 g t m. P8. The motion of a particle moving in the horizontal direction as shown in Fig.5 A particle moving in the horizontal direction. (b) Use your results in part (a) to sketch y(t) for 0 ≤ t ≤ 9 s.8 m/s2 is the acceleration due to gravity. dt a(t) = d2 y(t) . (c) x(t) = 10 sin(10 𝜋t) + 5 e3 t m.Problems ( ) ( ) 2 3 𝜋t + 3 sin 𝜋t m. The height of the particle is given by y(t) = 2 t3 − 15 t2 + 24 t + 8 m (a) Find the values of the position and acceleration when the velocity is zero.7 Motion of a particle in the vertical plane. 15. 8-9. If C = 2 𝜇F and v(t) = t2 e−10 t V.9 Voltage and current in an inductor. (c) Use the above information to sketch i(t). The current flowing through the capacitor shown in Fig. The current flowing through a capacitor shown in Fig. P8. sketch the (a) Knowing that a(t) = dv dt acceleration a(t). P8. y(t) = 2 t3 − 15 t2 + 24 t + 8 m Figure P8. If i(t) = t3 e−2 t A and L = 0. 8-14.12 is given by dv(t) i(t) = C dt . If C = 500 𝜇F and v(t) = 250 sin(200 𝜋t) V.16. 8-16. (a) Find the current i(t). Determine the voltage v(t). The velocity of the vehicle for the next 8 seconds is shown in Fig.268 Chapter 8 Derivatives in Engineering (a) Find the current i(t). (b) Find the value of the current when the voltage is zero.9 is given by v(t) = L dt . P8.25 H and i(t) = t2 e−t A. At time t = 0. a vehicle located at position x = 0 is moving at a velocity of 10 m/s. 8-13. sketch the dt position x(t) if the maximum position is x = 30 m and the final position is x = 10 m. (b) Find the power p(t) = v(t) i(t) and its maximum value pmax . .12 Current flowing through a capacitor. P8.8 Motion of a particle in the vertical i(t) v(t) + − C plane. (b) Knowing that v(t) = dx . (b) Find the value of the voltage when the current is zero. The voltage across the inductor of di(t) Fig. A vehicle starts from rest at position x = 0. sketch the dt position x(t) if the minimum position is x = −16 m and the final position is x = 0.125 H. (b) i(t) = 20 cos(2 𝜋60 t) A. i(t) v(t) + − L Figure P8. . Repeat problem P8-12 if C = 40 𝜇F and v(t) = 500 cos(200 𝜋t) V. 8-12. The velocity of the vehicle for the next 8 seconds is shown in Fig. 8-10. and the maximum power transfered if the inductance is L = 2 mH and the current i(t) is given by (a) i(t) = 13 e−200 t A. 8-11. The voltage across an inductor is given di(t) by v(t) = L dt . . (c) Use the given information to sketch v(t). 8-15. sketch the (a) Knowing that a(t) = dv dt acceleration a(t).12 is given by dv(t) i(t) = C dt . (b) Knowing that v(t) = dx . Figure P8. Repeat problem P8-9 if L = 0. (a) Find the voltage. v(t). P8. the power p(t) = v(t) i(t). x(t) v(t). s −10 Figure P8. 8-18. m/s2 3 0 0 2 4 6 t. A vehicle starting from rest at position x = 0 is subjected to the acceleration a(t) shown in Fig. (b) Knowing that v(t) = dx . (b) Knowing that v(t) = dx . sketch the dt velocity v(t) if the initial velocity is 12 m/s. sketch the (a) Knowing that a(t) = dv dt velocity v(t). s 8 0 0 2 4 6 8 t. s −8 Figure P8. P8. . sketch the dt position x(t) if the final position is x = 48 m. Clearly indicate both its maximum and final values.17. 8-19. x(t) v(t). P8. m/s 8 x(t) a(t). m/s 10 0 0 2 4 6 8 t. sketch the dt position x(t) if the final position is x = 60 m.17 Acceleration of a vehicle for problem P8-17. A vehicle starting from rest at position x = 0 is subjected to the following acceleration a(t): . At time t = 0.18. (a) Knowing that a(t) = dv . Figure P8.Problems 269 8-17. a moving vehicle is located at position x = 0 and subjected to the acceleration a(t) shown in Fig.15 Velocity of a vehicle for problem −3 P8-15.16 Velocity of a vehicle for problem P8-16. 8-21. m/s2 8-20. dv . dt 0 2 4 6 t. 5 (a) Knowing that a(t) = velocity v(t).270 Chapter 8 Derivatives in Engineering (a) Knowing that a(t) = velocity v(t). sketch the (b) Knowing that v(t) = dx . Clearly indicate both its maximum and final values. i(0) = 0 A. Assume L = 2 H. P8. A vehicle starting from rest at position x = 0 is subjected to the following acceleration a(t): 10 0 dv . The voltage across an inductor is given di(t) in Fig. . sketch the dt position x(t) if the final position is x = 15 m. x(t) a(t). sketch the graphs of i(t) and p(t).21. m/s2 Figure P8.18 Acceleration of a vehicle for problem 10 P8-18. sketch the dt position x(t) if the final position is x = 260 m. m/s2 Figure P8. dt sketch the (b) Knowing that v(t) = dx .19 Acceleration of a vehicle for problem P8-19. s −10 x(t) a(t). s x(t) −10 a(t). Knowing that v(t) = L dt and p(t) = v(t) i(t). 0 0 2 13 15 t.20 Acceleration of a vehicle for problem P8-20. 0 0 2 5 6 t. and p(0) = 0 W. s −5 Figure P8. Clearly indicate both its maximum and final values. q(0) = 0 C. V v(t) 6 + − C i(t). A 0 0 1 2 3 4 5 t. sketch the graphs of i(t) and p(t). sketch the graphs of the dt stored charge q(t) and the voltage v(t). and v(0) = 0 V. The current flowing through a capacitor is given in Fig. 8-23. P8.25.22. i(0) = 0 A.25 H. i(t) + − 1 2 t. v(t) 0 L Figure P8. ms −1 Figure P8. The voltage across an inductor is given di(t) in Fig. P8. Knowing that v(t) = L dt and p(t) = v(t) i(t).23. and p(0) = 0 W. ms 3 −10 problem P8-21. Knowing that i(t) = dq(t) dv(t) = C dt . P8.24 Current flowing through a capacitor for problem P8-24. Assume L = 0.24. Knowing that v(t) = L dt and p(t) = v(t) i(t). Knowing that . i(t) v(t). and p(0) = 0 W.23 Voltage across an inductor for problem P8-23. 8-25. V 10 6 0 271 0 1 2 3 4 0 t. 0 1 2 3 4 5 6 t. sketch the graphs of i(t) and p(t). Assume L = 2 mH.21 Voltage across an inductor for 8-22. 8-24. The current applied to a capacitor is given in Fig. The voltage across an inductor is given di(t) in Fig. i(0) = 0 A.Problems i(t) i(t) + − v(t) L + − v(t) L v(t). s 2 −6 0 Figure P8.22 Voltage across an inductor for problem P8-22. Assume C = 250 𝜇F. s 5 Figure P8. V v(t). P8. 272 Chapter 8 Derivatives in Engineering dq(t) dv(t) i(t) = dt = C dt , sketch the graphs of the stored charge q(t) and the voltage v(t). Assume C = 250 𝜇F, q(0) = 0 C, and v(0) = 0 V. 8-27. A simply supported beam is subjected to a load P as shown in Fig. P8.27. The deflection of the beam is given by l Pbx 2 (x + b2 − l2 ), 0 ≤ x ≤ 6 EI l 2 where EI is the flexural rigidity of the beam. Find the following: (a) The location and the value of maximum deflection ymax . y(x) = i(t) v(t) + − C i(t), A (b) The value of the slope 𝜃 = the end x = 0. 2 y 1 0 0 1 2 3 4 5 dy(x) dx at P t, ms a b A B C −2 x Figure P8.25 Current flowing through a capacitor for problem P8-25. l 8-26. The current flowing through a 500 𝜇F capacitor is given in Fig. P8.26. Knowdv(t) ing that i(t) = C dt , plot v(t) for 0 ≤ t ≤ 4 sec if v(0) = −4 V, v(2) = 2 V, v(4) = 2 V, and the maximum voltage is 4 V. i(t) v(t) + − C 1 2 3 4 5 6 −2 Figure P8.26 Current flowing through a capacitor for problem P8-26. problem P8-27. 8-28. A simply supported beam is subjected L as shown in Fig. to a load P at x = 4 P8.28. The deflection of the beam is given by L P (7 L2 x − 16 x3 ), 0 ≤ x ≤ 128 EI 4 where EI is the flexural rigidity of the beam. Find the following: dy(x) (a) The equation of the slope 𝜃 = dx . 2 0 Figure P8.27 A simply supported beam for y(x) = − i(t), mA 0 x=0 t, s (b) The location and magnitude of maximum deflection ymax . (c) Evaluate both the deflection and L slope at x = 0 and x = . 4 (d) Use the results of parts (b) and (c) to sketch the deflection y(x) for L . 0 ≤ x ≤ 4 Problems y P L 4 A 8-30. A simply supported beam is subjected to a sinusoidal distributed load, as shown in the Fig. P8.30. The deflection y(x) of the beam is given by ) ( wo L4 2𝜋 x sin y(x) = − L 16 𝜋 4 EI 3L 4 B C x where EI is the flexural rigidity of the beam. Find the following: L x=0 y Figure P8.28 A simply supported beam for problem P8-28. (b) The location and the value of maximum deflection ymax . (c) Evaluate both the deflection and slope at the points A and B (x = 0 and x = L2 ). (d) Use the results of parts (a)–(c) to sketch the deflection y(x) from x = 0 to x = L2 . y B Mo x A x L Figure P8.30 A simply supported beam for problem P8-30. (a) The equation for the slope 𝜃 = dy . dx (b) The value of the slope where the deflection is zero. (c) The value of the deflection at the location where the slope is zero. (d) Use the results of parts (b) and (c) to sketch the deflection y(x). 8-31. Consider a beam under a linear distributed load and supported as shown in Fig. P8.31. The deflection of the beam is given by y(x) = w0 (−x5 + 2 L2 x3 − L4 x) 120 EI L where L is the length and EI is the flexural rigidity of the beam. Find the following: (a) The location and value of the maximum deflection ymax . dy(x) L Figure P8.29 A simply supported beam for problem P8-29. 2π ( L x) w(x) = w0 sin 8-29. A simply supported beam is subjected to an applied moment Mo at its center x = L2 as shown in Fig. P8.29. The deflection of the beam is given by ( ) Mo 3 L x2 2 3 L x− −x , y(x) = 6 EI L 4 L 0 ≤ x ≤ 2 where EI is the flexural rigidity of the beam. Find the following: dy (a) The equation for the slope 𝜃 = dx . L/2 273 (b) The value of the slope 𝜃 = dx at the ends x = 0 and x = L. (c) Use your results in (a) and (b) to sketch the deflection y(x). 274 Chapter 8 Derivatives in Engineering y(x) q(x) = w0 x L location and value of the maximum deflection. 8-33. Consider the buckling of a pinnedpinned column under a compressive axial load P as shown in Fig. P8.33. x y(x) L P x Figure P8.31 A beam under linear distributed load L for problem P8-31. Figure P8.33 A pinned-pinned column under a 8-32. A fixed-fixed beam is subjected to a sinusoidal distributed load, as shown in Fig. P8.32. y w(x) = w0 cos ( 2Lπ x) x L Figure P8.32 A fixed-fixed beam subjected to a sinusoidal distributed load. The deflection y(x) is given by )] [ ( w0 L4 2𝜋 y(x) = − x . 1 − cos L 16 𝜋 4 EI (a) Determine the equation for the dy slope 𝜃 = dx . (b) Evaluate both the deflection and the slope at the points x = 0 and x = L. (c) Determine both the location and value of the maximum deflection. (d) Use your results of parts (b) and (c) to sketch the deflection y(x), and clearly indicate both the compressive axial load. The deflection y(x) of the second buckling mode is given by ) ( 2 𝜋x , y(x) = −A sin L where A is an undetermined constant. (a) Determine the equation for the dy . slope 𝜃 = dx (b) Determine the value of the slope at the locations where the deflection is zero. (c) Determine the value of the deflection at the locations where the slope is zero. (d) Use your results of parts (b) and (c) to sketch the buckled deflection y(x). 8-34. Consider the buckling of a pinned-fixed column under a compressive load P as shown in Fig. P8.34. The deflection y(x) of the buckled configuration is given by ) ] [ ( 0.97616 4.4934 x + x y(x) = −A sin L L where A is an undetermined constant. (a) Determine the equation for the dy . slope 𝜃 = dx Problems P 275 (a) The location and value of the maximum deflection ymax . dy(x) (b) The value of the slope 𝜃 = at dx the ends x = 0 and x = L. (c) Use your results in (a) and (b) to sketch the deflection y(x). y(t) L 8-36. A cantilever beam is pinned at the end x = L, and subjected to an applied moment Mo as shown in Fig. P8.36. The deflection y(x) of the beam is given by x Figure P8.34 A pinned-fixed column under a y(x) = − compressive load. (b) Evaluate both the deflection and the slope at the points x = 0 and x = L. (c) Determine the location and magnitude of the maximum deflection. (d) Use your results of parts (b) and (c) to sketch the buckled deflection y(x). 8-35. Consider a beam under a uniform distributed load and supported as shown in Fig. P8.35. The deflection of the beam is given by −w (2 x4 − 5 x3 L + 3 x2 L2 ) y(x) = 48 EI where L is the length and EI is the flexural rigidity of the beam. Find the following: Mo (x3 − L x2 ) 4 EI L where L is the length and EI is the flexural rigidity of the beam. Find the following: (a) The equation of the slope 𝜃 = dy(x) . dx (b) The location and magnitude of the maximum deflection ymax . (c) Evaluate both the deflection and slope at the points A and B (x = 0 and x = L). (d) Use the results in parts (b) and (c) to sketch the deflection y(x) and clearly indicate the location of the maximum deflection on the sketch. y y(x) Mo B A w x L x L Figure P8.35 A beam under uniform distributed load for problem P8-35. Figure P8.36 A cantilever beam for problem P8-36. 8-37. Consider a shaft subjected to an applied torque T, as shown in Fig. P8.37. The internal normal and shear stresses at the surface vary with the angle relative 276 Chapter 8 Derivatives in Engineering to the axis and are given by equations (8.71) and (8.72), respectively. 𝜎𝜃 = 32 T sin 𝜃 cos 𝜃 𝜋d3 (8.71) 𝜏𝜃= 16 T (cos2 𝜃 − sin2 𝜃) 𝜋d3 (8.72) Find (a) the angle 𝜃 where 𝜎𝜃 is maximum (b) the angle 𝜃 where 𝜏𝜃 is maximum T θ d τθ y = 13,000 ft y = 0 ft v(t), ft/s 0 6 0 −18 24 27 162 180 t, s (180, 0) (162, −18) (27, −18) σθ −193.2 (6, −193.2) (24, −193.2) T Figure P8.38 Velocity of a skydiver jumping from a height of 13,000 ft. Figure P8.37 Applied torque and internal forces in a shaft. highway by a parabolic arc y(x) = a x2 + b x, 8-38. The velocity of a skydiver jumping from a height of 13,000 ft is shown in Fig. P8.38. (a) Find the equation of the velocity v(t) for the five time intervals shown in Fig. P8.38. dv , sketch the (b) Knowing that a(t) = dt acceleration a(t) of the skydiver for 0 ≤ t ≤ 180 s. (c) Use the results of part (b) to sketch the height y(t) for 0 ≤ t ≤ 180 s. 8-39. A proposed highway traverses a hilltop bounded by uphill and downhill grades of 15% and −10%, respectively. These grades pass through benchmarks A and B located as shown in Fig. P8.39. With the origin of the coordinate axes (x, y) set at benchmark A, the engineer has defined the hilltop segment of the which is tangent to the uphill grade at the origin. (a) Find the slope of the line for the uphill grade and the value of b for the parabolic arc. (b) Find the equation of the line ŷ = c x + d for the downhill grade. (c) Given that at the downhill point of tangency (̄x), both the elevation and the slope of the parabolic arc are equal to their respective values of the downhill line, for example, y(̄x) = ŷ (̄x) d̂y dy |x=̄x = dx dx x=̄x Problems 277 y, m Downhill point of tangency (x¯, y¯) Uphill grade 15% 0 A 0 x, m 8 300 B Downhill grade −10% Figure P8.39 Parabolic arc traversing highway hilltop. determine the point of tangency (̄x, ȳ ) of the parabolic arc with the downhill grade. Also, compute its elevation. (d) Find the equation of the parabolic arc. CHAPTER 9 Integrals in Engineering This chapter will discuss what integration is and why engineers need to know it. It is important to point out that the objective of this chapter is not to teach techniques of integration, as discussed in a typical calculus course. Instead, the objective of this chapter is to expose students to the importance of integration in engineering and to illustrate its application to the problems covered in core engineering courses such as physics, statics, dynamics, and electric circuits. 9.1 INTRODUCTION: THE ASPHALT PROBLEM An engineering co-op had to hire an asphalt contractor to widen the truck entrance to the corporate headquarters, as shown in Fig. 9.1. The asphalt extends 50 ft in the x- and y-directions and has a radius of 50 ft. Thus, the required asphalt is the area under the circular curve is given by (x − 50)2 + (y − 50)2 = 2500. (9.1) r = 50 ft 50 ft New asphalt 50 ft Figure 9.1 Driveway of corporate headquarters. The asphalt company charges by the square foot and provides an estimate based on “eyeballing” the required area for new asphalt. The co-op asks a young engineer to estimate the area to make sure that the quote is fair. The young engineer proposes to estimate the area as a series of n inscribed rectangles as shown in Fig. 9.2. The area 278 9.1 Introduction: The Asphalt Problem 279 A is given by A≈ n ∑ f (xi )Δ x i=1 50 is the width of each rectangle and f (xi ) is the height. The equation n of the function f (x) is obtained by solving equation (9.1) for y, which gives √ y = f (x) = 50 − 2500 − (x − 50)2 . (9.2) where Δ x = y (x − 50)2 + (y − 50)2 = 2500 (xi, f(xi)) x xi . . . xn x1 50 Δx = n Figure 9.2 Division of asphalt area into n inscribed rectangles. Suppose, for example, that n = 4, as shown in Fig. 9.3. Here Δ x = the area can be estimated as A≈ 4 ∑ 50 = 12.5 ft and 4 f (xi )Δ x i=1 = 12.5 ∗ [f (x1 ) + f (x2 ) + f (x3 ) + f (x4 )]. y f(x) = 50 −√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 2500 − (x − 50)2 f(x1) 12.5 x1 x2 x3 Figure 9.3 Calculation of area using four rectangles. x4 x (9.3) . b while ∫a f (x) dx is the definite integral of f (x) between x = a and x = b.. This result clearly underestimates the actual value of the area.e.4).e.2) at the corresponding values of x: f (x1 ) = f (12. then f (x) is the derivative of F(x).59 + 0) = 315.5) where F(x) is an antiderivative of f (x). lim n→∞ ∑n n ∑ b f (xi ) Δ x = i=1 i=1 f (xi ) ∫a f (x) dx.59 f (x4 ) = f (50. its derivative backward).e. In the case of the asphalt problem.70 + 1.93 f (x2 ) = f (25.5) = 1.5 × (16. The value of the integral is obtained from the fundamental theorem of calculus. in comes the old engineer.4) Δ x is the area under f (x) between x = a and x = b. A = lim n→∞ n ∑ f (xi ) Δ x. (9. Hence. Based on the knowledge of derivatives. i. who recognizes this as the definition of the definite integral.4 ft2 .0) = 6. lim n→∞ In equation (9.70 f (x3 ) = f (37.. f (x) = d F(x). dx (9.0) = 0. i.280 Chapter 9 Integrals in Engineering The values of f (x1 ) … f (x4 ) are obtained by evaluating equation (9.93 + 6. i. a = 0 and b = 50 since these are the limits of the asphalt in the x-direction. i=1 However.3) gives A ≈ 12.6) Hence.0.5) = 16. Substituting these values in equation (9. the definite integral of a function over an interval is the area under the function over that same interval. If F(x) is the antiderivative of f (x). evaluating the integral of a function amounts to finding its antiderivative (i. the antiderivatives . b ∫a f (x) dx = [F(x)]ba = F(b) − F(a) (9..e. The young engineer claims he would need an ∞ number of rectangles to get it right. F(x) = ∫ f (x) dx. F(x) = ∫ f (x) dx c f (x) 1 cos(𝜔 x) + C 𝜔 1 sin(𝜔 x) + C 𝜔 1 ax e +C a 1 xn+1 + C n+1 c ∫ f (x) dx f1 (x) + f2 (x) ∫ f1 (x) dx + ∫ f2 (x) dx sin(𝜔 x) cos(𝜔 x) ea x xn − Note that in Table 9. d d F(x) = f (x) dx = f (x) dx dx ∫ d F(x) = f (x) dx = F(x) dx = F(x). With this background.1 Introduction: The Asphalt Problem 281 (integrals) of sin (x) and xn can. Since F(x) is the antiderivative of f (x). . for example.7) show that differentiation and integration are inverse operations. (9.6) and (9.9. i. as shown in Fig. and 𝜔 are constants as they do not depend on x.1 The antiderivatives of some common functions in engineering. Function. 9. the young engineer finds the total area as the sum of all elemental areas dA. xn dx = ∫ n+1 x2 dx = The previous integrals are called indefinite integrals since there are no limits a and b. f (x) Antiderivative. a.4.. n. c.e.7) Equations (9. n+1 F(x) = sin(x) dx = −cos x + C x3 +C ∫ 3 xn+1 + C.1. be written as ⇒ f (x) = sin (x) f (x) = x2 ⇒ f (x) = xn ⇒ ∫ F(x) = −cos (x) + C x3 +C 3 xn+1 F(x) = + C. ∫ ∫ dx f (x) = TABLE 9. (9.2) gives √ 50 A= 2500 − (x − 50)2 ) dx (50 − ∫0 50 = 50 dx − ∫0 ∫0 √ 2500 − (x − 50)2 dx 50 √ 2500 − (x − 50)2 dx ∫0 50 = 50 50 dx − − = 50 [x]50 0 ∫0 ∫0 50 √ 2500 − (x − 50)2 dx 50 = 50 (50 − 0) − ∫0 √ 2500 − (x − 50)2 dx or A = 2500 − I. which gives I = 625 𝜋. who notes that the result can be calculated without calculus! The total area is simply the area of a square (of dimension 50 × 50) minus . In comes the oldest engineer.8) √ 50 The integral I = ∫0 2500 − (x − 50)2 dx is not easy to evaluate by hand.8) gives A = 2500 − 625 𝜋 A = 536. The total area is thus 50 A= 50 dA = ∫0 ∫0 y(x) dx.5 ft2 .282 Chapter 9 Integrals in Engineering y dA = y(x) dx y(x) = f(x) 0 x dx x = 50 x Figure 9. However.4 Asphalt area with elemental area dA. Substituting I into equation (9. Substituting the value of y(x) from equation (9. this integral can be evaluated using MATLAB (or other engineering programs). A = (50 × 50) − ] 1 [ 𝜋 (50)2 4 1 (2500 𝜋) 4 = 2500 − 625 𝜋 A = 536. If the object is moved by a constant force as shown in Fig.7. as shown in Fig. 9. as shown in Fig. Indeed. Therefore.2 CONCEPT OF WORK Work is done when a force is applied to an object to move it a certain distance.2 Concept of Work 283 the area of a quarter-circle of radius r = 50 ft. If the force F is constant.6. 9. 9.6 Force F moving an object a distance d. the work done is just the force times the distance.5.5 ft2 . .9. W =F×d Distance d Force F W=F×d Figure 9. the work done is the area under the force-displacement curve W = F ×d = F × (x2 − x1 ) where d = x2 − x1 is the distance moved.5 Calculation of area without calculus. one of the most important things about calculus in engineering is understanding when you really need to use it! 9. = 2500 − y Quarter−circle r = 50 ft 50 ft x 50 ft Figure 9. the area under the curve (i.9).8.7 Work as area under a constant force curve. the work) must be determined by integration: x2 W= ∫x1 F(x) dx d = ∫0 F(x) dx. as shown in Fig. If d = 1.9)) are demonstrated in the following examples.284 Chapter 9 Integrals in Engineering F Work x = x1 x = x2 Distance d = x2 − x1 x Figure 9.8 Work as area under a variable force curve. Calculations used to find the work done by a variable force (equation (9. find the work done for the following forces: (a) f (x) = 2 x2 + 3 x + 4 N ( ) ( ) 𝜋 𝜋 (b) f (x) = 2 sin x + 3 cos x N 2 2 (c) f (x) = 4 e𝜋x N . 9.e. 9. N Work = area under the curve 0 d x. If the force is not constant but is a function of x.0 m. m Figure 9. (9..9) F. Example 9-1 The work done on the block shown in Fig.7 is defined by equation (9. 17 N-m.9.2 Concept of Work Solution d (a) W= f (x) dx ∫0 1 = ∫0 (2 x2 + 3 x + 4) dx 1 =2 x2 dx + 3 1 1 x dx + 4 ∫0 ∫0 ∫0 [ 2 ]1 [ 3 ]1 x x +3 + 4 (x)10 =2 3 0 2 0 1 dx 2 3 (1 − 0) + (1 − 0) + 4 (1 − 0) 3 2 2 3 = + +4 3 2 = = 37 6 or W = 6. (b) d W= f (x) dx ∫0 1 = ∫0 [ ( ) ( )] 𝜋 𝜋 2 sin x + 3 cos x dx 2 2 ( 1 =2 ∫0 sin ( ) ) 1 𝜋 𝜋 cos x dx + 3 x dx ∫0 2 2 ( ) 1 ) 1 ( ⎡ cos 𝜋 x ⎤ ⎡ sin 𝜋 x ⎤ ⎢ ⎥ ⎥ ⎢ 2 2 = 2 ⎢− ⎥ +3 ⎢ ⎥ 𝜋 𝜋 ⎢ ⎥ ⎥ ⎢ 2 2 ⎣ ⎦0 ⎦0 ⎣ [ [ ( )]1 ( )]1 2 2 𝜋 𝜋 = 2 − cos +3 x sin x 𝜋 2 𝜋 2 0 0 [ ( [ ( )]1 )]1 𝜋 𝜋 4 6 cos sin =− + x x 𝜋 2 𝜋 2 0 0 ( ) ( ) 6 4 𝜋 𝜋 − cos(0)] + [sin − sin(0)] = − [cos 𝜋 2 𝜋 2 4 6 = − (0 − 1) + (1 − 0) 𝜋 𝜋 4 6 10 = + = 𝜋 𝜋 𝜋 285 . A x .0 m.10) Ai y y = y(x) = f(x) x¯ G y¯ Figure 9.1 Center of Gravity (Centroid) The centroid.9. Note: In all three cases. the centroid of a two-dimensional object bounded by a function y = f (x) is given by a point G = (̄x. The average x-location x̄ of the material is given by ∑ x̄ i Ai x̄ = ∑ (9.9 Centroid of a two-dimensional object. 9. ȳ ) as shown in Fig. For example.18 N-m.3 APPLICATION OF INTEGRALS IN STATICS 9. the distance moved by the object is 1. of an object is a point within the object that represents the average location of its mass.286 Chapter 9 Integrals in Engineering or W = 3.3. or center of gravity.2 N-m. but the work (energy) expended by the force is completely different! 9. (c) d W= f (x) dx ∫0 1 = ∫0 (4 e 𝜋x ) dx 1 e 𝜋x dx ∫0 [ ]1 1 𝜋x =4 e 𝜋 0 =4 ] 4 [ 𝜋 e − e0 𝜋 4 = [e 𝜋 − 1] 𝜋 = or W = 28. if x̄ i = x.10) and (9.11. Find the location of the centroid. ȳ i = 2 .12) Ai ∫ dA ∫ y(x) dx and y(x) 1 ∑ dA (y(x))2 dx ∫ ȳ i Ai 2 2 ∫ = ȳ = ∑ = .11). as shown in Fig. ȳ i ) = x. Ai ∫ dA ∫ y(x) dx (9. .13) y(x) y ( (x¯i . y(x) Now. y¯i) = x.11) can be written as ∑ ∫ x dA ∫ x y(x) dx x̄ A x̄ = ∑ i i = = (9. y(x) 2 ) A x Ai = dA = y(x) dx x y(x) 2 dx Figure 9.11) To evaluate the summation in equations (9. y y=− h x¯ G h x+h b 1 A = hb 2 y¯ x b Figure 9.10) and (9. 9.10 Two-dimensional object with elemental area dA.11 Centroid of triangular section. and Ai = dA = y(x) dx is the elemental area.10. )consider a rectangular ( y element of the area of width dx and centroid (̄xi .9.3 Application of Integrals in Statics while the average y-location ȳ of the area is given by ∑ ȳ i Ai ȳ = ∑ . equations (9. Ai 287 (9. 2 . 