257054874-Solution-Manual-Mechanics-J-L-Mariam.docx

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University ofLahore. Cell#03338189587 By Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 What is Mechanics? Mechanics is the physical science which deals with the effects of forces on objects. The subject of mechanics is logically divided into two parts: statics,which concerns the equilibrium of bodies under action of forces, and dynamics, which concerns the motion of bodies. BASIC CONCEPTS The following concepts and definitions are basic to the study of mechanics, and they should be understood at the outset. Space is the geometric region occupied by bodies whose positions are described by linear and angular measurements relative to a coordinate system. For three-dimensional problems, three independent coordinates are needed. For two-dimensional problems, only two coordinates are required. Time is the measure of the succession of events and is a basic quantity in dynamics. Time is not directly involved in the analysis of statics problems. Mass is a measure of the inertia of a body, which is its resistance to a change of velocity. Mass can also be thought of as the quantity of matter in a body. The mass of a body affects the gravitational attraction force between it and other bodies. This force appears in many applications in statics. Force is the action of one body on another. A force tends to move a body in the direction of its action. The action of a force is characterized by its magnitude, by the direction of its action, and by its point of application. Thus force is a vector quantity.. A particle is a body of negligible dimensions. In the mathematical sense, a particle is a body whose dimensions are considered to be near zero so that we may analyze it as a mass concentrated at a point. We often choose a particle as a differential element of a body. We may treat a body as a particle when its dimensions are irrelevant to the description of its position or the action of forces applied to it. Rigid body. A body is considered rigid when the change in distance between any two of its points is negligible for the purpose at hand. For instance, the calculation of the tension in the cable which supports the boom of a mobile crane under load is essentially unaffected by the small internal deformations in the structural members of the boom. For the purpose, then, of determining the external forces which act on the boom, we may treat it as a rigid body. Statics deals primarily with the calculation of external forces which act on rigid bodies in equilibrium. Determination of the internal deformations belongs to the study of the mechanics of deformable bodies, which normally follows statics in the curriculum. Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 PROBLEMS 2/2 The magnitude of the force F is Introductory Problems 600 N. Express F as a vector in terms 2/1 The force F has a magnitude of of the unit vectors i and j. Identify both 800 N. Express F as a vector in terms the scalar and vector components of F. of the unit vectors i and j. Identify the x and y scalar components of F. Soln. Step 1: Magnitude of force F=600 lb Soln. Step 2: Free body diagram Step1: Free body Diagram Step3: Force vector F=600cos30i - 600sin30j Step2:Magnitude of force 800N F=520i – 300j x component of force, Fx=-Fsin35o Step 4:Scalar components of force Fx=-800sin35o Along x-axis Fx=-459 N Fx=520 lb y component of force, Fy=Fcos35o Along y-axis Fy=800 cos35o Fy= -300 lb Fy=655 N Force vector, F=(-459i-655j)N Step5:Vector components of force Along x-axis Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Fx=520i lb Along y-axis Fy= -300j lb 2/3 The slope of the 4.8-kN force F is specified as shown in the figure. Express F as a vector in terms of the unit vectors i and j. Soln. Step1: Magnitude of force F=4800 lb Position of point A= -15i-20j Position of point B=30i+10j Step2: Free body diagram Soln. Step1:Free body diagram Stpe3: Position vector of AB, AB=OB – OA Step2: AB=45i+30j Magnitude of force, F=4.8 kN Magnitude of AB, AB =√452 + 302 3 4 Unit vector of force, n= - i- j = 54.08 in 5 5 Force vector F= F n 𝐴𝐵 3 4 Unit vector of AB, n= 𝐴𝐵 F=4.8(- i- j) 5 5 =(-2.88i-3.84j)kN 2/4 The line of action of the 9.6-kN force 45𝐢+30𝐣 = F runs through the points A and B as 54.08 shown in the figure. Determine the x and y scalar components of F. 2/5 A cable stretched between the fixed supports A and B is under a tension T of 900 lb. Express the tension as a vector using the unit vectors i and j, first, as a forceTA acting on A and second, as a force TB acting on B. Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Soln. Soln. Stpe1: Step1: Free body Diagram Magnitude of force F=1800 N 3 4 Unit vector of force, n= - i- j 5 5 Force vector F= F n 3 4 F=1800(- i- j) 5 5 =-1080iN-1440jN 2/7 The two structural members, one of which is in tension and the other in compression, exert the indicated Step2: forces on joint O. Determine the Magnitude of tension in cable AB, T = magnitude of the resultant R of the two forces and the angle θ which R makes 900 lb with the positive x-axis. 2 Unit vector of AB=nAB= 𝒊 - √22 +32 3 j √22 +32 nAB = 0.832i - 0.55j TA= TA nAB TA=900(0.832i - 0.55j) TA=(749i-499j)lb But TA= - TB =-(749i-499j)lb =(-749i+499j)lb Soln. Step1: Free body Diagram 2/6 The 1800-N force F is applied to the end of the I beam. Express F as a vector using the unit vectors i and j. Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 x-components of resultant force Rx= ∑Fx Soln. Rx= -3cos60o-2cos30o Step1: Free body Diagram Rx= -3.23 kN y-components of resultant force Ry= ∑Fy Ry= 3sin60o+2 sin30o Ry= -1.598 kN Magnitude of the resultant force R= √R2x + R2y R= √(−3.23)2 + (−1.598)2 R=3.6 kN Step3: Step2: Angle θ made by ‘R’ F1=800 lb 𝑅𝑥 F2=425 lb θ=tan-1( ) Given that the resultant force is normal to 𝑅𝑦 −3.23 x-axis.Therefore the x-component of =tan-1( ) resultant force is zero. −1.598 Rx= ∑Fx = 0 =26o Rx= -425cosθ+800cos70o = 0 Angle θ made by ‘R’ with positive x-axis 𝑜 θ=180o+26o 800𝑐𝑜𝑠70 Cosθ= =206o 425 θ=49.90 2/8 Two forces are applied to the y-components of resultant force construction bracket as shown. Ry= ∑Fy Determine the angle θ which makes Ry= -425sin49.9-800 sin70o the resultant of the two forces vertical. Ry= -1070 lb Determine the magnitude R of the Magnitude of the resultant force resultant. R= √R2x + R2y R= √02 + (−1077)2 R=1077 lb Representative Problems 2/9 In the design of a control mechanism, it is determined that rod Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 AB transmits a 260-N force P to the Scalar components of force P along ‘n’ crank BC. Determine the x and y Pn=260 cos(30o-22.6o) scalar components of P. Pn=258 N Soln. Scalar components of force P along ‘t’ Step1: Free body Diagram Pt=260 sin(30o-22.6o) Pt=33. 5N 2/11 The t-component of the force F is known to be 75N. Determine the n- component and the magnitude of F. Step2:Magnitude of force P=260 N 12 5 Unit vector of force, n= - i- j 13 13 Soln. Force vector P= P n Step1: Free body Diagram 12 5 P=260(- i- j) 13 13 =-240iN-100jN Scalar component of P along x Px=-240 N Scalar component of P along y Py=-100 N 2/10 For the mechanism of Prob. 2/9, determine the scalar components Pt and Pn of P which are tangent and normal, respectively, to crank BC. Soln. Step2: The t-component of the force Step1:Free body diagram F,Ft=75 N.Let Fn be the n-component of the force F Ft=Fcos40o (1) Fn=-Fsinθ (2) Dividing equation (2) by (1) 𝐹𝑛 =-tan40o 𝐹𝑡 Fn=-Ft tan40o Fn=-75 tan40o Fn=-62.9o By equation (1) Ft=Fcos40o Step2:magnitude of force F=260 N 75=F cos40o 5 F=97.9 N 𝜃 =tan-1(12) 2/12 A force F of magnitude 800 N is θ=22.6o applied to point C of the bar AB as Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 shown. Determine both the x-y and the 2/13 The two forces shown act at point A n-t components of F. of the bent bar. Determine the resultant R of the two forces. Step1: Free body diagram Step1: Free body diagram Step2: Components of force F along ‘x’ Fx= -800 cos200 Step2: Fx= -752 lb x-components of resultant force Components of force F along ‘y’ Rx= ∑Fx Fy= 800 sin200 Rx= 7cos45o-3cos30o Fy= 274 lb Rx= 2.35 kips y-components of resultant force Step3: Ry= ∑Fy Components of force along ‘t’ Ry= -7sin60o+3 sin30o Ft= -800 cos400 Ry= -3.45 kips Ft= -613 lb The resultant of the two forces is Components of force along ‘n’ R=Rxi+Ryj Fn= -800 sin400 R= (2.35i-3.45j)kips Fn= -514 lb 2/14 To satisfy design limitations it is necessary to determine the effect of the 2- Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 kN tension in the cable on the shear, Soln. tension, and bending of the fixed I-beam. Step1:Free body diagram For this purpose replace this force by its equivalent of two forces at A, Ft parallel and Fn perpendicular to the beam. Determine Ft and Fn. Step2:Magnitude of force F=120 lb Soln. Tensile spring force be Fs Step1:Free body diagram Since the horizontal component of ‘R’ is zero F=Fs cos60o F=120 cos60o F=60 lb The resultant force R=Fsin60o R=120sin60o R=103.9 lb 2/16 The ratio of the lift force L to the Step2: drag force D for the simple airfoil is L/D Components of force along ‘t’ =10. If the lift force on a short section of the airfoil is 200 N, compute the Ft= 2 cos(200+300) magnitude of the resultant force R and the Ft=1.286 kN angle θ which it makes with the Components