Sciloperações com vetoresab - Operacoes Com Vetores

Vector Operations with SCILABBy Gilberto E. Urroz, Ph.D., P.E. Distributed by i nfoClearinghouse.com ©2001 Gilberto E. Urroz All Rights Reserved A "zip" file containing all of the programs in this document (and other SCILAB documents at InfoClearinghouse.com) can be downloaded at the following site: http://www.engineering.usu.edu/cee/faculty/gurro/Software_Calculators/Scil ab_Docs/ScilabBookFunctions.zip The author's SCILAB web page can be accessed at: http://www.engineering.usu.edu/cee/faculty/gurro/Scilab.html Please report any errors in this document to: [email protected] VECTORS WITH SCILAB 2 Operations with vectors 2 Vectors in Cartesian coordinates 5 Vector operations in SCILAB 6 Calculating 2× ×2 and 3× ×3 determinants 9 Cross product as a determinant 12 Polynomials as vector components 13 Applications of vector algebra using SCILAB Example 1 - Position vector Example 2 - Center of mass of a system of discrete particles Example 3 - Resultant of forces Example 4 – Equation of a plane in space Example 5 – Moment of a force Example 6 – Cartesian and polar representations of vectors in the x-y plane Example 7 – Planar motion of a rigid body Solution of equations – one at a time Solution of a system of linear equations using matrices 16 16 17 18 23 24 25 26 30 31 Exercises 32 Download at InfoClearinghouse.com 1 © 2001 - Gilberto E. Urroz item (a). the negative of a vector is a vector of the same magnitude. as shown in the figure below. Operations with vectors Vectors in two.com 2 © 2001 . The vector sum.. The figure below shows a vector A. subtracting vector B from vector A is equivalent to adding vectors A and (–B). respectively. velocity. is the diagonal of the parallelogram that starts at the common origin of vectors A and B. In other words. Urroz . Subtraction is accomplished by using the definition: A – B = A + (–B). item (b).e. also known as the resultant.Vectors with SCILAB A vector in two. To illustrate vector addition and subtraction refer to the figure below.. and the arrowhead indicates the direction of the vector. Download at InfoClearinghouse. and completing the parallelogram resulting from drawing lines parallel to vectors A and B at the tips of vectors B and A. Vectors in the plane or in space can be used to represent physical quantities such as the position. but with opposite sense.Gilberto E. as shown in the figure. There are two ways that you can construct the vector sum. item (d). item (c). angular velocity and acceleration or a rotating body. and acceleration of a particle. The length of the arrow represents the magnitude of the vector. and its negative –A.or three-dimensions are represented by arrows. i. (2) By showing the vectors A and B with a common origin. forces and moments. Consider two vectors in the plane or in space. displacement. This operation is illustrated in the figure below. A+B: (1) By attaching the origin of the vector B to the tip of vector A. A and B. Vectors can be added and subtracted.or three-dimensions represents a directed segment. In this case. the vector sum is the vector extending from the origin of vector A to the tip of vector B. a mathematical object characterized by a magnitude and a direction in the plane or in space. As you can see. as shown in the figure below. |A| 3 © 2001 . |A| By definition. |eA| = 1. The magnitude of a vector A is represented by |A|. The unit vector corresponding to a vector A is shown in the figure below. The projection of vector B onto vector A. The unit vector along the direction of A is defined by eA = A .A vector A can be multiplied by a scalar c. item (b).com | A | ⋅ | B | ⋅ cosθ .0. Urroz . A unit vector in the direction of A is a vector of magnitude 1 parallel to A. we see from the figure that PB/A = |B|⋅cos θ = |eA|⋅|B|⋅cos θ . The figure below illustrates the case in which c = 2. parallel and oriented in the same direction as A. or PB/A = Download at InfoClearinghouse. then the orientation of the vector c⋅A will be opposite that of A. is shown in the figure below. If θ represents the angle between the two vectors. PB/A.Gilberto E. resulting in a vector c⋅A. and whose magnitude is c times that of A. item (a). If the scalar c is negative. if A is perpendicular (or normal) to B (A⊥B). From the definition of the dot product it follows that the angle between two vectors can be found from  A•B  . such that |C| = |A|⋅ |B|⋅ sin θ. from A to B). for the right-hand rule indicates that B×A = . where θ is the angle between the vectors. of two vectors in space is defined as the vector C = A×B.e. This expression is known as the vector triple product.We will define the dot product. as shown in the figure below. C is perpendicular to both A and B). i. Notice that A•B = B•A. Obviously. cos θ = 0. the right hand thumb will point towards the orientation of C. where θ is the angle between the vectors when they have a common origin.. in space. Download at InfoClearinghouse. or vector product. B. having a common origin.. θ = arccos | A | ⋅| B | From the same definition it follows that if the angle between the vectors A and B is θ = 90o = π/2 rad. if A•B = 0. of two vectors A and B as the scalar quantity A•B = | A|⋅|B|⋅cos θ. The cross product is illustrated in the figure below. the order of the factors in a cross product affects the sign of the result. Since there could be two orientations for a vector C perpendicular to both A and B.. then A⊥B. The cross product.Gilberto E. It can be proven that the volume