9. Example 9-2 Centroid of Triangular Section: Consider a triangular section of width b and height h. as shown in Fig. 12) as x̄ = ∫ x y(x) dx ∫ y(x) dx ( ) b h x − x + h dx ∫0 b = 1 bh 2 ) b( h − x2 + hx dx ∫0 b =2 bh ( )( [ 2 ]b ) [ ]b 2 x h x3 = +h − bh b 3 0 2 0 ( )( h h = − (b3 − 0) + (b2 − 0) 3b 2 ( )( ) 2 hb2 hb2 = − + bh 3 2 ( ) ( 2) 2 hb = bh 6 2 bh or x̄ = b . the x-location of the centroid is calculated from equation (9. 9.14) and (9. The above result can also be obtained by h integrating y(x) = − x + h with respect to x from 0 to b. 2 (9.11 is the area bounded by the line h y(x) = − x + h. b (9.15).15) which is simply the area of the triangle.288 Chapter 9 Integrals in Engineering Solution The two-dimensional triangular section shown in Fig. Using the information b in equations (9. 3 ) .14) The area of the section is given by b A= ∫0 y(x) dx or A= 1 bh. 9. These coordinates can also be calculated using the horizontal rectangles.13) as 1 y2 (x) dx 2 ∫ ȳ = ∫ y(x) dx ) )2 ( )( b ( 2 h 1 dx − x+h = 2 b h ∫0 b ( ) ) b [( 2 ) ] ( 2h h 1 x h + h2 dx x2 − = h b ∫0 b b2 ) ( ( ) [ ]b ( 2 ) [ 2 ]b 2h 1 h2 x3 x b 2 − + h (x)0 = hb b 2 0 b2 3 0 ( ) )( 2 1 h2 2 h 3 2 = (b − 0) − (b − 0) + h (b − 0) hb b 3 b2 ( ) )( 2 1 h b 2 2 = −h b+h b hb 3 ( )( 2 ) 1 h b = bh 3 or ȳ = h . 3 Thus. the y-location of the centroid is calculated from equation (9. In the above example. 9. the coordinates of the centroid were found by using .12. 3 3 vertical rectangles. as shown in Fig. ȳ ) = ( b h . the ) centroid of a triangular section of width b and height h is given by (̄x. y) Ai = dA = x dy y x b Figure 9.12 Evaluation of centroid using horizontal elemental areas. y¯i) = h dy x x ( 2 . The area of the horizontal element is given by Ai = dA = x dy = g(y)dy . y h y=− x+h b (x¯i.3 Application of Integrals in Statics 289 Similarly. (c) Determine the x-coordinate of the centroid by integration with respect to x.290 Chapter 9 Integrals in Engineering b where x = g(y) = − y + b is obtained by solving the equation of the line for x. (b) Determine the area of the cooling fin by integration with respect to x. the elemental area of the horizontal rectangle is given by ( ) b Ai = dA = − y + b dy.16) ( ) ] b [y2 ]h0 2 ( ) ] [( ) b b 3 2 − (h − 0) + (h − 0) 3h 2 [ ] b h2 b h2 − + 3 2 [( − b 3h ) [y3 ]h0 + which gives h . 3 This is the same result previously found using the vertical rectangles! ȳ = Example 9-3 The geometry of a cooling fin is described by the shaded area bounded by the parabola shown in Fig. (d) Determine the y-coordinate of the centroid by integration with respect to x.13. h Therefore. (a) If the equation of the parabola is y(x) = −x2 + 4. determine the height h and the width b of the fin. . 9. h The y-coordinate of the triangular section can be calculated as ∑ ∫ y dA ȳ A ȳ = ∑ i i = A Ai or ( ) b h ∫0 y − y + b dy h ȳ = 1 bh 2 ) h ( 2 b 2 = − y + b y dy b h ∫0 h ] [( ) h h 2 b = y2 dy + [b] y dy − ∫0 bh h ∫0 = 2 bh = 2 bh = 2 bh (9. (a) The equation of the parabola describing the cooling fin is given by y(x) = −x2 + 4.3 Application of Integrals in Statics 291 y.17) The height h of the cooling fin can be found by substituting x = 0 in equation (9. . which gives y(x) = −x2 + 4 = 0 ⇒ x2 = 4 ⇒ x = ± 2. it follows that b = 2′′ . b Figure 9.9. in. y(x) = −x2 + 4 h x. (9. (b) The area A of the fin is calculated by integrating equation (9.17) as h = y(0) = −02 + 4 = 4′′ .17) from 0 to b as b A= y(x) dx ∫0 2 (−x2 + 4) dx ∫0 [ 3 ]2 x = − + 4x 3 0 [( 3 ) ] 2 = − + 4 (2) − (0 + 0) 3 = or A= 16 2 in.13 Solution Geometry of a cooling fin. 3 . The width b of the fin can be obtained by setting y(x) = 0. in. Since the width of the fin must be positive. 14. where x̄ i = x and Ai = dA = y(x) dx. in. the y-coordinate of the centroid can be determined by integration with respect to x as ∑ ȳ A ȳ = ∑ i i Ai Therefore. y¯i) = x. Thus. x̄ = . dx y(x) = −x2 + 4 h ( (x¯i. ∑ x̄ A x̄ = ∑ i i Ai y.292 Chapter 9 Integrals in Engineering (c) The x-coordinate of the centroid can be found using the vertical rectangles as illustrated in Fig. 4 (d) Similarly. b Figure 9. 9. in. By definition. x ) dA = y(x) dx y(x) 2 x. 16 3 in. x̄ = = ∫ xy(x) dx A 2 ∫0 x (−x2 + 4) dx 16 3 2 3 (−x3 + 4 x) dx 16 ∫0 [ 4 ]2 3 x x2 = − +4 16 4 2 0 ) ] [( 4 3 2 2 = − + 2 (2 ) − (0 + 0) 16 4 = or x̄ = 12 in.14 y(x) 2 Determination of centroid using vertical rectangles. (9. 5 9.15.3. then the x.18) = 0.3 Application of Integrals in Statics where ȳ i = 293 y(x) and Ai = y(x) dx. 2 1 y2 (x) dx 2 ∫ ȳ = A 2 1 (−x2 + 4)2 dx 2 ∫0 = 16 3 2 3 (x4 − 8 x2 + 16) dx 32 ∫0 [ 5 ]2 3 x x3 = −8 + 16 x 32 5 3 0 [( 5 ) ] 3 3 2 2 = −8 + 16 (2) − (0 + 0 + 0) 32 5 3 [ ] 3 32 64 = − + 32 32 5 3 [ ] 1 2 =3 − +1 5 3 [ ] 3 10 15 =3 − + 15 15 15 = = 24 15 ȳ = 8 .9. ȳ = 8 in. 9.2 Alternate Definition of the Centroid If the origin of the x-y coordinate system is located at the centroid as shown in Fig.19) .and y-coordinates of the centroid are given by x̄ = ȳ = ∫ x dA A ∫ y dA A =0 (9. Thus. 5 or Therefore. 9. there is no first moment about the origin).17 Rectangular section.294 Chapter 9 Integrals in Engineering y x G Figure 9.and y-axes is zero: First moment of area about x-axis = Mx = ∫ y dA = 0 First moment of area about y-axis = My = ∫ x dA = 0. this means that the first moment of the area about both the x. y y dA dA x y x G G x Figure 9. Hence. As shown in Fig. 9.e.17 are x̄ = ȳ = 0. 9-4 y b/2 h x G h/2 b Figure 9. . an alternative definition of the centroid is the location of the origin such that ∫ x dA = ∫ y dA = 0 (i.16. Example Show that the coordinates of the centroid of the rectangle in Fig.15 Triangular section with origin at centroid. The above definition of the centroid is illustrated for a rectangular section in the following example..16 First moment of area. Hence. y b dy ∫− h y2 2 ( ] h2 − h2 h2 h2 − 4 4 ) . which gives ∫ y dA = h 2 2 [ =b = b 2 = 0.18.19.18 x-coordinate of the centroid using vertical rectangles. x̄ = ∫ xdA A = 0. Hence. the first moment of area about the x-axis can be calculated using horizontal rectangles. which gives ∫ x dA = b 2 x hdx ∫− b 2 [ =h h = 2 x2 2 ( ]b 2 − b2 b2 b2 − 4 4 ) = 0. ȳ = ∫ y dA A = 0. as shown in Fig.3 Application of Integrals in Statics Solution 295 The first moment of area about the y-axis can be calculated using vertical rectangles. y dx dA = h dx h x G b Figure 9. Similarly. 9. 9. as shown in Fig.9. as well as the location of that force required for statically equivalent loading. Since pressure is force/unit area. 9..20). 9. These are among the primary applications of integrals in statics. The resultant force acting on the wall is calculated by adding up (i.22. integrating) all the differential forces dF shown in Fig.1 Hydrostatic Pressure on a Retaining Wall Consider a retaining wall of height h and width b that is subjected to a hydrostatic pressure from fluid of density 𝜌 (Fig.4.4 DISTRIBUTED LOADS In this section.20 Hydrostatic force acting on the rectangular retaining wall.e. 9. The pressure acting on the wall satisfies the linear equation p(y) = 𝜌 g y where g is the acceleration due to gravity.19 y-coordinate of the centroid using horizontal rectangles.296 Chapter 9 Integrals in Engineering y dA = b dy dy h x G b Figure 9. the differential force is found by multiplying the value of the pressure at any depth y by an elemental area of the wall as dF = p(y) dA where dA = b dy. integrals are used to find the resultant force due to a distributed load. Retaining wall p(y) = ρ g y h g b y Figure 9. 9. . Since this is a triangular load.4 Distributed Loads 297 dy p(y) dA dF b y Figure 9.21 Forces acting on the retaining wall.23. The resultant force acting on the wall is obtained by integration as F= dF ∫ h = p(y) b dy ∫0 h = ∫0 𝜌 g y b dy h = 𝜌gb y dy ∫0 [ 2 ]h y = 𝜌gb 2 0 or F= 𝜌 g b h2 .9. h h Note that ∫0 p(y) b dy = b ∫0 p(y) dy is simply the width b times the area of the triangle shown in Fig. Therefore. the area can be calculated .22 Forces acting on the retaining wall. 9. the resultant force can be obtained from the area under the distributed load. 2 dy p(y) dA dF b y Figure 9. the equivalent load R is the area under the distributed load. (9.23 Area under hydrostatic pressure. The distributed load w(x) varies in intensity with position x and has units of force per unit length (lb/ft or N/m). 2 2 9. 9.2 Distributed Loads on Beams: Statically Equivalent Loading Fig.4. To find the location of the statically equivalent force.25. The equivalent load R and location l are shown in Fig. and is given by A= 𝜌 g h2 1 (𝜌 g h)(h) = .24 shows a simply supported beam with a distributed load applied over the entire length L. the resultant load shown in Fig.24 Distributed load on a simply supported beam. The resultant force is obtained by multiplying the area with the width b. 2 2 p(y) h ρgh y Figure 9.25 must have the same moment about every point as the distributed load . 9. The goal is to replace the distributed load with a statically equivalent point load. and is given by L R= ∫0 w(x) dx. which gives F=b 𝜌 g b h2 𝜌 g h2 = .20) y w(x) x L Figure 9.298 Chapter 9 Integrals in Engineering without using integration. 9. As concluded in the previous section. 9. The moment M0 for the distributed load can be calculated by summing moments due to elemental loads dw. 9. Hence. For example.21) and (9. the moment (force times distance) about the point x = 0 must be the same for both the distributed and equivalent loads. as shown in Fig. The moment M0 about the x = 0 point of the equivalent load R is given by M0 = R l. (9.4 Distributed Loads y 299 R l x L Figure 9.25 Beam with an equivalent point load. ∫0 (9.9. M0 = xdw ∫ L = xw(x) dx.22) Equating the two moments in equations (9.24. or L M0 = l ∫0 w(x) dx.26 Beam with a small elemental load.22) gives L l ∫0 L w(x) dx = ∫0 xw(x) dx or L l= ∫0 xw(x) dx L ∫0 w(x) dx . shown in Fig.26.21) dw = w(x) dx y x x L Figure 9. (9.23) . 9.23) is identical to equation (9. Example 9-5 Find the magnitude and location of the statically equivalent point load for the beam of Fig.28).28 Solution R2 Reaction forces acting at the supports. which is the x-coordinate of the centroid of the area under the load! Thus. y w(x) = w0 x L w0 x L Figure 9. a distributed load can always be replaced by its resultant force acting at its centroid. which is given by L R= w(x) dx ∫0 L (w ) 0 xdx L ( w ) ( 2 )L x 0 = L 2 0 ( ) (w ) L2 0 = −0 L 2 = ∫0 or R= 1 w L. Use your results to find the reactions at the support (Fig.27 Simply supported beam with linear distributed load.300 Chapter 9 Integrals in Engineering Equation (9.27.12) but with w(x) instead of y(x). y w0 x L R1 Figure 9. Hence. 9. it can be concluded that l = x̄ . The resultant R is the area under the triangular load w(x). 2 0 . for the purpose of statics. 9.29.4 Distributed Loads 301 Note that the above result is just the area under the triangle defined by w(x). the location l could have been determined without any further calculus. 2 0 (9. The location l of the statically equivalent load is the x-coordinate of the centroid of the area under w(x) and can be calculated using equation (9.24) .e. Hence.. i. 3 Note that this is two-thirds of the way from the base of the triangular load (i. R1 + R2 = R or R1 + R2 = 1 w L.29 Statically equivalent loading of the beam.9. R l R1 x 2L 3 L 3 R2 Figure 9. For equilibrium.23) as L l= ∫0 x w(x) dx L ∫0 w(x) dx (w ) L 0 ∫0 x x dx L = 1 w L 2 0 L 2 x2 dx L2 ∫0 ( 3 )L x 2 = 3 0 L2 = = 2 (L3 − 0) 3 L2 or 2L ..e. l= The statically equivalent loading is shown in Fig. the sum of the forces in the y-direction must be zero. the centroid of the triangle). 26) APPLICATIONS OF INTEGRALS IN DYNAMICS It was discussed in Chapter 8 that if the position of a particle moving in the x-direction as shown in Fig.302 Chapter 9 Integrals in Engineering Also. 3 2 R1 = w0 L w0 L − 2 3 Solving for R1 gives or R1 = 9.25) in equation (9.25) Substituting equation (9.30 is given by x(t). the velocity v(t) is the first derivative of the position. 3 (9. = dt dt2 v(t). 6 (9.5 w0 L . a(t) x(t) Figure 9. Taking the moments about x = 0 gives ( ) 1 2L R2 L − w L =0 2 0 3 which gives R2 L = w0 L2 3 or R2 = w0 L . 9. .24) gives R1 + w0 L w0 L = . the sum of moments about any point on the beam must be zero. and the acceleration is the first derivative of the velocity (second derivative of the position): v(t) = dx(t) dt a(t) = dv(t) d2 x(t) .30 A particle moving in the horizontal direction. Beginning with dv(t) = a(t) dt integrating both sides between t = t0 and any time t gives t ∫t0 t By definition. so that equation (9.31) can be written as 0 dt [x(t)]tt = 0 t ∫t0 v(t) dt which gives t x(t) − x(t0 ) = v(t) dt ∫t0 or x(t) = x(t0 ) + t ∫t0 v(t) dt.5 Applications of Integrals in Dynamics 303 Now. (9.31) dx(t) dt = [x(t)]tt . ∫t0 t dx(t) v(t) dt.32) .28) can be written as 0 dt [v(t)]tt = 0 t a(t) dt ∫t0 which gives t v(t) − v(t0 ) = a(t) dt ∫t0 or v(t) = v(t0 ) + t a(t) dt. so that equation (9. if the acceleration a(t) of the particle is given.29) Thus.9. Beginning with dx(t) = v(t) dt integrating both sides between t = t0 and any time t gives t ∫t0 t By definition.30) (9. dt = ∫t0 dt (9. dt = ∫t0 dt (9. ∫t0 t dv(t) a(t) dt. both the velocity v(t) and the position x(t) can be determined by integrating with respect to t.27) (9. given the velocity v(t).28) dv(t) dt = [v(t)]tt . the velocity of the particle at any time t is equal to the velocity at t = t0 (initial velocity) plus the integral of the acceleration from t = t0 to the time t. Now. ∫0 (9. the position x(t) can be determined by integrating with respect to t. and v(0) = 0 in equation (9. the position of the ball at any time t is given by y(t) = 1. a(t) = −9. Now.81 t m/s.29). Find v(t). t=0 1. and the time it takes for the ball to hit the ground.81 2 0 = y(0) − ) 9. y(t).32).0 m y(t) Figure 9.905 t2 .31 Solution a = −9.304 Chapter 9 Integrals in Engineering Thus. substituting v(t) into equation (9.0 − 4. the position y(t) of the ball at any time t can be obtained as t y(t) = y(0) + ∫0 −9. v(0) = 0 m/sec. Example 9-6 A ball is dropped from a height of 1. Since the initial height is y(0) = 1 m. the velocity at any time t can be obtained as t v(t) = 0 + −9.81 m/s2 A ball dropped from a height of 1 m.81 dt ∫0 = −9. 9.81 [t]t0 = −9. Substituting t0 = 0. Since the ball is dropped from rest at time t = 0 sec. the position of the particle at any time t is equal to the position at t = t0 (initial position) plus the integral of the velocity from t = t0 to the time t.81 t dt [ 2 ]t t = y(0) − 9.81 ( 2 t −0 2 y(t) = y(0) − 4. . as shown in Fig.31.81 m/s2 .0 m at t = t0 = 0.81 (t − 0) or v(t) = −9.905 t2 m. 43 − 9.43 − 9. and a(t) = −9. Find v(t) and y(t). 9. as shown in Fig.81 m/s2 Figure 9. the position of the ball is given as t y(t) = y(0) + ∫0 (4.905 or timpact = 0.33) Now substituting the velocity v(t) into equation (9.452 s. 1.43 (t − 0) − 2 = y(0) + 4.32.81 t m/s. y(t) v0 = 4.905 timpact which gives 2 4.43 − 9.43 m/s a = −9. (9.81 m/s2 into equation (9.9.5 Applications of Integrals in Dynamics 305 The time to impact is obtained by setting y(t) = 0 as 2 = 0. the velocity of the ball at any time t is given by t v(t) = 4. v(0) = 4.43 − 9.43 m/s.0 4. Solution Substituting t0 = 0.81 (t − 0) or v(t) = 4.905 timpact = 1. Example 9-7 Suppose that a ball is thrown upward from ground level with an initial velocity v(0) = v0 = 4.81 dt = 4.29).43 [t]t0 − 9.81 [t]t0 = 4.81 ( 2 t −0 = y(0) + 4.32 A ball thrown upward with an initial velocity.81 t) dt [ 2 ]t t 2 0 ) 9.81 .32). Solving for timpact gives √ timpact = 1.43 m/s.0 − 4.43 + ∫0 −9. as shown in Fig.33.33 A stone thrown from the top of a building.35) and the position is t y(t) = y(0) + ∫0 (a) Find and plot the velocity v(t).43 t − 4.81 (t − 0) .81 m/s2 into equation (9.81 [t]t0 = 10 − 9. Example 9-8 A stone is thrown from the top of a 50 m high building with an initial velocity of 10 m/s.81 m/s2 y(t) = 0 m Figure 9. (b) Find and plot the position y(t). Since the initial position is y(0) = 0 m.306 Chapter 9 Integrals in Engineering or y(t) = y(0) + 4.905 t2 m. v(0) = 10 m/s y(0) = 50 m 50 m a = −9. the position of the ball at any time t is given by y(t) = 4.34) as t v(t) = v(0) + ∫0 a(t) dt t = 10 + ∫0 −9.43 t − 4.905 t2 . Solution (a) The velocity of the stone can be calculated by substituting v(0) = 10 m/s and a(t) = −9. 9.34) v(t) dt (9. (c) Determine both the time and the velocity when the stone hits the ground. Knowing that the velocity is t v(t) = v(0) + ∫0 a(t) dt (9.81 dt = 10 − 9. (9.371 t.37) The plot of the position is as shown in Fig. m/s 10 0 −32.0194 sec.9.097 m.0194) − 4.81 = 50 + 10 (t − 0) − or [ 2 ]t t 2 0 9.81 m/s2 as shown in Fig.0194)2 or ymax = 55. 9.34 Velocity of the stone.0194 4.36) The plot of the velocity is a straight line with y-intercept vo = 10 m/s and slope −9. s 0 Figure 9. (b) The position of the stone can be calculated by substituting y(0) = 50 m and v(t) from equation (9.88 1. Solving dt for t gives tmax = 1. (9.35) as t y(t) = y(0) + ∫0 v(t) dt t = 50 + ∫0 (10 − 9.34.81 t m/s.81 t) dt = 50 + 10 [t]t0 − 9. The maximum height can be dy determined by setting = v(t) = 0. 9. . which gives v(t) = 10 − 9.905 (1.81 2 (t − 0) 2 y(t) = 50 + 10 t − 4. v(t).36) into equation (9.5 Applications of Integrals in Dynamics 307 or v(t) = 10 − 9. The maximum height is thus ymax = 50 + 10 (1.35.905 t2 m.81 t = 0. 3516 t = 1.38) can be solved by using one of the methods described in Chapter 2.905 t2 = 0 or t2 − 2.039 2 = 10.371) = −32. we can complete the square as t2 − 2.3516 or t = 4. m ymax 50 0 0 1. The velocity when the stone hits the ground is obtained by evaluating v(t) at t = 4.332 s) in equation (9. it takes t = 4.35 Position of the stone.039 t − 10.38) The quadratic equation (9.194 + t − 2.194 = 0.81 (4.371 t.39) Since the negative time (t = −2.371.332 s.233)2 t− 2 t − 1.0194 = ± 3. (9.0194 4.194 ( ( )2 )2 2. For example.308 Chapter 9 Integrals in Engineering y(t).371) = 10 − 9.039 2.371 s for the stone to hit the ground. − 2.039 = (± 11.039 t + 2 2 )2 ( √ 2. (9. . (c) The time it takes for the stone to hit the ground can be calculated by setting the position y(t) equal to zero as y(t) = 50 + 10 t − 4. which gives v(4.88 m/s.0194 ± 3. s Figure 9.39) is not a possible solution.039 t = 10.371. 9. In other words. It follows that the change in velocity can be determined from the area under the graph of a(t). Integratdt dv(t) dt dt t = [v(t)]t2 1 or t2 ∫t1 a(t) dt = v2 − v1 . we get t2 t2 a(t) dt = ∫t1 ∫t1 dv(t) . 9. This can be shown by considering the definition of acceleration.1 Graphical Interpretation The velocity v(t) can be determined by integrating the acceleration. the area under a(t) between t1 and t2 equals the change in v(t) between t1 and t2 .5 Applications of Integrals in Dynamics 309 9.5. a(t) = ing both sides from time t1 to time t2 . as shown in Fig.37. t2 t . 9. as shown in Fig.37 Change in velocity from time t1 to t2 . v(t) v2 t v2 − v1 = ∫ t2 a(t) dt 1 v1 t1 Figure 9.36. a(t) t A = ∫ t2 a(t) dt 1 t1 t t2 Figure 9. The change in velocity v2 − v1 can be added to the initial velocity v1 at time t1 to obtain the velocity at time t2 .36 Velocity as an area under the acceleration graph. v(t) = both sides from time t1 to time t2 gives t2 t2 v(t) dt = ∫t1 ∫t1 dx(t) . In other words. 9. This can be shown by considering the definition of velocity. the area under v(t) between t1 and t2 equals the change in x(t) between t1 and t2 .38 Position as an area under the velocity graph.38. x(t) x2 t x2 − x1 = ∫ t2 v(t) dt 1 x1 t1 Figure 9. dt Integrating dx(t) dt dt t = [x(t)]t2 1 or t2 ∫t1 v(t) dt = x2 − x1 . as shown in Fig. t2 t .39.310 Chapter 9 Integrals in Engineering Similarly. v(t) t A = ∫ t2 v(t) dt 1 t1 t t2 Figure 9.39 Change in position from time t1 to t2 . The change in position x2 − x1 can be added to the initial position x1 at time t1 to obtain the position at time t2 . as shown in Fig. It follows that the change in position can be determined from the area under the graph of v(t). 9. the position x(t) can be determined by integrating velocity. 2 < t ≤ 4 : Since a(t) = 0 m/s2 . Since v0 = 0. which gives v2 = v0 + 4.40 Acceleration of a vehicle for Example 9-9. Therefore. Thus. 9. m/s2 2 0 x(t) 0 2 4 t. v2 − v0 = (2) (2) = 4 (area of a rectangle). v(t) is a straight line with a slope of 2 m/s2 .40. s 6 −2 Figure 9. v4 = v2 + 0 = 4+0 .5 Applications of Integrals in Dynamics Example 9-9 311 The acceleration of a vehicle is measured as shown in Fig. Thus.9. v2 = 0 + 4 = 4 m/s. sketch the velocity v(t) and position x(t) using integrals. the change in velocity is 2 v2 − v0 = ∫0 a(t) dt or v2 − v0 = area under a(t) between 0 and 2 s. Solution (a) Velocity: t Knowing v(0) = 0 and v(t) − v(t0 ) = ∫t a(t) dt. v(t) is constant. Also. a(t). 4 v4 − v2 = ∫2 a(t) dt = area under a(t) between 2 and 4 s = 0. Also. 0 ≤ t ≤ 2 : 0 a(t) = 2 m/s2 = constant. Knowing that the vehicle starts from rest at position x = 0. . 0 0 ≤ t ≤ 2 : v(t) is a linear function (straight line) with a slope of 2 m/s. Also.41 Velocity of the vehicle for Example 9-9. 4 < t ≤ 6 : a(t) = −2 m/s2 = constant. (b) Position: Now use v(t) to sketch x(t) knowing that x(0) = 0 and x(t) − x(t0 ) = t ∫t v(t) dt. v6 = v4 − 4 = 4−4 or v6 = 0 m/s. s Figure 9. Thus.312 Chapter 9 Integrals in Engineering or v4 = 4 m/s. x(t) is a quadratic function with increasing slope (concave up). m/s 4 0 0 2 4 6 t. The graph of the velocity obtained above is as shown in Fig. Therefore. Also.41. v(t). Therefore. the change in x(t) is 2 x2 − x0 = ∫0 v(t) dt = area under v(t) between 0 and 2 = or 1 (2) (4) (area of a triangle) 2 x2 − x0 = 4. v(t) is a straight line with a slope of −2 m/s2 . 9. the change in v(t) is 6 v6 − v4 = a(t) dt ∫4 = area under a(t) between 4 and 6 = (−2) (2) (area of a rectangle) or v6 − v4 = −4. x6 = x4 + 4 = 12 + 4 or x6 = 16 m. x2 = x0 + 4. x2 = 0 + 4 or x2 = 4 m. Therefore. Therefore. Also. Since x0 = 0. Thus. . x4 = x2 + 8 = 4+8 or x4 = 12 m.5 Applications of Integrals in Dynamics 313 Thus. x(t) is a straight line with a slope of 4 m/s. Also. 2 < t ≤ 4 : v(t) has a constant value of 4 m/s. 4 < t ≤ 6 : v(t) is a linear function (straight line) with a slope of −2 m/s.9. the change in x(t) is 4 x4 − x2 = v(t) dt ∫2 = area under v(t) between 2 and 4 = (2) (4) (area of a rectangle) or x4 − x2 = 8. the change in position is 6 x6 − x4 = ∫4 v(t) dt = area under v(t) between 4 and 6 = or 1 (2) (4) (area of a triangle) 2 x6 − x4 = 4. Thus. x(t) is a quadratic function with a decreasing slope (concave down). a current i(t) = 24 e−40 t mA is applied to a 3 𝜇 F capacitor.42 Position of the particle for Example 9-7. s Figure 9. Example 9-10 For t ≥ 0. integrals are used to obtain the voltage across a capacitor when a current is passed through it (charging and discharging of a capacitor).6. dt dw(t) (b) Given that p(t) = v(t) i(t) = . 2 Assume that the capacitor is initially completely discharged. and Energy Stored in a Capacitor In this section. x(t). m 16 Quadratic 12 8 Linear 4 0 Quadratic 0 2 4 6 t.1 Current. the initial voltage across the capacitor and initial energy stored are zero. 9. as shown in Fig.43 Current applied to a capacitor. find the voltage v(t) across the capacitor.314 Chapter 9 Integrals in Engineering The graph of the position obtained above is shown in Fig.e. i. dv(t) .42.6 APPLICATIONS OF INTEGRALS IN ELECTRIC CIRCUITS 9. as well as the total stored energy. (a) Given that i(t) = C + i(t) ↑ C = 3 μ F v(t) − Figure 9.. find the stored energy w(t) and show that dt 1 w(t) = C v2 (t). 9. . 9. Voltage.43. 41) Substituting v(0) = 0 V. C ∫0 (9.024 e−40 t dt 3.42) gives t ∫0 t dw(t) p(t) dt dt = ∫0 dt [w(t)]t0 = t p(t) dt ∫0 t w(t) − w(0) = ∫0 p(t) dt (9.6 Applications of Integrals in Electric Circuits Solution (a) Given i(t) = C dv(t) dt 315 . it follows that dv(t) 1 = i(t). C = 3 × 10−6 F.024 e−40 t dt = 3.0 × 10−6 ∫0 t 0. and i(t) = 0.0 × 10−6 ∫0 [ ]t 1 −40 t = 8000 − e 40 0 ( ) 8000 −40 t 0 e =− −e 40 = −200 ( e−40 t − 1) v(t) = or v(t) = 200 ( 1 − e−40 t ) V. dt C (9.41). (b) By definition.40) Integrating both sides gives t ∫0 t dv(t) 1 dt = i(t) dt ∫0 C dt [v(t)]t0 = 1 C ∫0 t i(t) dt t v(t) − v(0) = 1 i(t) dt C ∫0 or v(t) = v(0) + t 1 i(t) dt.024 e−40 t A into equation (9.9. the power supplied to a capacitor is given by p(t) = dw(t) dt Integrating both sides of equation (9.42) . the voltage across the capacitor at any time t is given by t 1 0. 