of force along ‘n’ horizontal. Fn=2 sin(200+300) Fn=1.532 kN 2/15 Determine the magnitude Fs of the tensile spring force in order that the resultant of Fs and F is a vertical force. Determine the magnitude R of this vertical resultant force. Soln. Step1: Free body diagram Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Step2: Step2: Lift force L=50 lb By the law of sine’s 𝐿 𝐹𝑎 2 Ratio of the lift force to drag force =10 = 𝐷 𝑠𝑖𝑛15𝑜 𝑠𝑖𝑛120𝑜 50 𝑠𝑖𝑛15𝑜 =10 Fa= ×2 𝐷 𝑠𝑖𝑛120𝑜 D=5 lb Fa=0.598 kN 𝐹𝑏 2 Magnitude of resultant force R =√𝐿2 + 𝐷2 = 𝑠𝑖𝑛45 𝑠𝑖𝑛120𝑜 𝑜 R=√52 + 502 𝑠𝑖𝑛45𝑜 R=50.2 lb Fb= ×2 𝑠𝑖𝑛120𝑜 Step3: Fb=1.633 kN Angle made by the resultant with ‘D’ Step3: 𝐿 𝜃 =tan-1( ) 𝐷 50 𝜃 =tan-1( ) 5 θ =84.3o 2/17 Determine the components of the 2-kN force along the oblique axes a and b. Determine the projections of F onto the a- and b-axes. Let Pa and Pb be the projections of force P along a and b. Pa=2 cos45o Pa=1.414 kN Pb=2 cos15o Pb=1.932 kN 2/18 Determine the scalar components Ra and Rb of the force R along the non rectan -gular axes a and b. Also determine the orthogonal projection Pa of R onto axis a. Soln. Step1:Free body diagram Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Soln. Step1: Free body diagram Soln. Step1: Free body diagram Step2: By the law of sine 𝑅𝑏 800 = 𝑠𝑖𝑛30 𝑠𝑖𝑛40 Rb= 622 N 𝑅𝑎 800 = 𝑠𝑖𝑛110 𝑠𝑖𝑛40 Step2:(a) Ra= 1170 N From the law of cosine’s Step3:Let Pa be the orthogonal R2=4002+6002 – 2(400)(600)cos1202 projection of P onto a-axis R2=760000 R2=872 N Let θ be the angle made with the vertical,then by the law of sine’s 600 872 = 𝑠𝑖𝑛𝜃 𝑠𝑖𝑛120𝑜 𝜃=36.6o Step3:(b) x-components of resultant force Rx= ∑Fx Step4: Rx=600cos30o Pa=R cos300 Rx=520 N =800 cos300 y-components of resultant force =693 N Ry= ∑Fy 2/19 Determine the resultant R of the Ry= -400-600sin30o two forces shown by (a) applying the Ry=-700 N parallelogram rule for vector addition Magnitude of the resultant force and (b) summing scalar components. Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 6 R= √R2x + R2y θ2=tan-1( ) 8 R= √(520)2 + (−700)2 θ2=36.87o R=872 N α=180o – (θ1+θ2) Angle θ made by ‘R’ α=180o – (26.57o+36.87o) 𝑅𝑥 α=116.56o θ=tan-1( ) By the law of sine 𝑅𝑦 𝑃 400 700 𝑠𝑖𝑛𝜃2 =𝑠𝑖𝑛𝜃1 =tan-1( ) 520 =53.4o P= 537 lb So the angle made by resultant with the 𝑇 𝑃 vertical = 𝑠𝑖𝑛𝛼 𝑠𝑖𝑛𝜃2 θ=90o-53.4o T= 800.541 lb θ=36.6o 2/21 At what angle θ must the 800-N 2/20 It is desired to remove the spike from force be applied in order that the the timber by applying force along its resultant R of the two forces has a horizontal axis. An obstruction A prevents magnitude of 2000 N? For this direct access, so that two forces, one 1.6 condition, determine the angle θ kN and the other P, are applied by cables between R and the vertical. as shown. Compute the magnitude of P necessary to ensure a resultant T directed along the spike. Also find T. Soln. Step1: Free body diagram Soln. Step1: Free body diagram Step2: Magnitude of the resultant force R=2000 lb From the law of cosine’s 20002=14002+8002 – Step2: From figure 2(1400)(800)cos(180o-θ) 4 θ1=tan-1( ) But cos(180o-θ)=-cosθ 8 Therefore θ1=26.57o Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 20002=14002+8002 + Position of point A=0.06i+0.04j 2(1400)(800)cosθ Position of point P=0.08sin30i-0.08cos30j θ=51.3o P=0.04i-0.0693j From the law of sine’s Position vector PA=OA-OP 800 𝑅 PA=(0.06-0.04)i-[(0.04)-(-0.0693)]j = 𝑠𝑖𝑛𝛽 sin(180−𝜃) PA=0.02i+0.1093j Magnitude of PA=√0.022 + 0.10932 sin(180−𝜃) PA=0.1111m 𝑠𝑖𝑛𝛽 =R× 𝑃𝐴 800 Unit vector nPA= β=18.19o 𝑃𝐴 0.02𝐢+0.1093𝐣 2/22 The unstretched length of the n PA= √(0.02)2 +(0.1093)2 spring of modulus k =1.2 kN/m is 100 mm.When pin P is in the position n PA=0.18i+0.984j θ=30o determine the x- and y- Magnitude of spring force F= Kx components of the force which the Where x is change in length. spring exerts on the pin. (The force in x=PA- l a spring is given by F=kx, where x is the deflection from the unstretched x =0.1111-0.1 length.) x=0.0111m F=1.2 kN×0.0111 F=13.32 N Step3: Spring force vector F=F nPA =13.32(0.18i+0.984j) =(2.4i+13.1j) N x and y components of force are Fx=2.4 N Fy=13.1 N 2/23 Refer to the statement and figure of Prob. 2/22. When pin P is in the position θ=20o,determine the n- and t- Soln. components of the force F which the Step1: Free body diagram spring of modulus k =1.2 kN/m exerts on the pin. Step1:Given that Spring constant of the spring k=1.2 kN/m Unstrectched length of the spring, l=0.1m The angle made by the pin with the vertical θ=20o Let the centre of the coordinate system be positioned at 0. Step2: Givent hat spring modulus k=1.2 Step2:The arrangement of the spring kN/m system at the present instant is as Unstretched length of spring l =0.1m shown by the free body diagram Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Step3: Soln. The coordinates of points A and P are Step1: Free body diagram A=(0.06i,0.04j) P=(0.08 sin20oi-0.08cos20oj) P=(0.0274i-0.0752j) Hence the position vector PA is given by PA=OA-OP PA=(0.06-0.0274)i + [0.04-(-0.0752]j PA=0.0326i+0.1152j The magnitude of the vector is given by PA= √(0.0326)2 + (0.1152)2 PA=0.1197m The magnitude of the spring force is Step2: given by Tension in the cable T=750 N F=kx Let Tn and Tt be the components of force Where x=PA - 0.01 ‘T’ along ‘n’ and ‘t’axes respectively x=0.1197-0.01 By the cosine law x=0.0197m AB= F=1.2×0.0197 √OA2 + OB2 − 2(OA)(OB)cos120o F=0.0237 kN AB= F=23.7 kN √(1.5)2 + (1.2)2 − 2(1.5)(1.2)(−0.866) AB=2.34 m 2/24 The cable AB prevents bar OA from Step3: rotating clockwise about the pivot O. If the Applying sine rule for triangle BOA cable tension is 750N.Determine the n- 𝑠𝑖𝑛𝜃 𝑠𝑖𝑛120o and t-components of this force acting on = 1.2 𝐴𝐵 point A of the bar. 𝑠𝑖𝑛120o Sin𝜃= 2.34 ×1.2 𝜃=26.37o Step3: Components of force T along ‘n’ Tn=T Sin𝜃 Tn=750 Sin26.37 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Tn=333.12 N sin(180−51.32𝑜 ) 𝑠𝑖𝑛𝛽=400× 1000 Components of force T along ‘t’ β=18.19o Tt=-T cos𝜃 2/26 In the design of the robot to insert the Tn=-750 cos26.37 small cylindrical part into a close-fitting Tn=-672 N circular hole, the robot arm must exert a 2/25 At what angle θ must the 400-N 90-N force P on the part parallel to the force be applied in order that the axis of the hole as shown. Determine the resultant R of the two forces have a components of the force which the part magnitude of 1000 N? For this exerts on the robot along axes (a) parallel condition what will be the angle θ and perpendicular to the arm AB, and (b) between R and the horizontal? parallel and perpendicular to the arm BC. Soln. Step1: Given Force exerted by robot P=90 N Step1:Free body diagram Let parallel force = Pt Let perpendicular force = Pn Step2: (a) Components of force which the part exerts on the robot along parallel and perpendicular to the angle arm AB. Step2: Magnitude of the resultant force R=1000 lb Let θ be the angle between the resultant force ‘R’ and horizontal force 700 lb. From the law of cosine’s 10002=4002+7002 – 2(400)(700)cos(180o-θ) cosθ=0.625 By resolving forces, θ=51.32o Pt=-90cos30 By the law of sine’s Pt=-77.9 N (parallel force) 400 1000 𝑠𝑖𝑛𝛽 =sin(180−𝜃) Pn=90sin30 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Pn=45 N (perpendicular force) Step3: (b) Parallel and perpendicular forces to the arm BC. Step2: By resolving forces Tension in the cable AC,Tac=8 kN Pt=90sin45 Let the tension in the cable AB be Tab Pt=63.6 N (parallel force) Let the magnitude of the resultant Pn=90cos45 force be R Pn=63.6 N (perpendicular force) From the figure 2/27 The guy cables AB and AC are 50 attached to the top of the transmission θ1=tan-1( ) 40 tower. The tension in cable AC is 8 kN. θ1=51.3o Determine the required tension T in 40 cable AB such that the net effect of the θ2=tan-1( ) 60 two cable tensions is a downward θ2=37.3o force at point A. Determine the θ3=180o - θ1 - θ2 magnitude R of this downward force. θ3=95o By the law of sine 𝑇𝑎𝑐 𝑇𝑎𝑏 = sinθ1 sinθ2 sinθ2 Tab=400× 𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛37.7o Tab=400× 𝑠𝑖𝑛51.3𝑜 Tab=5.68 kN 𝑅 𝑇𝑎𝑐 = Soln. sinθ3 sinθ1 Step1: Free body diagram 𝑠𝑖𝑛𝜃3 R=8× sinθ1 𝑠𝑖𝑛95o R=8×𝑠𝑖𝑛51.3𝑜 R=10.21 kN 2/28 The gusset plate is subjected to the two forces shown. Replace them by two equivalent forces, Fx in the x-direction and Fa in the a-direction. Determine the Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 magnitudes of Fx and Fa. Solve 180o=β+45o+65o+ α geometrically or graphically. 