of the parallelepiped is obtained through the expression A•(B×C). determine a solid figure called a parallelepiped. Three vectors A. The reverse statement is also true.com 4 © 2001 . Urroz . and C. and A•C = 0 and B•C = 0 (i. i. and A•B = 0. The so-called right-hand rule indicates that if we were to curl the fingers of the right hand in the direction shown by the curved arrow in the figure (i.. or internal product.e.e.e. we need to refine the definition of C = A×B by indicating its orientation.C. j. Urroz . Ay. y-. k×i = j. These unit vectors are such that i•j = i•k = j•k =0. i×k = -j. and z-directions. j×i = -k. n terms of these unit vectors. and i×j = k. are called the Cartesian components of the vector A. respectively. and Az.Vectors in Cartesian coordinates The mathematical representation of a vector typically requires it to be referred to a specific coordinate system. we introduce the unit vectors i. and k.com 5 © 2001 . any vector A can be written as A = Ax⋅i + Ay⋅j + Az⋅k. j. k×j = -i. and k. Using a Cartesian coordinate system. Download at InfoClearinghouse.Gilberto E. j×k = i. where the values Ax. corresponding to the x-. The unit vectors i. are shown in the figure below. (Ax⋅i + Ay⋅j + Az⋅k ) = -Ax⋅i . You can enter these three-dimensional vectors as row vectors in SCILAB.com 6 © 2001 . -5. A×B = (Ax⋅i + Ay⋅j + Az⋅k )×(Bx⋅i + By⋅j + Bz⋅k) = (Bz⋅Ay – By⋅Az)⋅i + (Az⋅Bx– Bz⋅Ax)⋅j + (By⋅Ax – Ay⋅Bx)⋅k. i. ! -->A-B ans = ! 1. therefore. ! 2.By)⋅j + (Az .Bx)⋅i + (Ay . negative of vector: addition: A+B = (Ax+ Bx)⋅i + (Ay+ By)⋅j + (Az+ Bz)⋅k. cross product: Vector operations in SCILAB Consider the vectors A = 3i+5j-k and B = 2i+2j+3k.Using Cartesian components.Gilberto E. 3] A = ! B ! 3. ! --> -B Download at InfoClearinghouse. 5. The negative of these vectors are calculated as: --> -A A = ! -3. ! Addition and subtraction of vectors is straightforward: -->A+B ans = ! 5.e.1. we can also write B = Bx⋅i + By⋅j + Bz⋅k.Bz)⋅k.Az⋅k. 2. -1].Ay⋅j . dot product: magnitude: |A|2 = A•A = Ax2 + Ay2 + Az2. 2. = 5. ! 3. unit vector: eA = A/|A| = (Ax⋅i + Ay⋅j + Az⋅k)/(Ax2 + Ay2 + Az2)1/2. -->A = [3. B = [2. multiplication by a scalar. 7. A•B = (Ax⋅i + Ay⋅j + Az⋅k)•(Bx⋅i + By⋅j + Bz⋅k) = Ax⋅Bx + Ay⋅By + Az⋅Bz . subtraction: A-B = (Ax . . Urroz . and define the following vector operations: -A = . |A| = √(A•A) = (Ax2 + Ay2 + Az2)1/2. c: c⋅A = c⋅(Ax⋅i + Ay⋅j + Az⋅k ) = c⋅Ax⋅i + c⋅Ay⋅j + c⋅Az⋅k. 1..4. 3. . 2. the magnitudes of vectors A and B are calculated as: -->norm(A) ans = 5. 10.8451543 - .15. -2.10. Download at InfoClearinghouse. -->norm(eB) ans = 1. ! = ! . ! The product of the vectors by a constant is also straightforward: -->2*A. ! The following is a linear combination of the two vectors: -->5*A-3*B ans = ! 9. For example.4850713 .1690309 ! -->eB = B/norm(B) eB = ! . -5*B ans = ! 6. eA and eB) are given by: -->eA = A/norm(A) eA = ! .com 7 © 2001 . .9160798 -->norm(B) ans = 4. .Gilberto E.B = ! -2.5070926 .e.4850713 . 19. ans .7276069 ! You can check that these are indeed unit vectors by using norm: -->norm(eA) ans = 1. -3.14. ..2. For example. unit vectors parallel to A and B (i. Urroz .10. ! The magnitude of a vector is obtained by using the function norm.1231056 Unit vectors are calculated by dividing a vector by its magnitude. will produce error messages: -->A.3.*B ans = ! 6.*B' !--error 9999 inconsistent element-wise operation -->A'. i. the product of a column vector times a row vector produces a matrix. 9.. 10. ! -->B'*A ans = ! ! ! 6. -->A'*B ans = ! 6. ! . -->B*A' ans = 13. ! . ! When the multiplication symbol between two row (or column) vectors is preceded by a dot.Gilberto E. . ! -->B.The matrix multiplication of a row vector times the transpose of a second row vector (i. 15. 10...*A ans = ! 6. ! 10. -->A*B' ans = 13.3. The function sum. when applied to a vector. 9.e. e. ! 15. Following the rules of matrix multiplication (see Chapter… ).e. Urroz .2. the result is a vector whose components are the product of the corresponding components of the factors. .*B !--error 9999 inconsistent element-wise operation The dot product of two row or column vectors can be obtained by combining a term-by-term multiplication with the function sum.2. for example: Download at InfoClearinghouse. for example. 10. This is referred to as a term-by-term multiplication. 6. produces the sum of the vector components. a column vector) which will produce a dot product. . The following term-by-term multiplications. Thus. For example.g. .3. Term-by-term multiplication requires that the factors have the same dimensions. 6.com 8 © 2001 . ! .3. the term-by-term multiplication of a row vector and a column vector is not defined. ! .2. ! 10. try: -->A.2. 10. While there is a general rule to obtain the determinant of any square Download at InfoClearinghouse.com 9 © 2001 .e.*B) ans = 13. the cosine of the angle between vectors A and B is given by -->costheta = A*B'/(norm(A)*norm(B)) costheta = . To convert to degrees use: -->theta = 180/%pi*acos(costheta) theta = 57. The cosine of the angle between two vectors is obtained from the dot product of the vectors divided by the product of their magnitudes. Urroz . the result is in radians. __________________________________________________________________________________ Calculating 2×2 and 3×3 determinants The calculation of a cross product can be simplified if the cross product is written as a determinant of a matrix of 3 rows and 3 columns (also referred to as a 3×3 matrix).. i.Gilberto E. the natural angular units. a matrix with the same number of rows and columns.. the function sum is used to calculate the dot product of two vectors: -->sum(A.1974011 The calculation of cross products requires the use of determinants.0087155 Of course. A determinant is a number associated with a square matrix. i.e.