Vector Operations with SCILAB. Gilberto E. Urroz, Ph.D., P.E. Distributed by. i nfoclearinghouse.com Gilberto E. Urroz All Rights Reserved

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1 Vector Opertions with SCILAB By Gilberto E. Urroz, Ph.D., P.E. Distributed by i nfocleringhouse.com 2001 Gilberto E. Urroz All Rights Reserved

2 A "zip" file contining ll of the progrms in this document (nd other SCILAB documents t InfoCleringhouse.com) cn be downloded t the following site: b_docs/scilbbookfunctions.zip The uthor's SCILAB web pge cn be ccessed t: Plese report ny errors in this document to: gurro@cc.usu.edu

3 VECTORS WITH SCILAB 2 Opertions with vectors 2 Vectors in Crtesin coordintes 5 Vector opertions in SCILAB 6 Clculting 2 2 nd 3 3 determinnts 9 Cross product s determinnt 12 Polynomils s vector components 13 Applictions of vector lgebr using SCILAB 16 Exmple 1 - Position vector 16 Exmple 2 - Center of mss of system of discrete prticles 17 Exmple 3 - Resultnt of forces 18 Exmple 4 Eqution of plne in spce 23 Exmple 5 Moment of force 24 Exmple 6 Crtesin nd polr representtions of vectors in the x-y plne 25 Exmple 7 Plnr motion of rigid body 26 Solution of equtions one t time 30 Solution of system of liner equtions using mtrices 31 Exercises 32 Downlod t InfoCleringhouse.com Gilberto E. Urroz

4 Vectors with SCILAB A vector in two- or three-dimensions represents directed segment, i.e., mthemticl object chrcterized by mgnitude nd direction in the plne or in spce. Vectors in the plne or in spce cn be used to represent physicl quntities such s the position, displcement, velocity, nd ccelertion of prticle, ngulr velocity nd ccelertion or rotting body, forces nd moments. Opertions with vectors Vectors in two- or three-dimensions re represented by rrows. The length of the rrow represents the mgnitude of the vector, nd the rrowhed indictes the direction of the vector. The figure below shows vector A, nd its negtive A. As you cn see, the negtive of vector is vector of the sme mgnitude, but with opposite sense. Vectors cn be dded nd subtrcted. To illustrte vector ddition nd subtrction refer to the figure below. Consider two vectors in the plne or in spce, A nd B, s shown in the figure below, item (). There re two wys tht you cn construct the vector sum, A+B: (1) By ttching the origin of the vector B to the tip of vector A, s shown in the figure, item (b). In this cse, the vector sum is the vector extending from the origin of vector A to the tip of vector B. (2) By showing the vectors A nd B with common origin, nd completing the prllelogrm resulting from drwing lines prllel to vectors A nd B t the tips of vectors B nd A, respectively, s shown in the figure below, item (c). The vector sum, lso known s the resultnt, is the digonl of the prllelogrm tht strts t the common origin of vectors A nd B. Subtrction is ccomplished by using the definition: A B = A + ( B). In other words, subtrcting vector B from vector A is equivlent to dding vectors A nd ( B),. This opertion is illustrted in the figure below, item (d). Downlod t InfoCleringhouse.com Gilberto E. Urroz

5 A vector A cn be multiplied by sclr c, resulting in vector c A, prllel nd oriented in the sme direction s A, nd whose mgnitude is c times tht of A. The figure below illustrtes the cse in which c = 2. If the sclr c is negtive, then the orienttion of the vector c A will be opposite tht of A. The mgnitude of vector A is represented by A. A unit vector in the direction of A is vector of mgnitude 1 prllel to A. The unit vector corresponding to vector A is shown in the figure below, item (). The unit vector long the direction of A is defined by A e A =. A By definition, e A = 1.0. The projection of vector B onto vector A, P B/A, is shown in the figure below, item (b). If θ represents the ngle between the two vectors, we see from the figure tht P B/A = B cos θ = e A B cos θ, or P B/A A B cosθ =. A Downlod t InfoCleringhouse.com Gilberto E. Urroz