06 e−80 t − 0.024 e−40 t ) or p(t) = 4. the quantity C v2 (t) can be calculated as 2 2 1 1 C v2 (t) = (3 × 10−6 ) (200 − 200 e−40 t )2 2 2 ] [ = (1.45) and (9.8 e−40 t − 4.12 + 0.8 −80 t =− − 1) + − 1) (e (e 40 80 = −0. or v(t) = v(t0 ) + voltage v(t) across the capacitor knowing that i(t) = C dt t 1 i(t) dt.024 e−40 t ) = (200) (0. w(0) = 0. Comparison of equations (9.44) Substituting w(0) = 0 and p(t) from equation (9.12 e−40 t + 0.43) Since the energy stored in the capacitor at time t = 0 is zero.44) into equation (9.024) e−40 t − (200 e−40 t ) (0.43) gives t (4. C ∫t0 v(0) = 0 V).8 − − 4.46) reveals that w(t) = Example 9-11 (9.. The power p(t) is given by p(t) = v(t) i(t) = 200 (1 − e−40 t ) (0.8 e−80 t W. Assume that the initial voltage across the capacitor is zero (i. (9. (9.8 −40 t 4.06 e−80 t − 0.e.06 J.8 e−80 t ) dt ∫0 [ [ ]t ]t 1 −40 t 1 −80 t = 4. To show that w(t) = (9.06 e−80 t − 0.45) 1 1 C v2 (t).5 × 10−6 ) 2002 − 2(200)(200)e−40t + (200e−40t )2 = (1. 2 The current i(t) shown in Fig.5 × 10−6 )(4 × 104 − 8 × 104 e−40 t + 4 × 104 e−80 t ) = 0.06 J. .12 e−40 t + 0.12 e−40 t + 0.06 w(t) = 0 + or w(t) = 0. Sketch the dv(t) .316 Chapter 9 Integrals in Engineering or w(t) = w(0) + t ∫0 p(t) dt.44 is applied to a 20 𝜇F capacitor.8 e−40 t − 4. 9.8 − e e 40 80 0 0 4.46) 1 C v2 (t). 001 − 0) or v1 = 200 V.e.6 Applications of Integrals in Electric Circuits 317 i(t).001 0 = 200..001 v1 = v0 + 50. and i(t) = 4 A = constant. The change in voltage across the capacitor between 0 and 1 ms can also be calculated without evaluating the integral (i. Since C = 20 𝜇F. ms −4 Figure 9. 1 v(t) = i(t) dt is a straight line with positive slope.000 ∫0 4 dt = 0 + 200.44 Solution Current applied to a capacitor. A 4 i(t) ↑ C = 20 μ F + 2 v(t) 0 − −2 0 1 2 4 5 t. v(t0 ) = v0 = 0 V. 1 1 = C 20 × 10−6 106 20 102 × 104 = 20 = or 1 = 50. from geometry) as 1 × area under the current between 0 and 1 ms C = 50. The voltage at time C∫ t = 1 ms can be calculated as 0. Therefore. C The voltage during each time interval can now be calculated as follows: (a) 0 ≤ t ≤ 1 ms: t0 = 0.001)] (area of a rectangle) v1 − v0 = .9.000 (0.000 [ t ]0.000 [(4) (0.000 F−1 . Therefore. The voltage at t = 4 ms can be calculated as 0. (b) 1 ms < t ≤ 2 ms: t0 = 1 ms.002 2 dt = 100 + 100.002) ] v4 − v2 = . v(t) = C1 ∫ i(t) dt is a straight line with positive slope.002) = 100 + 100.001) ] v2 − v1 = v2 − v1 = −100 V Therefore. v1 = v0 + 200 = 200 V. v(t0 ) = v1 = 200 V.318 Chapter 9 Integrals in Engineering or v1 − v0 = 200 V.002 0. and i(t) = −2 A = constant.004 v4 = v2 + 50.000 [ (−2) (0.001 (−2) dt = 200 − 100. 1 Therefore.000 [t]0.000 ∫0.002 v2 = v1 + 50.001 = 200 − 100. The change in voltage across the capacitor between 1 ms and 2 ms can also be calculated from geometry as 1 × area under the current waveform between 1 and 2 ms C = 50.004 0.000 ∫0. The voltage C∫ at t = 2 ms can be calculated as 0.000 [ (2) (. v(t0 ) = v2 = 100 V.000 [ t ]0. The change in voltage across the capacitor between 2 ms and 4 ms can also be calculated from geometry as 1 × area under the current waveform between 2 and 4 ms C = 50.000 (0. v(t) = i(t) dt is a straight line of negative slope.002 − 0.004 − 0.001) = 200 − 100 or v2 = 100 V. Therefore. (c) 2 ms < t ≤ 4 ms: t0 = 2 ms.000 (0.002 = 100 + 100.002) = 100 + 200 or v4 = 300 V. v2 = v1 − 100 = 200 − 100 = 100 V.000 (0. and i(t) = 2 A = constant. 000 [t]0. The voltage at t = 5 ms can be calculated as 0. v(t).000 (0. v(t) = C1 ∫ i(t) dt is a straight line with negative slope.005 − 0.45.001) ] v5 − v4 = or v5 − v4 = −200 V Therefore.000 ∫0. V v4 300 v1 200 0 Figure 9. Therefore. and i(t) = −4 A = constant. ms . 9.6 Applications of Integrals in Electric Circuits 319 or v4 − v2 = 200 V.45 v5 v2 100 0 1 2 4 5 Voltage across the capacitor in Example 9-11.000 (0. v(t0 ) = v4 = 300 V. (d) 4 ms < t ≤ 5 ms: t0 = 4 ms.000 [ (−4) (.001) = 300 − 200 or v5 = 100 V.004 = 300 − 200.004) = 300 − 200.004 (−4) dt = 300 − 200.005 0. t. The sketch of the voltage across the capacitor is shown in Fig. Therefore.005 v5 = v4 + 50. v4 = v2 + 200 = 100 + 200 = 300 V. v5 = v4 − 200 = 300 − 200 = 100 V.9. The change in voltage across the capacitor between 4 ms and 5 ms can also be calculated from geometry as 1 × area under the current waveform between 4 and 5 ms C = 50. 5 (−2) = −4 V/s. Sketch the voltage v(t) across the capacitor knowing that i(t) = C dt 1 or v(t) = i(t) dt. i(t).5 F v(t) 0 0 1 2 4 5 t. therefore.46 is applied to a 0.e.5[ ] 1 (1)(−2) (area of a triangle) =2 2 or v1 − v0 = −2. The voltage across the capacitor during different time intervals can be calculated as follows: (a) 0 ≤ t ≤ 1 s: i(t) is a straight line.320 Chapter 9 Integrals in Engineering Example 9-12 The sawtooth current i(t) shown in Fig. dt time t. s − −2 Figure 9. the slope of the voltage is . with a zero slope at t = 1 s. since dt = C i(t) is the slope of v(t) at 1 at t = 0 is dv = 0. v(t) = C1 ∫ i(t) dt is a quadratic function.5 F capacidv(t) tor. v1 = v0 + (−2) = 0−2 or v1 = −2 V. At t = 1 s. 9.5 Therefore. Solving for v1 gives. dt 0. Assume that the capacitor is completely discharged at C∫ t = 0 (i. v(t) is a decreasing quadratic function starting at v(0) = 0 V and ending at −2 V.46 Solution Sawtooth current applied to a capacitor. v(0) = 0 V). the slope of the voltage dv 1 = (0) = 0 V/s. dv(t) 1 Also.. A 2 + i(t) ↑ C = 0. The change in voltage across the capacitor between the time interval 0 to 1 s can be calculated as t 1 i(t) dt v1 − v0 = C ∫0 1 = (area under the current waveform between 0 and 1 s) 0. Since the voltage at t = 2 s is zero and the current supplied to the capacitor between 2 and 4 s is the same as the current applied to the capacitor from 0 to 2 s.5 Therefore. The same is true for remaining intervals.6 Applications of Integrals in Electric Circuits 321 (b) 1 < t ≤ 2 s: i(t) is a straight line.47. The change in voltage across the capacitor during the time interval 1 to 2 s can be calculated as 1 (area under the current waveform between 1 and 2 s) v2 − v1 = 0. dt 0. Also. v(t). v(t) = C1 ∫ i(t) dt is a quadratic function. s v(t) −2 Figure 9.9. since dv(t) dt = 1 C i(t) is the slope of v(t). .47 Voltage across a capacitor subjected to a sawtooth current. dv(t) = 0 V/s at t = 1 s. 9. therefore. V 2 0 0 i(t) 1 2 3 4 5 6 t. dt dv(t) 1 = (2) = 4V/s at t = 2 s. A sketch of the voltage across the capacitor is shown in Fig.5 ] [ 1 (1)(2) (area of a triangle) =2 2 or v2 − v1 = 2. v(t) is an increasing quadratic function starting with a zero slope at v1 = −2 V and ending at v2 = 0 V with a slope of 4 V/s. the voltage across the capacitor between 2 and 4 s is identical to the voltage between 0 and 2 s. Solving for v2 gives v2 = v1 + 2 = −2 + 2 or v2 = 0 V. 2 (a) The voltage/current relationship for an inductor is given by L di(t) = v(t) dt or di(t) 1 = v(t). (9.48 + − L = 100 mH Voltage applied to an inductor. Also. (a) Suppose the initial current flowing through the inductor is i(0) = 10 A. Knowdi(t) ing that v(t) = L dt . find the power p(t) = v(t) i(t) supplied to the inductor. integrals are used to find the current flowing through an inductor when it is connected across a voltage source.48) from an initial time t0 to time t gives t ∫t0 t di(t) 1 v(t) dt = dt L ∫t0 t [i(t)]tt = 1 v(t) dt L ∫t0 i(t) − i(t0 ) = 1 v(t) dt L ∫t0 0 t . find the stored energy t w(t) = w(0) + and show that w(t) = Solution ∫0 p(t) dt.48.47) 1 2 L i (t).7 CURRENT AND VOLTAGE IN AN INDUCTOR In this section. 9. as shown in Fig. (9. If the initial energy stored in the inductor is w(0) = 5 J.322 Chapter 9 Integrals in Engineering 9. Example 9-13 A voltage v(t) = 10 cos(10 t) V is applied across a 100 mH inductor. (b) Given your results in part (a). i(t) v(t) Figure 9. integrate both sides of the equation to determine the current i(t).48) dt L Integrating both sides of equation (9. plot the current for 0 ≤ t ≤ 𝜋/5 s. the plot of i(t) is simply the sinusoid 10 sin (10 t) shifted upward by 10 amps as shown in Fig.50) is a periodic function with frequency 𝜔 = 10 rad/s.9.1 ∫0 [ ]t 1 = 10 + 100 sin(10 t) 10 0 i(t) = 10 + = 10 + 10 ( sin 10 t − 0 ) or i(t) = 10 + 10 sin 10 t A. L = 0.50) The current i(t) obtained in equation (9. 5 Thus.1 H. The period is 2𝜋 𝜔 2𝜋 = 10 T= or T= 𝜋 s.49. i(0) = 10 A. A 20 10 0 0 Figure 9. and v(t) = 10 cos(10 t) V in equation (9. (9.49) gives t 1 10 cos(10 t) dt 0.7 Current and Voltage in an Inductor 323 or t i(t) = i(t0 ) + 1 v(t) dt. L ∫t0 (9. t. 9. i(t). s .49 π /20 π /10 3π /20 π /5 Current flowing through the inductor in Example 9-13.49) Substituting t0 = 0. 324 Chapter 9 Integrals in Engineering (b) The power p(t) supplied to the inductor is given by p(t) = v(t) i(t) = (10 cos (10 t)) (10 + 10 sin (10 t)) = 100 cos (10 t) + 100 sin (10 t) cos (10 t) = 100 cos (10 t) + 50 (2 sin (10 t) cos (10 t)) = 100 cos (10 t) + 50 sin (20 t) or p(t) = 100 (cos (10 t) + 0.5 sin (20 t)) W. (9.51) The energy stored in the inductor is given by t w(t) = w(0) + ∫0 p(t) dt. (9.52) Substituting w(0) = 5 J and p(t) calculated in equation (9.51) gives t w(t) = 5 + ∫0 = 5 + 100 100 (cos (10 t) + 0.5 sin (20 t)) dt [ sin (10 t) 10 [ ]t + 50 0 cos 20 t − 20 ]t 0 = 5 + 10 (sin (10 t) − 0) − 2.5 (cos (20 t) − 1) or w(t) = 7.5 + 10 sin (10 t) − 2.5 cos (20 t) J. (9.53) 1 1 L i2 (t), the quantity L i2 (t) can be calculated as 2 2 1 1 L i2 (t) = (0.1) (10 + 10 sin (10 t))2 2 2 ] [ = 0.05 102 + 2(10)(10) sin (10 t) + (10 sin (10 t))2 To show that w(t) = = 0.05 (100 + 200 sin (10 t) + 100 sin2 (10 t)). ( ) 1 − cos (20 t) 2 , we get Noting that sin (10 t) = 2 ( ) 1 − cos (20 t) 1 L i2 (t) = 5 + 10 sin (10 t) + 5 2 2 = 5 + 10 sin (10 t) + 2.5 − 2.5 cos (20 t) = 7.5 + 10 sin (10 t) − 2.5 cos (20 t) J, which is the same as equation (9.53). (9.54) 9.7 Current and Voltage in an Inductor Example 9-14 325 A voltage v(t) is applied to a 500 mH inductor as shown in Fig. 9.50. Knowing that di(t) v(t) = L dt (or i(t) = L1 ∫ v(t) dt), plot the current i(t) using integrals. Assume the initial current flowing through the inductor is zero (i.e., i(0) = 0 A). v(t), V 9 i(t) v(t) + − L = 500 mH 0 0 2 4 6 8 t, s −9 Figure 9.50 Voltage applied to an inductor. Solution Using equation (9.49), the current i(t) flowing through the inductor during each time interval can be determined as follows: (a) 0 ≤ t ≤ 2 s: v(t) = 9 V = constant. Therefore, i(t) = line with positive slope. Also, the change in current is 1 L ∫ v(t)dt is a straight 2 1 v(t)dt L ∫0 1 = (area under the voltage waveform between 0 and 2 s) 0.5 1 = [ (2) (9) ] 0.5 i 2 − i0 = or i2 − i0 = 36 A. Solving for i2 is i2 = i0 + 36 = 0 + 36 or i2 = 36 A. Note that the equation for the current flowing through the inductor at any time t between 0 and 2 s can also be calculated as t i(t) = i(0) + 1 v(t) dt L ∫0 t 1 9 dt 0.5 ∫0 = (2) (9)[t]t0 = 0+ 326 Chapter 9 Integrals in Engineering or i(t) = 18 t. (b) 2 < t ≤ 4 s: v(t) = −9 V = constant. Therefore, i(t) = L1 ∫ v(t) dt is a straight line with a negative slope. Also, the change in current is 1 (area under the voltage waveform between 2 and 4 s) 0.5 1 = [ (2) (−9) ] 0.5 i 4 − i2 = or i4 − i2 = −36 A. Solving for i4 gives i4 = i2 − 36 = 36 − 36 or i4 = 0 A. Note that the equation for the current flowing through the inductor at any time t between 2 and 4 s can also be calculated as t i(t) = i(2) + 1 v(t) dt L ∫2 t = 36 + 1 −9 dt 0.5 ∫2 = 36 − 18 [t]t2 = 36 − 18 (t − 2) or i(t) = −18 t + 72. Since the current at t = 4 s is zero (the same as at t = 0) and the voltage applied to the inductor between 4 and 8 s is the same as the voltage from 0 to 4 s, the current flowing through the inductor between 4 and 8 s is identical to the current between 0 and 4 s. The resulting current waveform (triangle wave) is shown in Fig. 9.51. i(t), A i(t) = −18t + 72 36 0 0 2 4 6 8 t, s Figure 9.51 Current flowing through the inductor in Example 9-14. 9.8 Further Examples of Integrals in Engineering 9.8 Example 9-15 327 FURTHER EXAMPLES OF INTEGRALS IN ENGINEERING A biomedical engineer measures the velocity profiles of a belted and unbelted occupant during a 35 mph (≈ 16 m/s) frontal collison, as shown in Fig. 9.52. t (a) Knowing that x(t) = x(0) + ∫0 v(t)dt, find and plot the displacement x(t) of the belted occupant for time 0 to 50 ms. Assume that the initial displacement at t = 0 is 0 m (i.e., x(0) = 0 m). (b) Find and plot the displacement x(t) of the unbelted occupant for time 0 to 50 ms. Assume x(0) = 0 m. (c) Based on the results of parts (a) and (b), how much farther did the unbelted occupant travel compared to the belted occupant? v(t), m/s 16 v(t) = 16 cos(50 π t) Belted Occupant: +X v(t) = 0 0 0 v(t), m/s 0.01 t, s 0.05 16 v(t) = 16 cos(10 π t) Unbelted Occupant: 0 +X 0 t, s 0.05 Figure 9.52 Velocities of the belted and unbelted occupants during frontal collison. Solution (a) The displacement of the belted occupant can be calculated using the expression t x(t) = x(0) + (i) for 0 ≤ t ≤ 0.01 s t x(t) = 0 + ∫0 [ 16 cos(50 𝜋t)dt = 16 8 = sin(50 𝜋t) m. 25 𝜋 Therefore, x(0.01) = 8 25 𝜋 sin ∫0 v(t)dt sin(50 𝜋t) 50 𝜋 ]t = 0 16 [sin(50 𝜋t) − 0] 50 𝜋 (9.55) ( ) 𝜋 2 = 0.102 m. 328 Chapter 9 Integrals in Engineering (ii) 0.01 ≤ t ≤ 0.05 s t x(t) = 0.102 + ∫0 0 dt = 0.102 m. (9.56) The displacement of the belted occupant during a frontal collison is shown in Fig. 9.53. x(t), m 0.102 0 0 0.01 0.05 t, s Figure 9.53 Displacement of the belted occupant during a frontal collison. (b) The displacement of the unbelted occupant for 0 ≤ t ≤ 0.05 s can be calculated as t 16 cos(10 𝜋t)dt ∫0 [ ] sin(10 𝜋t) t = 16 10 𝜋 0 x(t) = 0 + 16 [sin(10 𝜋t) − 0] 10 𝜋 8 sin(10 𝜋t) m. (9.57) = 5𝜋 ( ) Therefore, x(0.05) = 58𝜋 sin 𝜋2 = 0.509 m. The displacement of the unbelted occupant is shown in Fig. 9.54. (c) To find how much farther the unbelted occupant traveled during collison as compared to the belted occupant, the total distance traveled by the belted occupant in 50 ms is subtracted from the total distance traveled by the unbelted occupant as = Δ x = xunbelted (0.05) − xbelted (0.05) = 0.509 − 0.102 = 0.407 m. 9.8 Further Examples of Integrals in Engineering 329 x(t), m 0.509 0 0 Figure 9.54 Example 9-16 0.05 t, s Displacement of the unbelted occupant during a frontal collison. A biomedical engineer is evaluating an energy-absorbing aviation seat on a vertical deceleration tower, as shown in Fig. 9.55. The acceleration profile of the drop cage is described by a(t) = 500 sin(40 𝜋t) m/s2 . t (a) Knowing that v(t) = v(0) + ∫0 a(t)dt, find and plot the velocity v(t) of the drop cage. Assume the drop cage starts from rest at t = 0 s. (b) What is the impact velocity vimpact of the drop cage if it takes 25 ms to hit the ground? (c) The total impulse I is equal to the change in momentum. For example I = Δ p = pf − pi = mvimpact − mv0 , where m is the mass of the system, vimpact is the final velocity, v0 is the initial velocity, p is the momentum, and I is the impulse. Find the total impulse after 25 ms. Assume that the total mass of the drop cage, seat, and crash test dummy is 1,000 kg. Figure 9.55 Energy-absorbing aviation seat. 330 Chapter 9 Integrals in Engineering Solution (a) The velocity of the cage can be calculated as t v(t) = v(o) + ∫0 a(t)dt t 500 sin(40 𝜋t)dt ∫0 [ ] cos(40 𝜋t) t = 500 − 40 𝜋 0 = 0+ 500 [cos(40 𝜋t) − 1] 40 𝜋 25 = (9.58) [1 − cos(40 𝜋t)] m/s. 2𝜋 (b) The velocity of the cage when it impacts the ground can be found by substituting t = 25 ms = 0.025 s in equation (9.58) as =− vimpact = = = = = 25 [1 − cos ( 40 𝜋(0.025) )] 2𝜋 25 [1 − cos(𝜋)] 2𝜋 25 [1 − (−1)] 2𝜋 50 2𝜋 7.96 m/s. (c) The total impulse I can be found as I = m vimpact − m v0 = (1000) (7.96) − (1000) (0) kg m . = 7960 s Example 9-17 A civil engineer designs a building overhang to withstand a parabolic snow loading ) ( x 2 per unit length p(x) = p̂ 1 − , as shown in Fig. 9.56. L L (a) Compute the resulting force V = ∫0 p(x)dx. L (b) Compute the corresponding moment M = ∫0 x p(x)dx. L (c) Locate the centroid of the loading x̄ = ∫o x p(x) dx L ∫0 p(x) dx = M . V 9.8 Further Examples of Integrals in Engineering 331 p(x) pˆ V V M x L Figure 9.56 Solution Snow loading of a building overhang. (a) The resulting force per unit width V can be calculated as ) ( x 2 dx p̂ 1 − ∫0 ∫0 L ( ) ) [ ( )]L L ( 2 x2 2 x x2 1 x3 = p̂ 1 − + + dx = p̂ (x) − ∫0 L L 2 L2 L2 3 0 [ ] ] [ ) ) 1( 2 1 ( 3 L L −0 + L − 0 = p̂ L − L + = p̂ (L − 0) − L 3 3 L2 L L p(x)dx = V= or V = p̂ L . 3 (9.59) (b) The corresponding moment M can be calculated as [ ( ) ] x 2 x p(x)dx = p̂ x 1 − dx M= ∫0 ∫0 L ( ) ] [( 2 ) ( )]L L [ 2 x2 x3 2 x3 1 x4 x p̂ x − + − + dx = p̂ = ∫0 L 2 L 3 L2 L2 4 0 [( ) ] ( ( ) ) 1 2 1 L3 − 0 + L4 − 0 = p̂ (L2 − 0) − 2 3L 4 L2 ] [ 2 2 L2 L2 L − + = p̂ 2 3 4 L or M = p̂ L2 . 12 L Example 9-18 p(x) p0 V Upper face plate h V M x L (a) Cantilever overhang. Figure 9. ) ( x L into equation (9. and G is the shear modulus.60) for the deflection y(x). (9.60) gives and p(x) = p0 1 − 2 L ] ) ( x [ x 1 x L dx − p0 y(x) = p 1− dx ∫0 0 h G ∫0 L 2 ]x ] x [ [ L x2 1 − p0 dx p0 x − = h G ∫0 2L 0 2 (a) Substituting V = p0 .57. This sandwich beam construction deforms primarily due to shear deformation with deflection y(x) satisfying ] x [ x 1 p(x) dx − V dx.57 Solution Core Lower face plate (b) Sandwich construction. The overhang is constructed from two metal face plates separated by a nonmetallic core of thickness h. as shown in Fig. 4 A building overhang is subjected to a triangular snow loading p(x). where V = p0 .332 Chapter 9 Integrals in Engineering (c) The location of the centroid x̄ can be found as x̄ = M V p̂ L2 = 12 p̂ L 3 or x̄ = L .60) y(x) = ∫0 h G ∫0 ) ( x L . p(x) = p0 1 − 2 L (a) Evaluate (9. (b) Find the location and value of maximum deflection. 9. Triangular snow loading of sandwich-constructed building overhang. (9.9.8 Further Examples of Integrals in Engineering x [ ) } ] {( L x2 − (0 − 0) − p0 dx x− 2L 2 = 1 h G ∫0 = ] x [ p0 L x2 − dx x− h G ∫0 2L 2 p = 0 hG p = 0 hG [ ]x x2 x3 L − − x 2 6L 2 [( p x = 0 2hG p0 333 x2 x3 L − − 2 6L 2 ( 0 ) ] − (0 − 0 − 0) ) x2 x− −L 3L or ( )] p0 x [ x L−x 1− .61) 2hG 3L (b) The location of the maximum value of the deflection can be found by equating the derivative of y(x) to zero as y(x) = − dy(x) =0 dx ( p x [ ( )]) x d − 0 L−x 1− =0 dx 2hG 3L ( ) p0 3 x2 − L − 2x + =0 2hG 3L x2 − 2 x L + L2 = 0 (x − L)2 = 0. 6hG Therefore. the deflection is maximum at x = L.61) as ( )] p0 L [ L L−L 1− 2hG 3L ( ) p L L =− 0 2hG 3 y(L) = − =− p0 L2 . The value of the maximum deflection can now be obtained by substituting x = L in equation (9. Therefore. 6hG . the maximum value of deflection is p0 L2 . 9.2 Profile of a gear tooth for problem P9-2.1 Area of a gear tooth for problem P9-1. A 0 l 0 x 0 (b) The gear tooth area inscribed by six rectangles. i.3.e.2 is approximated by ( the trigonomet( )) ric equation y(x) = k 2 1 − cos 2 𝜋x l . P9. 0 y k (a) The profile of a gear tooth. ∫0 9-2. A = l y(x) dx.. The profile of a gear tooth shown in Fig. A= 8 ∑ y(xi ) Δ x. as shown in Fig. y(x) k Area. The acceleration is constant during the first 4 seconds of motion. The velocity of an object as a function of time is shown in Fig.334 Chapter 9 Integrals in Engineering PROBLEMS 9-1. i=1 y(x) k (b) Calculate the exact area by integration.1(a) is approximated by the quadratic equation y(x) = − 4lk x (x − l). (a) Estimate the area A using six rectangles of equal width (Δ x = l∕6). The velocity is constant during the last 6 s.e. Δ x = l∕8.. (a) Estimate the total distance covered (the area. 9-3. (b) Calculate the exact area by evaluating the definite integral. A 0 l 0 x ∫ y(x) dx. (b) Now estimate the total distance covered using 10 rectangles of equal width. l A= Area. x . The profile of a gear tooth shown in Fig. Figure P9. (a) Estimate the area A using eight rectangles of equal width 0 Δx l Figure P9. A) under the velocity curve using five rectangles of equal width (Δ t = 10∕5 = 2 s). i. P9.1(b). so the velocity is a linear function of time with v(t) = 0 at t = 0 and v(t) = 100 ft/s at t = 4 s. P9. 4. The total work done on the object is given by 100 5 W= 0 4 10 t. 9-7. as shown in Fig.7(a).4. b y(x) dx. (c) f (x) = 2 sin 𝜋x l l 9-6. it travels a distance of 5 m. P9. Figure P9. Determine the work done if the force is given by (a) f (x) = (x + 1)3 N. as shown in Fig.Problems (c) Calculate the exact area under the velocity curve. ȳ = A A (d) Now solve for x as a function of y. (b) Find the area A by integration. ft/s 335 9-5. ∫0 (c) Find the centroid G by integration with vertical rectangles.7(b). s Figure P9. (b) f (x) = 4 e−2 x N. (b) f (x) = 10 sin 10 Hint: Use the double angle formula. f (x) dx N-m. A particle is accelerated along a curved path of length l under the action of an applied force f (x). f(x) ∫0 x x̄ = ∫ x dA A = ∫ x y(x) dx A and y 1 dA (y(x))2 dx ∫ 2 2∫ = .0 m under the action of an applied force f (x). find A= If l = 4. ∫0 (d) Calculate the exact area by adding the area of the triangle and the area of the rectangle bounded by the velocity curve.. ( ) 𝜋x N.4 A particle moving on a curved path. ( ) ( ) 𝜋x + 3 cos N.e. and recalculate the y-coordinate of the centroid G by integration with . (c) f (x) = 1 − 2 sin2 2 Hint: Use a trigonometric identity. (b) f (x) = e−2 x (1 + e4 x ) N. The total work done on the particle is l W= ∫0 f (x) dx N-m. P9. P9. i. as shown in Fig. ) ( ) ( 𝜋 𝜋 x cos 10 x N. (a) Find the equation of the line y(x). definite integral Δ x = v(t). determine the work done for (a) f (x) = 8 x3 + 6 x2 + 4 x + 2 N.3 The velocity of an object. Determine the work done for (a) f (x) = 2 x4 + 3 x3 + 4 x − 1 N. as shown in Fig. in other words. When a variable force is applied to an object. P9. A particle is accelerated along a curved path of length l = 3. find the total distance traveled by evaluating the 10 v(t) dt. A triangular area is bounded by a straight line in the x-y plane. The total work done on the particle is l W= ∫0 f (x) dx.0 m. 9-4. Area. cm 9 dA = x(y) dy 6 x/2 dy 3 x(y) 0 0 3 6 9 12 x.. P9. (a) Find the equation of the line y(x). as illustrated in Fig.7 Centroid of a triangular cross section.8 Area bounded by a curved surface.7(c). Figure P9. A y = kxn G 0 b x Figure P9. y horizontal rectangles. .336 Chapter 9 Integrals in Engineering y. cm 9 9 y(x) dx 6 6 Area. (a) Determine the area A by integration with respect to x. P9.9. ȳ = ∫ y dA A = ∫ kbn y x(y) dy A . cm 12 0 0 (a) Area bounded by a straight line. 9-8. Consider the shaded area under the line y(x). as shown in Fig. i. cm (c) Horizontal rectangles.8. An area in the x-y plane is bounded by the curve y = k xn and the line x = b. A 3 dA = y(x) dx G x¯ 3 y(x) y¯ 0 0 3 y(x)/2 6 9 x. (c) Evaluate your answer to part (b) for the case n = 1. P9. cm y. 3 6 9 x. 9-9. cm 12 (b) Vertical rectangles. (b) Determine the coordinates of the centroid G by integration with respect to x.e. y. as shown in Fig. Problems (b) Determine the area under the line by integration with respect to x. (d) Determine the y-coordinate of the centroid by integration with respect to x. (b) Determine the area of the cooling fin by integration with respect to x. (a) Given the above equation for y(x). (c) Determine the x-coordinate of the centroid by integration with respect to x. 9-14. h h 2 b y. 3) 3 y x. Repeat problem P9-10 if the shaded area of the cooling fin is y(x) = −x2 + 4.13. P9. Repeat problem P9-10 if the geometry of a cooling fin is described by the shaded area shown in Fig. ft 9-11. y(x) = −x2 + 16 h x. The geometry of a cooling fin is defined by the shaded area that is bounded by the parabola y(x) = −x2 + 16. . as illustrated in Fig.9 Shaded area for problem P9-9. 6 9-12. determine the height h and width b of the fin. 1 (16 − x) 2 x Figure P9. (c) Determine the x-coordinate of the centroid by integration with respect to x. The cross section of an airfoil is described by the shaded area that is bounded by a cubic equation y(x) = −x3 + 9 x. in. P9. as shown in Fig. 337 (b) Determine the area of the cooling fin by integration with respect to x.13 Geometry of the cooling fin for problem P9-13. P9. 0 0 b Figure P9. (6. ft 0 0 3 6 y(x) = Figure P9.14. 9-10. (d) Determine the y-coordinate of the centroid by integration with respect to x. Repeat problem P9-10 if the shaded area of the cooling fin is y(x) = 9 − x2 . determine the height h and width b of the fin. y(x) 9-13. (a) Given the above equation for y(x). (c) Determine the x-coordinate of the centroid by integration with respect to x. in.10 Geometry of a cooling fin. y.10. and calculate the width b.16 Geometry of an airfoil. (cm) y. P9. 