-->sum(A) ans = 7. -->acos(costheta) ans = 1. -->PB_A = A*B'/norm(A) PB_A = 2. For example. In the following examples.795141 The projection of vector B over vector A is calculated by dividing the dot product of the two vectors by the magnitude of A.5329480 The corresponding angle is obtained by using the function acos. as described in the following section. com 10 © 2001 . Next. we concentrate our attention on 2×2 and 3×3 determinants. a 22 A 3×3 determinant is calculated by augmenting the determinant. The elements of a matrix are identified with two sub-indices. the first representing the row and the second the column..Gilberto E. The 2×2 determinant is. a 33  The determinants corresponding to these matrices are represented by the same arrangement of elements enclosed between vertical lines. a 2×2 and a 3×3 matrix will be represented as:  a11 a  21 a12  . Urroz . The diagram also shows the elements to be multiplied with the corresponding sign to attach to their product. and placing them to the right of column 3. a11 a 21 a12 . Download at InfoClearinghouse.e. a 22 a11 a 21 a31 a12 a 22 a 32 a13 a 23 .matrix. therefore. a 33 A 2×2 determinant is calculated by multiplying the elements in its diagonal and adding those products accompanied by the positive or negative sign as indicated in the diagram shown below. as shown in the diagram below. a11 a 21 a12 = a11 ⋅ a 22 − a12 ⋅ a 21 . we present a simple way to calculate the determinant for 2×2 and 3×3 matrices. an operation that consists on copying the first two columns of the determinant. After multiplication the results are added together to obtain the determinant. in a similar fashion as done earlier for a 2×2 determinant. i. a 22   a11 a  21  a31 a12 a 22 a32 a13  a 23 . Therefore. 5. Urroz . . The following commands shown examples of 2x2 and 3x3 determinant calculations in SCILAB. -1.com 11 © 2001 . 2.3. 5]) ans = 18. 2.1. -2 --> 4. -->det([2. -->A3x3 = [2.3. -1 --> 5. ! . 5] A2x2 = ! ! 3.Gilberto E. 4. ! 1.5. ! -->det(A3x3) ans = . -3.Therefore. 5. -->A2x2 = [3.-2. ! 5. 1] A3x3 = ! ! ! 2.1. Determinants can be calculated with SCILAB by using the function det. ! -->det(A2x2) ans = 17. a 3×3 determinant produces the following result: a11 a 21 a12 a 22 a13 a 23 = a11 ⋅ a 22 ⋅ a 33 + a12 ⋅ a 23 ⋅ a 31 + a13 ⋅ a 21 ⋅ a 32 a 31 a32 a 33 − (a13 ⋅ a 22 ⋅ a31 + a11 ⋅ a 23 ⋅ a 31 + a12 ⋅ a 21 ⋅ a 33 ). __________________________________________________________________________________ Download at InfoClearinghouse. . . 3. 4.2.6. com 12 © 2001 . 1. p = [px. v(2)]. u(1). 3. will produce the result A×B = (Bz⋅Ay – By⋅Az)⋅i + (Az⋅Bx– Bz⋅Ax)⋅j + (By⋅Ax – Ay⋅Bx)⋅k.mu] = size(u). calculates the cross product of two three-dimensional vectors: function [p] = CrossProd(u. py = det(A2). v = [-3. CrossProd. v(3). Urroz .1. SCILAB does not provide a cross-product function of its own. A2 = [ u(3). v(1). The following function.v) ans = Download at InfoClearinghouse. [nv. 4. -1].Cross product as a determinant The cross product A×B in Cartesian coordinates can be expressed as a determinant if the first row of the determinant consists of the unit vectors i. 4] u = ! 2. u(3). and k. py. = 3. v(2). . Bz Evaluation of this determinant. 1.v) //Calculates the cross-product of two vectors u and v. end A1 = [ u(2). //Vectors u and v can be columns or row vectors.3. v(3)]..Gilberto E. v(1)]. px = det(A1).mv] = size(v). as indicated above. pz = det(A3). i. if nu*mu <> 3 | nv*mv <> 3 then error('Vectors must be three-dimensional only') abort. but //they must have only three elements each. u(2). A3 = [ u(1). ! v -->CrossProd(u. ! ! .e. j. pz] //end function An application of the function CrossProd follows: -->getf('CrossProd') -->u = [2. [nu. The components of vector A and B constitute the second and third rows of the determinant. i A × B = Ax Bx j Ay By k Az . 'c') c = c With the symbolic variable c defined above. 2. 3] A = ! B ! C ! 2. you can produce the following vector operations: -->c*A. c*B. -1]. 3. = 3. . C = [1.C)') ans = 21. 2.com 13 © 2001 . ! 5. For example. In this section we present simple applications of polynomials as vector components. 2. 11. v = -3i+j+4k. the following statement creates the symbolic variable c: -->c=poly(0. . ! These operations are interpreted as u = 2i+3j-k. Urroz . ! 1. 1]. Polynomials are a special type of SCILAB objects useful in linear system analysis. = 2.5. They are presented in more detail elsewhere in this book. __________________________________________________________________________________ Polynomials as vector components Although SCILAB is not a symbolic environment. 3. some pseudo-symbolic calculations are possible using polynomials. B = [5. Simple “symbolic” variables can be defined using the following call to the SCILAB function poly. and C.c = = Download at InfoClearinghouse. As an example in SCILAB.Gilberto E. is given by the magnitude of the vector triple product A•(B×C). B.! 13. 1. ! -->abs(A*CrossProd(B. Earlier in this chapter we indicated that the volume of the parallelepiped defined by vectors A.1. try the following calculation: -->A = [2. c*(A+B). c*(A-B) ans = ! 3c ans ! 5c 2c ans ! ! 2c 3c ! 