180o=β+45o+65o+48.12o β =21.88o By the law of sine 𝑠𝑖𝑛𝛽 𝑠𝑖𝑛45 = Fx 1037.93 Fx=547.02 N Step3: Now considering triangle ‘OCa’ By the law of sine sin(𝛼+65) 𝑠𝑖𝑛45 = Fa 1037.93 sin(48.12+65) 𝑠𝑖𝑛45 = Fa 1037.93 sin(113.12) 𝑠𝑖𝑛45 = Fa 1037.93 Soln. Fa=1349.963 N Step1:Geometric solution Step2: Let Fx and Fa be the forces along x and a-axes respectively. Let R be the resultant force. From law of cosine R2=9002+8002-2(900)(800)cos75o R2=1077300.575 R=1037.93 Step3: By the law of sine 𝑠𝑖𝑛𝛼𝑠𝑖𝑛75 = 800 1037.93 𝑠𝑖𝑛𝛼=0.744 α=48.12o. From triangle ‘OCD’ Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 PROBLEMS Mo=8×5 - 6×5 Introductory Problems Mo=10 kNm 2/29 The 10 kN force is applied atpoint Step3:Considering similar triangles A. Determine the moment of F about point O. Determine the points on the x- and y-axes about which the moment of F is zero. 3 5 = 4 4+𝑥 20 4+x= 3 x=2.67 m 𝑦 3 = 𝑥 4 3 y=2.67× Soln. 4 Step1: Free body diagram y=2 m point B(0,2) C(2.67,0) 2/30 Determine the moment of the 800-N force about point A and about point O. Step2: The magnitude of force F=10 kN Let B and C be the points on y and x- axes respectively. Components of force along x and y- axes. Soln. 4 Step1: Free body diagram Fx=10× 5 Fx=8 kN 3 Fy=10× 5 Fy=6 kN Taking moment about ‘O’ (CW +) Mo=Fx×5 - Fy×4 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Force acting along line AB,F=50 kN From figure. Position vector of A,OA=-15i-20j Position vector of B,OB=40i+10j Position vector of C,OC =25j Vector form of AB,AB=OB-OA AB=(40+15)i+(10+20)j AB=55i+30j 𝐀𝐁 Step2:Magnitude of force F=200 lb Force vector along line AB,F=F ǀ𝐀𝐁ǀ Taking moment about ‘A’ F=F[ 55𝐢+𝟑𝟎j ] MA=200cos30o×35 √552 +302 55𝐢+𝟑𝟎j MA=6062.2 lb-in (CW) F=50[ ] Taking moment about ‘O’ 62.65 F=43.89i+23.94j MO = 200cos30o×35 – 200sin30o×35 MO = 6062.2 – 2500 Step3:(a) MO = 3562.2 lb –in (CW) Taking moment about ‘O’ 2/31 Determine the moment of the 50N Mo=OA×F force (a) about point O by Varignon’s Mo=(-15i-20j)×( 43.89i+23.94j) theorem and (b) about point C by a Mo=(-15×23.94)k+(-20×43.89)(-k) vector approach. Mo=518.7k Magnitude of moment about ‘O’ is Mo=518.7 Nmm(CCW) Step4:(b) Moment about ‘C’ Mc=CA×F Position vector of CA,CA=OA-OC CA=-15i+(-20-25)j CA=-15i-45j Mc=(-15i+(-20-25)j)×( 43.89i+23.94j) Mc=(-15×23.94)k+(-45×43.89)(-k) Mc=1615.95k Soln. Magnitude of moment about ‘C’ is Step1: Free body diagram Mc=1615.95 Nmm(CCW) 2/32 The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Soln. Step2: Given that Step1: Free body diagram Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Magnitude of force F=120 N Moment of force about ‘O’ (CW +) Mo=F cos(20o+15o)×0.15 Mo=120cos35o×0.15 Mo=14.74 Nm 2/34 The throttle-control sector pivots freely at O. If an internal torsional spring Step2: exerts a return moment M =1.8 on the Point A(0,h) sector when in the position shown, for Point B(b,0) design purposes determine the necessary Vector AB=bi-hj throttle-cable tension T so that the net b𝐢−h𝐣 moment about O is zero. Note that when T Unit Vector AB= nAB= √𝑏2 +ℎ2 is zero, the sectorrests against the idle- Vector OA=hj control adjustment screw at R. Moment about ‘O’ MO= OA × F Fb𝐢−Fh𝐣 MO= hj ×[ √𝑏2 +ℎ2 ] Fbh MO= - k √𝑏2 +ℎ2 Fbh MO=√𝑏2+ℎ2 (CW) 2/33 In steadily turning the water pump,a person exerts the 120N force on the handle as shown. Determine the moment of this force about point O. Soln. Step1: Free body diagram Soln. Step1: Free body diagram Step2: Let the tension required be ‘T’ Taking moment about ‘O’ (Anticlock wise +) MO=0 1.8 - T×0.05=0 1.8 T= 0.05 T=36 N Step2: Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 2/35 A force F of magnitude 60 N is applied to the gear. Determine the moment of F about point O. Soln. Step1: Free body diagram Soln. Step1: Free body diagram Step2: Given Magnitude of force F=250 N x and y components of force ‘F’ Fy=250 cos15o Fy=241.48 N Fx=250 sin15o Fx=64.7 N Step3: Taking moment about ‘O’ Mo – Fy(0.2)+Fx(0.03)=0 Mo – 241.48(0.2)+64.7(0.03)=0 Step2: Mo - 48.296+1.92 =0 Magnitude of force F=60 N Mo=46.36 Nm Fy=F cos cos20o 2/37 A mechanic pulls on the 13-mm Fy=60 cos cos20o combination wrench with the 140-N Fy=56.38 N force shown. Determine the moment of Taking moment about ‘O’ (CW+) this force about the bolt center O. Mo=Fy×0.1 Mo=56.38×0.1 Mo=5.64 Nm 2/36 Calculate the moment of the 250-N force on the handle of the monkey wrench about the center of the bolt. Soln. Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Step1: Free body diagram Mo= 120 sin30o ×1.5+120cos30o × 11 Mo=90+1143.15 Mo=1233.15 lb-in (CW) 2/39 A portion of a mechanical coin sorter works as follows:Pennies and dimes roll down the 20o incline,the last triangular portion of which pivots freely about a horizontal axis through O. Dimes are light enough (2.28 grams each) so that the triangular portion remains stationary, and the dimes roll Step2: into theright collection column. Magnitude of force F=140 N Pennies, on the other hand, are heavy Taking moment about ‘O’ (CCW+) enough (3.06 grams each) so that the Mo=Fcosθ×0.095 triangular portion pivots clockwise, and Mo=Fcos(25o-10o)×0.095 the pennies roll into the left collection Mo=140cos10o×0.095 column. Determine the moment about Mo=13.1 Nm O of the weight of the penny in terms 2/38 As a trailer is towed in the forward of the slant distance s in millimeters. direction, the force F =500 N is applied as shown to the ball of the trailer hitch. Determine the moment of this force about point O. Soln. Step1: Free body diagram Soln. Step1: Free body diagram Step2: mgcos200 and mgsin200 are Step2: Given that force F=120 lb components of weight perpendicular Let Fx and Fy be the components of force parallel to the inclined plane along x and y-axes respectively. respectively. Taking moment about ‘O’ Taking moment about pivot ‘O’ (CW +) (Anticlock wise +) MO=Fx ×1.5+Fy × 11 Mo= mgcos200×s+ mgsin200(9.5+3.5) Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Mo=3.06×10-3×9.81cos200×s+3.06×10-3 Mo=104+24.57 ×9.81sin200(9.5+3.5) Mo=128.57 lb-in (CW) Mo=(0.0282s+0.1335) Nmm (s in mm) The combined moment due to the all 2/40 Elements of the lower arm are shown forces is zero. in the figure.The mass of the forearm is 2.3 (Clock wise +)∑ Mo=0 kg with mass center at G. Determine the 128.57-T×2=0 combined moment about the elbow pivot T=64.29 lb O of the weights of the forearm and the 3.6kg homogeneous sphere. What must the biceps tension force be so that the overall 2/41 A 32 lb pull T is applied to a moment about O is zero? cord,which is wound securely around the inner hub of the drum.Determine the moment of T about the drum center C. At what angle θ should T be applied so that the moment about the contact point P is zero? Soln. Step1: Free body diagram Soln. Step1: Free body diagram Step2: Magnitude of force T=32 lb Taking moment about ‘C’(CW+) Mc=T×5 Mc=32×5 Step2: Given that Mc=160 lb-in Weight of the fore arm W1=5 lb For the moment about the contact Weight of the sphere W2=8 lb point P to be zero,the applied force Let ‘T’ be the tension in the bicep. should pass through point P. Moment about ‘O’ due to the weights 5 lb Let θ be the angle of T with the horizontal such that and 8 lb. (Clock wise +) Mo=8×13+5×6sin55o Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 (Clock wise +) Mo=F×OA Mo=200×391.27 Mo=78253.84 Nmm 2/43 In order to raise the flagpole OC, a light frame OAB is attached to the pole and a tension of 780 lb is developed in the hoisting cable by the power winch D. Calculate the moment Mo of this tension about thehinge point From the triangle PCB O. 𝐵𝐶 Cosθ= 𝐶𝑃 5 Cosθ= 8 θ=51.3o 2/42 A force of 200 N is applied to the end of the wrench to tighten a flange Soln. bolt which holds the wheel to the axle. Step1: Free body diagram Determine the moment M produced by this force about the center O of the wheel for the position of the wrench shown. Step2: Magnitude of tenion T=780 lb Taking moment about ‘O’ (CCW +) Mo=780cos20o×10cos30o-780sin20o×5 Mo=6347.62-1333.88 Mo=5013.74 lb-ft 2/44 The uniform work platform, which has a mass per unit length of 28 kg/m, is simply supported by cross rods A Soln. and B. The 90-kg construction worker starts from point B and walks to the Step1: Free body diagram right. At what location ‘s’ will the combined moment of the weights of the man and platform about point B be zero? Step2: Magnitude of force F=200 N OA=450-62.5 cos20 OA=391.27 mm Moment about ‘O’ Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Soln. Step1: Free body diagram Soln. Step1: Free body diagram Step2: Given that Mass of the platform M=28 kg/m Mass of the worker m=90 kg Weight of the platform W=28×(1+4+3)×9.81 W=2197.44 N Moment of all force at ‘B’ is zero. MB=0 Step2: mg×S -W×1=0 Given,moment about ‘O’ Mo=72 kN mg×S=W Let ‘T’ be the tension in the cable Considering triable OBC 𝑊 S= ℎ 𝑚𝑔 sin60o= 30 2197.44 h=25.98 m S= 𝑥 90×9.8 cos60o= 30 S=2.49 m x=15 m 2/45 In raising the pole from the Now considering triangle ‘ABC’ position shown, the tension T in the ℎ cable must supply a moment about O tanθ= 12+𝑥 of 72kNm . Determine T. 25.98 tanθ= 12+15 25.98 tanθ= 27 θ=tan-10.9622 θ=43.897o From ΔOAC 1800=α+θ+120o Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 1800=α+43.897o +120o α=16.103o Step3: Step2: Given that Force exerted by the plunger F=40 N 100 Taking moment about ‘O’ tanθ= ∑Mo=0 400 θ=14.036o Mo-Tsinα×30=0 Taking moment about point ‘O’ Tsin16.103o×30=72 Mo=40cos14.036o(75)+ 40sin14.036o(425) T=8.653 kN Mo=2910.43+4123.035 2/46 The force exerted by the plunger of Mo=7033.465 Nmm (CCW) cylinder AB on the door is 40 N directed Let ‘Fc’ be the reaction force at ‘C’ along the line AB, and this force tends to Taking moment about point ‘O’ keep the door closed. Compute the ∑Mo=0 moment of this force about the hinge O. Fc (825) - Mo=0 What force FC normal to the plane of the Fc (825) - 7033.465 =0 door must the door stop at C exert on the door so that the combined moment about Fc (825) = 7033.465 O of the two forces is zero? Fc =8.525 N 2/47 The 2lb force is applied to the handle of the hydraulic control valve as shown. Calculate the moment of this force about point O. Soln. Step1: Free body diagram Soln. Step1: Free body diagram Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Considering the force applied at ‘D’ CD=400tan15o Step2: CD=107.18 mm Given,applied force F=2 lb DA=AC+CD Taking moment about ‘O’ (CCW +) DA=280+107.18 ∑Mo=-2cos20o×(10sin60o+1.5)- DA=387.18 mm 2sin20o×(1ocos60o) Taking moment about ‘A’ ∑Mo=-19.095-3.42 MA= 200cos15o×DA ∑Mo=-22.5 lb-in MA= 200cos15o×387.18 2/48 Calculate the moment MA of the 200- MA=74797.4 Nm (CW) N force about point A by using three scalar Step4: Method3 methods and one vector method. Finding the perpendicular distance AE DA=387.18 mm DE=DAsin15o DE=387.18 sin15o DE=100.21 mm AE=√𝐷𝐴2 + 𝐷𝐸 2 AE=√387.182 + 100.212 AE=374 mm Taking moment about ‘A’ MA=200×374 MA=74797.4 Nm Soln. Step5: Vector method Step1: Free body diagram r=200i+480j F=-200 cos15oi+200 sin15oj M A= r × F MA=(200i+480j) × (=-200 cos15oi+200 sin15oj) MA=74797.4k Nm 2/50 (a) Calculate the moment of the 90-N force about point O for the Step2: Method1 condition θ=15o. Also, determinethe Taking moment about ‘A’ value of θ for which the moment about MA=Fcos15o×280+ Fsin15o×400 O is (b) zero and (c) a maximum. MA=200cos15o×280+ 200sin15o×400 MA=74797.4 Nm (CW) Step3: Method2 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Soln. Given that Soln. Applied force F=90 N Step1: Free body diagram a) θ=15o Step1: Free body diagram Step2: Elastic modulus of band k=60 N/m Step2: Unstretched length of band xo=0.74m (Anticlock wise +) OC=OA+AC Mo=Fcosθ×0.6 - Fsinθ × 0.8 OC=0.635+0.74 Mo=90cos15o×0.6+ 90sin15o×0.8 OC=1.375 m Mo=33.52 Nm 1.375 θ=tan-1 Step3: 0.635 b) Moment of the force about ‘O’ is zero. θ=65.2o Mo=0 Step3: Fcosθ×0.6 - Fsinθ × 0.8=0 Change in the length of band x=BC-xo cosθ×0.6=sinθ × 0.8 x=1.515-0.74 0.6 Deflection of spring x=0.775m tanθ= Spring force F=kx 0.8 F=60×0.775 θ=36.9o F=46.5 N Step4: Taking moment about ‘O’ (CW +) Mo=Fsinθ×OB Mo=46.5sin62.5o×0.535 Mo=26.8 Nm 2/50 (a) Calculate the moment of the 90-N force about point O for the condition θ=15o. Also, determine the value of θ for which the moment about O is (b) zero and (c) a maximum. Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Step2: c) For the moment to be maximum the Let the angle made by the force F with applied force should be perpendicular. the horizontal be α 0.6 360+340sin40−110cos40 tanα= tanα= 0.8 340cos40+110sin40 494.283 tanα= α=36.9o 331.162 α=56.2o θ=90+α Taking moment about ‘O’ (CW +) θ=90+36.9o Mo=Fcosα×OB θ=126.9o Mo=4.5cos56.2o ×0.36 2/51 The small crane is mounted along Mo=0.902 kNm the side of a pickup bed and facilitates 2/52 Design criteria require that the robot the handling of heavy loads. When the exert the 90-N force on the part as shown boom elevation angle is θ=40o,the while inserting a cylindrical part into the force in the hydraulic cylinder BC is 4.5 circular hole. Determine the moment about kN, and this force applied at point C is points A, B, and C of the force which the in the direction from B toC (the cylinder part exerts on the robot. is in compression). Determine the moment of this 4.5-kN force about the boom pivotpoint O. Soln. Soln. Step1: Free body diagram Step1: Free body diagram Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Step2: Let T be the tension in the string. Taking moment about ‘O’ ∑Mo=0 5cos30o×90+5sin30o×90- Step2: Given that 5 2×60 T ×120- T =0 Force exerted by the robot at ‘D’ P=90 N √22 +52 √22 +52 Taking moment about ‘C’ 389.71+150-11.42T-22.28T=0 ∑MC=0 133.7T=539.71 MC – P(150)=0 T=4.04 kN MC=13500 Nmm 2/54 The piston, connecting rod, and crankshaft of a diesel engine are shown in MC=13.5 Nm the figure. The crank throw OA is half the Taking moment about ‘B’ stroke of 8in, and the length AB of the rod ∑MB=0 is 14in. For the position indicated, the rod MB – P(EF+FB)=0 is under a compression along AB of MB = P(EF+FB) 3550lb. Determine the moment M of this MB =90(150+450sin30o) force about the crankshaft axis O. MB =90(375) MB =33750 Nmm MB =33.75 Nm Step3: Taking moment about ‘A’ ∑MA=0 MA – P(EF+FB+BG)=0 MA = P(EF+FB+BG) MA =90(150+450sin30o+550sin45o) MA =90(3763.91) MA =68751.9 Nmm MA =68.752 Nm 2/53 The masthead fitting supports the two forces shown. Determine the magnitude of T which will cause no Soln. bending of the mast (zero moment) at Step1: Free body diagram point O. Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Step2: OA=8 in AB=14 in Compression in the rod AB, T=3550 lb. AD=OA cos30o Soln. AD=4(0.866) Step1: Free body diagram AD=3.464 in OD=OA sin30o OD=4 sin30 OD=2 in BD=√𝐴𝐵 2 − 𝐴𝐷2 BD=√142 − 3.4642 BD=13.56 in OB=BD+OD OB=13.56+2 OB=15.56 in AB Step2:Applied force F=120 N θ=tan-1 Taking moment about ‘O’(CW+) AD Mo=120cos30o(70+150+70)+ 3.464 θ=tan-1 120cos30o(25+70+70+25) 13.56 Mo=41537.68Nmm Step3: θ=14.33o For maximum moment Mo the force F Step3: should be perpendicular to the line Taking moment about ‘O’ joining AB. (Clockwise +) Mo=Fsinθ×OB 25+70+25+70 Mo=3550 sin14.33o×15.56 tanα= 70+150+70 Mo=13671.81 lb-in α=33.2o Mo=1139 lb-ft Step4:For this condition 2/55 The 120-N force is applied as shown to one end of the curved wrench. If α=30o, calculate the moment of F about the center O of the bolt. Determine thevalue of α which would maximize the moment about O; state the value of this maximum moment. OA=√(25 + 70 + 25 + 70)2 + (70 + 150 + 70)2 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 OA=346.7mm Rx=-111.6 lb Taking moment about ‘O’ ∑Fy=Ry Mo=120×346.7 Ry=100 sin60o Mo=41603.84 Nmm Ry=86.6 lb 2/56 If the combined moment of the two R=√𝑅𝑥2 + 𝑅𝑦2 forces about point C is zero, determine R=√(−111.6)2 + (86.6)2 (a) the magnitude of the force P. R=141.3 lb (b) the magnitude R of the resultant of the Step3: two forces. 𝐹𝑦 (c) the coordinates x and y of the point A c) θ=tan-1 𝐹𝑥 on the rim of the wheel about which the 86.6 combined moment of the two forces is a θ=tan -1 143.3 maximum. θ=37.81o (d) the combined moment MA of the two Step4: forces about A. x=8sinθ x=8sin37.81o x=4.9 in y=8cosθ Soln. x=8cos37.81o Step1: Free body diagram x=6.32 in Step5: Moment about ‘A’ MA=R×BA BA=OA+OB BA=8+4 cosθ BA=8+4cos37.81o BA=11.61 in Since MA=R×BA MA=143.3×11.61 MA=15771 lb-in (CW) Step2: a) Taking moment about ‘C’. ∑MC=0 P×8+100cos60o×4-100sin60o×8=0 P=61.6 lb Step3: b) ∑Fx=Rx Rx= - 100cos60o-61.6 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected] Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. PROBLEMS Introductory Problems 2/57 The caster unit is subjected to the pair of 400-Nforces shown. Determine the moment associated with these forces. 2/60 The indicated force–couple system is applied to a small shaft at the center of the plate.Replace this system by a single force and specify the coordinate of the point on the x- axis through which the line of action of this resultalt force passes. 2/58 A force F=60 N acts along the line AB. Determine the equivalent force– couple system at point C. 2/61 The bracket is spot welded to the end of the shaft at point O.To show the effect of the 900-N force on the weld, replace the force by its equivalent of a force and couple M at O. Express M in vector notation. 2/59 The top view of a revolving entrance door is shown. Two persons simultaneously approach the door and exert force of equal magnitudes as shown. If the resulting moment about the door pivot axis at O is 25 Nm, determine the force magnitude F. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. replacement is frequently done in the design of structures. 2/62 As part of a test, the two aircraft engines are revved up and the propeller pitches are adjusted so as to result in the fore and aft thrusts shown. What force F must be exerted by the 2/64 Each propeller of the twin-screw ground on each of the main braked ship develops a fullspeed thrust of 300 wheels at A and B to counteract the kN. In maneuvering the ship,one turning effect of the two propeller propeller is turning full speed ahead thrusts? Neglect any effects of the and the other full speed in reverse. nose wheel C, which is turned 90N What thrust P must each tug exert on and unbraked. the ship to counteract the effect of the ship’s propellers? Representative Problems 2/65 A lug wrench is used to tighten a square-head bolt. If 250-N forces are applied to the wrench as shown, determine the magnitude F of the equal forces exerted on the four contact points on the 25-mm bolt head 2/63 Replace the 10-kN force acting so that their external effect on the bolt on the steel column by an equivalent is equivalent to that of the two 250-N force–couple system at point O.This forces. Assume that the forces are Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. perpendicular to the flats of the bolt head. 2/66 During a steady right turn, a 2/68 A force F of magnitude 50 N is person exerts the forces shown on the exerted on the automobile parking - steering wheel. Note that each force brake lever at the position x =250mm. consists of a tangential component Replace the force by an equivalent and a radiallyinward component. force–couple system at the pivot point Determine the moment exerted about O. the steering column at O. 2/69 The tie-rod AB exerts the 250-N force on the steering knuckle AO as shown. Replace this force by an equivalent force–couple system at O. 2/67 The 180-N force is applied to the end of body OAB. If θ= 50o, determine the equivalent force–couple system at the shaft axis O. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 2/72 Calculate the moment MB of the 2/70 The combined drive wheels of a 900-N force about the bolt at B. front-wheel-drive automobile are acted Simplify your work by first replacing on by a 7000-N normal reaction force the force by its equivalent force-couple and a friction force F, both of which are system at A. exerted by the road surface. If it is known that the resultant of these two forces makes a 15o angle with the vertical, determine the equivalent force–couple system at the car mass center G. Treat this as a two dimensional problem. 2/71 The system consisting of the bar 2/73 The bracket is fastened to the OA,two identicalpulleys, and a section girder by means of the two rivets A of thin tape is subjected to the two and B and supports the 2-kN force. 180N tensile forces shown in the Replace this force by a force acting figure. Determine the equivalent force– along the centreline between the rivets couple system at point O. and a couple. Then redistribute this force and couple by replacing it by two forces,one at A and the other at B, and ascertain the forces supported by the rivets. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 2/76 The device shown is a part of an automobile seatback-release mechan- ism. The part is subjected to the 4-N force exerted at A and a restoring moment exerted by a hidden torsional spring. Determine the y-intercept of the line of action of the single equivalent force. 2/74 The angle plate is subjected to the two 250-N forces shown. It is desired to replace these forces by an equivalent set consisting of the 200-N force applied at A and a second force applied at B. Determine the y- coordinate of B. 2/75 The weld at O can support a maximum of 2500 N of force along each of the n- and t-directions and a maximum of 1400 Nm of moment. Determine the allowable range for the direction θ of the 2700-N force applied at A. The angle θ is restricted to 0 ≤θ≥90o. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Resultant Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. PROBLEMS replace this force–couple system with Introductory Problems a stand-alone force. 2/77 Calculate the magnitude of the tension T and the angle θ for which the eye bolt will be under a resultant downward force of 15 kN. 2/80 Determine the height h above the base B at which the resultant of the three forces acts. 2/78 Determine the resultant R of the four forces acting on the gusset plate. Also find the magnitude of R and the angle θx which the resultant makes with the x-axis. 2/81 Where does the resultant of the two forces act? 2/82 Under nonuniform and slippery road conditions, thetwo forces shown are exerted on the two rear-drive 2/79 Determine the equivalent force– wheels of the pickup truck, which has couple system at the origin O for each a limited-slip rear differential. of the three cases of forces being Determine the y-intercept of the applied to the edge of a circular disk. If resultant of this force system. the resultant can be so expressed, Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 2/86 A commercial airliner with four jet 2/83 If the resultant of the two forces engines, each producing 90 kN of and couple M passes through point O, forward thrust, is in a steady, level determine M. cruise when engine number 3suddenly fails.Determine and locate the resultant of the three remaining engine thrust vectors. Treat this as a twodimensional problem. 2/84 Determine the magnitude of the force F applied to the handle which will make the resultant of the three forces pass through O. 2/87 Replace the three forces acting on the bent pipe by a single equivalent force R. Specify the distance x from point O to the point on the x-axis through which the line of action of R passes. 2/85 Determine and locate the resultant R of the two forces and one couple acting on the I-beam. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 2/88 The directions of the two thrust vectors of an experimental aircraft can be independently changed from the 2/90 The gear and attached V-belt conventional forward direction within pulley are turning counterclockwise limits. For the thrust configuration and are subjected to the tooth load of shown, determine the equivalent 1600 N and the 800-N and 450-N force–couple system at point O. Then tensions in the V-belt. Represent the replace this force–couple system by a action of these three forces by a single force and specify the point on resultant force R at O and a couple of the x-axis through which the line of magnitude M. Is the unit slowing down action of this resultant passes. These or speeding up? results are vital to assessing design performance. 2/89 Determine the resultant R of the three forces acting on the simple truss. 2/91 The design specifications for the Specify the points on the x- and y-axes attachment at A for this beam depend through which R must pass. on the magnitude and location of the applied loads. Represent the resultant of the three forces and couple by a single force R at A and a couple M. Specify the magnitude of R. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 2/92 In the equilibrium position shown, the resultant of the three forces acting on the bell crank passes through the bearing O. Determine the vertical force P. Does the result depend on θ? 2/94 While sliding a desk toward the doorway, three students exert the forces shown in the overhead view. Determine the equivalent force–couple system at point A. Then determine the equation of the line of action of the resultant force. 2/93Two integral pulleys are subjected to the belt tensions shown.If the resultant R of these forces passes through the center O, determine T and 2/95 Under nonuniform and slippery the magnitude of R and the road conditions, the four forces shown counterclockwise angle θ it makes are exerted on the four drive wheels of with the x-axis. the all-wheel-drive vehicle. Determine the resultant of this system and the x- and y-intercepts of its line of action. Note that the front and rear tracks are equal (i.e.AB=CD). Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Treating the problem as two- dimensional, determine the equivalent force–couple system at the car center of mass G and locate the position x of the point on the car centreline through which the resultant passes. Neglect all forces not shown. 2/96 The rolling rear wheel of a front- wheel-drive automobile which is accelerating to the right is subjected to the five forces and one momentshown. The forces Ax =240 N and Ay= 2000 N are forces transmitted from the axle to 2/98 An exhaust system for a pickup the wheel, F =160 N is the friction truck is shown in the figure. The force exerted by the road surface on weights Wh, Wm, and Wt of the the tire, N =2400 N is the normal headpipe,muffler, and tailpipe are 10, reaction force exerted by the road 100, and 50 N, respectively, and act at surface, and W = 400 N is the weight the indicated points. If the exhaust- of the wheel/tire unit. The couple M=3 pipe hanger at point A is adjusted so Nm is the bearing friction moment. that its tension FA is 50 N, determine Determine and locate the resultant of the required forces in the hangers at the system. points B, C, and D so that the force– couple system at point O is zero. Why is a zero force–couple system at O desirable? 2/97 A rear-wheel-drive car is stuck in the snow between other parked cars as shown. In an attempt to free the car, three students exert forces on the car at points A, B, and C while the driver’s actions result in a forward thrust of 200 N acting parallel to the plane of rotation of each rear wheel. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Ch#3 Equilibrium Step1:Free body diagram Step2: Step3: Step4: Step5: 1.5' Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. PROBLEMS Introductory Problems 3/1 Determine the force P required to maintain the 200-kg engine in the position for which θ=30o. The diameter of the pulley at B is negligible. Soln. Step1:Free body diagram Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Step2: Mass of car m=1400kg Step2: Weight W=1400×9.81 Mass m=200kg W=13734 N Weight W=200×9.81 ∑Fy=0 =1962 N Since there are two front and two rear DC=2sin30o tyres of a car,therefore DC=1 m 2RA+2RB-13734=0 (i) AD=2cos30o Taking moment about ‘A’ AD=1.732 m ∑MA=0 BD=2-AD 2RB(1.386+0.964)-13734×1.386=0 BD=0.27 m 4.7RB=19035.324 𝐷𝐶 RB=4050 N (ii) α=tan-1 𝐵𝐷 Using the value of RB in (i) 1 α= tan-1 2RA+2(4050)-13734=0 0.27 α=75o 2RA=2817 N Step3: 3/3 A carpenter carries a 12 lb uniform Applying sine’s law board as shown.What downward force 𝑃 𝑊 does he feel on his shoulder at A? = sin(90 +30 ) sin(180 −30𝑜 −75𝑜 ) 𝑜 𝑜 𝑜 𝑠𝑖𝑛120𝑜 P=1962× 𝑠𝑖𝑛75𝑜 P=1759 N 3/2 The mass center G of the 1400-kg rear-engine car is located as shown in the figure. Determine the normal force under each tire when the car is in equilibrium.State any assumptions. Step1:Free body diagram Soln. Step1:Free body diagram Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Step2:Weight of the board W=12 lb Let NA and NB be the reactions at points A and B respectively.Taking moment about B, ∑MB=0 W×(2+1)-NA×2=0 12×3-2NA=0 Step2: 2NA=36 Weight of TV, W 1=70×9.81 NA=18 lb W 1=686.7 N Weight of the cabinet W 2=24×9.81 3/4 In the side view of a 70kg W 2=235.44 N television resting on a24kg cabinet, the Taking moment about A, mass centers are labeled G1 and G2. ∑MA=0 Determine the force reactions at A and NB ×0.7-W 1×0.6-W 2×0.35=0 B. (Note that the mass center of most 0.7NB -686.7×0.6-235.44×0.35=0 televisions is located well forward 0.7NB-412.02-82.404=0 because of the heavy nature of the 0.7NB=706 N front portion of picture tubes.) Now ∑Fy=0 NA+NB-W 1-W 2=0 NA+706-686.7-235.44=0 NA=216 N 3/5 The roller stand is used to support portions of long boards as they are being cut on a table saw. If the board exerts a 25-N downward force on the roller C, determine the vertical reactions at A and D. Note that the connection at B is rigid, and that the feet A and D are fairly lengthy horizontal tubes with a nonslip coating. Soln. Step1:Free body diagram Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. RA+RD-25=0 RA+16.55-25=0 RA=8.45 N 3/6 The 450-kg uniform I-beam supports the load shown.Determine the reactions at the supports. Soln. Step1:Free body diagram Soln. Step1:Free body diagram Step2: Weight of the I-beam W 1=450×9.81 W1=4414.5 N Weight of the drum W 2=220×9.81 W2=2158.2 N RA+RB=4414.5+2158.2 RA+RB=6572.7 (i) Taking moment about A, ∑MA=0 (CW+) 4414.5×4+2158.2×5.6-RB×8=0 RB=3718 N Putting this value in (i) Step2: RA+3718=6572.7 Downward force at ‘C’ is 25N RA=2854.7 N Let RA and RB be the reactions at A and B. 3/7 Calculate the force and moment Taking moment about A, reactions at the bolted base O of the ∑MA=0 (CCW+) overhead traffic-signal assembly. Each RD×0.355-25×0.235=0 traffic signal has a mass of 36 kg,while RD=16.55 N the masses of members OC and AC Now ∑Fy=0 are 50 kg and 55 kg, respectively. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. shown.Determine the contact forces at A and B. Soln. Step1:Free body diagram Soln. Step1:Free body diagram Step2: Step2: Mass of the sphere, m=20kg Weight of each traffic signal is 80 lb. Weight of sphere, W=20×9.81 Weight of members OC and AC are W=196.2 N 110 lb and 120 lb respectively. Considering forces along x-axis Let Ox,Oy and Mo=0 be the reactions ∑Fx=0 and reactive moment at ‘O’ RA cos150-RB sin30o=0 respectively. 0.966NA-0.5NB=0 (i) Considering forces along x-axis Considering forces along y-axis ∑Fx=0 ∑Fy=0 Ox =0 RA sin150+RB cos30o-196.2=0 Considering forces along y-axis 0.966NA+0.5NB=196.2 ∑Fy=0 0.259NA=196.2-0.866NB Oy-80-80-110-120=0 NA=758.06-3.344NB (ii) Oy=390 lb Putting above value in (i) Taking moment about ‘O’ 0.966(758.06-3.344NB)-0.5NB=0 ∑Mo=0 (CCW+) 732.29-3.23NB-0.5NB=0 80(15+3+12)+80(3+12)+120×12-Mo=0 3.73NB=732.29 Mo=5040 lb-ft (CW) NB=196.3 Using this value in (i) 0.966NA-0.5(196.3)=0 0.966NA=98.15 3/8 The 20-kg homogeneous smooth NA=101.6 N sphere rests on the two inclines as 3/9 A 120 lb crate resets on the 60 lb pickup tailgate. Calculate the tension T Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. in each of the two restraining cables, G. Determine the upward force F one of which is shown. The centers of necessary to reduce the normal force gravity are at G1 and G2. The crate is at A to onehalf its nominal (F = 0) located midway between the two value. cables. Soln. Soln. Step1:Free body diagram Step1:Free body diagram Step2: Step2: Weight of the crate W 1=120 lb Weight of the generator W=160×9.81 Weight of the tailgate W 2=60 lb W=1569.6 N 12 Let F be the force required. tanθ= 9.5+2.75 Let RA and RO be the reactions at A θ=tan-10.9796 and O respectively. θ=44.4o For F=0 Taking moment about ‘O’ Taking moment about ‘O’ ∑Mo=0 (CCW+) ∑Mo=0 (CCW+) 120(14)+60(9.5)-2Tsin44.4o(12.25)=0 -RA×0.46+1596.6×0.135=0 1680+570-17.14T=0 RA=460.64 N T=131.27 lb 1 3/10 A portable electric generator has For RA=230.32 N 2 a mass of 160 kg with mass center at Taking moment about ‘O’ Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. ∑Mo=0 (CCW+) T=100 lb 1569.6(0.135)-230.32(0.46)-F(0.7)=0 3/12 The device shown is designed to aid -0.7F=-105.95 in the removal of pull-tab tops from cans. If F=151.36 N the user exerts a 40-N force at A, 3/11 With what force magnitude T must determine the tension T in the portionBC the person pull on the cable in order to of the pull tab. cause the scale A to read 2000 N? The weights of the pulleys and cables are negligible. State any assumptions. Soln. Step1:Free body diagram Soln. Step1:Free body diagram Step2: Force applied F=40 N Let the tension in portion ‘BC’ be ‘T’ Let Ro be reaction at ‘O’ Taking moment about ‘O’ ∑Mo=0 (CCW+) Tcos45o(32)+Tsin45o(36) - 40cos10o(78+32)-40sin10o(27)=0 Step2: T=94.06 N Weight of the block is 1000 lb and reading 3/13 A woodcutter wishes to cause the is 500 lb. tree trunk to fall uphill, even though the Let T be the tension in the cable. trunk is leaning downhill.With the aid of Considering forces along y-axis the winch W, what tension T in the cable ∑Fy=0 will be required? The 1200lb trunk has a 5T+500-1000=0 center of gravity at G. The felling notch at Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. O is sufficiently large so that the resisting moment there is negligible. Soln. Step1:Free body diagram Soln. Step1:Free body diagram Step2: Step2: Weight of the block W=300×9.81 Weight of the trunk W=1200 lb W=2943 N Let ‘T’ be the tension in the cable Let RA and RB be reactions at A and B. Taking moment about ‘O’ Taking moment about ‘A’ ∑Mo=0 (CCW+) ∑MA=0 (CCW+) 1200(12sin5o)-T(10+4)cos15o=0 2943×0.4 - RB×0.6=0 T=81.21 lb RB=1962 N 3/14 To facilitate shifting the position of a Considering forces along y-axis lifting hook when it is not under load, the ∑Fy=0 sliding hanger shown is used. The RA-RB-W=0 projections at A and B engage the flanges RA=1962+2923 of a box beam when a load is supported, RA=4905 N and the hook projects through a horizontal 3/15 Three cables are joined at the slot in the beam. Compute the forces at A junction ring C. Determine the tensions in and B when the hook supports a 300-kg cables AC and BC caused by the weight mass. of the 30-kg cylinder. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. contact points B and E is sufficient to prevent slipping;friction at the pulley contact points C and F is negligible. Soln. Step1:Free body diagram Soln. Step1:Free body diagram Step2: Let TAC,TAB and TDC be the tension in the cables AB,BC and CD respectively. Weight of the cylinder W=30×9.81 W=294.3 N Here Step2: TDC=294.3 N Total axial force ‘F’ requires is 700N ∑Fx=0 Since two prybars are acting at both -TACcos45o+294.3cos15o-TBCcos60o=0 sides of the pully,the axial force by 0.707TAC+0.5TBC=284.27 (i) each prybar would be 350N. ∑Fy=0 Taking moment about ‘E’ TACsin45o+294.3sin15o-TBCsin60o=0 ∑ME=0 (CCW+) 𝑭 0.707TAC=0.866TBC-76.17 -𝟐×0.038+Fcos5(0.25)+Fsin5(0.031)=0 TAC=1.225 TBC-107.74 (ii) -13.3+0.249F+0.0027F=0 Using (ii) in (i) 0.252F=13.3 0.707(1.225TBC-107.74)+0.5TBC=284.27 F=52.84N 0.866TBC-76.17+0.5TBC=284.27 3/17 The uniform beam has a mass of 50 1.366TBC=360.44 kg per meter of length. Compute the TBC=263.87 N reactions at the support O.The force loads Using above value in (ii) shown lie in a vertical plane. TAC=1.225(263.87)-107.74 TAC=215.49 N 3/16 A 700-N axial force is required to remove the pulley from its shaft. What force F must be exerted on the handle of each of the two prybars? Friction at the Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 3/18 A pipe P is being bent by the pipe bender as shown. If the hydraulic cylinder applies a force of magnitude F = 24 kN to the pipe at C, determine the magnitude of the roller reactions at A and B. Soln. Step1:Free body diagram Soln. Step1:Free body diagram Soln. Step1:Free body diagram Step2: Let Ox and Oy be the reactions at ‘O’ along x and y directions respectively. Let Mo,be the reaction moment at ‘O’ Self load of the beam portions.i.e F1=50×9.81×(1.8+0.6) Step2: F1=1177.2 N Applied force 24 kN F1=1.117 kN Let RA and RB be the reactions at A and B. F2=50×9.81×(0.6+0.6) Here RA=RB F2=588.6 N ∑Fy=0 F2=0.589 kN 24-RAcos15o-RBcos15o=0 Step3: 2RAcos15o=24 Taking moment about ‘O’ RA=12.42 kN ∑Mo=0 RA= RB=12.42 kN Mo-F1×1.2-3×2.8-F2(2.4+0.6cos30o)+4 – 3/19 The uniform 15m pole has a mass of 1.4(2.4cos30o+1.2)=0 150 kg and is supported by its smooth Mo-1.177×1.2-5.4-0.589×2.916+4- ends against the vertical walls and by the 1.4×3.278=0 tension T in the vertical cable.Compute Mo-1.41-5.4-1.72+4-4.59=0 the reactions at A and B. Mo=9.12 kNm ∑Fx=0 Ox+1.4sin30o=0 Ox+0.7=0 Ox=-0.7 kN ∑Fy=0 Oy-F1-F2-3-1.4cos30o=0 Oy-1.177-0.589-3-1.21=0 Oy=5.98 kN Representative Problems Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 𝐴𝐶 12 = 5 15 AC=4 m Now consider trianglesAFE and ABO 𝐴𝐸 𝐴𝑂 = 𝐴𝐹 𝐴𝐵 𝐴𝐸 12 = 7.5 15 AE=6 m Taking moment about ‘A’ (CCW+) ∑MA=0 RB×9-1471.5×6+1471.5×4=0 RB=327 N Soln. RA=RB=327 N Step1:Free body diagram 3/20 Determine the reactions at A and E if P = 500 N. What is the maximum value which P may have for static equilibrium? Neglect the weight of the structure compared with the applied loads. Step2: Length of the pole, L=15m Weight of the pole, W=150×9.8 W=1471.5 N Let ‘T’ be the tension in the cable. Let RA and RB be the horizental reactions at A and B respectively. OB2+122=152 Soln. Step1:Free body diagram OB=√152 + 122 OB=9 m ∑Fx=0 RA=RB ∑Fy=0 T-1471.5=0 T=1471.5 N Step3: Step2: Let Ax,Ay and Ex be the reactions at A and E respectively. ∑Fx=0 From similar triangles ADC and ABO 𝐴𝐶 𝐴𝑂 Ax+Ex-4000sin30o=0 = Ax+Ex=2000 (i) 𝐴𝐷 𝐴𝐵 ∑Fy=0 Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Ay-4000cos30o+500=0 Ay=3464.1-500=0 Ay=2964.1 N Taking moment about ‘A’ ∑MA=0 500×8+Ex×3-4000cos30o(4)=0 3Ex=16000cos30o-4000 Ex=3285.46 N Using above value in (i) Ax+3285.46=2000 Ax=-1285.46 N Step2: For maximum P, Ex=0 Let (Fc)x and (Fc)y be the horizontal and Taking moment about ‘A’ vertical reactions at ‘C’. ∑MA=0 Let FB be the force exerted by rock at ‘B’ P×8-4000cos30o(4)=0 (a) Considering the weight of the prybar. 