9-15. . y w(x) = − w0 2 (x − l2) l2 w0 x Figure P9.e. the maximum value of y(x)). (cm) Figure P9. P9. (a) Determine the values of a and b for the heat-sink fin. (e) Determine the y-coordinate of the centroid by integration with respect to x. The geometry of a gear tooth is approximated by the following quadratic equation. l R= ∫0 w(x) dx. (c) Determine the x-coordinate of the centroid by integration with respect to y. (mm) y(x) = −2x2 + 12x − 16 h x1 y(x) = −x3 + 9x x2 x.17. (mm) b h Figure P9.338 Chapter 9 Integrals in Engineering (d) Determine the y-coordinate of the centroid by integration with respect to x. (c) Determine the area of the gear tooth by integration with respect to x. as shown in Fig. y. (cm) b Figure P9. as shown in Fig.16.15 Geometry of an airfoil. A cantilever beam is subjected to a quadratic distributed load. 9-17.17 Cantilever beam subjected to a quadratic distributed load. (cm) x = f(y) = y2 + 2 2 1 a b x.. (a) Determine the height h of the tooth (i. (d) Determine the y-coordinate of the centroid by integration with respect to y.14 Geometry of an airfoil. y. The geometry of a decorative heat-sink fin of height 2 cm and length b (in cm) is approximated. (a) Determine the total resultant force. (d) Determine the x-coordinate of the centroid by integration with respect to x. 9-16. x.15. P9. (b) Determine the coordinates x1 and x2 where y(x) = 0. as shown in Fig. (b) Determine the area of the cooling fin by integration with respect to y. and clearly indicate both its maximum and final values. (b) a(t) = 5 e−5 t + 𝜋4 cos ( 𝜋4 t) m/s2 . l x̄ = ∫0 x w(x) dx R . and clearly indicate both its maximum and final values. P9. plot the position x(t) of the vehicle. m/s2 10 0 0 Figure P9. y w(x) = − 339 9-20. (b) Given your results of part (a). A simply supported beam is subjected to a quadratic distributed load. ∫0 (b) Determine the x-location of the resultant R.21 The acceleration of an automobile. A vehicle starting from rest at position x(0) = 0 is subjected to the acceleration shown in Fig. P9. l x̄ = x w(x) dx ∫0 .23. (b) a(t) = 4 sin 2 t cos 2 t m/s2 . R 9-18..e.18. 9-21. determine the centroid of the area under the distributed load.e. 9-22.21. determine the centroid of the area under the distributed load.18 Simply supported beam subjected 2 14 16 t. i. w0 x(x − 2l ) l2 x(t) w0 x l a(t). Determine the velocity v(t) and the position y(t) of a vehicle that starts from rest at position y(0) = 0 and is subjected to the following accelerations: (a) a(t) = 40 t3 + 30 t2 + 20 t + 10 m/s2 . s to a quadratic distributed load.. If the automobile starts from rest at position x(0) = 0. −10 (a) Determine the total resultant force. 9-23. Find the velocity v(t) and position x(t) if the particle is subjected to the following accelerations: (a) a(t) = 6 t3 − 4 t2 + 7 t − 8 m/s2 .22. . A vehicle starting from rest at a position x(0) = 0 is subjected to the acceleration shown in Fig. i.Problems (b) Determine the x-location of the resultant R. P9. (a) Plot the velocity v(t) of the vehicle. sketch the velocity v(t) and position x(t) of the automobile. P9. and clearly indicate both its maximum and final values. w(x) dx. as shown in Fig. Hint: Use a trigonometric identity. A particle starts from rest at a position x(0) = 0. The acceleration of an automobile is measured as shown in Fig. (a) Plot the velocity v(t) of the vehicle. 9-19. l R= Figure P9. .26 has R = 10 Ω.5 F. A vehicle starting from rest at a position x(0) = 0 is subjected to the acceleration shown in Fig. find the voltage v(t) supplied by the voltage source. If the current i(t) flowing through the circuit is i(t) = 10 sin (240 𝜋t) A.22 A vehicle subjected to a given acceleration. plot the position x(t) of the vehicle. dq(t) Knowing that i(t) = dt . 3 6 9 12 Figure P9. 9-26. x(t) a(t). plot the position x(t) of the vehicle. 2 4 6 8 −5 −10 Figure P9. P9. and C = 0. P9. −5 9-25.24 A vehicle subjected to a given acceleration. and clearly indicate both its maximum and final values. x(t) a(t).340 Chapter 9 Integrals in Engineering 9-24. and clearly indicate both its maximum and final values. (b) Given your results of part (a). (m/s2) 20 x(t) 10 t(s) 0 1 2 3 4 a(t). find the equation of the charge of (t) if q(0) = 0. The RLC circuit shown in Fig. The current flowing in a resistor is given by −10 i(t) = (t − 1)2 A.24. (m/s2) 10 −10 5 −20 t(s) 0 Figure P9. and clearly indicate both its maximum and final values. (b) Given your results of part (a). (m/s 2) 10 5 t(s) 0 (a) Plot the velocity v(t) of the vehicle.23 A vehicle subjected to a given acceleration. L = 2 H. 62) to determine the voltage vo (t).5 F 9-29. C = 0.31. P9. Integrate both sides of equation (9. A current i(t) = 10 e−10 t mA is applied to a 100 𝜇F capacitor. Also. and that the initial stored energy is w(0) = 0.26 A series RLC circuit.Problems which is given by: v(t) = i R + L (b) Suppose that the instantaneous power absorbed by the capacitor is p(t) = 100(1 − cos(10 t))sin (10 t) mW. t di(t) 1 i(t) dt + dt C ∫0 i(t) R = 10 Ω v(t) + − L=2H 9-28.63) Determine the voltage vo (t). integrate both sides of the equation to determine the total stored energy w(t). P9. +vcc − + + −vcc + − P9-27 9-30. integrate both that i(t) = C dt sides of the equation to determine .1 dvo (t) . Knowing dvo (t) . Figure P9. R = 10 kΩ vin = 5 cos (20t) + − − + R = 20 kΩ +vcc + −vcc v0 − (a) The relationship between the input and output of the Op–Amp is given by vin = −0. dt (9. (9. The relationship between the input and output is given by C = 10 μ F R = 10 kΩ 341 C = 20 μ F v0 − Figure P9.27.62) Suppose that the initial output voltage of the Op–Amp is zero.30 An Op–Amp circuit for problem P9-30. sketch the output voltage vo (t) for one cycle. Repeat problem P9-27 if vin = 10 (1 − cos (100 t)) V.27 An Op–Amp circuit for problem P9-27. An input voltage vin = 10 sin(10 t) V is applied to an Op–Amp circuit as shown in Fig. 9-31. 9-27. (a) Suppose the initial voltage across the capacitor is vo = 10 V. P9. Knowing that dw(t) p(t) = dt . vin = 10 sin (10t) if vin = ( vo = − 2 vin + 5 ) t ∫0 vin (t) dt. An input voltage vin = 5 cos(20 t) V is applied to an Op–Amp circuit as shown in Fig. Repeat problem 10 e−10 t V. Figure P9. sketch the output voltage vo (t) for one cycle. Also. as shown in Fig.30. e. mA 10 + C = 500 μ F v(t) − i(t) ↑ 0 0 1 2 3 4 5 t. i(t) ↑ C + v(t) − 0 −50 1 2 3 4 t (ms) −100 Figure P9. 9-34. dv(t) (a) Knowing that i(t) = C dt .33 Sawtooth current applied to a capacitor in problem P9-33. i(0) = 0 A). i(t) (mA) 100 Figure P9.2 e−20 t W.32. 9-32. as shown in the Figure 9. i(t) v(t) + − L = 100 mH 9-35. the voltage is v(t) = 5 cos(5 t) V. determine the total stored energy. Integrate both sides of the equation p(t) = ddwt to determine the total stored energy w(t). .35. Evaluate the voltage at times t = 0.. A current i(t) is applied to a capacitor of C = 100 𝜇F. i(t). + i(t) ↑ C vo (t) − Figure P9. and 0.e. Sketch the voltage v(t) across dv(t) the capacitor knowing that i(t) = C dt or v(t) = C1 ∫ i(t) dt. s −10 Figure P9. For the circuit shown in Fig.32 A voltage applied to an inductor.34 Sawtooth voltage applied across an inductor. P9.34 is applied across a 100 mH inductor. and the total power is p(t) = 25 sin(10 t) W. and use your results to sketch v(t). The sawtooth current i(t) shown in Fig.5 s.342 Chapter 9 Integrals in Engineering the voltage vo (t).31 A current applied to a capacitor.1.33 is applied to a 500 𝜇F capacitor. 9. the current is i(t) = 10 sin(5 t) A.005 J. If the initial stored energy is w(0) = 0 J. v(0) is 0 V). Sketch the current i(t) passing through the inductor knowing di(t) that v(t) = L dt or i(t) = L1 ∫ v(t) dt. and that the initial stored energy is w(0) = 0. (b) Suppose the stored power is p(t) = 0. 0.35 A current applied to a 100 𝜇F capacitor. Assume that the current flowing through the inductor at t is zero (i. V 10 i(t) L =100 mH 0 0 v(t) + − 1 2 3 4 5 t. Assume that the capacitor is completely discharged at t = 0 (i. 50 9-33. 9. v(t). ms −10 Figure P9.2.. t w(t) = w(0) + ∫0 p(t) dt. and plot one cycle of w(t). 0. The sawtooth voltage v(t) shown in Fig. sketch the voltage across the capacitor v(t). p is the momentum. m/s 20 t. and crash test dummy is 1200 kg. as shown in Fig. . Find the total impulse after 20 ms. I = Δ p = pf − pi = mvimpact − mv0 . Also note that the time t is measured in ms and the current i(t) is measured in mA. (a) Knowing that v(t) = v(0) + t ∫0 a(t)dt. m/s 20 20 cos(62. 9-36. (b) What is the impact velocity vimpact of the drop cage if it takes 20 ms to hit the ground? (c) The total impulse I is equal to the change in momentum. s Figure P9. vimpact is the final velocity. sketch the power p(t) = v(t)i(t). as shown in Fig. where m is the mass of the system. A biomedical engineer measures the velocity profiles of a belted and unbelted occupant during a 45 mph (≈ 20 m/s) frontal collison. P9. Assume the drop cage starts from rest at t = 0 s. find and plot the displacement x(t) of the belted occupant for time 0 to 40 ms. (c) Based on the results of parts (a) and (b).5 π t) 0 0 0. P9.5 π t) 0 0 0.e. x(0) = 0 m).36.04 t. vo is the initial velocity.04 20 cos(12. find and plot the velocity v(t) of the drop cage.37. i. Assume that the total mass of the drop cage.37 Velocities of the belted and unbelted Figure P9. 343 9-37. (b) Given results of part (a).. Assume that the initial displacement at t = 0 is 0 m (i. Assume x(0) = 0 m.36 Energy-absorbing aviation seat. occupants during frontal collison.Problems Assume v(0) = 0 V.. and I is the impulse. how much farther did the unbelted occupant travel compared to the belted occupant? Belted Occupant: +X Unbelted Occupant: +X v(t). A biomedical engineer is evaluating an energy-absorbing aviation seat on a vertical deceleration tower. seat. (b) Find and plot the displacement x(t) of the unbelted occupant for time 0 to 40 ms. s 0.008 v(t). t (a) Knowing that x(t) = x(0) + ∫0 v(t) dt.e. The acceleration profile of the drop cage is given by a(t) = 400 sin(50 𝜋t) m/s2 . Figure P9. The overhang is constructed from two metal face plate separated by a nonmetallic core of thickness h.344 Chapter 9 Integrals in Engineering 9-38. .39 Parabolic snow loading of sandwich-constructed building overhang. A civil engineer designs a building overhang to withstand a triangular snow ) per unit length p(x) = ( loading x .38. This sandwich beam construction deforms primarily due to shear deformation with deflection y(x) satisfying ] x [ x 1 p(x) dx − V dx. P9. as shown in Fig. (a) Show that the deflection of the sandwich beam is given by { [ ( )]} p̂ x L x 1 x 1− −x − . y(x) = ∫0 h G ∫0 h Core Lower face plate (b) Sandwich construction. po 1 − L (a) Compute the resulting force V = L ∫0 p(x)dx. as shown in Fig. y(x)= − hG 3 2 3L 4L (b) Find the location and value of maximum deflection y(x). P9. Figure P9. (c) Locate the position of the centroid L ∫o x p(x) dx M x̄ = L = . p(x) pˆ V p(x) po V V M V M L x L x (a) Cantilever overhang. p(x) = p̂ 1 − Lx G is the shear modulus. 9-39. A building overhang is subjected to a parabolic snow loading p(x).38 Triangular snow loading on a building Upper face plate overhang. (b) Compute the corresponding L moment M = ∫0 x p(x)dx. V ∫0 p(x) dx )2 ( and where V = p̂ L3 .39. A h(t) Volume. V K Qout = K h(t) Figure 10.1 INTRODUCTION: THE LEAKING BUCKET Consider a bucket of cross-sectional area A being filled with water at a volume flow rate Qin . circuits.and second-order linear differential equations with constant coefficients. 345 . and dynamics.1 A leaking bucket with a small hole. These are the most common type of differential equations found in undergraduate engineering classes. 10. A differential equation relates an output variable and its derivatives to an input variable or forcing function. There are several different types of differential equations. This chapter discusses first. 10. the rate of change of the volume is given by dh(t) dV =A . If h(t) is the height and V = A h(t) is the volume of water in the bucket.Differential Equations in Engineering CHAPTER 10 The objective of this chapter is to familiarize engineering students with the solution of differential equations (DEQ) as needed for first.1.and second-year engineering courses such as physics. as shown in Fig.1) Qin Area. dt dt (10. 4) Equation (10. y(t). dt (10. y(t) = dy(t) ̈ . velocity dt . functions of y. dt2 equation (10. equation dt2 .2 DIFFERENTIAL EQUATIONS An nth order linear differential equation relating an output variable y(t) and its derivatives to some input function f (t) can be written as An dn y(t) dn−1 y(t) dy(t) + A + … + A1 + A0 y(t) = f (t) n−1 n n−1 dt dt dt (10. the volume of water in the bucket is given by dV = Qin − Qout . or function of t. dt or (10. as well as the properties of the fluid.1) and (10. in other words.2) into equation (10.2) where K is a constant.5) takes the form A2 d2 y(t) dt2 + A1 dy(t) + A0 y(t) = f (t). but it is assumed here for simplicity. An−1 . a general discussion of differential equations and the solution of linear differential equations with constant coefficients is given. Qout is not a linear function of h(t). and acceleration d2 y(t) . dy(t) For a second-order system involving position y(t).346 Chapter 10 Differential Equations in Engineering Suppose the bucket has a small hole on the side through which water is leaking at a rate Qout = K h(t) (10.6) Note that engineers often use a dot notation when referring to derivatives with ̇ respect to time.3) Substituting equations (10. and so on. for example. dt (10. A0 can be constants.5) where the coefficients An . Before presenting the solution of this equation. 10. By conservation of volume.3) gives A dh(t) = Qin − K h(t) dt A dh(t) + K h(t) = Qin . In reality. The solution of the differential equation is the output variable. …. y(t) dt = d2 y(t) . The input function f (t) (also called the forcing function) represents everything on the right-hand side of the differential equation. The objective is to solve the differential equation. The constant K is an engineering design parameter that depends on the size and shape of the hole.4) is a first-order linear differential equation with constant coefficients. determine the height h(t) of the water when an input Qin and the initial condition h(0) are given. In this case. 8) where m is the mass and k is the spring constant. cn are determined later from the initial conditions of the system. (10. and so on.9). the solution can be made general by multiplying one of the roots by t. in the case of constant coefficients. Assume a transient solution of the form y(t) = cest . the differential equation given by equation (10. the solution y(t) can be obtained by following the step-by-step procedure outlined below.. A0 are functions of y or t. This makes the right-hand side of equation (10.8) is a second-order differential equation given by ̈ + k y(t) = f (t) m y(t) (10..7) In many engineering applications. The roots are the n values of s that make the characteristic equation equal to zero. Call these values s1 .6) can be written as ̈ + A1 y(t) ̇ + A0 y(t) = f (t). sn . ytran (t) (also called the Homogeneous or Complementary Solution): The transient solution is obtained using the following steps: a. so that each term will contain cest . n dt dt dtn−1 b. In many cases. c2 . dn y(t) dn−1 y(t) dy(t) + An−1 + … + A1 (10. in the case of a spring-mass system subjected to an applied force f (t). s2 . i. ….10) c. canceling the cest will result in a polynomial in s: An sn + An−1 sn−1 + ⋯ + A1 s + A0 = 0.e. and the solution y(t) must be obtained numerically (e. Since the right-hand side of equation (10. An−1 . which is known as the characteristic equation. A2 y(t) (10. the transient solution of the differential equation has the general form ytran (t) = c1 es1 t + c2 es2 t + … + cn esn t where the constants c1 . and substitute it into (10. A0 are constants (not functions of y or t). Set the forcing function f (t) = 0. 10. …. For the case of n distinct roots.9) + A0 y(t) = 0. Transient Solution. 1. exact solutions can be difficult to obtain.5) zero.3 Solution of Linear DEQ with Constant Coefficients 347 (10. For example.. the total solution for the output variable y(t) is the sum of two solutions: the transient solution and the steady-state solution. For example.9) is zero. However. dt2 = cs2 est .10. e. two of the roots are the same). equation (10. For the special case of repeated roots (i. An−1 . Solve for the roots of the above equation.3 SOLUTION OF LINEAR DEQ WITH CONSTANT COEFFICIENTS In general. …. using the differential equation solvers in MATLAB). An dy(t) d2 y(t) Note that dt = csest .g. d.e. If the coefficients An .5) is known as a linear differential equation with constant coefficients. In this case. exact solutions do not exist. the coefficients An . . …. B. This will usually have the same general form as the forcing function and its derivatives but will contain unknown constants (i. c2 .. ….e. Substitute the assumed steady-state solution yss (t) and its derivatives into the original differential equation. Example guesses are shown in Table 10.12) + K h(t) = Qin . B. where K. 4. This can usually be done by equating the coefficients of like terms on the left. If input f (t) is Assume yss (t) K Kt K t2 K sin 𝜔 t or K cos 𝜔 t A At + B A t2 + B t + C A sin 𝜔 t + B cos 𝜔 t b.1. undetermined coefficients). (10. A. Assume that the initial height of the water is zero (i. yss (t) (also called the Particular Solution): The steady-state solution can be found using the Method of Undetermined Coefficients: a. c.. Apply the initial conditions on y(t) and its derivatives.348 Chapter 10 Differential Equations in Engineering for a second-order system with s1 = s2 = s. A differential equation of order n must have exactly n initial conditions. 3. C. Find the total solution. y(t) = ytran (t) + yss . which will result in an n × n system of equations for n constants c1 . y(t): The total solution is just the sum of the transient and steady-state solutions.1. yss .).1 Assumed solutions yss (t) for common input finctions f (t).e. the transient solution is ytran (t) = c1 est + c2 t es t . dt Find the total solution of h(t) if the input Qin = B is a constant. cn . Assume (guess) the form of the steady-state solution. Solve for the unknown (undetermined) coefficients (A. Example 10-1 The Leaking Bucket Problem Consider again the leaking bucket of Section 10. A .11) 2. which satisfies the following first-order differential equation: dh(t) (10. Steady-State Solution.3 to a variety of first-order differential equations in engineering. 10.and righthand sides of equations. and C are constants.4 FIRST-ORDER DIFFERENTIAL EQUATIONS This section illustrates the application of the method described in Section 10. etc. TABLE 10. h(0) = 0). 13).19) dt According to the method of undetermined coefficients (Table 10. (10.17) A Substituting the above value of s into equation (10. (10. Since the given input is Qin = B. Thus.14) and (10.14): A htran (t) = cest .18) The constant c depends on the initial height of the water and cannot be determined until the initial condition is applied to the total solution in part (d).4 First-Order Differential Equations Solution 349 (a) Transient (Complementary or Homogeneous) Solution: Since the transient solution is the zero-input solution. the steadystate solution will have the same general form as the input and its derivatives.15) in equation (10. (b) Steady-State (Particular) Solution: The steady-state solution of a differential equation is the solution to a particular input.14). the transient solution for the leaking bucket is given by s=− K htran (t) = ce− A t (10.10.12) can be written as dh(t) + Kh(t) = B.1).16) is the characteristic equation for the leaking bucket. (10. the differential equation (10.16) Equation (10.13) dt Assume that the transient solution of the height htran (t) is of the form given by equation (10. (10. Factoring out cest gives cest (As + K) = 0 Since cest ≠ 0. A . the homogeneous differential equation of the leaking bucket is given by dh(t) + Kh(t) = 0. (10.16) for s gives K . Solving equation (10.14) The constant s is determined by substituting htran (t) and its derivative into the homogeneous differential equation (10. The derivative of htran (t) is given by dhtran (t) d = (cest ) dt dt = csest .13) yields A(csest ) + K(cest ) = 0. (10. the input on the right-hand side (RHS) of equation (10.15) Substituting equations (10.12) is set to zero. it follows that A s + K = 0. 22) (c) Total Solution: The total solution for h(t) is obtained by adding the transient and the steady-state solutions as K h(t) = c e− A t + B . The value of E can be determined by substituting hss (t) and its derivative into equation (10. K (10.25) . K which gives c=− B .24) Substituting the above value of c into equation (10.23) yields h(t) = − or h(t) = B K B − KA t B + e K K ( K ) 1 − e− A t . (10. Solving for E gives E= B .21) into equation (10. The derivative of hss (t) is dhss (t) d = (E) dt dt = 0.19). the steady-state solution of the leaking bucket subjected to a constant input Qin = B is given by hss (t) = B . (10.350 Chapter 10 Differential Equations in Engineering Since the input in this example is constant.19) gives A(0) + KE = B.20) and (10. the steady-state solution is assumed constant to be hss (t) = E. K (10.20) where E is a constant. K Therefore.23) (d) Initial Conditions: The constant c can now be obtained by substituting the initial condition h(0) = 0 into equation (10. (10. K (10.23) as K h(0) = c e− A (0) + B =0 K or c (1) + B = 0.21) Substituting equations (10. reached its steady-state value.4 First-Order Differential Equations 351 B Note that as t → ∞. in other words. K At time t = 5 A∕K s. K At time t = A K dh(t) dt = 0).632 B .368) K or ( h A K ) = 0. 10. B K = Qin .26) The plot of the height h(t) for input Qin = B is shown in Fig.9933 . the leaking bucket) can generally be written as ( t ) y(t) = steady-state solution 1 − e− 𝜏 .10. It can be seen from this figure that after t = 5𝜏 s. it takes t = A∕K s for the height to reach 63. the total solution reaches the steady-state solution.. h 5 K K Thus.2% of the steady-state value and t = 5A∕K s to reach 99. Thus. the water level has.e. The response of a first-order system (for example. the bucket continues to fill until the pressure is great enough that Qout = Qin (i. the bucket fills to a height of ( A h 5 K ) B = K ( −K A 1−e ( 5A K = B (1 − e−5 ) K = B (1 − 0. (10. h(t) → K . That value depends only on sec. for all practical purposes. . at steady state the height h(t) reaches a constant B value of K . The time t = A∕K s is known as the time constant of the response and is usually denoted by the Greek letter 𝜏.33% of the steady-state value.2.0067) K )) or ) ( B A = 0. the bucket fills to a height of ( h A K ) B = K ( −K A ( )) A K 1−e = B (1 − e−1 ) K = B (1 − 0. Physically speaking. s A K Figure 10. find the time it takes for the water to completely leak out of the bucket. A Qin = 0 Area. V K Qout = K h(t) Figure 10. The height h(t) of the water is governed by the firstorder differential equation dh(t) + Kh(t) = 0.3). 10. . (10. Example 10-2 Leaking Bucket with No Input Suppose now that Qin = 0.632 B K 0 0 A K 5 t. Also.2 Solution for h(t) for Qin = B and h(0) = 0.27) dt Determine the total solution for h(t).352 Chapter 10 Differential Equations in Engineering h(t) B K 0. and that the initial height of the water is h0 (Fig. A h(0) = h0 Volume.3 Leaking bucket with no input for Example 10-2. the bucket empties to a height of ( ) ( ) A −K A A K = h0 e h K = h0 e−1 or ( h At time t = A K ) = 0. the total solution for h(t) is K h(t) = h0 e− A t .29) is a decaying exponential function with time constant 𝜏 = A∕K. (c) Total Solution: The total solution for the height h(t) is given by h(t) = htran (t) + hss (t) K = c1 e− A t + 0 or K h(t) = c1 e− A t .28) as K h(0) = c1 e− A (0) = h0 or c1 (1) = h0 . (10. which gives c1 = h0 . the input is zero). the bucket empties to a height of K ( ) ) ( 5A −K 5A A K = h0 e h K = h0 e−5 = 0.e. At time t = A∕K s.368 h0 . (b) Steady-State Solution: Since the RHS of the differential equation (10.27) is zero (i.10. 