7c 2c ! = 5c ans . and u×v = 13i+j+4k. 6c .4c ! It should be kept in mind that these are not symbolic results in the most general sense of the word (i.c ! -->u. For example. as those obtained in a symbolic mathematical environment such as Maple or Mathematica).3c ! -->u +3*A.4c .*u ans = ! ! 2 10 + 10c 16 + 32c + 16c 2 3 ! . Urroz .2c ! Some operations.*u ans = ! ! 2 1 . The results obtained above.*B ans = ! ! 2 2 + 2c 2 3 ! 6c + 3c ! 2 + 4c + 2c -->A-B. c^3+2*c^2 ] u = ! ! 2 1 + c 2 3 ! 2c + c ! 1 + 2c + c -->u+c*A+3*c*B ans = ! ! 2 1 + 10c 1 + 13c + c 2 3 ! 8c + 2c + c ! Operations such as term-by-term multiplication.1 . are permitted with vectors whose components are polynomials. such as norm.2c . are threedimensional vectors whose components are polynomials.com 14 © 2001 .2c 3 . Other examples of vectors with polynomial components would be: -->u = [ 1+c.4c .e.! c 3c . in terms of the polynomial c. (c+1)^2. are not defined when a vector has polynomial components: -->norm(u) !--error 4 undefined variable : %p_norm -->norm(c*A) !--error 4 undefined variable : %p_norm Download at InfoClearinghouse. or linear combinations..*u ans = ! ! 2 3 + 3c 5 + 10c + 5c 2 3 ! . -->A.Gilberto E.2c 2 3 ! . the following command defines a new polynomial “symbolic” variable r: -->r=poly(0.1 3. ! ! .'c') c = c -->A = [2. i.. From which we can find c = 8/3. suppose that you want to determine the value of c such that vector A = 2i+cj-k is normal to vector B = -5i+3j-2k.2.5. For example. B = [-5. operations that attempt to combine two polynomial variables (i. gets redefined as a constant: -->c=2 c = 2. 3. . symbolic operations) are not allowed. which so far has been used as a polynomial variable. We will use the fact that if vectors A and B are perpendicular then their dot product is zero. -2] A = ! 2 = B .8 + 3c This translates into the equation -8 + 3c = 0.Also. __________________________________________________________________________________ Download at InfoClearinghouse.2. ! As an application of polynomials as components of vectors.e.Gilberto E. A•B = 0. c ! -->A*B' ans = . c. Urroz . 10.. -1]. Using SCILAB: -->c = poly(0.com 15 © 2001 . . in keeping with SCILAB’s numerical nature. In the next example. With the new value of c the following operation produces a constant vector: -->c*A ans = ! 6.e. the value of c.'r') r = r The following linear combination of the polynomial variables c and r fails to produce a result: -->c*A+r*B !--error 4 undefined variable : %p_a_p Assignment statements can replace the values of polynomial variables. . -->abs_r = norm(r) abs_r = 6. 5).Applications of vector algebra using SCILAB In this section we present examples of operations with 2. 5.y.164414.2. then the position vector is r = x⋅i+y⋅j+z⋅k. If the current position of the particle is P(x. -2. the position vector for this particle is r = 3i -2j+5k.z). The examples are taken from applications in different physical sciences.164414 In paper. In SCILAB this position vector is written as [3 –2 5]. -->r = [3. you can write |r| = 6. use: Download at InfoClearinghouse.-2. Example 1 .com 16 © 2001 . To determine the unit vector.Gilberto E.and 3-dimensional vectors using SCILAB functions. If a particle is located at point P(3. as illustrated in the figure below. ! To determine the magnitude of the position vector the function norm.Position vector The position vector of a particle is a vector that starts at the origin of a system of coordinates and ends at the particle’s position. Urroz .5] r = ! 3. 81111].3244428 .-->e_r = r/abs_r e_r = ! . We can write position vectors for each particle as ri = xi⋅i+yi⋅j+zi⋅k.Gilberto E. located at position Pi (xi. n ∑m i =1 i A system of four particles is illustrated in the figure below. yi. with i = 1.com 17 © 2001 . Urroz . i xi yi zi mi 1 2 3 4 5 2 -1 3 5 5 3 6 -1 4 3 5 4 2 -7 2 12 15 25 10 30 Download at InfoClearinghouse. Example 2 .4866 -0.3244 0. …. zi).4866643 - . we can write: er = r/|r| = [0.Center of mass of a system of discrete particles Consider a system of discrete particles of mass mi. Determine their center of mass. The center of mass of the system of particles will be located at a position rcm defined by n rcm = ∑m i =1 i ⋅ ri . 3.8111071 ! In paper. The following table shows the coordinates and masses of 5 particles. 2. n. y cm = i ∑ y i ⋅mi i =1 n ∑m i =1 i n . 3. ! 4. we enter vectors containing the coordinates xi. ! Then.-1.Resultant of forces Download at InfoClearinghouse.4.25.-1. i Using SCILAB. yi. 5.2. 25. (xcm.7.0869565 -->y_cm = sum(y.-7.6. 5. 2.15.First.5108696 -->z_cm = sum(z. -->m = [12.*m)/sum(m) x_cm = 3.1.*m)/sum(m) y_cm = 2. xcm = ∑ z ⋅m i =1 n i ∑m i =1 i . ycm. . 30. 6. as well as the masses mi: -->x = [2.5.com 18 © 2001 .5] x = ! 2.3] y = ! 3.*m)/sum(m) z_cm = 1.2] z = ! 5.Gilberto E. 10. ! -->y = [3.30] m = ! 12. .4. and zi. we proceed to calculate the coordinates of the center of mass. -->z = [5. 3.3. 2. 4. ! . Urroz .7391304 Example 3 . zcm) using the formulas: n xcm = ∑ x i ⋅mi i =1 n ∑m i =1 n . 15.1. these formulas are calculated using the function sum: -->x_cm = sum(x.10. we will use --> rEA = rA . 0. where ei is a unit vector in the direction of the cable where the tension acts. where rA is the position vector of point A (the tip of the vector rEA). enter rE and rA : --> rE = [ 0 0 2. are T1 = 150 N. rA = [-1 -1. Using SCILAB we would calculate the vector rEA as follows. The magnitude of the tensions in each of those cables. rEA = rA – rE. To obtain. 