8P=13856.4 Taking moment about ‘C’ ∑MC=0 P=1732 N 8 3/21 While digging a small hole prior to 50×68+40×34tan20o-FB× =0 𝑐𝑜𝑠20𝑜 planting a tree, a homeowner encounters 3400+495-8.51FB=0 rocks. If he exerts a horizontal 225-N force 8.51FB=3895 on the prybar as shown, what is the FB=457.7 lb horizontal force exerted on rock C? Note Taking force along x-axis that a small ledge on rock C supports a (Fc)x+50- FBcos20o=0 vertical force reaction there. Neglect (Fc)x= 457.7cos20o-50 friction at B. Complete solutions (a) (Fc)x=380 lb including and(b)excluding the weight of Step3: the 18kg prybar. (b) Excluding the weight of the prybar. Taking moment about ‘C’ ∑MC=0 8 50×68-FB× =0 𝑐𝑜𝑠20𝑜 FB=399 lb Taking forces along x-axis. ∑Fx=0 FC- FBcos20o+50=0 FC- 399cos20o+50=0 FC=325 lb 3/22 Determine the force P required to begin rolling the uniform cylinder of mass m over the obstruction of height h. Soln. Step1:Free body diagram Soln. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Step1:Free body diagram Soln. Step1:Free body diagram Step2:considering the triangle ABC Step2: Let ‘F’ be the force applied at handle. Let Ox and Oy be the horizontal and BC=√𝑟 2 + (𝑟 − ℎ)2 vertical reactions at point ‘O’. BC=√𝑟 2 − 𝑟 2 − ℎ2 + 2𝑟ℎ Given that 35N of force is the reactive BC=√2𝑟ℎ − ℎ2 force at ‘B’ 𝐵𝐶 Taking moment about ‘O’ sinθ= 𝐴𝐶 ∑Mo=0 √2𝑟ℎ−ℎ2 35×18-Fcos10o(38)- Fsin10o(44)=0 sinθ= 𝑟 45.06F=630 Taking moment about ‘C’ F=13.98 N ∑MC=0 Ste3: P(r-h)-mgrsinθ=0 ∑Fx=0 √2𝑟ℎ−ℎ2 Ox-Fcos10o-35=0 P(r-h)-mgr =0 𝑟 Ox-13.98cos10o-35=0 √2𝑟ℎ−ℎ2 Ox=48.8 N P=mg 𝑟−ℎ ∑Fy=0 3/23 A 35-N axial force at B is required to Oy-Fsin10o=0 open the springloaded plunger of the Oy=2.43 N water nozzle. Determine the required force Magnitude of reaction at ‘O’ F applied to the handle at A and the magnitude of the pin reaction at O. Note O=√𝑂𝑥 2 + 𝑂𝑦 2 =√48.82 + 2.432 that the plunger passes through a O=48.86 N vertically-elongated hole in the handle at 3/24 A person holds a 30-kg suitcase by B, so that negligible vertical force is its handle as indicated in the figure. transmitted there. Determine the tension in each of the four identical links AB. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Step1:Free body diagram Soln. Step1:Free body diagram Step2: Force applied on the handle F=50 lb Let ‘T’ be the tension in the nail. Let Ax and Ay be the horizontal and vertical reactions at ‘A’ respectively. Taking moment about ‘A’ ∑MA=0 Step2: 50×8-2T=0 Weight of the suitcase W=60 lb T=200 lb Number of links is 4. ∑Fx=0 Let ‘T’ be the tension in each link,so there Fcos20o -Ax=0 are 2T force in the left and right. Ax = 50cos20o ∑Fy=0 Ax = 46.98 lb 60-4Tsin35o=0 ∑Fy=0 Tsin35o=60 Fsin20o +Ay-T=0 T=26.15 lb Ay =T- 50sin20o 3/25 A block placed under the head of the Ay =200- 50sin20o claw hammer as shown greatly facilitates Ay =182.9 lb the extraction of the nail. If a 200-N pull on The magnitude of the force ‘A’ exerted by the handle is required to pull the nail, the hammer head of the block. calculate the tension T in the nail and the magnitude A of the force exerted by the A=√𝐴𝑥 2 + 𝐴𝑦 2 hammer head on the block. The A=√46.982 + 182.92 contacting surfaces at A are sufficiently A=188.8 lb rough to prevent slipping. 3/26 The indicated location of the center of mass of the 3600 lb pickup truck is for the unladen condition. If a load whose center of mass is x = 400 mm behind the rear axle is added to the truck, determine the mass mL for which the normal forces under the front and rear wheels are equal. Soln. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 3/27 The wall-mounted 2.5-kg light fixture has its mass center at G. Determine the reactions at A and B and also calculate the moment supported by the adjustment thumbscrew at C. (Note that the lightweight frame ABC has about 250 mm of horizontal tubing, directed into and out of the paper, at both A and B.) Soln. Step1:Free body diagram Soln. Step1:Free body diagram Step2: Let the load weight be WL Weight of the truck W=3600 lb Let the normal forces at A and B be RA and RB respectively. Given that the normal forces under the front and rear wheels are equal. RA=RB Taking moment about ‘A’ ∑MA=0 Step2: 3600(45)+ W L(45+67+16)- RB(45+67)=0 Weight of the light W=2.5×9.81 162000+ 128W L-112RB =0 (i) W=24.53 N Considering forces along y-axis. Let Ax,Ay and Bx be the reactions at A and ∑Fy=0 B respectively. RA+RB-3600-W L=0 Taking moment about ‘A’ 2RB =3600+ W L (ii) ∑MA=0 (CW+) Using (ii) in (i) Bx×230-24.53×300=0 162000+ 128W L-56(3600+W L) =0 Bx=32 N 162000+128W L-210600-56W L=0 Considering fores along x-axis 72W L=39600 ∑Fx=0 WL=550 lb Ax-Bx=0 Putting above value in (ii) Ax-32=0 2RB =3600+550 Ax=32 N 2RB =4150 Step3:Free body diagram of fixture only RA=RB=2075 lb Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 3/29 The chain binder is used to secure loads of logs,lumber, pipe, and the like. If the tension T1 is 2 kN when θ=30o, determine the force P required on the lever and the corresponding tension T2 for this position.Assume that the surface under A is perfectly smooth. Taking moment about ‘C’due to weight ‘W’ ∑MC=0 MC=24.53×100 MC=2453 Nmm MC=24.53 Nm 3/28To test the validity ofaerodynamic Soln. assumptions made in the design of the Step1:Free body diagram aircraft, its model is being tested in a wind tunnel. The support bracket is connected to a force and moment balance, which is zeroed when there is no airflow. Under test conditions,the lift L, drag D, and pitching moment MG act as shown. The force balance records the lift, drag, and a moment MP. Determine MG in terms ofL, D, and MP. Step2: Tension T1=2 kN θ=30o Let Ay be the reaction at support ‘A’ Taking moment about ‘A’ ∑MA=0 (CW+) P×600-2×100sin30o=0 P=0.1667 kN Soln. Considering fores along x-axis Step1:Free body diagram ∑Fx=0 T2-T1+P sin30o=0 T2-25+0.1667×0.5=0 T2=1.92 kN 3/30 The device shown is designed to apply pressure when bonding laminate to each side of a countertop near an edge. If a 120-N force is applied to the handle, Step2: determine the force which each roller Taking moment about ‘P’ exerts on its corresponding surface. ∑MP=0 MP-MG-Ld-Dh=0 MG=MP-Ld-Dh=0 Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Taking moment about ‘C’ ∑MC=0 B×3.5-30×5.5cos45o=0 3.5B=116.67 B=33.43 lb Considering fores along y-axis ∑Fy=0 C-B-30=0 C-33.34-30=0 C=63.34 lb 3/31 The two light pulleys are fastened together and form an integral unit.They are prevented from turning about their bearing at O by a cable wound securely Soln. around the smaller pulley and fastened to Step1:Free body diagram point A.Calculate the magnitude R of the force supported by the bearing O for the applied 2kN load. Step2: Force applied at handle F=30 lb Soln. Let the reaction forces at B and C are RB Step1:Free body diagram and RC respectively. Considering triangle BCO Step2: Weight acting on the pulley W=2 kN OB=3.5tan45o Let ‘T’ be the tension in the cable at ‘A’ OB=3.5 Let Rx and Ry be the reactions at ‘O’ BC=√𝑂𝐵2 + 𝑂𝐶 2 Considering the triangle BC=√3.52 + 3.52 BC=4.95 Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 125 sinθ= 325 θ=22.62o Taking moment about ‘O’ ∑Mo=0 (CCW+) 2×200-T×125=0 Soln. T=3.2 kN Step1:Free body diagram Considering fores along x-axis ∑Fx=0 3.2cos22.62o+Rx=0 Rx=-2.95 kN Considering fores along y-axis ∑Fy=0 Ry-2-3.2sin22.62o=0 Ry=3.23 kN Magnitude of the reaction force ‘R’ R=√𝑅𝑥 2 + 𝑅𝑦 2 R=√(−2.95)2 + (3.23)2 R=4.37 kN 3/32 In a procedure to evaluate the Step2: strength of the triceps muscle, a person Weight of the lower arm is 3.2 lb pushes down on a load cell with the palm Let ‘F’ be the force in the triceps muscle of his hand as indicated in the figure. If the load at the palm is 35 lb load-cell reading is 160 N, determine the Taking moment about ‘O’ vertical tensile force F generated by the ∑Mo=0 (CCW+) triceps muscle. The mass of the lower arm 35×(6+6)-3.2×6-F×1=0 is 1.5 kg with mass center at G. State any -F= - 401 assumptions. F=401 lb 3/33 A person is performing slow arm curls with a 10-kg weight as indicated in the figure. The brachialis muscle group (consisting of the biceps and brachialis muscles) is the major factor in this exercise. Determine the magnitude F of the brachialis-musclegroup force and the magnitude E of the elbow joint reaction at point E for the forearm position shown in the figure. Take the dimensions shown to locate the effective points of application of the two muscle groups; these points are 200 mm directly above E and 50 mm Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. directly to the right of E. Include the effect Ey=-126.4 lb of the 1.5-kg forearm mass with mass Resultant force at ‘E’ center at point G. State any assumptions. E=√𝐸𝑥 2 + 𝐸𝑦 2 E=√(37.4)2 + (126.4)2 E=131.82 lb 3/34 A woman is holding a 3.6-kg sphere in her hand with the entire arm held horizontally as shown in the figure. A tensile force in the deltoid muscle prevents the arm from rotating about the shoulder joint O; this force acts at the 21o angle shown.Determine the force exerted by the deltoid muscle on the upper arm at A and the x- and y-components of the force reaction at the shoulder joint O. The mass of the upper arm is mU =1.9 kg, the mass of the lower arm is mL = 1.1 kg, and the Soln. mass of the hand is mH = 0.4 kg; all the Step1:Free body diagram corresponding weights act at the locations shown in the figure. Step2: Weight of the lower arm is 3.2 lb Soln. Let ‘F’ be the force is the Brichiates Step1:Free body diagram muscle load in the palm is 20 lb. tanθ=2⁄8 θ=14.04o Taking moment about ‘E’ ∑ME=0 (CCW+) Fcos14.04o(2)-G×6-20×14=0 1.94F-3.2×6-280=0 F=154.23 lb Considering fores along x-axis ∑Fx=0 Step2: Ex-Fsin14.04o=0 Weight of the upper arm Wu=1.9×9.81 N Ex-154.23sin14.04o=0 Weight of the lower arm Wl=1.1×9.81 N Ex=37.4 lb Weight of the hand W h=0.4×9.81 N Considering fores along y-axis Weight of the sphere W=3.6×9.81 N ∑Fy=0 Let ‘T’ be the tensile force acting in the Ey-3.2-20+Fcos14.04o=0 deltoid muscle. Let Ox and Oy be the reactions at joint ‘O’ Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Taking moment about ‘O’ ∑Mo=0 (CCW+) Fsin21o(125)-1.9×9.81×130-1.1×9.81 ×412- (3.6+0.4) ×9.81×6.35=0 31786.362 F=125sin21o F=710 N Considering fores along x-axis ∑Fx=0 Ox-Fcos21o=0 Ox =662.8 N Considering fores along y-axis ∑Fy=0 Oy+710sin21o-1.9×9.81-1.1×9.81- 0.4×9.81-3.6×9.81=0 Step2: Oy=-185.7 N Let ‘F’ be the force in the patellar tendon. 