5A s.0067 h0 . Thus.28) (d) Initial Conditions: The constant c1 is determined by substituting the initial height h(0) = h0 into equation (10.. the steady-state solution is also zero: hss (t) = 0.29) The height h(t) given in equation (10. (10.4 First-Order Differential Equations Solution 353 (a) Transient Solution: The transient solution is identical to that of the previous example and is given by K htran (t) = c1 e− A t . v(0) = 0).5 RC circuit with constant input for Example 10-3. 10.e.8% of the initial value in one time constant 𝜏 = A∕K. It can be seen from this figure that the height starts from the initial value h0 and decays to 36. and is approximately zero after five time constants. Example 10-3 Voltage Applied to an RC Circuit Find the voltage v(t) across the capacitor if a constant voltage source vs (t) = vs is applied to the RC circuit shown in Fig.4. Assume that the capacitor is initially completely discharged (i.4 Solution for h(t) with Qin = 0 and h(0) = h0 . R + vR(t) − vs(t) + − i(t) C + v(t) − Figure 10.5.368 h0 0 0 A K 5 t. sec A K Figure 10. 10.. The plot of the height is shown in Fig. h(t) h0 0.354 Chapter 10 Differential Equations in Engineering or ( h 5A K ) ≈ 0. . Therefore.31) is a first-order differential equation with constant coefficients. Since the resistor and capacitor are connected in series.36) which gives 1 . (10. vR (t) = RC dt .33) and assuming a solution of the form vtran (t) = cest . The derivative of vtran (t) is given by dvtran (t) d = (cest ) = csest .33). It follows that RCs + 1 = 0. (10.30) can be written as dv(t) (10.32) The goal is to solve the voltage v(t) if vs (t) = vs is constant and (a) Transient Solution: The transient solution is obtained by setting the righthand side of the differential equation equal to zero as ̇ + v(t) = 0.4 First-Order Differential Equations 355 The governing equation for v(t) follows from Kirchhoff’s voltage law (KVL). RC Substituting the above value of s into equation (10. Solution dv(t) .33) yields (10. dt Equation (10.10. dt (10.34) and (10.34) gives s=− 1 vtran (t) = ce− RC t (10.35) in equation (10. the same current flows through dv(t) dv(t) the resistor and capacitor.38) . and equation (10. RCv(t) ̇ where v(t) = v(0) = 0. Factoring out cest gives est (RCs + 1) = 0. RCv(t) (10. which gives vR (t) + v(t) = vs (t).34) The constant s is determined by substituting vtran (t) and its derivative into equation (10. (10. i(t) = C dt . dt dt Substituting equations (10.37) (10.35) RC(csest ) + (cest ) = 0. This equation can also be written as RC ̇ + v(t) = vs (t).31) + v(t) = vs (t).30) From Ohm’s law. the voltage across the resistor is given by vR (t) = Ri(t). 43) Substituting the above value of c into equation (10.41) as 1 v(t) = ce− RC t + vs ..44) Note that as t → ∞.41) (c) Total Solution: The total solution for v(t) is obtained by adding the transient and the steady-state solutions given by equations (10. (10.39) Since the input to the RC circuit is constant. RC v(t) (10.42) (d) Initial Conditions: The constant c can now be obtained by applying the initial condition as 1 v(0) = ce− RC (0) + vs = 0 or c (1) + vs = 0. ̇ + v(t) = vs .39). Solving for E yields E = vs .e. the capacitor is fully charged to a voltage equal to the . (10. the steady-state solution for the output voltage is vss (t) = vs . (b) Steady-State Solution: For vs (t) = vs . which is applied to the total solution in part (d). The value of E can be determined by substituting vss (t) and its derivative in equation (10. Solving for c gives c = −vs . (10. the steady-state solution of the output voltage v(t) is assumed to be vss (t) = E.42) yields 1 v(t) = −vs e− RC t + vs or ( 1 ) v(t) = vs 1 − e− RC t .38) and (10. v(t) → vs (i.40) where E is a constant. At steady state. Thus. (10. (10. which gives RC(0) + E = vs .356 Chapter 10 Differential Equations in Engineering The constant c depends on the initial voltage across the capacitor. the total solution reaches the steady-state solution). 368) or v = 0.10.632 vs 0 0 RC 5RC t.2% of the input voltage and at t = 5 RC s. . which is a measure of the time required for the capacitor to fully charge. V vs 0.6 The voltage across the capacitor to a constant voltage in an RC circuit of Example 10-3.4 First-Order Differential Equations 357 input voltage. Also. the voltage across the capacitor is given by ( ) 1 v(5 RC) = vs 1 − e− RC (5 RC) = vs (1 − e−5 ) = vs (1 − 0. it takes t = RC s for the voltage to reach 63. v(t). It can be seen from this figure that it takes the response approximately 5𝜏 to reach the steady state.33% of the input value. at time t = 5 RC s. The time t = 𝜏 = RC s is the time constant of the RC circuit.6. s Figure 10. to reduce the charge time of the capacitor. the voltage across the capacitor at time t = RC s is given by ( ) 1 v(RC) = vs 1 − e− RC (RC) = vs (1 − e−1 ) = vs (1 − 0. the resistance value of the resistor is reduced. Typically.632 vs . The plot of the voltage v(t) is shown in Fig. Thus.9933 vs or v ≈ vs . 10.0067) = 0. the voltage reaches 99. which is identical to the result obtained for the leaking bucket with constant Qin . While the capacitor is charging. (10. Thus. (b) Steady-State Solution: Since the input applied to the RC circuit is 10 V. (10.49) Since the input is constant. 0.45) can be written as ̇ + v(t) = 10.358 Chapter 10 Differential Equations in Engineering Example 10-4 For the circuit shown in Fig.50) . 0. the steady-state solution of the output voltage v(t) is assumed to be vss (t) = E. Assume that the initial voltage across the capacitor is zero (i.5 v(t) (10. the transient solution of the output voltage is given by vtran (t) = ce−2 t (10.45) Find the output voltage v(t) across the capacitor if the input voltage vs (t) = 10 V.5 s + 1 = 0 s = −2.5(csest ) + (cest ) = 0 cest (0. the differential equation relating the output v(t) and input vs (t) is given by ̇ + v(t) = vs (t).e. Solution (a) Transient Solution: The transient solution is obtained by setting the righthand side of the differential equation to zero as ̇ + v(t) = 0.46) and solving for s gives 0.47) Substituting the transient solution and its derivative into equation (10. v(0) = 0).46) and assuming a solution of the form vtran (t) = cest . 0.5 s + 1) = 0 0.7. 10. equation (10.5 v(t) (10.7 RC circuit with input voltage vs (t).48) where c will be obtained from the initial condition.5 v(t) (10.. R = 5 kΩ + vR(t) − vs(t) + − i(t) + v(t) − C = 100 μ F Figure 10. (10.51) (c) Total Solution: The total solution for v(t) is obtained by adding the transient and steady-state solutions given by equations (10.5 s to reach 63.49) as 0. (10.5 s to fully charge to approximately 10 V.53) Substituting the value of c from equation (10. The value of E can be determined by substituting vss (t) and its derivative into equation (10.4 First-Order Differential Equations 359 where E is a constant.5 (0) + E = 10 which gives E = 10 V. The plot of the output voltage v(t) is shown in Fig. the steady-state solution of the output voltage is given by vss (t) = 10 V.8.5 s.53) into equation (10. 10. Solving for c gives c = −10 V. (10.2% of the input voltage and approximately 5(0.51) as v(t) = ce−2 t + 10.48) and (10. .52) yields v(t) = −10 e−2 t + 10 or v(t) = 10 (1 − e−2 t ) V.5) = 2. (10.54) Since the time constant is 𝜏 = 1∕2 = 0. Thus. it takes the capacitor 0.10.52) (d) Initial Conditions: The constant c can now be obtained by applying the initial condition (v(0) = 0) as v(0) = ce−2 (0) + 10 = 0 or c (1) + 10 = 0. Example 10-5 The differential equation for the capacitive circuit shown in Fig.5 0. s Figure 10.48) as vtran (t) = ce−2 t . 10. Solution (a) Transient Solution: Since the left-hand side of the governing equation is the same as that for the previous example.360 Chapter 10 Differential Equations in Engineering v(t).32 0 0 2. 0.55) Find the output voltage v(t) across the capacitor C as it discharges from an initial voltage of v(0) = 10 V.9 is given by ̇ + v(t) = 0. V 10 6.5 t.8 The voltage across the capacitor in Example 10-4. R = 5 kΩ + vR(t) − C = 100 μ F i(t) + v(t) − v(0) = 10 V Figure 10.9 Discharging of a capacitor in an RC circuit. . the transient solution of the output voltage is given by equation (10.5 v(t) (10. the response of a capacitor discharging in an RC circuit is identical to the response of a leaking bucket with initial fluid height h0 ! . at time t = 2. The plot of the output voltage.10.5) = 10 e−2 (2.68 V. (d) Initial Conditions: The constant c can now be obtained by applying the initial condition (v(0) = 10 V) as v(0) = ce−2 (0) = 10. the voltage across the capacitor is given by v(2. the voltage across the capacitor at time t = 0. which gives c = 10 V.5 s (five time constants). Also. (c) Total Solution: The total solution for v(t) is obtained by the adding the transient and the steady-state solutions. v(t) is shown in Fig.4 First-Order Differential Equations 361 (b) Steady-State Solution: Since there is no input applied to the circuit. which gives v(t) = ce−2 t . While the capacitor is discharging. 10. Thus. Mathematically speaking.10.5) = 10 e−2 (0. the output voltage is given by v(t) = 10 e−2 t V. the steadystate value of the output voltage is zero: vss (t) = 0.5) = 10 e−5 = 0.067 or v ≈ 0.5 s (one time constant) is given by v(0.5) = 10 e−1 or v = 3. (10.5 2. Thus. Applying KVL yields R vs(t) + − + vR(t) − + vL(t) i(t) L − Figure 10. find the total solution for the current i(t).11.56) di(t) is the dt voltage across the inductor.57) If the applied voltage source is vs (t) = vs = constant.68 0 0 0.5 t. sec Figure 10. Assume the initial current is zero (i(0) = 0). vR (t) + vL (t) = vs (t). Example 10-6 Consider a voltage vs (t) applied to an RL circuit. 10. dt (10.10 The voltage across the capacitor in Example 10-5. as shown in Fig.56) can be written in terms of the current i(t) as where vR (t) = R i(t) is the voltage across the resistor and vL (t) = L L di(t) + R i(t) = vs (t). V 10 3.11 Voltage applied to an RL circuit. equation (10. .362 Chapter 10 Differential Equations in Engineering v(t). 61) The constant c depends on the initial current flowing through the circuit and is found in part (d). dt and assuming a transient solution of the form L itran (t) = cest . R E= (10.57) can be written as di(t) (10. The value of E can be determined by substituting iss (t) and its derivative into equation (10.59) The constant s is determined by substituting itran (t) and its derivative into equation (10. which gives L(0) + R (E) = vs . (10. (b) Steady-State Solution: Since the input vs (t) = vs . L Substituting the above value of s into equation (10. R Thus.59). L dt Because the voltage applied to the RL circuit is constant.62) + Ri(t) = vs . Factoring out est gives cest (Ls + R) = 0.4 First-Order Differential Equations Solution 363 (a) Transient Solution: The transient solution is obtained by setting the right-hand side of the differential equation to zero as di(t) + R i(t) = 0.62). equation (10.58) (10. Solving for E gives vs . (10. (10. the steady-state solution of the current i(t) is assumed to be iss (t) = E. which implies Ls + R = 0.64) .58). Solving for s gives R (10.63) where E is a constant. the steady-state solution of the current is given by v iss (t) = s . which gives L (csest ) + R (cest ) = 0.10. the transient solution of the output voltage is given by R itran (t) = ce− L t .60) s=− . A vs R 0. i(t) → vs ∕R (i. i(t).65) i(t) = ce− L t + s .364 Chapter 10 Differential Equations in Engineering (c) Total Solution: The total solution for the current i(t) is obtained by adding the transient and the steady-state solutions given by equations (10.632 vs R 0 0 L R 5 L R Figure 10. R or v c (1) + s = 0 R Solving for c gives v (10.65) gives R v v i(t) = − s e− L t + s R R or R v (10.12.65) as R v i(0) = ce− L (0) + s = 0. It can be seen that the current i(t) takes approximately 5𝜏 to reach the steady-state value. s . The plot of the current i(t) is shown in Fig. the steady-state solution). t.66) c = − s.2% of its steady-state value vs ∕R. as obtained for both the charging of a capacitor and the filling of a leaking bucket. R (d) Initial Conditions: The constant c can now be obtained by applying the initial condition (i(0) = 0) to equation (10.61) and (10.12 The current flowing through an RL circuit.64) as R v (10. 10.e.67) i(t) = s (1 − e− L t ) A. It takes the current t = 𝜏 = L∕R s to reach 63. R Substituting the above value of c into equation (10. R As t → ∞.. dt (10.1s + 100) = 0 0. (10. The circuit is described by the following differential equation: 0. R = 100 Ω i(t) vs(t) = 10 V + − L = 100 mH Figure 10. the transient solution of the current is given by itran (t) = ce−1000 t .13 RL circuit for Example 10-7.68) Find the current i(t) if the initial current is 50 mA (i. (10. dt (10.1 di(t) + 100 i(t) = 0. 10.1 (csest ) + 100 (cest ) = 0 cest (0.1 di(t) + 100 i(t) = 10. Solution (a) Transient Solution: The transient solution is obtained by setting the righthand side of the differential equation to zero as 0. (10. (b) Steady-State Solution: Because the voltage applied to the RL circuit is constant (the right-hand side of equation (10.71) Substituting the value of s into equation (10.72) The constant c depends on the initial current flowing through the circuit and will be obtained by applying the initial condition to the total solution in part (d)..68) is constant). (10.10.4 First-Order Differential Equations Example 10-7 365 A constant voltage vs (t) = 10 V is applied to the RL circuit shown in Fig.70) The constant s is determined by substituting itran (t) and its derivative into equation (10.69) and solving for s as 0.73) . the steady-state solution is assumed to be iss (t) = E.69) and assuming a solution of the form itran (t) = cest .1s + 100 = 0 s = −1000.13.e.70). i(0) = 50 × 10−3 ). 366 Chapter 10 Differential Equations in Engineering where E is a constant. (10.14.75) (d) Initial Conditions: The constant c can now be obtained by applying the initial condition i(0) = 50 mA into equation (10. (10. the current reaches its steady-state solution).0816 A or 81.1 A.5 e−1000 t A.05 e−1000 t + 0..72) and (10. It takes the current t = 𝜏 = 1∕1000 = 1 ms to reach 0.75) gives i(t) = −0.1 or ) ( i(t) = 0. Solving for c gives c = −0.1(0) + 100 (E) = 10.632 × (Steady-state value − Initial value) or 50 + 0.75) as i(0) = ce−1000 (0) + 0.1 = 100 mA (i.76) Substituting the value of c into equation (10. . which gives 0. (10. i(t) → 0.1(1 − 0.e. the steady-state solution for the current is given by iss (t) = 0.74) as i(t) = ce−1000 t + 0.1 = 0. Solving for E gives E = 0.68).6 mA.368) = 0.05. (10. The value of E can be determined by substituting iss (t) and its derivative into equation (10.05 or c (1) + 0.1 A.74) (c) Total Solution: The total solution is obtained by adding the transient and the steady-state solutions given by equations (10.1 A.77) Note that as t → ∞.1 = 0.6 mA. It can be seen from this figure that the current i(t) takes approximately 5𝜏 = 5 ms to reach the final value. The value of the current at t = 𝜏 can also be found from the expression: Initial value + 0. Thus.5 × 0. 10. The plot of the current i(t) is shown in Fig.05.1 1 − 0.632 × (100 − 50) = 81. 14 The current flowing through the RL circuit of Example 10-7. and C = 0.78) dt RC C where Q̇ in is the volumetric blood flow. 10. ms 5 Figure 10.6 50 0 1 0 t. Also. Q˙in P Artery Figure 10.15 Windkessel model. and C is cm3 . the arterial pressure P(t) satisfies the following first-order differential equation: Q̇ dP(t) 1 + P(t) = in . (10. If the volumetric blood flow Q̇ in is 80 s (a) Find the transient solution Ptran (t) for the arterial pressure.5 7 mmHg.15 to investigate the relationship between arterial blood flow and blood pressure in a single artery. In this model. and sketch the solution of P(t) for 0 ≤ t ≤ 5 𝜏. (b) Determine the steady-state solution Pss (t) for the arterial pressure. The unit for P(t) is mmHg. arterial compliance.10. mA 100 81. Example 10-8 A biomedical engineering graduate student uses the Windkessel model shown in Fig. Q˙out . (c) Determine the total solution P(t) assuming that the initial arterial pressure is mmHg cm3 . assume R = 5 mmHg (cm3 ∕s) (d) Evaluate P(t) after one time constant 𝜏. R is the peripheral resistance.4 First-Order Differential Equations 367 i(t). (10.80) The constant s is determined by substituting Ptran (t) and its derivative into equation (10. The value of K can be determined by substituting Pss (t) and its derivative into equation (10. (10. (10.84) . Thus. RC C Solving for K gives K = 80 R mmHg.368 Chapter 10 Differential Equations in Engineering Solution (a) Transient Solution: The transient solution is obtained by setting the righthand side of the differential equation (10.79) and solving for s as 1 (cest ) = 0 RC (csest ) + cest (s + 1 )=0 RC s+ 1 =0 RC 1 .78) is constant). which gives (0) + Q̇ 1 (K) = .83) where K is a constant.78) to zero as dP(t) 1 + P(t) = 0 dt RC (10. (10. the transient solution of the arterial pressure is given by s=− 1 Ptran (t) = ce− RC t .82) The constant c depends on the initial pressure of the blood flowing through the artery and will be obtained by applying the initial condition to the total solution in part (d).78). (b) Steady-State Solution: Because the volumetric blood flow in the artery is constant (the right-hand side of equation (10. the steady-state solution is assumed to be Pss (t) = K (10.81) RC Substituting the value of s from equation into equation (10.79) and assuming a solution of the form Ptran (t) = cest . the steady-state solution for the arterial pressure is given by Pss (t) = 80 R mmHg.80). 10.84) as 1 P(t) = ce− RC t + 80 R mmHg.4 t + 400 or ) ( P(t) = 400 1 − 0.5 s to reach 400(1 − 0. s . 10 12.5 s to reach its final value..85) (d) Initial Conditions: The constant c can now be obtained by applying the initial condition P(0) = 7 mmHg into equation (10.82) and (10.16.87) Note that as t → ∞. Solving for c gives c = −393.16 The blood pressure in a single artery.9825 e−0. (10.4 mmHg. The plot of the current P(t) is shown in Fig.632 × (Steady-state value − Initial value) or 7 + 0.5 gives c (1) + 400 = 7.4 mmHg.9825(0.4 t mmHg.4 First-Order Differential Equations 369 (c) Total Solution: The total solution is obtained by adding the transient and the steady-state solutions given by equations (10.86) Substituting the value of c into equation (10.5 Figure 10. It can be seen from this figure that it takes the pressure P(t) approximately 5𝜏 = 12. (10.4 = 2. It takes the pressure t = 𝜏 = 1∕0.632 × (400 − 7) = 255. P(t). mmHg 400 255.10.85) gives P(t) = −393 e−0.85) as 1 P(0) = ce− RC (0) + 80 R = 7. P(t) → 400 mmHg (i.368)) = 255. (10. Substituting R = 5 and C = 0.5 t. the pressure reaches its steady-state solution).e. The value of the pressure at t = 𝜏 can also be found from the expression: Initial value + 0.5 5 7.4 0 0 2. RC v(t) (10. which gives v̇ ss (t) = A𝜔 cos 𝜔 t − B 𝜔 sin 𝜔 t.94) Comparing the coefficients of sin 𝜔 t on both sides of equation (10.94) gives −RCB𝜔 + A = V.93) yields (−RCB𝜔 + A) sin 𝜔 t + (RCA𝜔 + B) cos 𝜔 t = V sin 𝜔 t.1. (10.95) Similarly. (b) Steady-State Solution: Since vs (t) = V sin 𝜔 t.89) The constant c will be determined by applying the initial condition to the total solution in part (d).92) into equation (10.88) Find the output voltage v(t) if vs (t) = V sin 𝜔 t and the initial voltage is zero (i. (10.92) Substituting equations (10.5 is given by RC dv(t) + v(t) = vs (t).94) gives RCA𝜔 + B = 0.96) . Solution (a) Transient Solution: The transient solution vtran (t) for the differential equation (10. (10..93) Grouping like terms in equation (10.88) is the same as that found in Example 10-3 and is given by 1 vtran (t) = ce− RC t .90). (10.91) where A and B are constants to be determined. the steady-state solution of the output voltage v(t) has the form vss (t) = A sin 𝜔 t + B cos 𝜔 t (10.90) gives RC(A𝜔 cos 𝜔 t − B 𝜔 sin 𝜔 t) + A sin 𝜔 t + B cos 𝜔 t = V sin 𝜔 t.88) can be written as ̇ + v(t) = V sin 𝜔 t.91).90) According to Table 10. The derivative of vss (t) is obtained by differentiating equation (10. (10. equation (10. The values of A and B can be found by substituting vss (t) and its derivative into equation (10. v(0) = 0).370 Chapter 10 Differential Equations in Engineering Example 10-9 The differential equation for the RC circuit of Fig. dt (10.91) and (10. 10.e. comparing the coefficients of cos 𝜔 t on both sides of equation (10. (10. 99) As discussed in Chapter 6. 1 + (RC𝜔)2 (10.97) A= (10. These equations can be solved using one of the methods discussed in Chapter 7 and are given by V 1 + (RC𝜔)2 −RC𝜔 V B= .102) v(t) = ce 1 + (RC𝜔)2 (d) Initial Conditions: The constant c can now be obtained by applying the initial condition v(0) = 0 to equation (10.102) as ) ( V sin 𝜙 = 0 v(0) = c (1) + √ 1 + (RC𝜔)2 or ( c=− ) V √ 1 + (RC𝜔)2 sin 𝜙.95) and (10.92) yields ( ) ( ) V −RC𝜔V sin 𝜔 t + cos 𝜔 t vss (t) = 1 + (RC𝜔)2 1 + (RC𝜔)2 or vss (t) = V (sin 𝜔 t − RC𝜔 cos 𝜔 t). (10.10.96) represent a 2 × 2 system of equations for the two unknowns A and B.98) Substituting A and B from equations (10.89) and (10. Substituting equation (10.4 First-Order Differential Equations 371 Equations (10.97) and (10. (10. (10.101) (c) Total Solution: The total solution is obtained by adding the transient and the steady-state solutions given by equations (10. 1) = −tan−1 (RC𝜔).100) where 𝜙 = atan2(−RC𝜔. (10.101) as ( ) 1 V − RC t + √ sin (𝜔 t + 𝜙).103) .100) into equation (10. summing sinusoids of the same frequency gives √ sin 𝜔 t − RC𝜔 cos 𝜔 t = 1 + (RC𝜔)2 sin(𝜔 t + 𝜙).99) gives the steady-state solution as ( ) ) (√ V 2 1 + (RC𝜔) sin (𝜔 t + 𝜙) vss (t) = 1 + (RC𝜔)2 or ( vss (t) = ) V √ 1 + (RC𝜔)2 sin (𝜔 t + 𝜙). 1 + (RC𝜔)2 (10.98) into equation (10. Thus. (10.106) Note that as t → ∞.107) |v(t)| = √ 1 + (RC𝜔)2 The amplitude of the input voltage vs (t) = V sin(𝜔 t) is given by |vs (t)| = V. (10.17 as −RC𝜔 sin 𝜙 = √ 1 + (RC𝜔)2 (10.104) y 1 x ϕ √1⎯ ⎯+ ⎯⎯(⎯ R⎯C⎯ω ⎯⎯)⎯2 ⎯ −RCω Figure 10. (10. the value of sin 𝜙 can be found from the fourthquadrant triangle shown in Fig. Substituting sin 𝜙 from equation (10. 2 1 + (RC𝜔) 1 + (RC𝜔)2 (10.372 Chapter 10 Differential Equations in Engineering Since 𝜙 = −tan−1 (RC𝜔).105) 1 + (RC𝜔)2 Substituting the value of c from equation (10.104) into equation (10.108) Dividing the amplitude of the output at steady-state (equation (10.17 Fourth-quadrant triangle to find sin 𝜙.108)) gives V √ 1 + (RC𝜔)2 |v(t)| = V |vs (t)| or |v(t)| 1 . (10.105) into equation (10.107)) by the amplitude of the input (equation (10. the total solution reaches the steady-state solution. the amplitude of the output voltage as t → ∞ is given by V . 10.103) yields ( )( ) V −RC𝜔 c=− √ √ 1 + (RC𝜔)2 1 + (RC𝜔)2 or RC𝜔V c= .102) gives v(t) = 1 RC𝜔V V e− RC t + √ sin (𝜔 t + 𝜙).109) =√ |vs (t)| 1 + (RC𝜔)2 . 1 rad/s and 10 rad/s.5 𝜔).1 and 10 rad/s are shown in Figs.9988. the ratio is given by |v(t)| 1 = 0.5 and V = 10 V.19. the amplitude of the output is close to zero. |vs (t)| This means that for high-frequency input. =√ |vs (t)| 1 + (0. respectively.10. as 𝜔 → ∞. However. |v(t)| → 0.1)2 For 𝜔 = 10 rad/s. This will be further illustrated in Example 10-10. the ratio (10.988 V to 10 ∗ (0. 10.1961.106) as v(t) = 10 5𝜔 e−2 t + √ sin (𝜔 t − tan−1 0.1961) = .5𝜔)2 (b) For 𝜔 = 0.5𝜔)2 1 + (0. =√ |vs (t)| 1 + (0. 10.109) as |vs (t)| |v(t)| 1 = 0. |vs (t)| This means that for low-frequency input.5 ∗ 0. |v(t)| as 𝜔 → ∞. the amplitude of the output is about the same as the input. because it passes the low-frequency inputs but filters out the high-frequency inputs.5 and V = 10 into equation (10.1 rad/s.1 to 10 rad/s. Solution (a) The total solution for the output voltage is obtained by substituting RC = 0. (a) Find the total solution v(t). The RC circuit shown in Fig. |vs (t)| The plots of the output for 𝜔 = 0.110) |v(t)| can be found from equation (10. 1 + (0. the ratio |v(t)| → 0.9988) = 9.5 ∗ 10)2 As 𝜔 → ∞. |v(t)| → 1. It can be seen that as 𝜔 increases from 0.4 First-Order Differential Equations 373 Note that as 𝜔 → 0. Also plot the steady-state output for both (b) Find the ratio |vs (t)| 𝜔 = 0.5 is known as a low-pass filter.18 and 10. the amplitude of the steady-state output decreases from 10 ∗ (0. Example 10-10 Consider the low-pass filter of the previous example with RC = 0. 