2. The coordinates of these points.Gilberto E.1 0]. First. T2 = 300 N. 0) and E(0. To determine a unit vector along cable EA.1j) m.5].5k) m. Download at InfoClearinghouse. Therefore. and rA = (-i –1. T3 = 200 N. EC.5 m). and T4 = 150 N.The figure below shows a vertical pole buried in the ground and supported by four cables EA. EB. are A(-1. as shown in the figure.) Writing out tension T1 Tension T1 acts along cable EA. To find the vector resultant of all those forces you need to add the four tensions written out as vectors.1 m. The tension in cable i will be given by Ti = Ti⋅ei. obtained from the figure.rE. we can write rE = (2. and rA is the position vector of point A (the origin of the vector rEA). you need to write the vector rEA = rA – rE. where the cables are attached to the pole. (The tension vectors act so that the cables pull away from point E. Urroz .0 m. -1.com 19 © 2001 . and ED. e.860k. --> T1 = 150*eEA T1 = ! -51. where rA = xA⋅i+yA⋅j+zA⋅k is the position vector of point A.344 . To find the unit vector along which T1.Gilberto E.378 -0.860] = -0. rEA = -i –1.344i . Writing the vector that joins two points in space The vector that joints points A and B. and rB = xB⋅i+yB⋅j+zB⋅k is the position vector of point B. This operation is illustrated in the figure below. yA.com 20 © 2001 .5k. eEA = rEA/|rEA| = [-0. zA) is the origin and point B(xB.73 -128.73j -128. the unit vector. (|rEA| = 2..0. is calculated as --> eEA = rEA/rEA_abs eEA = ! -0. yB.0.rEA = ! -1..93k) N.73 -128.57 -56.344 ..rA. times the unit vector eEA.1 0.57 -56. zB) the tip of the vector. -1. Finally.860 ! i.yA)⋅j+(zB . where A(xA. Download at InfoClearinghouse.0. (To verify that this is indeed a unit vector.xA)⋅i+(yB .9086079). --> rEA_abs = norm(rEA) i.= (xB . is written as rAB = rB .1j-2.e. The tension vector is calculated by multiplying the magnitude of the tension T1 = 150. ! i. Urroz ..93].e.378j -0.000).378 -0.zA)⋅k. first we find the magnitude. use the command --> norm(eEA) . You should get as a result 1. or T1 = (-51. i.57i -56.93 ! The result is the vector [-51.e. ! .5] rA = ! . zB) with respect to point A(xA. These points are: A(-1. rB = [0. |rEC|.). zA).0).yA)⋅j+(zB .1. 0. and rED. T3 = T3⋅eEC.1. C(-1. 0. -1. 1. 2.Relative position vector The previous operation can also be interpreted as obtaining the relative position vector of point B(xB. in order to obtain T2.0).5 m). T3.2 m.8 = . we can proceed to write out the vectors representing the tensions T2. and E: -->rA =[-1. First. -1. 0. 0. 0. find their magnitudes. yB.5 m).com 21 © 2001 . and written as rB/A = rB .Gilberto E. we need to write out the vectors rEB. T3 = 200 N.5 ! -->rD = [1. and calculate the unit vectors eEB = rEB/|rEB|.8. and T4 Having developed a procedure to determine the unit vector along any of the cables.5. eEC = rEC/|rEC|.zA)⋅k.rA. Thus. Urroz .1.1. ! rC . D. and T4. rE = [0. Writing out tensions T2. 0) B(0. rB = .xA)⋅i+(yB . will be written as T2 = T2⋅eEB.2.3m.1. and T4.= (xB . (Recall that T2 = 300 N. we determine the coordinates of relevant points from the figure describing the problem. The tensions. T3. we enter the position vectors for points A. and T4 = T4⋅eED. |rEB|. First. 0]. C. and |rED|.0. T3. and T4 = 150 N. 1. and eED = rED/|rED|. 0. rEC. 0]. as vectors. B.5 0. yA.5] rD = Download at InfoClearinghouse.0 m.1m.3. rC = [-1. D(1. 2. -1.5 m.1 0.0. 0.8 m. 0. The following SCILAB commands perform the required calculations. and E(0.0 m. 0]. ! ! . -1. 2 rE 1.0626786 The unit vectors along the cables.8162789 ! Next.Gilberto E. eEC = rEC/abs_rEC. 0. eEC = rEC/|rEC|.5 abs_rED = 3.8089144 abs_rEC = 2.com 22 © 2001 .! 1.1. respectively: -->T2m = 300. ! ! = 0.5 ! Next. and |rED|.5 ! The magnitudes of the vectors.rE rEB = ! . T2 = 300 N. rEC = rC .4244650 - . rED = rD .2..8 = . T2m. ! . = T3m 200.2. T4m = 150 T2m = 300. Urroz . and rED: -->rEB = rB . and eED = rED/|rED|.3 0. .3560094 - .6 eED = ! 0. rEC.3918139 - .2848075 = eEC ! . abs_rEC = norm(rEC).3 rEC 1. 2.rE.8 ! . These are identified in SCILAB as T1m. are calculated as: -->eEB = rEB/abs_rEB.1. and T4 = 150 N. T3m = 200. T3 = 200 N.5 ! . ! 1.8900236 ! - . ! . eEB = rEB/|rEB|. . are calculated next: -->abs_rEB = norm(rEB). abs_rED= norm(rED) abs_rEB = 2.rE. we enter the magnitudes of the tensions. |rEC|.2.5 rED = 0.2 . eED = rED/abs_rED eEB = ! . we calculate the relative position vectors. and T3m. rEB. |rEB|. Download at InfoClearinghouse. = T4m 150. The function PlaneEquation.com 23 © 2001 .67j –122.442263 = ! . can be obtained given a point on the plane A(xA.77i +63. Urroz .80j –267.yA.44i -106. Example 4 – Equation of a plane in space The equation of a plane in space. and T4 = T4⋅eED.y. T3 = (-120i –160k) N. namely.Gilberto E. ! 63. themselves. calculates the right-hand side of the equation.669757 .160. or (nx⋅i+ny⋅j+nz⋅k)•((x-xA)⋅i+(y-yA)⋅j+(z-zA)⋅k) = 0. We can form a relative position vector rP/A = rP – rA = (x-xA)⋅i+(y-yA)⋅j+(z-zA)⋅k. are calculated by using T2 = T2⋅eEB. which results in the equation nx⋅(x-xA) + ny⋅(y-yA) + nz⋅(z-zA) = 0. n•rA. T3 = T3⋅eEC. as shown in the figure below. T4 = ! 