3/35 With his weight W equally distributed Let Ox and Oy be the reactions at ‘O’. on both feet, a man begins to slowly rise Let 𝑊 ⁄2 be the weighton one leg. from a squatting position as indicated in Taking moment about ‘O’ the figure. Determine the tensile force F in ∑Mo=0 (CCW+) the patellar tendon and the magnitude of 𝑊 the force reaction at point O, which is the F×50 - 2 ×225=0 contact area between the tibia and the F=2.25W femur. Note that the line of action of the Considering fores along x-axis patellar tendon force is along its midline. ∑Fx=0 Neglect the weight of the lower leg. Fcos55o-Ox=0 2.25cos55o-Ox=0 Ox=1.29W Considering fores along y-axis ∑Fy=0 𝑊 Fsin55o+ 2 +Oy=0 2.25sin55o+ 0.5W+Oy=0 Oy=-2.34W Magnitude of the reaction at ‘O’ O= √𝑂𝑥 2 + 𝑂𝑦 2 O= √(1.29𝑊)2 + (−2.34𝑊)2 O=2.67W 3/36 The elements of an on-off Soln. mechanism for a table lamp are shown in Step1:Free body diagram the figure. The electrical switch S requires a 4N force in order to depress it. Whatcorresponding force F must be exerted on the handleat A? Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. ∑Mo=0 (CW+) 0.9×1.2- F×(2.4cos15o)-F×(3.6cos15o)=0 1.8-2.318-3.477F=0 F=0.186 lb 3/37 The uniform 18-kg bar OA is held in the position shown by the smooth pin at O and the cable AB. Determine the tension T in the cable and the magnitude and direction of the external pin reaction at O. Soln. Step1:Free body diagram Soln. Step1:Free body diagram Step2: The force at ‘S’ is 0.9 lb Let Ox and Oy be the reactions at ‘O’. The force has been replaced by a force- couple system at B. Sstep2: Weight of the bar ‘OA’ W=18×9.81 Let ‘T’ be the tension in the cable AB Let Ox and Oy be the reactions at ‘O’. 1.5𝑠𝑖𝑛60𝑜 tanθ= 1.2+1.5𝑐𝑜𝑠60𝑜 1.299 tanθ= 1.95 θ=33.7o Taking moment about ‘O’ ∑Mo=0 (CW+) Tsin33.7o×1.5cos60o-Tcos33.7o×1.5sin60o 1.5 + 18×9.81× 2 cos60o=0 0.42T-1.08T+66.22=0 Where M=F(2.4cos15o) 0.66T=66.22 Taking moment about ‘O’ T=100 N Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Step3: Let ‘F’ be the extended force. Considering fores along x-axis The reaction ‘R’ at B is zero at the point ∑Fx=0 trepping. Ox-100c0s33.7o=0 Taking moment about ‘A’ Ox=83.45 N ∑MA=0 Considering fores along y-axis 40×7-Fcos15o×(15+8)+Fsin15o×3=0 ∑Fy=0 280-21.44F=0 Oy-18×9.81-100sin33.7o=0 F=13.06 lb Oy=232 N 3/39 The exercise machine is designed Magnitude of the reaction at ‘O’ with a lightweight cart which is mounted on small rollers so that it is O= √𝑂𝑥 2 + 𝑂𝑦 2 free to move along the inclined ramp. Two O= √(83.45)2 + (232)2 cables are attached to the cart-one for O=246 N each hand. If the hands are together so 3/38 A person attempts to move a 20-kg that the cables are parallel and if each shop vacuum by pulling on the hose as cable lies essentially in a vertical plane, indicated. What force F will cause the unit determine the force P which each hand to tip clockwise if wheel A is against an must exert on its cable in order to maintain obstruction? equilibrium position.The mass of the person is 70 kg, the ramp angle θ is 15o, and the angle β is 18o. In addition, calculate the force R which the ramp exerts on the cart. Soln. Step1:Free body diagram Soln. Step1:Free body diagram Step2: Weight of the person W=70×9.81 N Inclination of the ramp θ=15o Let ‘R’ be the force exerted by the ramp Step2: on the cart. Weight of the cart W=40 lb Let ‘P’ be the force exerted by each hand. Let Ax and Ay be the horizontal and Considering the forces along the inclined vertical reactions at ‘A’ respectively. plane. ∑Fx=0 Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. 70×9.81×sin15o-2P-2Pcos18o=0 177.73-3.90P=0 P=45.6 N Considering the forces perpendicular to the plane. R-70×9.81×sin15o-2Psin18o=0 R=663.3+2×45.6 sin18o R=691N 3/40 The device shown is used to test automobile-engine valve springs. The torque wrench is directly connected to arm OB. The specification for the automotive intake-valve spring is that 370 N of force Taking moment at ‘O’ should reduce its length from 50 mm M=Fcos20o×15 (unstressed length) to 42 mm. What is the 498=14.095F corresponding reading M on the torque F=35.3 lb wrench, and what force F exerted on the torque-wrench handle is required to produce this reading? Neglect the small effects of changes in the angular position of arm OB. Soln. Step1:Free body diagram Step2: Force exerted by the spring on ‘OB’=83 lb Let Ox and Oy be the reactions at ‘O’. Taking moment about ‘O’ ∑Mo=0 M-83×6=0 M=498 l.in Considering free body diagram of ‘OA’ Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Commonly used bridge trusses. Truss:A framework composed of members joined at their ends to form arigid structure is called a truss. e.g; Trusses of roofing system, bridge trusses, transmission towers etc. Types of trusses: There are two types of trusses. i) Plane truss ii) Space truss Plane Truss: When the members of the truss lie essentially in a single plane, the truss is called a plane truss. Space Truss: When the members of the truss do not lie in the same plane, the truss is called space truss. Type of plane trusses depending upon the arrangement of members. SimpleTruss:A truss which is constructed from a basic triangular structure to such a manner that to increase new elements, two members and one joint is added, is known as simple truss. Compound Truss: A compound truss is formed by connecting two or more simple trusses. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Complex Truss: It is a truss which cannot be classified as simple or compound. Internal Determinacy: It can be checked if minimum number of reacting Assumptions in the analysis and design of components necessary for the external trusses. determinacy and stability are known. m+r=2j 1-Members are joined together by smooth where pins, although the members are riveted, m=number of members bolted and welded. r= 2-The centre line of all the members are j= concurrent at the joint. The structure will be stable and 3-All the loads are only applied at the determinate if joints and the weight of the members is m+r > 2j assumed negligible so there will be only Sample Problem 4/1 axial forces (tension or compression). Compute the force in each member of the loaded cantilever truss by the method of Stability and Determinacy of Trusses. joints. Stability: External Stability: A truss will be external unstable if all the reactions are concurrent or parallel. Internal Stability: A truss will be internally stable if it is not liable to collapse. A simple truss is always internally stable. Indeterminacy:A truss will be indetermina- te in which all support reactions and internal forces cannot be calculated only by available equilibrium equations for a given system of forces. Degree of indeterminacy: The access of total number of reactive components or access of members over the available Soln. equilibrium equations is known as degree Step1:Free body diagram of determinacy. It is convenient to consider stability and determinacy as follows: 1-With respect to reactions. e.g; external stability and determinacy. 2-With respect to members. e.g; internal stability and determinacy. 3-The combination of internal and external conditions. External Determinacy: A determinate structure should have at least three reactions. Step2: Ne=r-N Taking moment about E, Where ∑ME=0 r=number of reactions 5T-20(5)-30(10)=0 N=number of equilibrium equations 5T=400 available. T=80 kN Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. ∑FX=0 80cos30o-EX=0 EX=69.3 kN ∑FY=0 80sin30o+EY-30-20=0 EY=10 kN Step3: Considering the joint A ∑FY=0 CD Sin60o-BC Sin60o-20=0 CD(0.866)-34.6(0.866)-20=0 0.866CD-30-20=0 0.866CD=50 CD=57.7 kN (T) ∑FX=0 BC cos60o-CD cos60o-AC-CE=0 ∑FY=0 34.6(0.5)-57.7(0.5)-17.32-CE=0 AB Sin60o-30=0 CE=63.5 kN (C) 0.866AB=30 Step6: Finally consider joint E AB=34.6 kN (T) ∑FX=0 -AC+AB cos60o=0 -AC+34.6(0.5)=0 AC=17.32 kN (C) Step4: Considering joint B ∑FY=0 -DE Sin60o+10=0 0.866DE=10 ∑FY=0 DE=11.55 kN BC Sin60o-AB Sin60o=0 ∑FX=0 0.866BC-34.6(0.866)=0 DE cos60o+63.5-69.3=0 BC=34.6 kN(C) 11.55(0.5)-5.8=0 ∑FX=0 0=0 checks BD-AB cos600-BC cos60o=0 BD-34.6(0.5)-34.6(0.5)=0 BD=34.6 kN (T) Step5: Considering joint C Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. . Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. .