10. and y(t) is the position measured from equilibrium. 10. as shown in Fig.83.75 5 −2 Figure 10.25 2.374 Chapter 10 Differential Equations in Engineering 1. the external forces on the block are shown in the free-body diagram (FBD) of Fig. From equilibrium of forces in the y-direction.21.109) that if 𝜔 → ∞.5. m is the mass.5 SECOND-ORDER DIFFERENTIAL EQUATIONS 10.961 V. .18 Output voltage for 𝜔 = 0. V 3 2 t. v(t). and k 𝛿 is the restoring force in the spring.49 −10 Figure 10.66 t. V 10 0 62. 0 125. s 0 0 1. v(t).20.5 3. where 𝛿 is the equilibrium elongation of the spring. the amplitude of the output will approach zero. mg is the force due to gravity.1 Free Vibration of a Spring-Mass System Consider a spring-mass system in the vertical plane. s 188. k 𝛿 = mg. At the equilibrium position. 10.1 rad/s. It can be seen from equation (10.19 Output voltage for 𝜔 = 10 rad/s. where k is the spring constant. 111) This equilibrium elongation 𝛿 is also called the static deflection.10.5 Second-Order Differential Equations 375 k g m y(t) Figure 10. the FBD of the system is as shown in Fig.22 gives ̈ mg − k(𝛿 + y(t)) = m y(t) . kδ m mg Figure 10.20 Mass-spring system.22 Free-body diagram of the mass-spring system displaced from equilibrium. Since the system is no longer in equilibrium. k (10. 10.21 Free-body diagram of the mass-spring system with no motion. Summing the forces in Fig.22. law ( ∑ F = m a) can be used to write the equation of motion as ∑ Fy = m a = m ÿ (t). Now. which gives 𝛿= mg . Newton’s second k (δ + y(t)) m y(t) mg Figure 10. 10. if the mass is displaced from its equilibrium position by the amount y(t). Solving for s yields k m s2 = − which gives √ s=± − k m (10.112) Substituting 𝛿 from equation (10. (a) Transient Solution: Since the right-hand side of the equation is zero.111) gives ( mg ) ̈ − k y(t) = m y(t) mg − k k or ̈ −k y(t) = m y(t). m y(t) (10.20. Substituting the transient solution and its derivatives into equation (10.114) . Example 10-11 Solution Find the solution of equation (10.113) is a second-order differential equation for the displacement y(t) of the spring-mass system shown in Fig. 10.113) Equation (10. (10. The first and second derivatives of the transient solution are given by ẏ tran (t) = csest ÿ tran (t) = cs2 est . Factoring out est gives cest (ms2 + k) = 0 which implies that ms2 + k = 0. assume a transient solution of the form ytran (t) = cest .113) yields m(cs2 est ) + k(cest ) = 0. Note that the initial velocity is zero (y(0) = 0).113) if the mass is subjected to an initial displacė ment of y(0) = A and let go.376 Chapter 10 Differential Equations in Engineering or ̈ mg − k 𝛿 − k y(t) = m y(t). which gives ̈ + k y(t) = 0. 117) where c3 = c1 + c2 and c4 = j (c1 − c2 ) are real constants. equation (10. Therefore.116) can be written as ) ( √ √ k k ytran (t) = c1 cos t + j sin t + m m ) ( √ √ k k t − j sin t c2 cos m m or √ ytran (t) = (c1 + c2 ) cos This can be further simplified as k t + j (c1 − c2 ) sin m √ ytran (t) = c3 cos k t + c4 sin m √ k t m √ k t. Note that the constants c1 and c2 must be complex conjugates for ytran (t) to be real. m (10.10. yss (t) = 0. equa- where c1 and c2 tion (10. Using Euler’s formula ej 𝜃 = cos 𝜃 + j sin 𝜃.116) Since cos(−𝜃) = cos(𝜃) and sin(−𝜃) = −sin(𝜃). for example. (b) Steady-State Solution: Since the RHS of equation (10.114) are thus s1 = + j k m and s2 = − j k . the transient solution is given by ytran (t) = c1 es1 t + c2 es2 t or √ k m √ (10. m Therefore. The two √ roots of the characteristic equation (10. the steady-state solution is zero.5 Second-Order Differential Equations or √ s=0 ± j 377 k m √ where j √ = −1.118) .115) are constants.113) is already zero (no forcing function). (10. the transient solution of a mass-spring system √ can be written in terms of sines and cosines with natural frequency 𝜔n = k m rad/s. (10.115) can be written as ( √ ytran (t) = c1 k t + j sin cos m [ ( √ c2 cos − t −j k m t ytran (t) = c1 e j + c2 e √ ) k t m ) k t m + ( √ + j sin − )] k t m . the total solution for the displacement is given by √ k y(t) = A cos t m or y(t) = A cos 𝜔n t. Thus. (10.122) The constant c4 can now be found by substituting y(0) as √ √ k k ̇ sin(0) + c4 cos(0) = 0 y(0) = −A m m or (√ ) k = 0.117) and (10. . Thus. the displacement of the mass is given by √ √ k k y(t) = A cos t + c4 sin t. which gives c3 = A.122) y(t) = −A m m m m ̇ = 0 in equation (10. which gives √ √ k k t + c4 sin t.121) as √ √ √ √ k k k k ̇ sin t + c4 cos t. (10.121) The velocity of the mass can be found by differentiating y(t) in equation (10. −A (0) + c4 m which gives c4 = 0. Substituting y(0) = A in equation (10.120) or c3 (1) + c4 (0) = A.119) y(t) = c3 cos m m (d) Initial Conditions: The constants c3 and c4 are determined from using the ̇ initial conditions y(0) = A and y(0) = 0.119) gives y(0) = c3 cos(0) + c4 sin(0) = A (10.378 Chapter 10 Differential Equations in Engineering (c) Total Solution: The total solution for the displacement y(t) can be found by adding the transient and steady-state solutions from equations (10. m m (10.118). 5. .23 Displacement of the spring for Example 10-11.23.10. 10.5 Second-Order Differential Equations 379 The plot of the displacement y(t) is shown in Fig.24 Spring-mass system subjected to applied force. This is a general result for free vibration of mechanical systems. s −A Figure 10.. 10. y(t) ymax=A 0 π ωn 0 2π ωn t. and the k . 10. Note that the natural frequency is block oscillates at a frequency of 𝜔n = m proportional to the square root of the spring constant and is inversely proportional to the square root of the mass (i.2 Forced Vibration of a Spring-Mass System Suppose the spring-mass system is subjected to an applied force f (t). It can be seen that the amplitude of the displacement is simply √ the initial displacement A.24. as shown in Fig. the natural frequency increases with stiffness and decreases with mass). k g m y(t) f(t) Figure 10.e. Grouping like terms yields A(k − m𝜔2 ) sin𝜔 t + B(k − m𝜔2 ) cos𝜔 t = F cos(𝜔 t). √ if f (t) = F cos 𝜔 t and y(0) = y(0) investigate the response as 𝜔 → k .124) ytran (t) = c3 cos m m where c3 and c4 are real constants to be determined.123) gives m(−A 𝜔2 sin 𝜔 t − B 𝜔2 cos 𝜔 t) + k(A sin 𝜔 t + B cos 𝜔 t) = F cos 𝜔 t.380 Chapter 10 Differential Equations in Engineering In this case. the equation of motion of the system can be written as m ÿ (t) + k y(t) = f (t). and f (t) = F cos(𝜔 t) into equation (10. (10.123) is a second-order differential equation for the displacement y(t) of a mass-spring system subjected to a force f (t).113) for free vibration. (10.126) Substituting ÿ ss (t). (10. which gives ( A=0 √ provided 𝜔 ≠ ) k m . the derivation of the governing equation includes an additional force f (t) on the right-hand side. which gives ̈ + k y(t) = 0. (b) Steady-State Solution: Since the forcing function is f (t) = F cos 𝜔 t. Example 10-12 Solution ̇ Find the solution to equation (10.127) Equating the coefficients of sin𝜔 t on both sides of equation (10.117) as √ √ k k t + c4 sin t. Also.123) Equation (10.123) = 0. m y(t) This is the same as equation (10. (10.125) The first and second derivatives of the steady-state solution are thus ẏ ss (t) = A 𝜔 cos 𝜔 t − B 𝜔 sin 𝜔 t ÿ ss (t) = −A 𝜔2 sin 𝜔 t − B 𝜔2 cos 𝜔 t. Hence. the steadystate solution is of the form yss (t) = A sin 𝜔 t + B cos 𝜔 t. Thus. . yss (t).127) yields A(k − m𝜔2 ) = 0. m (a) Transient Solution: The transient solution is obtained by setting f (t) = 0. the transient solution is given by equation (10. (10. (10. (10. k − m𝜔2 which gives F .128) (c) Total Solution: The total solution for y(t) is obtained by adding the transient and steady-state solutions from equations (10.124) and (10. k − m𝜔2 (10.129) y(t) = c3 cos m m k − m𝜔2 (d) Initial Conditions: The constants c3 and c4 are determined from the initial coṅ ditions y(0) = 0 and y(0) = 0. The velocity of the mass can be obtained by differentiating equation (10. substituting y(0) = 0 in equation (10.130) yields √ ( ) k F ̇ cos(0) − 𝜔 sin(0) = 0 y(0) = −c3 (0) + c4 m k − m𝜔2 or √ ( ) k F c3 (0) + c4 (1) − 𝜔 (0) = 0 m k − m𝜔2 c3 = − .130) 𝜔 k − m𝜔2 Substituting y(0) = 0 in equation (10.129) as √ √ √ √ k k k k ̇ sin t + c4 cos t− y(t) = −c3 m m m m ) ( F sin 𝜔 t.5 Second-Order Differential Equations 381 Similarly.10. equating the coefficients of cos𝜔 t on both sides of equation (10. Therefore.127) yields B(k − m𝜔2 ) = F. which gives ( F B= k − m𝜔2 √ provided 𝜔 ≠ ) k m .129) gives ( ) F cos(0) = 0 y(0) = c3 cos(0) + c4 sin(0) + k − m𝜔2 or ( ) F c3 (1) + c4 (0) + (1) = 0. the steady-state solution is given by ( ) F yss (t) = cos 𝜔 t.128) as √ √ ( ) k k F t + c4 sin t+ cos 𝜔 t. k − m𝜔2 ̇ Similarly. m m ) ( √ √ ( ) F k k t − cos m t cos m y(t) → 0 0 = . √ k .27. However.25.132) But nevertheless we √ √ k k What is the response of y(t) as 𝜔 → ? As 𝜔 → . 0 This is an “indeterminate” form and can be evaluated by methods of calculus not yet available to all students. y(t) 15 10 5 0 −5 −10 −15 0 20 40 60 80 100 120 140 160 180 200 Figure 10. the result can be investigated by picking √ k m values of 𝜔 close to and plotting the results. Thus. let k = m = F = 1. and 10.9999 m .9 √ k . For example. 𝜔 = 0. sec .25 Displacement of the mass for 𝜔 = 0. cos 𝜔 t − cos y(t) = m k − m𝜔2 Note that the results obtained above assumed that 𝜔 ≠ √ k can investigate the behavior as 𝜔 gets very close to m .99 m . 10.9 m . the displacement of the mass is given by √ ( ( ) ) F k F y(t) = − t+ cos cos 𝜔 t m k − m𝜔2 k − m𝜔2 or ) √ ) ( ( k F t .131) (10.132) for these values are as shown in Figs. √ √ √ k k k and choose the values of 𝜔 = 0. m (10.26. m t. 10.382 Chapter 10 Differential Equations in Engineering which gives c4 = 0. and 𝜔 = 0. The plots of equation (10. 10. Figs. As 𝜔 is increased to 0.10. m t.99 √ t. m √ k respectively.5 Second-Order Differential Equations 383 y(t) 100 80 60 40 20 0 −20 −40 −60 −80 −100 0 20 40 60 80 100 120 140 160 180 200 Figure 10. s k .9 m in Fig. 10.25 shows the “beating” phenomenon typical of problems of the nat√ where the forcing frequency√𝜔 is in the neighborhood √ k k k ural frequency m .27 show y(t) increasing without bound.9999 √ k .26 and 10.9999 m . This is called resonance and is generally undesirable in mechanical systems. The plot for 𝜔 = 0. y(t) 100 80 60 40 20 0 −20 −40 −60 −80 −100 0 20 40 60 80 100 120 140 160 180 200 Figure 10.26 Displacement of the mass for 𝜔 = 0. sec .99 m and 0.27 Displacement of the spring for 𝜔 = 0. (b) Determine the steady-state solution yss (t) for the applied force shown in Fig. 10. (c) Determine the total solution.28 Spring-mass model of resistive training device. subject to the initial conditions.29. (a) Determine the transient solution ytran (t).384 Chapter 10 Differential Equations in Engineering Example 10-13 A biomedical engineer is designing a resistive training device to strengthen the latissimus dorsi muscle. The task can be represented as a spring-mass system. The displacement y(t) of the exercise bar satisfies the secondorder differential equation ̈ + k y(t) = f (t) m y(t) (10. 10. Solution (a) Transient Solution: The transient solution is obtained by setting the RHS of equation (10.134) . f(t) H 0 1 2 t Figure 10.133) equal to zero as m d2 y(t) dt2 + k y(t) = 0 and assuming a solution of the form ytran (t) = est .29 Applied force for resistive training device.133) ̇ subject to the initial condition y(0) = E and y(0) = 0. (10.28. k m f(t) Figure 10. as shown in Fig. (10.5 Second-Order Differential Equations 385 Substituting the transient solution and its second derivative into equation (10.135) are s1 = + j m transient solution is given by √ √ k k ytran (t) = c3 cos t + c4 sin t. the The two roots of equation (10.135) Solving for s yields s2 = − or k m √ s = ±j k . Factoring out est gives est (m s2 + k) = 0.134) yields m(s2 est ) + k (est ) = 0.137) Substituting the ÿ ss (t) and yss (t) into equation (10. 2k Similarly.138) yields kA = H 2 which gives A= H . where c3 and c4 are real constants and m (b) Steady-State Solution: Since the forcing function is f (t) = H t∕2. (10. Thus. (10. .138) Equating the coefficients of t on both sides of equation (10.10.138) yields B = 0.133) gives m × 0 + k (A t + B) = H t. which implies m s2 + k = 0. 2 (10. equating the constant coefficients on both sides of equation (10. the steadystate solution is of the form yss (t) = A t + B.136) m m √ k is the natural frequency 𝜔n . m √ √ k k and s2 = − j m . 3 Second-Order LC Circuit A source voltage vs (t) is applied to an LC circuit. √ 2k k∕m Thus.139) (c) Total Solution: The total solution of the displacement y(t) can be found by adding the transient and steady-state solutions from equations (10.136) and (10.386 Chapter 10 Differential Equations in Engineering Thus. which gives c3 = E.140) t + c4 sin t+ y(t) = c3 cos m m 2k The constants c3 and c4 are determined using the initial conditions y(0) = E ̇ and y(0) = 0. which gives √ √ ( ) k k H t. (10.141) m m m m 2k ̇ Substituting y(0) = 0 in equation (10. Substituting y(0) = E into equation (10. the steady-state solution is given by ( ) H t. as shown in Fig. The derivative of y(t) is obtained by differentiating equation (10.5. 10.141) yields √ ( ) k H 0 = −c3 (0) + c4 + m 2k which gives c4 = − H . sin √ m m 2k 2k k∕m 10. the displacement of the exercise bar is given by ( ) √ √ k k H H y(t) = E cos t− t+ t.142) .139).140) as √ √ √ √ k k k k H ̇ y(t) = −c3 sin t + c4 cos t+ (10.30. Applying KVL to the circuit gives vL (t) + v(t) = vs (10.140) yields ( ) H (0) = E y(0) = c3 cos(0) + c4 sin(0) + 2k or c3 (1) + c4 (0) + 0 = E. yss (t) = 2k (10. (10.30 is subjected to a voltage source vs (t) = V cos 𝜔 t. Factoring out est gives est (LC s2 + 1) = 0. the condt (a) Transient Solution: The transient solution is the solution obtained by setting the RHS of equation (10.143) is a second-order differential equation for an LC circuit subjected to forcing function vs (t). (10. 10. LC Example 10-14 Suppose the LC circuit of Fig. where vL (t) = L di(t) dt is the voltage across the inductor. which implies LC s2 + 1 = 0. Solve the resulting differential equation ̈ + v(t) = V cos 𝜔 t LC v(t) ̇ subject to the initial condition v(0) = v(0) = 0. Since the current flowing dv(t) through the circuit is given by i(t) = C dt . dt2 d2 v(t) + v(t) = vs (t). Substituting the transient solution and its second derivative into the equation (10. (10.144) yields LC(s2 est ) + (est ) = 0.10.5 Second-Order Differential Equations 387 L i(t) + vL (t) − + − vs(t) C + v(t) − Figure 10. Note that since i(t) = C ̇ dition v(0) = 0 means the initial current is zero. Solution dv .144) ytran (t) = est .143) equal to zero LC d2 v(t) dt2 and assuming a solution of the form + v(t) = 0.143) dt2 Equation (10. vL (t) can be written as vL (t) = LC Substituting vL (t) into equation (10.30 Voltage applied to an LC circuit.145) .142) yields d2 v(t) . Similarly.147) Substituting the v̈ ss (t) and vss (t) into equation (10. LC √ √ 1 1 The two roots of equation (10. the transient solution is given by √ √ 1 1 t + c4 sin t. the steady-state solution is given by ( ) V vss (t) = cos 𝜔 t.388 Chapter 10 Differential Equations in Engineering Solving for s yields s2 = − or 1 LC √ s = ±j 1 . (10. equating the coefficients of cos𝜔 t on both sides of equation (10. (10.145) are s1 = + j LC and s2 = − j LC . 1 − LC𝜔2 ) 1 LC . (b) Steady-State Solution: Since the forcing function is vs (t) = V cos 𝜔 t.146) vtran (t) = c3 cos LC LC √ 1 where c3 and c4 are real constants and is the natural frequency 𝜔n in LC rad/s.148) yields A(1 − LC𝜔2 ) = 0.148) yields B(1 − LC𝜔2 ) = V or ( V B= 1 − LC𝜔2 √ provided 𝜔 ≠ Thus. (10. (10. the steady-state solution is of the form vss (t) = A sin 𝜔 t + B cos 𝜔 t.148) Equating the coefficients of sin𝜔 t on both sides of equation (10. Grouping like terms yields A(1 − LC𝜔2 ) sin𝜔 t + B(1 − LC𝜔2 ) cos𝜔 t = V cos𝜔 t.143) gives LC(−A 𝜔2 sin 𝜔 t − B 𝜔2 cos 𝜔 t) + (A sin 𝜔 t + B cos 𝜔 t) = V cos𝜔 t. which gives ( A=0 √ provided 𝜔 ≠ ) 1 LC .149) . Thus. 5 Second-Order Differential Equations 389 (c) Total Solution: The total solution for the voltage v(t) can be found by adding the transient and steady-state solutions from equations (10.151) yields √ ( ) 1 V ̇ cos(0) − 𝜔 sin(0) = 0 v(0) = −c3 (0) + c4 LC 1 − LC𝜔2 or √ ( ) 1 V (1) − 𝜔 (0) = 0.150) v(t) = c3 cos LC LC 1 − LC𝜔2 (d) Initial Conditions: The constants c3 and c4 are determined using the initial ̇ conditions v(0) = 0 and v(0) = 0. which gives √ √ ( ) 1 1 V t + c4 sin t+ cos 𝜔 t.152) cos 𝜔 t − cos LC 1 − LC𝜔2 .146) and (10.10.150) yields ( ) V cos(0) = 0 v(0) = c3 cos(0) + c4 sin(0) + 1 − LC𝜔2 or ( ) V c3 (1) + c4 (0) + (1) = 0. 1 − LC𝜔2 which gives V . 1 − LC𝜔2 The derivative of v(t) is obtained by differentiating equation (10.150) as √ √ √ √ 1 1 1 1 ̇ v(t) = −c3 sin t + c4 cos t LC LC LC LC ) ( V sin 𝜔 t. the voltage across the capacitor is given by √ ( ) ) ( 1 V V t+ cos cos 𝜔 t v(t) = − LC 1 − LC𝜔2 1 − LC𝜔2 or ) √ ( ) ( V 1 v(t) = t V. Thus. Substituting v(0) = 0 into equation (10. c3 (0) + c4 LC 1 − LC𝜔2 which gives c4 = 0. (10.151) −𝜔 1 − LC𝜔2 c3 = − ̇ Substituting v(0) = 0 in equation (10.149). (10. (10. (c) Determine the total solution T(t). The initial temperature of the hot coffee cup shown in Fig. (a) Determine the transient solution. Ttran (t). the math is exactly the same. where h is the convective heat transfer Btu coefficient in . htran (t).1 Leaking tank for problem P10-1. The cup is placed in a room temperature of T∞ = 70◦ F. . If the container is initially empty. The temperature T(t) of the coffee at time t can be approximated by Newton’s law of cooling as dT + k T(t) = h T∞ dt h(t) Qout = kh(t) Figure P10. ft2 h◦ F (a) Find the transient solution. dt h(0) = 0 Qin (b) Suppose the faucet is turned on and off in a sinusoidal fashion. where k is a constant.390 Chapter 10 Differential Equations in Engineering Note: A comparison of Examples 10-12 (spring-mass) and 10-14 (LC circuit) reveals that the solutions are identical. P10. Deso that Qin = 2 termine the steady-state solution. 10-2. At the same time. PROBLEMS 10-1.1. then a little bit of math can go an awfully long way. the fluid leaks out the bottom at a rate Qout = k h(t). (c) Determine the total solution h(t). P10. Such is the case for a wide range of problems across all disciplines of engineering. A faucet supplies fluid to a container of cross-sectional area A at a volume flow rate Qin . as shown in Fig.2 is T(0) = 175◦ F. hss (t). subject to the initial condition. with the following corresponding quantities: Spring-Mass LC Circuit y(t) m k F v(t) LC 1 V Although the two physical systems are entirely different. Q (1 − sin 𝜔 t). Make no mistake … if you want to study engineering. the fluid height h(t) satisfies the following first-order differential equation and initial condition: A dh(t) + k h(t) = Qin . What is the time constant of the response? (b) Find the steady-state solution Tss (t). (d) Sketch the total solution T(t). For t ≥ 0. dt v(0) = 10V. P10. dt (a) Find the transient solution. Increasing the value of capacitance C will decrease the time the voltage v(t) reaches 99% of its steady-state value. What is the time constant of the response? (b) Find the steady-state solution vss (t). Assume that the switch has been in position 1 for a long time.2 A hot coffee cup placed in a room temperature of 70◦ F.3 RC circuit for problem P10-3. (c) Determine the total solution v(t). Doubling the value of resistance R will double the time constant of the response. How long does it take for the temperature to reach 99% of the room temperature? (e) Would increasing the value of h increase or decrease the time when the temperature of coffee reaches the room temp? (f) Would a lower or higher value of h be best for a cup of coffee? 10-3.6. . Repeat parts (a)–(d) of problem P10-3 if R = 20 kΩ and C = 100 𝜇F. Figure P10. P10.3. At t = 0.Problems (e) Mark each of the following statements as true (T) or false (F): Increasing the value of resistance R will increase the time the voltage v(t) reaches 99% of its steady-state value. 10-4. What is the time constant of the response? (b) Find the steady-state solution vss (t). How long does it take for the response to reach 99% of its steady-state value? 391 5 kΩ 1 2 t=0 vs(t) + − C + v(t) − R Figure P10. (d) Sketch the total solution v(t). 10-6. A constant voltage vs (t) = 10 V is applied to the RC circuit shown in Fig. the voltage v(t) across the capacitor satisfies the following differential equation and initial condition: RC dv(t) + v(t) = 0. (a) Find the transient solution. vtran (t). the switch is moved instantaneously to position 2. Repeat parts (a)–(d) of problem P10-3 if R = 10 kΩ and C = 50 𝜇F. The voltage v(t) across the capacitor satisfies the first-order differential equation RC dv(t) + v(t) = vs (t). A constant voltage vs (t) = 10 V is applied to the RC circuit shown in Fig. vtran (t). 10-5. Doubling the value of capacitance C will double the time constant of the response. (d) Calculate the voltage at times t = RC.6 RC circuit for problem P10-6. For t ≥ 0. Use your results to sketch v(t). A constant current i (t) = 100 mA is s applied to the RL circuit shown in Fig. the voltage across the capacitor satisfies the following firstorder differential equation and initial condition: dv(t) v(t) + = I. vss (t). P10. (c) If the initial voltage across the capacitor is v(0) = 0. Assume that the switch has been closed for a long time. C = 10 𝜇F. How long does it take for the response to reach 99% of its steady-state value? R i(t) vs(t) + − + C v(t) − Figure P10. Note: One of the two terms in the steady-state solution is small enough to be neglected. 10-9. Repeat problem P10-6 if R = 100 kΩ. (b) Determine the steady-state solution. C = 50 𝜇F. A sinusoidal voltage vs (t) = 10 sin (0. 10-14. vtran (t). the switch is opened instantaneously. and v(0) = 10 V. C dt R (a) Determine the transient solution. (c) Determine the total solution for v(t) if v(0) = 0 V. the current i(t) flowing . Repeat problem P10-9 if R = 20 kΩ and C = 20 𝜇F. and v(0) = 0 V. 2RC.01 t). determine the total solution v(t). Repeat problem P10-6 if R = 200 kΩ and C = 100 𝜇F. For time t ≥ 0.12 consists of a resistor and capacitor in parallel that are subjected to a constant current source I. and v(0) = 10 V. 4RC. What is the time constant of the response? (b) Find the steady-state solution vss (t) and plot one cycle of the response. Repeat problem P10-12 if R = 1 kΩ. 10-10.6. 10-13.15. P10. 10-7. 10-12. At t = 0. the initial voltage across the capacitor is zero. 10-15. The voltage v(t) across the capacitor satisfies the firstorder differential equation dv(t) + v(t) = 10 sin (0. The circuit shown in Fig. 10-11.01 t) V is applied to the RC circuit shown in Fig. t=0 + I ↑ R C v(t) − Figure P10. I = 5 mA.12 RC circuit for problem P10-12. determine the total solution v(t). (d) Sketch the total solution v(t). and as t → ∞. C = 100 𝜇F. Repeat problem P10-9 if R = 10 kΩ and C = 10 𝜇F. Repeat problem P10-12 if R = 2 kΩ. P10. At time t = 0. 10-8.392 Chapter 10 Differential Equations in Engineering (c) If the initial voltage across the capacitor is v(0) = 5 V. vs (t) = 20 V. I = 10 mA. RC dt (a) Find the transient solution vtran (t). and as R R R t → ∞. R dt 393 10-16. i(0) = 50mA. 10-21. dt L L If the input voltage vin (t) = 10 volts. vtran (t). (c) Determine the total solution for v(t). P10. How long does it take for the response to reach 99% of its steady-state value? (e) Mark each of the following statements as true (T) or false (F): Increasing the value of resistance R will increase the time the voltage v(t) reaches 99% of its steady-state value.Problems through the resistor satisfies the following differential equation and initial condition: L di(t) + i(t) = 0. an input voltage vin is applied to the RL circuit shown in Fig.18 RL circuit for problem P10-18. (a) Determine the transient solution. . s. L = 200 mH. the current i(t) flowing through the . Increasing the value of inductance L will decrease the time the voltage v(t) reaches 99% of its steady-state value. 