0.01k) N. which is contained in the plane.z) be a generic point on the plane of interest.00707 ! . turns out to be R = (-27. Because the vectors n and rP/A are perpendicular to each other. The resultant of these four forces. T3 = T3m*eEC.122. In SCILAB we use: -->T2 = T2m*eEB. 58.44k) N.zA).38k) N.120. we can write n•rP/A = 0.86j –678.106. and uses the function string Download at InfoClearinghouse. and a vector normal to the plane n = nx⋅i+ny⋅j+nz⋅k. in Cartesian coordinates. Let P(x. T4 = T4m*eED T2 = ! T3 85. To implement a SCILAB function to produce the equation of a plane in space we use the fact that the equation can be re-written as nx⋅x + ny⋅y + nz⋅z = n•rA. shown below.The tension vectors.44184 ! The results are: T2 = (85.772083 . then.267.80283 . R = T1 + T2 + T3 + T4. and for cable ED.36i –99. T4 = (58. = . eq = string(n(1))+'*x+'+string(n(2))+'*y+'+string(n(3))+'*z='+string(c) //end function The function requires as input the normal vector n and the point A (referred to as p in the function). Let r be the position vector of point P with respect to the Download at InfoClearinghouse.33. function [eq] = PlaneEquation(n. z = -2 x = y 3. y = 6. -2] -->myEquation = PlaneEquation(n. -->3*x-5*y+6*z ans = .A) myEquation = 3*x+-5*y+6*z=-33 To verify that the equation is satisfied by point A use: -->x = 3. = z 6. 6. A = [3. to find the equation of the plane with normal vector n = 3i-5j+6k passing through point A(3. The final result of the function is a string representing the equation. 6].p) //This function produces the equation of a plane //in space given the normal vector n and the point p.2. Suppose that the body is allowed to rotate about point O.Gilberto E. end c = n*p'.com 24 © 2001 .to convert numerical results to strings.mp] = size(p). -5. -2) use: -->getf('PlaneEquation') -->n = [3. both entered as SCILAB vectors. 6. [nn.mn] = size(n). [np. Urroz . For example. if nn*mn <> 3 | np*mp <> 3 then error('Vector n or point p in PlaneEquation must have 3 elements') abort. Example 5 – Moment of a force The figure below shows a force F acting on a point P in a rigid body. |M1| = M1 = 201.5] -->T1 = [-57.e.775 0].point of rotation O..825 i –143. To find the magnitude of the moment. and y/r = sin θ. If we use polar coordinates. the vector rOE = (2.5k) m. subtracted. we will use as the position vector r of the force’s line of action through the pole. Let its magnitude be r = |r|. The moment of the force F about point O is defined as M = r×F. about point O.954 m⋅N. as well as the resultant moment from all four tensions. thus.T1) Enter rOE Enter T1 Calculate M1 = rOE×T1 The result is [141.825 –143. Note: Moments are vector quantities and obey all rules of vectors. The moment M1 = rOE×T1. can be calculated as follows: -->rOE = [0 0 2.Gilberto E.com 25 © 2001 . we recognize x/r = cos θ. if we want to calculate the moment of tension T1 = (51.93] -->M1 = CrossProd(r0E. A unit vector along the direction of r is given by er = r/r = (x/r)⋅i+(y/r)⋅j. undergo internal and external vector products. the reader may want to calculate the moments corresponding to the other tensions in Example 3. As an exercise. Urroz . i. which we make coincide with the origin of our Cartesian coordinate system. multiplied by a scalar. Example 6 – Cartesian and polar representations of vectors in the xy plane A position vector in the x-y plane can be written simply as r = x⋅i+y⋅j.73j -128.775 j ) m⋅N. use [!][ABS]. thus. Referring to the figure of Example 3. Download at InfoClearinghouse.51 –56.57i -56. we can write the unit vector as er = r/r = cos θ⋅i+ sin θ⋅j. or M1 = (141.93k) N. they can be added.73 –128. we can easily put together the vector as r = r⋅er = r⋅ (cos θ⋅i+ sin θ⋅j). aA): vB = vA + ωAB × r B/A . aB = aA + αAB × r B/A . This result is illustrated in the figure below.7555802 ! Example 7 – Planar motion of a rigid body In the study of the planar motion of a rigid body the following equations are used to determine the velocity and acceleration of a point B (vB. if we are given the magnitude. you would use. aB) given the velocity and acceleration of a reference point A (vA. for example: -->r = 5. r.Thus. θ. you need to transform it to radians in the trigonometric functions. its polar coordinates (r.7040969 ! In this case the angle represents radians.ωAB2 ⋅ r B/A.e.75)] r = ! 1.5*[cos(0. and the direction. sin(0. for example: -->r = 2. of a vector (i.8292222 1.. If you want to use an angle in degrees.75). To enter a vector in SCILAB.4098501 2. Download at InfoClearinghouse.2*[cos(32*%pi/180).com 26 © 2001 . sin(32*%pi/180)] r = ! 4. given the magnitude r and the angle θ. Urroz .Gilberto E.θ)). Urroz . and the relative position vector of point B with respect to point A.The equations also use the angular velocity of the body connecting A and B.com 27 © 2001 . the angular acceleration of the body connecting A and B.Gilberto E. You are asked to determine the angular velocity and acceleration of bar BC and the linear velocity and acceleration of piston C. In this mechanism there are two bars AB and AC pin-connected at B. we use the data from the mechanism shown in the figure below. At the instant shown the angular velocity and acceleration of bar AB are 10 rad/s and 20 rad/s2 clockwise. and bar BC is attached through a pin to piston C. Bar AB is pin supported at A. The following figure illustrates the calculation of relative velocity and acceleration in planar motion of a rigid body. Piston C is allowed to move in the vertical direction only. ωAB. rB/A. αAB. whose magnitude is ωAB. To present applications of this equation using the HP 49 G calculator. Download at InfoClearinghouse. Urroz . ωAB = ±ωAB⋅k. alBC = [0. i.'alBC') wBCm = wBCm alBCm = alBC With the variables wBCm and alBCm we can define the vectors ωBC and αBC as: -->wBC = [0.4o. alBCm] wBC = ! 0 alBC ! 0 0 wBCm ! 0 alBC ! = Relative position vector The relative position vectors of interest in this problem are rB/A and rC/B. ! 0. alphaAB ! .Angular velocity and acceleration Angular velocities and accelerations in the x-y plane are represented as vectors in the zdirection.20. the angle θC/B = 90o – 13. The m in the polynomial variable names stands for ‘magnitude’: -->wBCm = poly(0. Thus. and αBC = (αBC⋅k). therefore. 0] rB_A = ! - . and αAB = ±αAB⋅k.Gilberto E. ωBC = (ωBC⋅k) rad/s.6o = 76. the angle corresponding to vector r B/A is θB/A = 90o + 45o = 135o. For the data in this problem we can write. In general. we will write: -->wAB = [0. we use the fact that for the xy coordinate system shown. alphaAB = [0. 0. . Using SCILAB. Thus. normal to the x-y plane. 0.com 28 © 2001 . For rC/B . which are obtained as follows: To obtain rB/A .1767767 0. r B/A = (-0. sin(135*%pi/180).10.177j)ft. in SCILAB. wBCm]. = The next statements define the polynomial variables wBCm (standing for ωBC) and alBCm (standing for αBC). we would write. therefore. 0. -20] wAB = ! 0. ! 0. αAB = (-20k)rad/s2. alBCm = poly(0. we can write -->rB_A = 0. Download at InfoClearinghouse. ! In paper. -10].1767767 .25*[cos(135*%pi/180).177i + 0.e. 0..'wBCm'). thus we can write. ωAB = (-10k) rad/s. 0. These vectors are positive if the angular velocity or acceleration is counterclockwise. this operation is written as: -->vC = vB + CrossProd(wBC.4*%pi/180).4*%pi/180). ! In paper. r C/B = (0. the velocity of point C can be written as vC = (vC⋅j) ft/s. vB = (1. we can write the system of equations: Download at InfoClearinghouse.com 29 © 2001 .175i + 0.1763566 . Since the x. sin(76.rC_B) vC = ! 1. we use vC = vB + ωBC × rC/B = (-1. ! Thus.729j)ft.175i + 0. Urroz .177i + 0.7289708wBCm 1.77j) ft/s.767767 0. rad⋅m = m. The figure above shows that the piston C is forced to move in the vertical direction.767767 + .177j)ft.75*[ cos(76. we would write: -->getf('CrossProd') -->vA = 0 vA = 0. vA = 0.77j) ft/s + (ω ωBC⋅k) rad/s × (0.767767 1.729⋅ωBC)⋅i + (1.rB_A) vB = ! 1. we write. -->vB = vA + CrossProd(wAB. Note: Radians are basically dimensionless units. To find the velocity of point C.Gilberto E.1763566wBCm 0 ! The result can be written in paper as: vC = ((1. 0] rC_B = ! . and we can write vB = ωAB× rB/A = (-10k) rad/s × (-0.729⋅ωBC)⋅i + (1.-->rC_B = 0. thus.77-0.77i+1.77+0.767767 - .and y-components of the two vectors in each side of the equal sign must be the same. Equating the two results presented immediately above for vC we get: (1. Velocity Since point A is fixed point.7289708 0. Using SCILAB.176⋅ωBC)⋅j) ft/s.77-0.77+0.176⋅ωBC) ⋅j = 0⋅⋅i + vC⋅j. thus rad⋅ft = ft.729j)ft.77i-1. With SCILAB. 4279835 2. Thus. -->aB = aA + CrossProd(alphaAB.77+ 0.43 rad/s.e.10. ωBC = 1.729.142 0. we can solve for the two unknowns (ωBC and vC) as follows: 1.213203 .14.176*2.77-0.77/0. -->vCm = 1.142j)ft/s2.729 wBCm = 2. The positive sign in vC means that point C is moving upwards in the vertical direction.20 ft/s. the solution of the system of equations is: ωBC = 2.19768 Thus.77+(0.77/0. i..729⋅ωBC = 0 1. ! The result is [21.com 30 © 2001 .e. Acceleration Again. The positive sign in ωBC means that the angular velocity is counterclockwise. From the second equation.24i -14. vC = 1.43). and vC = 2.177i + 0.176⋅ωBC = vC Solution of equations – one at a time Using SCILAB.77+0.ωAB2 ⋅ r B/A = (-20k)rad/s2 × (-0.. because A is a fixed point.000].177j)ft – (10 rad/s)2× (-0.177j)ft Using SCILAB we can obtain aB as follows: -->aA = 0 aA = 0. -->wABm = -10 wABm = .43 vCm = 2. aA = 0.Gilberto E.240 -14. Urroz .176)⋅(2. -->wBCm = 1.1.142136 0. i. From the first equation.rB_A)-wABm^2*rB_A aB = ! 21.177i + 0. or aB = (21. To calculate the acceleration of point C we use: Download at InfoClearinghouse. the acceleration of point B is given by aB = αAB × r B/A . as a matrix equation: 1 − 0. because the motion of point C is in the vertical direction.729  α  =  20.729].16j)ft/s2 +(αBCk)× (0.e. however. where A-1 is the inverse of matrix A. The equations are first re-written as ac .729j)ft .wBCm^2*rC_B aC = ! 20.176⋅αBC = aC.729⋅αBC)⋅i+(-18.175i + 0. However.729j)ft In Scilab: -->aC = aB + CrossProd(alphaBC.176 ! . This result follows from the original matrix equation. we would enter: -->A = [1. this “division” is not a regular arithmetic division.44 0. Urroz . is exactly the same.44.7289708alBC .729⋅αBC = 0.17] A = ! ! 1.44+0.ωBC2⋅rC/B = (21. Solution of a system of linear equations using matrices The system of linear equations in two unknowns (αBC and aC) obtained above can be re-written in matricial form as follows: 1. -0. b = [-18. 0. by “dividing” both sides of the equation by A.24i -14. The result.0. Typically.17-0.aC = aB + αBC × rC/B .729 ! Download at InfoClearinghouse.rC_B) .17 or.729⋅αBC = 20. - . provide the user with the “backward-slash” operator instead of using the inverse matrix. Using SCILAB.. such as SCILAB.44 0 0.175i + 0. -18. Equating the two expressions for the vector aC shown above. we get the following equations: 20.176  a C  − 18. i.42 rad/s)2⋅(0.44+0.(-2. Also.173563 - .176⋅αBC = -18. this operation would be represented as x = A-1b. x = A\b. Modern matrix-based numerical environments.439494 + . we can write aC = aC⋅j. 0.1763566alBC 0 ! The result is: aC = ((20. since A is a matrix. A⋅x = b.17-0.17     BC    2. 0.Gilberto E.176.com 31 © 2001 .176⋅αBC))⋅j) ft/s2.18. This matrix equation can be solved by using the backward-slash operator (\).20. -1] Download at InfoClearinghouse.5. i. r = (-2i+5j-3k) m (b) F = (i-4j-3k) lb. r = (-2i+5j-3k)/2 ft (b) F = 200(i-4j-k) N. Determine the resultant force from the four cables. v = -i-3k.13. s = -3i-2j-5k.-1).