10-18. (d) Calculate the output voltage v(t) L 2L 4L at times t = . Repeat problem P10-15 if R = 100 Ω and L = 10 mH. (d) Sketch the total solution i(t). R 10-20. vss (t). What is the time constant of the response? (b) Find the steady-state solution iss (t). Doubling the value of inductance L will double the time constant of the response. For t ≥ 0. 10-19. Repeat problem P10-15 if R = 100 Ω and L = 100 mH. At time t = 0. 10-17. t=0 i(t) ↑ is(t) 100 Ω L Figure P10. assuming the initial voltage is zero.18. The output voltage v(t) satisfies the following first-order differential equation: (a) Find the transient solution. Use your results to sketch v(t). Repeat problem P10-18 if R = 50 Ω and L = 500 mH. Doubling the value of resistance R will double the time constant of the response.15 RL circuit for problem P10-15. dv(t) R R + v(t) = vin (t). the switch is moved instantaneously to position 2. L + vin(t) + − R v(t) − Figure P10.21 has been in position 1 for a long time. P10. (b) Determine the steady-state solution. At t = 0. itran (t). The switch in the circuit shown in Fig. (c) Determine the total solution i(t). Repeat problem P10-18 if R = 10 Ω. and vin (t) = 20 V. vo. 1 t=0 5V (d) What is the time constant 𝜏 of the response? Plot the response vo (t) and give the values of vo at t = 𝜏.01 o + vo (t) = −vin (t). dt vo (0) = 0 V. The output voltage vo (t) satisfies the following first-order differential equation and initial conditions: 0. P10. − + +vcc + −vcc vo − Figure P10.21 RL circuit for problem P10-21. 10-23. (c) If the initial output voltage is vo (0) = 0 V. (b) Find the steady-state solution vo. 1A i(t) Figure P10.22 Op–Amp circuit for problem P10-22.22. 10-22. vo.394 Chapter 10 Differential Equations in Engineering inductor satisfies the following differential equation and initial condition: 0. What is the time constant of the response? (b) Find the steady-state solution iss (t).23 Op–Amp circuit for problem P10-23. (d) Sketch the total solution i(t). R2 = 20 kΩ C = 10 μF R1 = 10 kΩ vin + − = 10 sin(10 t) V (a) Find the transient solution.ss if vin = 10 V.tran (t). (a) Find the transient solution. dt vo (0) = 0 V. The output voltage vo (t) satisfies the following firstorder differential equation and initial conditions: dv (t) 0.23.2 di(t) + 10 i(t) = 5.2 dvo (t) + vo (t) = −2vin (t). A constant voltage source vin (t) = 10 volts is applied to the Op–Amp circuit shown in Fig. How long does it take for the response to reach 99% of its steady-state value? 10 Ω 2 50 Ω 200 mH ↑ + − R2 = 10 kΩ C = 1 μF R1 = 10 kΩ vin = 10 V − + +vcc + −vcc + − vo − Figure P10. P10. dt i(0) = −1A. determine the total response. itran (t). (c) Determine the total solution i(t). 2 𝜏. and 5 𝜏. A sinusoidal voltage source vin (t) = 10 sin(10 t) volts is applied to the Op– Amp circuit shown in Fig. What is the time constant 𝜏 of the response? . (a) Find the transient solution.tran (t). Repeat problem P10-24 if the volucm3 and metric blood flow Q̇ in is 60 s the initial arterial pressure is 6 mmHg. What is the time constant of the arterial pressure? (b) Determine the steady-state solution Pss (t) for the arterial pressure. (a) Find the transient solution Ptran (t) for the arterial pressure.28 is given by ̈ + 25 y(t) = f (t). and t → ∞ and use your results to sketch P(t). 0. y(t) .25 y(t) (a) Find the transient solution. The displacement y(t) of a spring-mass system shown in Fig. (d) Sketch the total displacement y(t). 395 ̈ + 10 y(t) = 0. (c) Determine the total displacement y(t) if the initial displacement y(0) = 0. mmHg and C = Also. assume R = 4 (cm3 ∕s) cm3 . The displacement y(t) of a spring-mass system shown in Fig. 10-25.4 mmHg 10-26. (d) Evaluate P(t) at times t = RC. t = 4 RC. k = 10 N/m m = 0. (e) Mark each of the following statements as true (T) or false (F): Increasing the value of resistance R will increase the time it takes for the arterial pressure P(t) to reach 99% of its steady-state value.2 m and the initial veloċ ity y(0) = 0 m/s. determine the total response. Repeat problem P10-26 if the displacement y(t) of a spring-mass (m = 500 gm and k = 5 N/m) system shown ̈ + 5 y(t) = 0 and is given by 0. R is the peripheral resistance. ytran (t). and C is arterial compliance (all constant). (b) Find the steady-state solution of the displacement yss . (c) If the initial output voltage is vo (0) = 0 V. t = 2 RC. P10.26 is given by where Q̇ in is the volumetric blood flow. 10-28. The unit for P(t) is mmHg.25 kg y(t) Figure P10. (c) Determine the total solution P(t) assuming that the initial arterial pressure is 0.Problems (b) Find the steady-state solution vo. 10-24. The relationship between arterial blood flow and blood pressure in a single artery satisfies the following firstorder differential equation: Q̇ dP(t) 1 + P(t) = in dt RC C Doubling the value of capacitance C will double the time constant of the response.5 y(t) y(0) = 10 cm. Increasing the value of capacitance C will decrease the time it takes for the arterial pressure P(t) to reach 99% of its steadystate value. P10. 0.26 Mass-spring system for problem P10-26. 10-27. Doubling the value of resistance R will double the time constant of the response.ss if vin = 10 sin(10 t) V. g h t=0 k x(t) Figure P10.396 Chapter 10 Differential Equations in Engineering (a) Find the transient solution. 10-31. 10-30.30 Mass dropped on a spring.33 is measured from the equilibrium configuration of the spring. m x(t) √ ̇ x(0) = 2 g h. ytran (t). P10.30. 10-32. The displacement y(t) of the springmass system shown in Fig. and ̇ y(0) = 0. (c) Determine the total displacement y(t) if the initial displacement y(0) = 0 m and the initial velocity ̇ y(0) = 0 m/s. subject to the initial conditions. and g = 9. k = 25 N/m m = 1 kg y(t) f(t) (b) Determine the steady-state solution. and determine the frequency of oscillation. (c) Determine the total solution x(t). k = 20 N/m.28 Mass-spring systems for problem m P10-28. P10. Repeat parts (a)–(c) of problem P1030 if m = 500 g. Doubling the mass m will double the maximum value of x(t). Figure P10. Repeat problem P10-28 if the displacement y(t) of a spring-mass (m = 2 kg and k = 32 N/m) system is given by ̈ + 32 y(t) = f (t) 2 y(t) and f (t) = 6 N. deflection is 𝛿 = k . A block of mass m is dropped from a height h above a spring k. 10-33. where the static mg . Increasing the height h will increase the frequency of x(t).1 m/s. Increasing the mass m will decrease the frequency of x(t). 10-29. y(0) = 20 cm. k = 30 N/m.8 m/s2 . and g = 9. (b) Find the steady-state solution of the displacement yss if f (t) = 10 N. the position x(t) of the block satisfies the following secondorder differential equation and initial conditions: ̈ + k x(t) = m g x(0) = 0. xss (t).8 m/s2 . (d) Sketch the total displacement y(t). Repeat parts (a)–(c) of problem P1030 if m = 1 kg. (d) Mark each of the following statements as true (T) or false (F): Increasing the stiffness k will increase the frequency of x(t). Beginning at the time of impact (t = 0). Doubling the height h will double the maximum value of x(t). as shown in Fig. (a) Determine the transient solution xtran (t). 5 t).34 Rod under axial loading by a weight of mass m.Problems Equilibrium Configuration: Undeformed Configuration: g k k δ = mg/k m m y(t) Fcos(ω t) Figure P10. k = 1.5 t). 10-35.25 cos(0.33 Mass-spring system for problem 10-34.375 t) + 0.5 cos(0.5 cos(0. At time t = 0.375 t) √ + c4 sin( 0.5 y(t) = 0. 397 Dynamic Loading: m x(t) AE L Figure P10.5 cos(0. (b) Determine the steady-state solution. However.5 t). show that c3 = 0 and c4 = 0. If the mass m is initially at rest. (c) Determine the total solution for x(t).5 N/m. if the mass is applied suddenly (dynamic loading). where g is the acceleration 𝛿= AE due to gravity. ̈ + m x(t) If m = 4 kg. 4 y(t) (a) Find the transient response ytran (t) of the system. vibration of the mass will ensue. ̇ Given y(0) = 0. determine the minimum value of the deflection y(t) and the first time it reaches that value. L ̇ x(0) = 0. (d) Calculate the maximum value of the deflection x(t). xss (t). xtran (t).5 t) is applied at t = 0. a cart of mass m moving at an initial velocity v0 impacts a spring of stiffness k. . Under static loading by a weight of mass m. What is the natural frequency of the response? (b) Show that the steady-state solution is given by yss (t) = 0.5 t). (a) Determine the transient solution. AE x(t) = mg. the displacement y(t) satisfies the second-order differential equation ̈ + 1.25 cos(0. and a sinusoidal force of 0. (c) Assume the total solution is given mgL m δ= by AE √ L g AE y(t) = c3 cos( 0. P10. (d) Given the total solution y(t) = 0. a rod of length L and axial rigidity AE deforms by an amount mgL .35. subject to the given initial conditions. the deflection x(t) satisfies the following secondorder differential equation and initial conditions: P10-33. x(0) = 0. as illustrated in Fig. 𝜔 = 0.5 and y(0) = 0. How does your result compare to the static deflection 𝛿? Static Loading: Show all steps. Show all steps.5 rad/s. Increasing the inductance L will increase the frequency of i(t). (0) = s . Doubling the velocity v0 will double the amplitude of x(t). (a) Determine the total solution for i(t). and clearly label both its maximum value and the time it takes to get there. and clearly label both its maximum value and the time it takes to get there. x(0) ̇ m x(t) = v0 . (c) Mark each of the following statements as true (T) or false (F): 10-36. Increasing the initial velocity v0 will increase the frequency of x(t). (c) Mark each of the following statements as true (T) or false (F): Increasing the spring stiffness k will increase the frequency of x(t). i(0) = 0.398 Chapter 10 Differential Equations in Engineering vo t=0 x(t) L k m i(t) vs + − C Figure P10. V d2 i(t) di The voltage v(t) satisfies the + i(t) = 0. Increasing the capacitance C will increase the frequency of i(t). C = 500 𝜇F. subject to the given initial conditions. 10-38. Increasing the mass m will decrease the frequency of x(t). Increasing the capacitance C will increase the amplitude of i(t). (b) Plot one-half cycle of the current i(t). Increasing the inductance L will increase the amplitude of i(t). and Vs = 2 V. x(0) = 0. LC 2 dt L dt following second-order differential . An LC circuit is subjected to a constant voltage source Vs that is suddenly applied at time t = 0. Repeat parts (a) and (b) of problem P10-36 if L = 200 mH. Doubling the mass m will double the amplitude of x(t).36 LC circuit for problem P10-36. The current i(t) satisfies the following second-order differential equation and initial conditions: 10-37. Figure P10. The resulting deformation x(t) of the spring satisfies the following secondorder differential equation and initial conditions: ̈ + k x(t) = 0. An LC circuit is subjected to input voltage vin that is suddenly applied at time t = 0. Increasing the voltage Vs will increase the amplitude of i(t). subject to the initial conditions. (b) Plot one-half cycle of the deformation x(t).35 Moving cart for problem P10-35. (a) Determine the total solution for x(t). C = 400 𝜇F. m y(t) 0. v(0) = 0. 10-40. (c) Determine the total solution for v(t).43 Rod supported by a spring. A rod of mass m and length l is pinned at the bottom and supported by a spring of stiffness k at the top. subject to the initial conditions. m l 𝜃(t) 3 f(t) k θ m.0 V. as shown in Fig. (a) Determine the transient solution ytran (t). s 2 Figure P10. P10. What is the frequency of the response? (b) If vin = 10. (b) Determine the steady-state solution yss (t) for the applied force shown in Fig.01 m and y(0) = 0 m/s. the angle 𝜃(t) satisfies the second-order differential equation 1 ̈ + k l 𝜃(t) = f (t). 10-42. Repeat parts (a)–(c) of problem P10-38 if L = 40 mH.1 0 0 1 t.Problems equation and initial conditions: LC d2 v(t) dt2 ̇ + v(t) = vin . and vin = 10 sin(100 t) V. and determine the frequency of oscillation. as shown in Fig. subject to the given initial conditions. its position is described by the angle 𝜃(t).38 LC circuit for problem P10-38. C = 400 𝜇F. l Figure P10. vtran (t). When displaced by a force f (t). 10-41. Does your result depend on the values of L and C? 10-39. N Figure P10. Repeat parts (a)–(c) of problem P10-38 if L = 40 mH. 10-43.42 Applied force for resistive training device for problem P10-42.42. and vin = 10 sin(245 t) V. (a) Find the transient solution 𝜃tran (t). (c) Determine the total solution. L vin(t) + − C + − v(t) 399 subject to the initial condition y(0) = ̇ 0. .28. For relatively small oscillations. A biomedical engineer is designing a resistive training device to strengthen the latissimus dorsi muscle. (a) Determine the transient solution.43. f(t). P10. The task can be represented as a spring-mass system. vss (t). P10. C = 40 𝜇F. determine the steady-state solution. (d) Calculate the maximum value of v(t). and vin = 10 V. The displacement y(t) of the exercise bar satisfies the second-order differential equation ̈ + k y(t) = f (t). v(0) = 0. Repeat parts (a)–(c) of problem P10-38 if L = 100 mH. (d) Given your solution to part (c). mark each of the following statements as true or false: Increasing the length l would decrease the frequency of the oscillation. 18 Determine the total solution. Increasing the mass m would increase the amplitude of the oscillation. 𝜃(t). 𝜃ss (t). Increasing the stiffness k would increase the frequency of the oscillation. and that the initial conditions are 𝜋 ̇ 𝜃(0) = rad and 𝜃(0) = 0 rad/s. Increasing the stiffness k would decrease the amplitude of the oscillation.400 Chapter 10 Differential Equations in Engineering (b) Suppose f (t) = F cos 𝜔 t. . Increasing the mass m would increase the frequency of the oscillation. (c) Suppose now that f (t) = 0. Determine the steady-state solution. 8 T(◦ C) + 32◦ F (c) 68◦ F ≤ T ≤ 176◦ F P1-43 (a) p(x) = −12 x + 30 psi.6 V P1-31 (a) A(F) = 4. l = 2.0836 (b) c = 6.7 V P1-23 vo = 1.5 P + 160 lb P1-7 (a) v(t) = −9.667 V 𝜇g ml (c) a(0.17◦ C (b) v(t) = 10 m/s (c) v(t) = −10 t + 40 m/s P1-37 (a) R = 100 𝜀 + 50 Ω T(x) = 26.55 × 10−6 F V (c) A(200) = 9.48 x + 48 N P1-29 (a) F = 18 N P1-5 (a) Fb (P) = 0.00419) = 0. (b) p(1.5 vin − 4 V P1-25 ea = 2.75) = 50◦ F P1-17 (b) I =2A P1-19 (a) Vs = 3 I + 3 V P1-21 Vs = 50 I + 0.0 ft/s P1-11 (a) v(t) = −75 t + 150 ft/s P1-33 (a) T(V) = 0.0667 c + 0.04 V − 25◦ C P1-13 (a) v(t) = 10 t m/s P1-35 (a) T(V) = 18.0833 P1-15 (a) (b) T(0.091 × 10−4 V (c) ΔT = 65.8 t + 4 m/s P1-9 (a) v(t) = −32.2 t + 161.5 in.04 P1-41 (a) T(◦ F) = 1.Answers to Selected Problems Chapter 1 vo = −1000 iD + 12 V P1-1 k = 10 N/m P1-27 P1-3 (a) F = 3.6◦ C (b) V = 0.67 x + 30◦ F P1-39 (a) a(c) = 0.667 ia + 0.0) = 18 psi 401 .1 V + 4. 36◦ P3-9 h = 415 m and 𝜃 = 53.5 k2 − 3.061 m P3-13 x = −1. L2 = 200 mH (c) Length of tunnel = 50 m CO = 0.13 rad/s P2-23 (a) 𝜔2 − 30𝜔 − 400 = 0 (b) 𝜔 = 40 rad/s . CO2 = 0.061 m Maximum height = 81 ft P2-19 (a) 3 k22 + 0. H2 = 0.178 P2-11 (a) C12 − 100 C1 − 10000 = 0 (c) (b) C1 = 161.061 m and y = −1.9 P2 − 1.13◦ P3-11 x = −1.453 m2 P3-7 𝜃 = 77.9 P2 − 1.9 P + 0..4◦ P3-3 l = 163.402 ANSWERS TO SELECTED PROBLEMS P2-25 (a) R2 − 10 R − 200 = 0 Chapter 2 (b) R = 20 Ω P2-1 (a) I 2 + 2 I − 15 = 0 (b) I = 3 A or I = −5 A P2-27 (a) R2 + 40 R − 1200 = 0 (b) R = 20 Ω P2-3 (a) I 2 + 4 I − 21 = 0 (b) I = 3 A or I = −7 A P2-5 (a) t = 8 s (b) t = 9 s P2-7 (a) R21 − 1067 R1 − 266667 = 0 P2-29 s = −5.3125 = 0 (b) P = 0.061 m and y = 1.5 s height = 30 ft.5 = 0 (b) k1 = 3 lb/in. k2 = 1 lb/in.5 s and t = 2.562.5 s Time when rocket hit ground = 4 s 0. C2 = 261. C2 = 300 𝜇F (c) P2-35 (a) 0. and 𝜃 = 17. width = 20 ft (b) Time when ball hit ground = 4 s Chapter 3 (c) P3-1 h = 66.9178 = 0 P = 0.2 in.438 P2-39 (a) b2 + 10 b − 600 = 0 (b) b = 20 ft (d) Maximum height = 64 ft (c) P2-17 (a) t = 0 s and t = 3.748 P2-37 (a) x2 − 5 x + 2 = 0 (b) L1 = 600 mH. 000 P2-31 L = 3 m and W = 12 m or L = 12 m and W = 3 m P2-33 (a) x2 − 70 x + 600 = 0 (b) xA = 10 m and xB = 60 m (b) R1 = 1275 Ω.8 𝜇F.8 𝜇F P2-13 (a) L22 + 100 L2 − 60000 = 0 (b) x = 0.438.438 P2-15 (a) t = 1 s and t = 3 s (c) (b) t = 1.9 P + 0.14 m and 𝜃 = 41. R2 = 1775 Ω P2-9 (a) C12 − 140 C1 − 12000 = 0 (b) C1 = 200 𝜇F. 000 or s = −20. H2 O = 1.562. P2-21 (a) 𝜔2 + 𝜔 − 1000 = 0 𝜔 = 31.1◦ P3-5 A = 48. 0 Ω and 𝜃 = 26. and 𝜃 = 26. 4.75◦ (b) l = 8. and 𝜃 = 116. and 𝜃 = −45◦ 403 P3-19 𝜃1 = 16.13 ⃗j mph.0 ⃗j in.2 m P3-41 h = 16. y) = (3. (b) 𝜃1 = −60◦ and 𝜃2 = 30◦ Chapter 4 P4-1 ⃗ = 6 ⃗i + 10.04◦ P3-37 (a) 𝜃 = 11.14 lb and Fy = −14.5. or P ⃗ = 12 ∠60◦ in.4◦ ⃗ = 10 ⃗i + 5 ⃗j V. or P ⃗ = 10 ∠30◦ in.536.74% (b) Δx = 173.3◦ cm P P4-11 ⃗ = −1.18 ∠26.4 Ω and 𝜃 = P3-33 Xc = 500.67 ⃗j N T⃗2 = −424. or P ⃗ = 2..5 N (b) 𝜃 = 43. 3.41% (b) Δx = 489..0 ⃗j cm.5. y) = (−4.485 ⃗j mph V magnitude = 9.87◦ cm P l = 9.2◦ in.5 ∠126.33) cm (d) (x.56◦ P3-35 V = 20.014 in.9◦ P4-27 (a) (b) ⃗ = −2.8◦ P3-21 𝜃1 = −57.485 ⃗i − 8.25 ⃗j N P4-33 (a) 𝜃 = 30◦ 16.26◦ P4-15 P3-25 𝜃1 = 127.57◦ (b) l = 4. or P ⃗ = 1 ∠120◦ m P P4-5 ⃗ = 8.55 ⃗j in.3◦ and 𝜃2 = 73.61 ∠56.33) cm (c) (x.54◦ and grade = 20. or P ⃗ = 15 ∠−120◦ cm P (b) (x.8◦ P4-25 (a) (b) ⃗ = −4.62 ∠−14.77 mph and 𝜃 = 136.5 ⃗i + 2 ⃗j cm. 2.ANSWERS TO SELECTED PROBLEMS P3-15 (a) (x.57◦ V V P3-27 𝜃 = −60◦ and 𝜃2 = 90◦ P4-17 ⃗ = 20 ⃗i − 5 ⃗j V.28 m P3-43 (a) Resultant Force = 35.866 ⃗j m. or P ⃗ = 3.8◦ and 𝜃2 = 21. P . or V ⃗ = 11.04◦ V V P3-29 V = 8. −15.6 ⃗i − 14 ⃗j cm P ⃗ = 16.6 mph direction 𝜃 = −117.7◦ P3-31 Z = 104.5 ⃗i − 13.57◦ (c) P4-9 ⃗ = 2 ⃗i + 3 ⃗j cm.75 ⃗i + 245.05◦ P3-45 (a) P(x.75 ⃗i + 735.98 ∠−124. −4.472 in.5 ⃗j mph V magnitude = 10.5) cm P4-3 ⃗ = −0.66 ⃗i − 5 ⃗j V V P4-21 ⃗ = 246.33.9◦ mph V P4-23 (a) (b) ⃗ = −8 ⃗i + 7.97 mph direction 𝜃 = 136.39 ⃗j in.14 lb P3-23 𝜃1 = −145◦ and 𝜃2 = −63.28 ⃗i + 10. P P4-29 (a) (b) ⃗ = −9.6◦ P4-13 Fx = 14.8◦ P4-19 ⃗ = 23. y) = (−2. P ⃗ = −7.66 ⃗i + 5. y) = (18. and 𝜃 = 233.472 in. P ⃗ = 10.77 ⃗i + 192.97.536) cm P4-7 P3-17 (a) l = 4.75 ∠37.44) in. or V ⃗ = 20.4◦ cm P P4-31 T⃗1 = 424. y) = (2.8 ∠102.9 m P3-39 (a) 𝜃 = −30◦ and grade = −57.31◦ and 𝜃2 = 49.50 ⃗i + 0.485 in.62 V and 𝜃 = 14. or V ⃗ = 312. 404 ANSWERS TO SELECTED PROBLEMS (b) F⃗ = −0.866 F ⃗i + 0.5 F ⃗j lb ⃗ = 0.5 N ⃗i + 0.866 N ⃗j lb N ⃗ = 0 ⃗i − 2000 ⃗j lb W (c) F = 1000 lb N = 1732.1 lb P4-35 (a) 𝜃 = 30◦ (b) F⃗ = 0.866 F ⃗i + 0.5 F ⃗j N ⃗ = −0.5 N ⃗i + 0.866 N ⃗j N N (c) F = 250 N, N = 433 N P4-37 (a) F⃗1 = 0.7705 F1 ⃗i + 0.6374 F1 ⃗j lb F⃗2 = −0.4633 F2 ⃗i + 0.8862 F2 ⃗j lb ⃗ = 0 ⃗i − 346 ⃗j lb P (b) F1 = 163.8 lb, F2 = 272.5 lb P4-39 (a) F⃗1 = − F1 ⃗i + 0 ⃗j lb F⃗2 = 0.7466 F2 ⃗i + 0.6652 F2 ⃗j lb ⃗ = 0 ⃗i − 125 ⃗j lb W (b) F1 = 140.3 lb, F2 = 187.9 lb P4-41 (a) F⃗m = −43.3 ⃗i + 25 ⃗j lb W⃗ a = 0 ⃗i − 7 ⃗j lb W⃗p = 0 ⃗i − 3 ⃗j lb (b) Rx = 43.3 lb, Ry = −15 lb R = 45.82 lb, Direction, 𝜃 = −19.11◦ ⃗ = 0.319 ⃗i + 2.156 ⃗j ft P4-43 (a) P (b) P = 2.18 ft, Direction, 𝜃 = 81.58◦ Chapter 5 P5-1 (a) VR = 1 + j 0 V and VL = 0 + j 1 V (b) V = 1 + j 1 V, V = 1.414 ∠45◦ V P5-3 (a) VR = 8.05 − j 4.03 V VL = 2.02 + j 4.02 V (b) V = 10.07 + j 0 V, V = 10.06 ∠0◦ V (c) Re(V) = 10.06 V, Im(V) = 0 V P5-5 (a) VR = 9 + j 3 V, VC = 1 − j 3 V (b) V = 10 + j 0 V, V = 10 ∠0◦ V (c) Re(V) = 10 V, Im(V) = 0 V P5-7 (a) IR = 100 − j 0 mA and IL = 0 − j 200 mA (b) I = 100 − j 200 mA I = 223.6 ∠−63.43◦ mA (c) Re(I) = 100 mA, Im(I) = −200 mA P5-9 (a) IR = 75 + j 43.3 𝜇A and IL = 25 − j 43.3 𝜇A (b) I = 100 + j 0 𝜇A, I = 100 ∠0◦ 𝜇A (c) Re(I) = 100 𝜇A, Im(I) = 0 𝜇A P5-11 (a) IR = 0.5 + j 0 mA and IC = 0 + j 0.2 mA (b) I = 0.5 + j 0.2 mA I = 0.539 ∠21.8◦ mA (c) Re(I) = 0.5 mA, Im(I) = 0.2 mA ( ◦ ) P5-13 vo (t) = Re 15 ej(120 𝜋 t−60 ) V P5-15 (a) Z = 100 + j 82.4 Ω (b) Z = 129.6 ∠39.5◦ Ω (c) Z∗ = 100 − j 82.4 Ω Z Z∗ = 16, 789.9 P5-17 (a) I = 26.83 ∠56.57◦ A, or I = 14.75 + j 22.39 A (b) V = 31.3 ∠18.43◦ V V = 29.69 + j 9.9 V P5-19 (a) Z2 + Z3 = 0 − j 100.45 Ω Z2 + Z3 = 100.45 ∠−90◦ Ω (b) Z1 + Z2 + Z3 = 100 − j 100.45 Ω Z1 + Z2 + Z3 = 141.74 ∠−45.13◦ Ω (c) H = 0.502 − j 0.50 or H = 0.709 ∠−44.87◦ (d) H ∗ = 0.502 + j 0.50 H H ∗ = 0.503 ANSWERS TO SELECTED PROBLEMS P5-21 (a) ZR = 100 + j 0 Ω or ZR = 100.0 ∠0◦ Ω P5-25 (e) V = 100.0 − j 0 V or V = 100 ∠0◦ V ZL = 0 + j 60 Ω or ZL = 60.0 ∠90◦ Ω P5-27 (a) ZR = 100 + j 0 Ω or ZR = 100 ∠0◦ Ω ZC = 0 − j 50 Ω or ZC = 50.0 ∠−90◦ Ω ZL = 0 + j 188.5 Ω or ZL = 188.5 ∠90◦ Ω (b) Z = 100 + j 10 Ω Z = 100.5 ∠5.71◦ Ω ZC = 0 − j 106.1 Ω or ZC = 106.1 ∠−90◦ Ω (c) Z∗ = 100 − j 10 Ω, Z Z∗ = 10, 100 (d) VC = −5.45 − j 54.46 V VC = 54.73 ∠−95.71◦ V P5-23 (a) ZR = 9 + j 0 Ω or ZR = 9 ∠0◦ Ω VL = 92.56 + j 112.3 V or VL = 145.5 ∠50.5◦ V ZC = 0 − j 4 Ω or ZC = 4 ∠−90◦ Ω VC = −52.1 − j 63.2 V or VC = 81.91 ∠−129.5◦ V (b) ZLC = 0 + j 12 Ω or ZLC = 12 ∠90◦ Ω (e) V = 100.0 + j 0 V or V = 100 ∠0◦ V (c) Z = 9 + j 12 Ω or Z = 15 ∠53.13◦ Ω (d) = 9 − j 12 Ω, Z Z∗ (c) I = 0.596 − j 0.491 A or I = 0.772 ∠−39.5◦ A (d) VR = 59.57 − j 49.11 V or VR = 77.2 ∠−39.5◦ V ZL = 0 + j 3 Ω or ZL = 3 ∠90◦ Ω Z∗ (b) Z = 100 + j 82.4 Ω or Z = 129.6 ∠39.5◦ Ω = 225 P5-25 (a) ZR = 100 + j 0 Ω or ZR = 100 ∠0◦ Ω ZC = 0 − j 50 Ω or ZC = 50 ∠−90◦ Ω (b) Z = 100 − j 50 Ω or Z = 111.8 ∠−26.57◦ Ω (c) I = 0.800 + j 0.400 A or I = 0.894 ∠26.57◦ A P5-29 (a) ZR = 1000 + j 0 Ω or ZR = 1000 ∠0◦ Ω ZC = 0 − j 1000 Ω or ZC = 1000 ∠−90◦ Ω (b) I1 = 0.5 − j 0.5 mA or I1 = 0.7071 ∠−45◦ mA I2 = 0.5 + j 0.5 mA or I2 = 0.7071 ∠45◦ mA (c) I1 + I2 = 1 + j 0 mA or I1 + I2 = 1 ∠0◦ mA (d) VR = 80.0 + j 40.0 V or VR = 89.4 ∠26.57◦ V P5-31 (a) ZR = 1000 + j 0 Ω or ZR = 1000 ∠0◦ Ω VC = 20.0 − j 40.0 V or VC = 44.7 ∠−63.43◦ V ZL = 0 + j 1000 Ω or ZL = 1000 ∠90◦ Ω 405 406 ANSWERS TO SELECTED PROBLEMS P5-31 (b) I1 = 0.5 + j 0.5 mA or I1 = 0.7071 ∠45◦ mA P6-3 Amplitude = l = 20 cm Frequency f = 1 Hz, 𝜔 = 2 𝜋 rad/s Period = 1 s Phase angle 𝜙 = −𝜋∕4 rad Time shift = 1∕8 s (to right) P6-5 Amplitude = l = 15 cm Frequency f = 0.5 Hz, 𝜔 = 𝜋 rad/s Period = 2 s Phase angle 𝜙 = 3 𝜋∕4 rad Time shift = 3/4 s (to left) P6-7 Amplitude = A = 2 cm Frequency f = 2 Hz, 𝜔 = 4 𝜋 rad/s Period = 0.5 s Phase angle 𝜙 = 0 rad y(t) = 2 sin(4 𝜋 t) cm P6-9 Amplitude = A = 10 in. Frequency f = 1 Hz, 𝜔 = 2 𝜋 rad/s Period = 1 s Phase angle 𝜙 = 0.2 𝜋 rad x(t) = 10 sin(2 𝜋 t + 0.2 𝜋) in. I2 = 0.5 − j 0.5 mA or I2 = 0.7071 ∠−45◦ mA (c) I1 + I2 = 1 + j 0 mA or I1 + I2 = 1 ∠0◦ mA P5-33 (a) ZR = 1 + j 0 kΩ or ZR = 1 ∠0◦ kΩ ZC = 0 − j 2 kΩ or ZC = 2 ∠−90◦ kΩ (b) Vo = 2.5 + j 0 V or Vo = 2.5 ∠0◦ V P5-35 (a) ZR1 = 1.5 + j 0 = 1.5 ∠0◦ kΩ ZR2 = 1 + j 0 = 1 ∠0◦ kΩ ZC = 0 − j 0.5 = 0.5 ∠−90◦ kΩ (b) Vo = −0.3578 + j 0.1789 V or Vo = 0.4 ∠153.43◦ V P5-37 (a) Za = 20 + j 0 = 20 ∠0◦ Ω Zb = 0 − j 10 = 10 ∠−90◦ Ω Zc = 20 + j 50 = 53.85 ∠68.2◦ Ω (b) Z1 = −2.5 − j 2.5 = 3.54 ∠−135◦ Ω Z2 = 17.5 + j 7.5 = 19.04 ∠23.2◦ Ω Z3 = 3.75 − j 8.75 = 9.52 ∠−66.8◦ Ω ∠−45◦ Ω P5-39 (a) Z1 = 2.5 − j 2.5 = 3.53 Z2 = 17.5 + j 7.5 = 19.04 ∠23.2◦ Ω Z3 = 3.75 − j 8.75 = 9.52 ∠−66.8◦ Ω (b) Za = 72.4 − j 0 = 72.4 ∠0◦ Ω Zb = 5 − j 12.5 = 13.46 ∠−68.2◦ Ω Zc = 25 + j 10 = 26.92 ∠21.8◦ Ω Chapter 6 P6-1 Amplitude = l = 5 in. Frequency f = 1 Hz, 𝜔=2 𝜋 rad/s Period = 1 s Phase angle 𝜙 = 0 rad Time shift = 0 s P6-11 (a) Amplitude = A = 10 cm Frequency f = 2∕𝜋 Hz, 𝜔 = 4 rad/s Period = 𝜋∕2 s (b) t = 𝜋/4 s P6-13 (a) Amplitude = A = 8 cm Frequency f = 1 Hz, 𝜔 = 2 𝜋 rad/s Period = 1 s Time shift = 1/8 s (to left) (b) t = 1/8 s P6-15 (a) Amplitude = A = 5 cm Frequency f = 1/2 Hz, 𝜔 = 𝜋 rad/s Period = 2 s (b) t = 1 s P6-17 (a) Amplitude = A = 5 rad Frequency f = 1 Hz, 𝜔 = 2 𝜋 rad/s Period = 1 s (b) t = 0.25 s ANSWERS TO SELECTED PROBLEMS P6-19 (a) v(t) = 41.38 cos(120 𝜋 t) V (b) Amplitude = 41.38 V Frequency f = 60 Hz Period = T = 1/60 s Phase angle = 0◦ Time shift = 0 s P6-21 (a) Amplitude = 0.1 V Frequency f = 10 Hz Period = T = 1/10 s Phase angle = 𝜋/2 rad Time shift = 1/40 s (to left) (c) v(t) = 0.51 cos(20 𝜋 t − 78.7◦ ) V P6-23 (a) i(t) = 14.14 sin(120 𝜋 t − 45◦ ) A (b) Amplitude = 14.14 A Frequency f = 60 Hz Period = T = 1/60 s Phase angle = −𝜋/4 rad Time shift = 1/480 s (to right) P6-25 (a) i(t) = 4 sin(120 𝜋 t + 60◦ ) A (b) Amplitude = 4 A Frequency f = 60 Hz Period = T = 1/60 s Phase angle = 𝜋/3 rad = 60◦ Time shift = 1/360 s (to left) P6-31 (a) Amplitude = 8 in. Frequency f = 2 Hz Period = T = 1/2 s Phase angle = 𝜋/3 rad = 60◦ Time shift = 1/12 s (to left) (c) 𝛿(t) = 4.11 cos(4 𝜋 t − 76.918◦ ) in. P6-33 (a) Amplitude = 8 in. Frequency f = 2 Hz Period = T = 1/2 s Phase angle = 𝜋/4 rad = 45◦ Time shift = 1/8 s (to left) (c) 𝛿(t) = 11.79 cos(4 𝜋 t + 28.68◦ ) in. P6-35 (a) Amplitude = 164 V Frequency f = 60 Hz Period = T = 1/60 s Phase angle = 7 𝜋/18 rad = 70◦ Time shift = 7/2160 s (to left) (c) vT (t) = 220 cos(120 𝜋 t − 45◦ ) V P6-37 (a) Amplitude = 120 V Frequency f = 60 Hz Period = T = 1/60 s Phase angle = 2 𝜋/3 rad = 120◦ Time shift = 1/180 s (to left) (c) vbc (t) = 207.8 cos(120 𝜋 t + 0◦ ) V P6-27 (a) Amplitude = 10 V Frequency f = 50 Hz Period = T = 1/50 s Phase angle = −𝜋/4 rad = −45◦ Time shift = 1/400 s (to right) P6-39 (a) Amplitude = 200 N Frequency f = 5/2 Hz Period = T = 2/5 s Phase angle = 0 rad = 0◦ Vertical shift = 1000 N (c) vo (t) = 22.36 cos(100 𝜋 t + 63.4◦ ) V √ P6-29 (a) Amplitude = 10 2 V Frequency f = 250 Hz Period = T = 1/250 s Phase angle = −3 𝜋/4 rad = −135◦ Time shift = 3/2000 s (to right) Chapter 7 (c) vo (t) = 11.18 sin(500 𝜋 t − 153.4◦ ) V 407 P7-1 (a) I1 = 5.063 A, I2 = −3.22 A ] [ ][ ] [ 110 16 −9 I1 = (b) −110 −9 20 I2 (c) I1 = 5.063 A, I2 = −3.22 A (d) I1 = 5.063 A, I2 = −3.22 A 408 ANSWERS TO SELECTED PROBLEMS P7-3 (a) I1 = 0.0471 A, I2 = −0.0429 A [ ][ ] [ ] 1100 1000 I1 9 (b) = 1000 1100 I2 0 (c) I1 = 0.0471 A, I2 = −0.0429 A (d) I1 = 0.0471 A, I2 = −0.0429 A P7-5 (a) V1 = 7.64 V, V2 = 7.98 V [ ][ ] [ ] 17 −10 V1 50 (b) = −6 11 V2 42 (c) V1 = 7.64 V, V2 = 7.98 V (d) V1 = 7.64 V, V2 = 7.98 V P7-7 (a) V1 = 0.976 V, V2 = 1.707 V [ ][ ] [ ] 7 −4 V1 0 (b) = 2 −7 V2 −10 (c) V1 = 0.976 V, V2 = 1.707 V (d) V1 = 0.976 V, V2 = 1.707 V P7-9 (a) G1 = 0.0175 ℧, G2 = 0.0025 ℧ [ ][ ] [ ] 10 10 G1 0.2 (b) = 5 15 G2 0.125 (c) G1 = 0.0175 ℧, G2 = 0.0025 ℧ (d) G1 = 0.0175 ℧, G2 = 0.0025 ℧ [ ][ ] [ ] 0.6 0.8 T1 100 P7-11 (a) = 0.8 −0.6 T2 0 (b) T1 = 60 lb, T2 = 80 lb ] [ ][ ] [ 0 0.707 −1 F1 = P7-13 (a) 98.1 0.707 0 F2 (b) F1 = 138.7 N, F2 = 98.1 N (c) F1 = 138.7 N, F2 = 98.1 N [ P7-15 (a) ][ ] [ ] 0.866 0.707 F1 200 = 0.5 −0.707 F2 0 (b) F1 = 146.4 N, F2 = 103.5 N (c) F1 = 146.4 N, F2 = 103.5 N P7-17 (a) R1 = 900 lb, R2 = 2100 lb [ ][ ] [ ] 1 1 R1 3000 (b) = 7 −3 R2 0 (c) R1 = 900 lb, R2 = 2100 lb (d) R1 = 900 lb, R2 = 2100 lb [ ][ ] [ ] 0.9285 −0.3714 F 0 P7-19 (a) = 0.3714 0.9285 N 10 (b) F = 3.714 kN, N = 9.285 kN (c) F = 3.714 kN, N = 9.285 kN [ ][ ] [ ] 0.866 −0.5 F −30 P7-21 (a) = 0.5 0.866 N 110 (b) F = 24.0 lb, N = 101.6 lb (c) F = 24.0 lb, N = 101.6 lb P7-23 (a) NF = 415.4 N, a = 4 m/s2 ] [ ][ ] [ 565.6 0.4 100 NF = (b) a 415.4 1 0 (c) NF = 415.4 N, a = 4 m/s2 P7-25 (a) R1 + R2 = 9810 2 R1 − 1.5 R2 = 7500 (b) R1 = 6347.1 N, R2 = 3462.9 N [ ][ ] [ ] 1 1 R1 9810 (c) = 2 −1.5 R2 7500 ANSWERS TO SELECTED PROBLEMS P7-27 (a) R1 + R2 = 11772 2 R1 − 2 R2 = 8100 (b) R1 = 7911 N, R2 = 3861 N [ ][ ] [ ] 1 1 R1 11772 (c) = 2 −2 R2 8100 (d) R1 = 7911 N, R2 = 3861 N (e) R1 = 7911 N, R2 = 3861 N P7-29 (a) 0.707 Fm − 0.866 Wf = 49.5 0.707 Fm − 0.5 Wf = 80 (b) Wf = 83.29 N, Fm = 172.03 N [ ][ ] [ ] 0.707 −0.866 Fm 49.5 (c) = 0.707 −0.5 Wf 80 (d) Wf = 83.29 N, Fm = 172.03 N (e) Wf = 83.29 N, Fm = 172.03 N P7-31 (a) V1 + V2 = 400 0.2 V1 + 0.4 V2 = 100 (b) V1 = 300 L, V2 = 100 L [ ][ ] [ ] 1 1 V1 400 (c) = 0.2 0.4 V2 100 (d) V1 = 300 L, V2 = 100 L (e) V1 = 300 L, V2 = 100 L ( ) 50 100 1 + s2 A, P7-33 (a) I1 = 500 10 + 10s + s 100 A I2 = 500 10 + 10s + s ⎡(0.2s + 10) ⎤[ ] −0.2s ⎢ ( )⎥ I1 (b) ⎢ 10 ⎥ I = 2 0.2s + ⎢ −0.2s s ⎥⎦ ⎣ ⎡ 100 ⎤ ⎢ s ⎥ ⎥ ⎢ ⎣ 0 ⎦ P7-35 (a) X1 = 409 1.25(s2 + 65) , s(s2 + 25) 50 X2 = s(s2 + 25) ][ ] [ [ ] 40 −40 X1 50∕s (b) = 0 −40 s2 + 65 X2 P7-37 (a) XA = 0.415 lb, XB = 4.67 lb [ ][ ] [ ] 0.15 3.2 XA 15 (b) = 1.6 2.0 XB 10 (c) P7-39 (a) (c) XA = 0.415 lb, XB = 4.67 lb u = −0.0508 in., v = 0.0239 in. [ ][ ] [ ] 2.57 3.33 u 0.05 = 3.33 6.99 v 0 (d) u = −0.0508 in., v = 0.0239 in. Chapter 8 P8-1 (a) v(t) = 150 − 32.2 t ft/s (b) a(t) = −32.2 ft/s2 (c) tmax = 4.66, ymax = 399 ft P8-3 (a) v(t) = 250 − 9.81 t m/s (b) a(t) = −9.81 m/s2 (c) tmax = 25.5, ymax = 3185.5 m P8-5 (a) x(3) = 7.0 m v(3) = 14.14 m/s a(3) = −84.17 m/s2 (b) x(3) = 689.8 m v(3) = 1184.6 m/s a(3) = 1610.4 m/s2 (c) x(3) = 325.5 km v(3) = 1302.1 km/s a(3) = 5208.2 km/s2 P8-7 (a) y(2) = 52 m, y(6) = 20 m a(2) = −12 m/s2 , a(6) = 12 m/s2 (b) y(0) = 20 m, y(9) = 101 m 410 ANSWERS TO SELECTED PROBLEMS P8-9 (a) v(t) = 0.125 e−2 t (3 t2 − 2 t3 ) V (b) i(0) = 0 A and i(1.5) = 0.168 A P8-11 (a) v(t) = −5.2 e−200 t V p(t) = −67.6 e−400 t W pmax = 67.6 W at t = 0 s P8-19 (a) 0 ≤ t ≤ 2 s: linear with slope = 5 m/s2 , since v(0) =0, v(t) = 5 t, and v(2) = 10 m/s 2 < t ≤ 5 s: linear with slope = −5 m/s2 , since v(2) =10, v(t) = 10 − 5 (t − 2) = 20 − 5 t v(5) = −5 m/s (b) v(t) = −4.8 𝜋 sin(120 𝜋 t) V p(t) = −48 sin(240 𝜋 t) W 3 s pmax = 48 𝜋 W at t = 480 P8-13 (a) i(t) = −4 𝜋 sin(200 𝜋 t) A 5 < t ≤ 6 s: linear with slope = 5 m/s2 , since v(5)= −5 m/s, v(t) = −5 + 5(t − 5) = −30 + 5t v(6) = 0 m/s (b) p(t) = −1000 𝜋 sin(400 𝜋 t) W pmax = 1000 𝜋 W t > 6 s: v(t) = 0 m/s (b) 0 ≤ t ≤ 2 s: quadratic with increasing slope/concave up P8-15 (a) 0 ≤ t ≤ 2 s: a(t) = −4 m/s2 2 < t ≤ 6 s: a(t) = 4 m/s2 6 < t ≤ 8 s: a(t) = −4 m/s2 x(t) = 2.5 t2 , x(2) = 10 m. Also, (b) 0 ≤ t ≤ 2 s: quadratic with decreasing slope/concave down x(t) = −2 t2 m/s, x(2) = −8 m 2 < t ≤ 6 s: quadratic with increasing slope and zero slope at t = 4 s, therefore, xmin = −16 m at t = 4 s 1 x(6) = −16 + × 2 × 8 = −8 m 2 6 < t ≤ 8 s: quadratic with decreasing slope, zero slope at t = 8 s 1 x(8) = −8 + × 2 × 8 = 0 m 2 P8-17 (a) 0 ≤ t ≤ 4 s: v(t) = 12 − 3 t m/s 4 < t ≤ 8 s: v(t) = −12 + 3 t m/s t > 8 s: v(t) = 12 m/s (b) 0 ≤ t ≤ 4 s: quadratic with decreasing slope/concave down Slope = 0 at t = 4 s 1 x(4) = 0 + × 4 × 12 = 24 m 2 4 < t ≤ 8 s: quadratic with increasing slope/concave up 1 x(8) = 24 + × 4 × 12 = 48 m 2 1 × 4 × 10 = 10 m 2 2 < t ≤ 5 s: quadratic with decreasing slope/concave down zero slope at t = 4 s, therefore, x(2) = 0 + x(4) = 10 + (1∕2) × 2 × 10 = 20 m x(5) = 20 − (1∕2) × 1 × 5 = 17.5 m P8-21 5 < t ≤ 6 s: quadratic with increasing slope, zero slope at t = 5, 1 x(6) = 17.5 − × 1 × 5 = 15 m 2 Current: 0 ≤ t ≤ 1 s: linear with 1 slope = × 6 = 3 A/s 2 since i(0) = 0, i(t) = 3 t A i(1) = 3 A 1 < t ≤ 2 s: linear with slope = 0 A/s, since i(1) = 3 A, i(t) = 3 A, and i(2) = 3 A 2 < t ≤ 3 s: linear with slope = 3 A/s, since i(2) = 3, i(t) = 3 + 3(t − 2) = −3 + 3t i(3) = 6 A ANSWERS TO SELECTED PROBLEMS P8-21 3 < t ≤ 4 s: linear with slope = 0 A/s, since i(3) = 6 A, i(t) = 6 A, and i(4) = 6 A P8-25 Charge: ≤ t ≤ 1 ms: q(t) = t mC < t ≤ 2 ms: q(t) = 1 mC < t ≤ 3 ms:q(t) = −3 + 3 t mC < t ≤ 4 ms: q(t) = t mC < t ≤ 5 ms: q(t) = 12 − 2 t mC > 5 ms: q(t) = 2 mC 4 < t ≤ 5 s: linear with slope = 3 A/s, since i(4) = 6, i(t) = 6 + 3(t − 4) = −6 + 3t i(5) = 9 A 0 1 2 3 4 t Power: Voltage: 0 ≤ t ≤ 1 s: p(t) = 18 t W 1 < t ≤ 2 s: p(t) = 0 W 2 < t ≤ 3 s: p(t) = −18+ 18 t W 3 < t ≤ 4 s: p(t) = 0 W 4 < t ≤ 5 s: p(t) = −36+ 18 t W P8-23 Current: 0 ≤ t ≤ 1 ms: linear with 1 × 10 slope = 2 × 10−3 = 5000 A/s since i(0) = 0, i(t) = 5000 t A i(1 × 10−3 ) = 5 A 1 < t ≤ 2 ms: linear with 1 slope = (−10) 2 × 10−3 = −5000 A/s since i(1 × 10−3 ) = 5 A, i(t) = 5 − 5000(t − 1 × 10−3 ) = 10 − 5000 t A i(2 × 10−3 ) = 0 2 < t ≤ 3 ms: linear with 1 slope = × 10 = 5000 A/s 2 × 10−3 −3 since i(2 × 10 ) = 0, i(t) = 0 + 5000(t − 2 × 10−3 ) = −10 + 5000 t A i(3 × 10−3 ) = 5 A Power: 0 ≤ t ≤ 1 ms: p(t) = 50 × 103 t W 1 < t ≤ 3 ms: p(t) = − W −100 + 50 × 103 t 411 v(t) = 4000 q(t) V √ l2 − b2 P8-27 (a) x = 3 √ ( ) Pb l2 − b2 2 b2 2 l2 − ymax = 6EIl 3 3 3 Pb (b2 − l2 ) (b) 𝜃(0) = 6EIl Pb 𝜃(l) = (b2 + 2 l2 ) 6EIl P8-29 (a) 𝜃(x) = Mo 3 (L2 − L x − 3 x2 ) 6EI L 2 (b) x = 0.345 L 0.0358 Mo L2 ymax = EI Mo L (c) 𝜃(0) = , y(0) = 0 6 EI ( ) M L (L) =0 𝜃 L2 = − o , y 2 √ 12 EI P8-31 (a) x = L∕ 5 L4 ymax = −0.00239 EI Wo L3 (b) 𝜃(0) = − 120 EI 𝜃(L) = 0 ( ) 2𝜋 A 2𝜋 x P8-33 (a) 𝜃(x) = cos L L 2𝜋 A (b) 𝜃(0) = − L ( ) 2𝜋 A 𝜃 L2 = L 2𝜋 A 𝜃(L) = − L 61 in.2 (c) x̄ = 0.6481 k l P9-3 (a) Distance covered = A = 700 ft (b) Distance covered = A = 750 ft (c) Distance covered = A = 800 ft (d) A = 800 ft P9-5 (a) W = 172.4) m (d) y(x) = −0.15 x m Chapter 9 P9-1 (a) A ≈ 0.5 x + 6 ft (b) A = 27 ft2 (c) x̄ = 2.75 x + 9 cm (b) A = 54 cm2 (c) x̄ = 4 cm.5 t4 + t + 5 t2 m 3 1 (b) v(t) = (1 − cos(4 t)) m/s 2 1 1 y(t) = t − sin(4 t) m 2 8 P9-21 Velocity: P9-17 (a) R = 0 ≤ t ≤ 2 s: v(t) = 10 t m/s 2 < t ≤ 14 s: v(t) = 20 m/s 14 < t ≤ 16 s: v(t) = −10 (t − 16) t > 16 s: v(t) = 0 m/s .75 in.1 x + 22 m (c) (176.7 N-m (c) W = 3. b = 2 in. (d) ȳ = 1. (d) ȳ = 5.0054 W L4 EI (b) 𝜃(0) = 0 W L3 𝜃(L) = 48 EI 𝜋 P8-37 (a) 𝜃 = rad 4 (b) 𝜃 = 0 rad P8-39 (a) b = 0.67 cm2 (c) x̄ = 1. P9-13 (a) h = 8 in. (b) A = 48 in. b = 8 in. ȳ = 3 cm (d) ȳ = 3 cm P9-9 (a) y(x) = −0.00071 x2 + 0.333 ft P9-11 (a) h = 4 in.2 (c) x̄ = 3.412 ANSWERS TO SELECTED PROBLEMS P8-33 (c) L 3L and x = 4 4 ( ) L = −A y 4 ( ) 3L y =A 4 x= P8-35 (a) x = 0. P9-15 (a) a = 2 cm b = 6 cm (b) A = 6.82 N-m P9-7 (a) y(x) = −0.578 L ymax = −0.33 in.6 in.2 cm 2 w l 3 0 3 (b) x̄ = l 8 P9-19 (a) v(t) = 10 t4 + 10 t3 + 10 t2 + 10 t m/s 10 3 y(t) = 2 t5 + 2.667 ft (d) ȳ = 2.88 cm (d) ȳ = 1. 4.15 (b) ŷ = −0.56 in. (b) A = 5.95 N-m (b) W = 201. 04 s: xunbelted (t) = 0.04 s: xbelted (t) = 0.008 s: xbelted (t) = 0.01 1 − e−20 t + 0. v(2) = 10 V. v(4) = 5 V and v(t) = 5 V for t > 4 s Since i(t) is constant over each interval.5 𝜋 t) m (c) xunbelted (0. v(t) is linear between each interval 0 ≤ t ≤ 3 s: v(t) = 10 t m/s 3 < t ≤ 6 s: v(t) = −10(t − 6) 6 < t ≤ 12 s: v(t) = (t − 6) t > 12 s: v(t) = 30 m/s (b) 0 ≤ t ≤ 0.5 cos(20 t)− 10 cos(10 t) mJ ) ( P9-31 (a) vo (t) = 10 1 − e−10 t + 10 V ( ) (b) w(t) = 0.5 𝜋 t) m 0. x(14) = 260 m P9-23 14 < t ≤ 16 s: quadratic with decreasing slope/concave down 1 x(16) = 260 + (2)(20) = 280 m 2 Velocity: Then simply repeat every 2 s P9-35 (a) P9-37 (a) 3 < t ≤ 6 s: quadratic with decreasing slope/concave down 1 x(6) = 45 + (3)(30) = 90 m 2 6 < t ≤ 12 s: quadratic with increasing slope/concave up 1 x(16) = 90 + (30)(6) = 180 m 2 1 3 2 P9-25 q(t) = t − t + t C 3 P9-27 (a) vo (t) = 10 (cos(10 t) − 1) V 0 ≤ t ≤ 0. v(1) = −10 V and v(2) = 0 V 0 ≤ t ≤ 2 s: quadratic with increasing slope/concave up x(t) = 5 t2 . x(2) = 20 m 0 ≤ t ≤ 2 s: quadratic with increasing slope/concave up 2 < t ≤ 14 s: linear x(t) = 20 (t − 1). v(3) = 15 V. v(1) = 10 V.407 m p̂ x P9-39 (a) y(x) = − h G ( L3 − P9-39 (b) x=L ymax = x 2 + x2 3L − x3 ) 12 L2 p̂ L2 12 h G Chapter 10 (b) w(t) = 7.04) − xunbelted (0.04) = 0.ANSWERS TO SELECTED PROBLEMS P9-21 Position: P9-33 v(0) = 0 V.5 + 2. x(3) = 45 m v(0) = 0 V.005 J 413 k htran (t) = C e− A t Q )× A2 𝜔2 + k2 (k sin𝜔 t − A 𝜔 cos 𝜔 t) k Q (A 𝜔 e− A t 2 +k +k sin𝜔 t − A 𝜔 cos 𝜔 t) A2 𝜔2 .5 cos(20 t)− 10 cos(10 t) mJ P10-1 (a) P9-29 (a) vo (t) = sin(100 t) − 100 t V (b) hss (t) = ( (c) h(t) = (b) w(t) = 7.509 sin(12.5 + 2.008 < t ≤ 0.102 sin(62.102 m Position: 0 ≤ t ≤ 3 s: quadratic with increasing slope/concave up x(t) = 5 t2 . 02) = 8.ss (t) = 13.321 V v(0. 𝜏 = (e) 𝜏 = 0.647 V v(0.33 cos(10 t)+ 6.67 sin(10 t) V (c) v(t) = 10 sin(0.33 e−5 t +13.5 A (c) i(t) = −1.01t) V sin(0. 𝜏 = RC s (b) vss (t) = (c) v(t) = 10 (1+0. T P10-3 (a) P10-5 (a) 1 1 (b) iss (t) = 0 A − 12 t vtran (t) = C e (c) i(t) = 50 e−10.01t) V P10-11 (a) vtran (t) = C e−2. T (c) v(t) = 10 e− RC t V (e) T.04) = 9. T P10-7 (a) P10-17 (a) itran (t) = C e−10. F.5 A (c) vo (t) = −13. 𝜏 = 0. T.67 sin(10 t) V (b) vss (t) = 10 V ) ( (c) v(t) = 10 1 − e−100 t V 1 (d) 𝜏 = 0. T (b) vss (t) = 0 V (c) v(t) = 10 e− 2 t V (e) T.02) = 8. F. F.32 V v(0.5 e−50 t + 0.02 s (b) vo.2 s (b) vss (t) = 10 sin(0. F.82 V P10-25 (a) Ptran (t) = A e− RC t mmHg (b) Pss (t) = 60 R mmHg 1 L R s (c) P(t) = 240 − 234 e− 8 t mmHg √ P10-27 (a) ytran (t) = C3 cos( 10 √ t)+ C4 sin( 10 t) m . 𝜏 = 0.01) = 6.001 R2 C2 ) 10 (1+0.33 cos(10 t) + 6.414 ANSWERS TO SELECTED PROBLEMS R vtran (t) = C e− RC t V (c) i(t) = 50 e− L t mA (b) vss (t) = 0 V (e) F.000 t A. F.01t) V P10-15 (a) itran (t) = C e− L t A.000 t mA V (e) F.1 ms 1 P10-19 (a) vtran (t) = C e−100 t V (b) vss (t) = 10 V (c) v(t) = 10(1 − e−100 t ) V 1 vtran (t) = C e− 20 t V (b) vss (t) = 10 V (c) v(t) = 10 − 5 e− 20 t V P10-9 (a) 1 1 vtran (t) = A e− RC t V.001 R2 C2 ) sin(0.01 s v(0.817 V v(∞) = 10 V (b) iss (t) = 0 A P10-21 (a) itran (t) = C e−50 t A.01 s v(0.65 V v(0. T. 𝜏 = 0.tran (t) = C e−5 t V.5 t V. 𝜏 = 0.04) = 9.01) = 6.01t) V R v(∞) = 10 V P10-23 (a) vo.4 s P10-13 (a) vtran (t) = C e−100 t V (b) iss (t) = 0. F. T. F P10-37 itran (t) = C3 cos(100 t)+ C4 sin(100 t) A iss (t) = 0 A i(t) = 0.01 Hz P10-39 (a) (b) vss (t) = 10 V (c) v(t) = 10 (1 − cos(500 t)) V P10-41 (a) vtran (t) = C3 cos(250 t)+ C4 sin(250 t) V 𝜔n = 250 rad/s.245 cos( 40 t)+ √ √ 0.1 sin(100 t) A vtran (t) = C3 cos(500 t)+ C4 sin(500 t) V 𝜔n = 500 rad/s. f = 250∕𝜋 Hz (b) xss (t) = 0.5 cos(0. F .61 t)+ C4 sin(0.2 cos( 10 t) m P10-29 (a) ytran (t) = C3 cos(4 t)+ C4 sin(4 t) m (b) yss (t) = 3∕16 m (c) y(t) = 0.ANSWERS TO SELECTED PROBLEMS (b) yss (t) = 0 m √ (c) y(t) = 0.1 sin(245 t) V (c) v(t) = 5 sin(250 t)+ 5. F.107)+ 3∕16 m √ P10-31 (a) xtran (t) = C3 cos( 40 √ t)+ C4 sin( 40 t) m √ 𝜔n = 40 rad/s or f = 1.5 t) m (c) ymin = −0. f = 125∕𝜋 Hz (b) vss (t) = 5.61 t) m 𝜔n = 0.5 m at t = 2 𝜋 s (√ ) √ m k P10-35 (a) x(t) = Vo cos t m k m (c) T.7 h sin( 40 t) + 0.61 rad/s (b) yss (t) = 0. T. F.028 cos(4 t − 1. F.245 m √ (c) x(t) = −0.245 m 415 (b) (c) (d) F.1 sin(245 t) V (√ ) 3k 𝜃tran (t) = C3 cos t + m (√ ) 3k t rad C4 sin m √ 3k 𝜔n = rad/s m F cos(𝜔 t) rad 𝜃ss (t) = 1 kl − 3 m l 𝜔2 ( ) 𝜋 3k 𝜃(t) = cos t rad 18 m P10-43 (a) P10-33 (a) ytran (t) = C3 cos(0. T. . 280 Armature current in DC motor. and C as a. 294 of triangular section. 136–137 impedance of R. 159–162 frequency and period. 89 first moment of area.INDEX Acceleration. 137 Angular frequency. 145–147 Complex numbers. 296–302 on beams. 221 Addition of sinusoids of same frequency. 226–240 Acceleration due to gravity. 278–283 Average velocity. 140–141 Asphalt problem. 314–321 417 . 345–346 linear DEQ. 141–145 exponential form. 160–162 Antiderivative. 61 Cramer’s Rule. 348–374 leaking bucket. 225–240 applications in electric circuits. 133–134 subtraction of two complex numbers. 347 Complex conjugate. 251. 293–296 determination using vertical rectangles. See also separate entry Direct kinematics of two-link robot. 314–321 impedance of. 322–326 passing through a capacitor. 137 Cos(𝜃). 314–321 current passing through. 314–321 Definite integral. 139 polar form. 166–173. 280 Deflection. solution with constant coefficients. 314–321 energy stored in. 73–74 Distributed loads. 296–297 Division of complex number in polar form. 221–225 Differential equations (DEQ). 140–141 complex conjugate. 240 flowing in inductor. 135 voltage across. See also Material strength chain rule of. 134–135 multiplication of complex number in polar form. 286–293 alternate definition of. 225–240 integrals applications in. 261–266 maxima and minima. 136 impedance of a series RLC circuit. sinusoids. 161 Angular motion of one-link planar robot. 241 definition. 221. L. 240–250 applications in strength of materials. 132–156 addition of two complex numbers. 86 Chain rule of derivative. See also under Sinusoids Addition of two complex numbers. 347–348 second-order differential equations. 247–250. 136 position of one-link robot as. 345–400 first-order differential equations. 292 evaluation using horizontal elemental areas. statically equivalent loading. 250–260. 192 Current. 250 Capacitor. relations between. 374–390. 139–140 Dynamics derivatives applications in. 139–140 examples. 314–321 Center of Gravity (Centroid). 77–79 Electric circuits. 297–302 hydrostatic pressure on a retaining wall. 302–314 Elbow-up solution for 𝜃1 . 218–221 examples of. 218–277 applications in dynamics. 188–189. 251 Derivatives. 145–147 division of complex number in polar form. 137 armature current in a DC motor. 219 Bending moment. 247–250. 241 Complementary solution. 347–348 steady-state solution. 348–374 Force-displacement relationship. 348 Minima. derivatives applications in. 256–260 inclined section of the rectangular bar. 348 transient solution. 259 moment. 259 minimum value. 117. 226–240 Position of one-link robot as a complex number. 255. 298 Multiplication of complex number in polar form. 107–110 Position. 134. 157–159 Particular solution. 256 normal and shear stresses acting on the inclined cross section. 37–38 examples. 348 Period. 117 Newton’s second law. 251. 251 distributed load. 3. 314–321 Equivalent resistance. for two-loop circuit. 251 maximum stress under axial loading. 278–283 concept of work. 186–188 Maxima. 314–321 Energy stored in a capacitor. 278–344. 255 slope. 161 Phase angle. 258 rectangular bar under axial loading. 280 in dynamics. one-link planar robot. 134 Exponential form. 373 Material strength. 221–225 Moment. 184 One-link planar robot angular motion of. 347 Linear frequency. 243–247. 134–135 Integrals. See also Statics Inverse kinematics of one-link robot. 250 deflection. 347 Hydrostatic pressure on a retaining wall. 136 position vector in. 38–50 . 161 Low-pass filter. 286–296. 107 Power. 243 current and voltage in. 252 Matrix algebra method solution tension in cables. 6–7 Free-body diagram (FBD). 256 Method of undetermined coefficients. 133–134 Position vector in polar form. 36–37 equivalent resistance. 162–164 Phase shift. 107–110 in rectangular form. complex numbers. 68–72 Inverse kinematics of two-link robot. 221–225 Maximum stress under axial loading. 296–297 Impedance of capacitor. 3. 119 Graphical method solution. See also Work definite integral. 136–137 Indefinite integrals. 240–250 integrals applications in. 159–162 as a sinusoid. 281 Kirchhoff’s voltage law (KVL). 322–326 inverse operations. 302–314 in electric circuits. 281 inductor. solution with constant coefficients. 33 First-order differential equations. See also Distributed loads Asphalt problem. 240 Quadratic equations. 75–79 Inverse operations. 255 shear force. 139 Newton’s first law. 375 Ohm’s law. 283–286. 322–326 impedance of. 37–38 Euler’s formula. 191–192 for two-loop circuit. 281 Inductor as a circuit element. 134–135. 137–140 of resistor. 314–321 derivatives applications in. 136 Factoring.418 INDEX Electric circuits (continued ) current and voltage in a capacitor. sinusoids. 281 in statics. 162–164 Polar form complex numbers. 314–321 examples. 184 Linear DEQ. 327–333 indefinite integrals. 135 of inductor. current and voltage in. 185–186 Homogeneous solution. 250–260 bending moment. 32–59 current in a lamp. 137–140 of a series RLC circuit. 256–260 maximum value. 164–166 one-link planar robot as. 226–240 Voltage. 252 Slope-intercept method. 374–379 Static equilibrium. 219 Spring-mass system forced vibration of. solution of. 110–123 Two-loop circuit. 357 Time shift. 251. 166–173 amplitude of. 114 Repeated roots. 191–192 substitution method. 386–390 Second-order LC circuit. centroid of. 157–183 addition of sinusoids of same frequency. position vector in. voltage-current relationship in. 286–296 Center of Gravity (Centroid). 192 matrix algebra method. 1–3 voltage-current relationship in resistive circuit. 3–5 Substitution method solution tension in cables. 48–50 projectile in a vertical plane. 117. 184–189. 86 Trigonometry. 162–164 Transient solution. two cycles of. 6–7 slope-intercept method. 184–217 examples. values of. 190–191 Time constant. 162–164 phase shift. 61 Sinusoids. 5 vehicle during braking. 110–123 Velocity. 351. 386–390 Shear force. 37 Quadratic formula. 347 Triangular section. 379–386 free vibration of a spring-mass system. 3–5 Resistor. integrals application in. 284 419 . 240 in capacitor. 72–89 Two-dimensional vectors. 114 static equilibrium. 3–5 Work. 32–36 resistors in parallel. 160 general form of. 347 Resistive circuit. 33 Rectangular form. 1–31 examples. 379–386 free vibration of. 60–105 examples. 117. 157–159 phase angle. 383 Second-order differential equations. 37 Resonance. 159 time shift.INDEX pipeline through parabolic hill. See also individual entry Steady-state solution. 314–321 in inductor. 119 position vector relative velocity. 185 Subtraction of two complex numbers. 65 sine and cosine functions. 60–72 reference angle. 190–192 Cramer’s Rule. 283–286 as area under a constant force curve. 8–18 force-displacement relationship. 374–390 forced vibration of a spring-mass system. See also Tension in cables Tension in cables. 5 Speed at impact. 322–326 Voltage-current relationship in resistive circuit. 89–96 one-link planar robot. 162–164 Slope. 65 Relative velocity. 193–205 solution of a two-loop circuit. 190–191 for two-loop circuit. 348 Straight lines. 119 vector addition. 106–131 free-body diagram (FBD). 64 two-link planar robot. 286–293. 137 Systems of equations. impedance of. 107 Reference angle. 284 as area under a variable force curve. 64 trigonometric functions in four quadrants. 162–164 sine function. 184–189 Vector addition. 117. 255 Sin(𝜃). 297–302 Statics. 134 Resistors in parallel. 374–379 second-order LC circuit. 119 Statically equivalent loading. . then √ −b ± b2 − 4 a c x= 2a Lines Equation. y1 ) and P2 (x2 . k) x . Area and Circumference of Circle: Equation of line with slope m and yintercept b: (x − k)2 + (y − k)2 = r2 A = 𝜋 r2 C = 2𝜋 r y = mx + b Equation: y r (h. y2 ): √ d = (x2 − x1 )2 + (y2 − y1 )2 Area of a Triangle: 1 bh 2 1 = a b sin(𝜃) 2 A= a θ h b If a x2 + b x + c = 0. y1 ) with slope m: y − y1 = m (x − x1 ) Point-slope equation of line through point P2 (x2 . y2 ) with slope m: y − y2 = m (x − x2 ) Distance Formula Distance between points P1 (x1 . y1 ) and P2 (x2 .Useful Mathematical Relations Algebra and Geometry Arithmetic Operations a(b + c) = a c + = b d a+c = b ab + ac ad + bc bd a c + b b a b = a×d c b c d ad = bc Exponents and Radicals ( )n xn x = n y y √ 1∕n x = nx √ n xm xn = xm+n xm∕n = xm m ( )m √ x m−n n = x = x xn √ √ √ (xm )n = = xm n n xy = n x n y 1 √ √ n x−n = n x x x n = √ n n n n y (x y) = x y y Factoring Special Polynomials x2 − y2 = (x + y) (x − y) x3 + y3 = (x + y) (x2 − xy + y2 ) x3 − y3 = (x − y) (x2 + xy + y2 ) Quadratic Formula Slope of line through points P1 (x1 . y2 ): y − y1 m= 2 x2 − x1 Point-slope equation of line through point P1 (x1 . Trigonometry Angle Measurement cos(−𝜃) = cos(𝜃) ) ( cos 𝜃 − 𝜋2 = sin(𝜃) 𝜋 radians = 180◦ 𝜋 1◦ = rad 180 180 1 rad = deg 𝜋 s=r𝜃 r θ s r Double-Angle Formulas Right-Angle Trigonometry sin 𝜃 = sin (2 𝜃) = 2 sin( 𝜃) cos( 𝜃) cos (2 𝜃) = cos2 (𝜃) − sin2 (𝜃) = 2cos2 (𝜃) − 1= 1 − 2 sin2 (𝜃) opp hyp adj hyp opp tan 𝜃 = adj hyp cos 𝜃 = opp adj Addition and Subtraction Formulas P(x. y) sin(𝜃1 ± 𝜃2 ) = sin 𝜃1 cos 𝜃2 ± cos 𝜃1 sin 𝜃2 y x 𝜃 = atan2(y. x) x cos(𝜃1 ± 𝜃2 ) = cos 𝜃1 cos 𝜃2 ∓ sin 𝜃1 sin 𝜃2 Law of Sines Fundamental Identities 1 sin 𝜃 1 sec 𝜃 = cos 𝜃 sin 𝜃 tan 𝜃 = cos 𝜃 Half-Angle Formulas √ ( ) (1 − cos 𝜃) 𝜃 = ± sin 2 2 √ ( ) (1 + cos 𝜃) 𝜃 = ± cos 2 2 θ Trigonometric Functions y sin 𝜃 = r x cos 𝜃 = y r y tan 𝜃 = r x √ θ r = x2 + y2 csc 𝜃 = tan(−𝜃) = − tan(𝜃) ( ) tan 𝜃 − 𝜋2 = − cot(𝜃) 1 tan 𝜃 sin2 𝜃 +cos2 𝜃 = 1 sin A sin B sin C = = a b c cot 𝜃 = 1 + tan2 𝜃 = sec2 𝜃 1 + cot2 𝜃 = csc2 𝜃 sin(−𝜃) = − sin(𝜃) ) ( sin 𝜃 − 𝜋2 = − cos(𝜃) a B c A C b Law of Cosines a2 = b2 + c2 − 2 b c cos A . Differentiation Rules d (c) = 0 dt d [c f (t)] = c ḟ (t) dt d ̇ (c f (t) + c2 g(t)) = c1 ḟ (t) + c2 g(t) dt 1 d ̇ (c f (t) − c2 g(t)) = c1 ḟ (t) − c2 g(t) dt 1 d n (t ) = n tn−1 dt Product Rule d ̇ + ḟ (t) g(t) [f (t) g(t)] = f (t) g(t) dt Quotient Rule [ ] ̇ g(t) ḟ (t) − f (t) g(t) d f (t) = dt g(t) [g(t)]2 Chain Rule df dg d f (g(t)) = × dt dg dt Power Rule d n (t ) = n tn−1 dt Integration Rules ∫ y dx = y x − ∫ x dy ∫ xn dx = xn+1 + C. n+1 ∫ ea x dx = ea x +C a ∫ sin(a x) dx = − ∫ cos(a x) dx = n ≠ −1 cos(a x) +C a sin(a x) +C a ∫ c f (x) dx = c ∫ f (x) dx ∫ [c1 f (x) + c2 g(x)]dx = c1 ∫ f (x)dx + c2 ∫ g(x)dx Exponential Functions d at (e ) = a ea t dt Trigonometric Functions d [sin (a t)] = a cos (a t) dt d [cos (a t)] = − a sin (a t) dt . 3 or ft3 in./s = 0.-lb/s or ft-lb/s Volts (V) Amps (A) Ohms (Ω) Henrys (H) Farads (F) SI m m2 m3 m/s m/s2 N N/m2 (Pa) kg N-m (J) W (J/s) Volts (V) Amps (A) Ohms (Ω) Henrys (H) Farads (F) Commonly Used Prefixes in Engineering Nano (n) 10−9 Micro (𝜇) 10−6 Milli (m) 10−3 Centi (c) 10−2 Kilo (K) 103 Mega (M) 106 Giga (G) 109 Conversion Factor 1 in.45 N 1 psi = 6890 Pa 1 lbm = 0. = 0.ductance Capacitance English in.113 J 1 in.-lb = 0./s2 = 0./s or ft/s in.2 (psi) lbm in. or ft in.454 kg 1 in.2 or ft2 in.Commonly Used Units in Engineering Unit Length Area Volume Velocity Acceleration Force Pressure (Stress) Mass Energy Power Voltage Current Resistance in.113 W .0254m/s2 1 lb = 4.0254 m/s 1 in.-lb/s = 0.3 = 1 m3 1 in.2 = 1 m2 61024 in.-lb or ft-lb in.0254m 1550 in./s2 or ft/s2 lb lb/in.