-2). r.com 32 © 2001 . ev.17 ! -->A\b ans = ! .b = ! . The tensions in the four cables are: AE = 150 lb.w.0) on a structure.5. Determine the equation of the plane through point A with normal vector n. BE = 250 lb. B(2.57 ft/s2.-1). [3]. ew (k) u • v (n) u×v (q) (s×t) • r (t) (u×v)×(s×t) [2]. We have presented here two methods for solution of systems of linear equations.Gilberto E.3.4).5. and v.-5). n = [5. w = 5i-10j-3k. Exercises For the following exercises use the Cartesian vectors: u = 3i+2j-5k. and D(2.570425 ! ! 27.5). M = r×F: (a) F = (3i+2j-5k) N.44 ! ! 20. The negative sign in ac indicates that point C is decelerating.3.0. Urroz .2. r = 3(i+j-10k) m [4].-1.3.-1).-2.3. αBC = 27. r = (i+8j-13k) ft (a) F = 100(i+j-k) lb. These methods.67 rad/s2.-1] (c) A(2. n = [1. r = -8i+10j-2. Determine the result of the following operations: (a) |u| (b) |w| (d) |s|/|w| (e) a = u + v (g) c = 3r-2v (h) d = -t + 2s (j) angles between vectors u. 1. The four cables are anchored to points A(-1. -1. and others. Sketch the plane: (a) A(-2.5k.v. CE = 100 lb. Four different cables are attached to point E(0. DE = 50 lb.s. are presented in more detail in a different chapter. (m) s×t (l) w • r (o) r × w (p) u •(v×w) (r) w×(r×u) (s) (u×v)×t (c) |v||r| (f) b = r-t (i) unit vectors: eu. t = 6i-2j+15k. n = [3..18.3] (b) A(0.668038 ! The results are interpreted as aC = -13. C(-2. Determine the torque of the following forces F given the arm r. 2). The positive sign in αBC indicates that bar BC is accelerating angularly in the counterclockwise direction. [1]. n = [2. 1] (d) A(-1.e. define the vectors as follows: --> n = [2.2 ft/s and its acceleration is ac = -0.0. m = [5. determine the angular velocity and acceleration of bars AB and BC.-1] (b) n = [x.y]. e..5. m = [-1.y. for (a): -->y = poly(0.2 ft/s2.5.[5].-2].2. y.g. m = [5.1]. Calculate the dot product as a polynomial: --> p = n*m’ and solve for y using function roots: --> roots(p) [6].2] (d) n = [5.Gilberto E. m = [x. m = [3.3] Note: define the unknown variable as a SCILAB polynomial variable. Two vectors n and m are said to be orthogonal if n • m = 0.‘y’) Then. Determine the missing components in the following vectors m and n so that they are orthogonal: (a) n = [2.4].3.-2.5.2] (c) n = [4. Urroz . For the mechanism presented in Example 7.2].5. if the velocity of point C is vc = 1.com 33 © 2001 .-2]. Download at InfoClearinghouse. and G. Gullberg.S." John Wiley & Sons.." Dover Publications. Massachussetts. 1989." Brooks/Cole . A. Graphs.. and R."Random Functions and Hydrology. Friedman. "Partial Differential Equations for Scientists and Engineers. The University of Iowa.”Introduction to Geostatistics . New York. D. Inc." Dover Publications. 1971. J. "Advanced Engineering Mathematics . K. E. "Handbook of Mathematics and Computational Science.L.." Prentice Hall. Arora." The Mc-Graw Hill Companies. Brogan.B. Harman." Prentice Hall.Second edition. (editor). 1965. T.S.. P. J.J. C. Upper Saddle River.. and J. Upper Saddle River. "Principles and Techniques of Applied Mathematics. Statistics. 1969. 1999.” Prentice Hall.C.Towards a deterministic approach to turbulence. Pomeau.” Birkhäuser. McGraw-Hill. T.W. 1999.S..L.P. Azarm. Herold. and T. Upper Saddle River. "Schaum's Outline of Theory and Problems of Physics for Engineering and Science. Kreysig. Stocker." QPI series.. D. and C. 2000." Harcourt Brace Jovanovich College Outline Series.A. Reading. 2000. Balachandran. Middleton. New Jersey."Handbook of Mathematical Functions with Formulas. 1997. Cambridge CB2 2RU. J. J. New Jersey."Modern Control Theory." Dover Publications. New York. Dabney. Stegun (editors). "Data Analysis in the Earth Sciences Using Matlab®." Dover Publications Inc. 1998.. U.. Berge. 1983. G. G.Applications in Hydogeology. "Mathematical Handbook for Scientists and Engineers. W..M.. "Numerical Methods Using Matlab®. "Applied Fourier Analysis. Browne.. New York.K."Order within chaos . Washington.com 34 © 2001 . Stanley J. P. B. Australia. Kottegoda. 2000." Dover Publications Inc. Gomez. "Introduction to Optimum Design. 1985. and Reliability for Civil and Environmental Engineers. Thailand.L. J. Magrab. 1997. Walsh.”Hydrologic Analysis and Design .. New York. Duncan. Iowa. B.Y. and R. 1984. 1988. U.. Koch. New York. "Principles of Applied Mathematics .. Inc. W. P. Link. H. American Geophysical Union. Download at InfoClearinghouse." AIT .. Korn.From the Birth of Numbers." W. G. 1989. and H.." Addison-Wesley Publishing Company. and Mathematical Tables. Inc. 1982.A.. Farlow.. "Probability.. U.." Springer. Harris. New York. Kitanidis. E. Rosso. Julien. Y.” Cambridge University Press.P." Class notes.. "Hydraulic Laboratory Manual. R.. J." Addison-Wesley Publishing Company. New York. "Advanced Engineering Mathematics with MATLAB® . Hsu. Richert.. 1998.Thompson Learning.” Cambridge University Press. and N. Norton & Company.K. Iowa City. New York.com) Abramowitz. McCuen. Quantum Publisher Incorporated.. New York.”Erosion and Sedimentation.Volumes I and II. Washington. Publishers. "Fundamentals of Geostatistics in Five Lessons. “Engineering and Scientific Computing with Scilab. San Diego. Asian Institute of Technology.Bangkok. Harcourt Brace Jovanovich. Upper Saddle River. Keener. "An Engineer's Guide to MATLAB®".. 1997.V. 1984.H. Penny. M. Inc." Schaum's outlines.C. S. New York. New York. New York.second edition. California. Vidal. and I.REFERENCES (for all SCILAB documents at InfoClearinghouse. Prentice Hall.. Rodriguez-Iturbe. and I. G." 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