6 We will define the dot product, or internl product, of two vectors A nd B s the sclr quntity A B = A B cos θ, where θ is the ngle between the vectors when they hve common origin. Notice tht A B = B A. From the definition of the dot product it follows tht the ngle between two vectors cn be found from A B θ = rccos. A B From the sme definition it follows tht if the ngle between the vectors A nd B is θ = 90 o = π/2 rd, i.e., if A is perpendiculr (or norml) to B (A B), cos θ = 0, nd A B = 0. The reverse sttement is lso true, i.e., if A B = 0, then A B. The cross product, or vector product, of two vectors in spce is defined s the vector C = A B, such tht C = A B sin θ, where θ is the ngle between the vectors, nd A C = 0 nd B C = 0 (i.e., C is perpendiculr to both A nd B). The cross product is illustrted in the figure below. Since there could be two orienttions for vector C perpendiculr to both A nd B, we need to refine the definition of C = A B by indicting its orienttion. The so-clled right-hnd rule indictes tht if we were to curl the fingers of the right hnd in the direction shown by the curved rrow in the figure (i.e., from A to B), the right hnd thumb will point towrds the orienttion of C. Obviously, the order of the fctors in cross product ffects the sign of the result, for the right-hnd rule indictes tht B A = - C. Three vectors A, B, nd C, in spce, hving common origin, determine solid figure clled prllelepiped, s shown in the figure below. It cn be proven tht the volume of the prllelepiped is obtined through the expression A (B C). This expression is known s the vector triple product. Downlod t InfoCleringhouse.com Gilberto E. Urroz

7 Vectors in Crtesin coordintes The mthemticl representtion of vector typiclly requires it to be referred to specific coordinte system. Using Crtesin coordinte system, we introduce the unit vectors i, j, nd k, corresponding to the x-, y-, nd z-directions, respectively. The unit vectors i, j, nd k, re shown in the figure below. These unit vectors re such tht nd i j = i k = j k =0, i j = k, j k = i, k i = j, i k = -j, k j = -i, j i = -k. n terms of these unit vectors, ny vector A cn be written s A = A x i + A y j + A z k, where the vlues A x, A y, nd A z, re clled the Crtesin components of the vector A. Downlod t InfoCleringhouse.com Gilberto E. Urroz

8 Using Crtesin components, therefore, we cn lso write B = B x i + B y j + B z k, nd define the following vector opertions: negtive of vector: -A = - (A x i + A y j + A z k ) = -A x i - A y j - A z k. ddition: A+B = (A x + B x ) i + (A y + B y ) j + (A z + B z ) k. subtrction: multipliction by sclr, c: A-B = (A x - B x ) i + (A y - B y ) j + (A z - B z ) k. c A = c (A x i + A y j + A z k ) = c A x i + c A y j + c A z k. dot product: A B = (A x i + A y j + A z k) (B x i + B y j + B z k) = A x B x + A y B y + A z B z. mgnitude: A 2 = A A = A x 2 + A y 2 + A z 2 ; A = (A A) = (A x 2 + A y 2 + A z 2 ) 1/2. unit vector: e A = A/ A = (A x i + A y j + A z k)/(a x 2 + A y 2 + A z 2 ) 1/2. cross product: A B = (A x i + A y j + A z k ) (B x i + B y j + B z k) = (B z A y B y A z ) i + (A z B x B z A x ) j + (B y A x A y B x ) k. Vector opertions in SCILAB Consider the vectors A = 3i+5j-k nd B = 2i+2j+3k. You cn enter these three-dimensionl vectors s row vectors in SCILAB, i.e., -->A = [3, 5, -1], B = [2, 2, 3] A =! ! B =! ! Addition nd subtrction of vectors is strightforwrd: -->A+B! ! -->A-B! ! The negtive of these vectors re clculted s: --> -A A =! ! --> -B Downlod t InfoCleringhouse.com Gilberto E. Urroz

9 B =! ! The product of the vectors by constnt is lso strightforwrd: -->2*A, -5*B! !! ! The following is liner combintion of the two vectors: -->5*A-3*B! ! The mgnitude of vector is obtined by using the function norm. For exmple, the mgnitudes of vectors A nd B re clculted s: -->norm(a) >norm(b) Unit vectors re clculted by dividing vector by its mgnitude. For exmple, unit vectors prllel to A nd B (i.e., e A nd e B ) re given by: -->ea = A/norm(A) ea =! ! -->eb = B/norm(B) eb =! ! You cn check tht these re indeed unit vectors by using norm: -->norm(ea) 1. -->norm(eb) 1. Downlod t InfoCleringhouse.com Gilberto E. Urroz

10 The mtrix multipliction of row vector times the trnspose of second row vector (i.e., column vector) which will produce dot product, i.e., -->A*B' >B*A' 13. Following the rules of mtrix multipliction (see Chpter ), the product of column vector times row vector produces mtrix, e.g., -->A'*B! !! !! ! -->B'*A! !! !! ! When the multipliction symbol between two row (or column) vectors is preceded by dot, the result is vector whose components re the product of the corresponding components of the fctors. This is referred to s term-by-term multipliction. For exmple, try: -->A.*B! ! -->B.*A! ! Term-by-term multipliction requires tht the fctors hve the sme dimensions. Thus, the term-by-term multipliction of row vector nd column vector is not defined. The following term-by-term multiplictions, for exmple, will produce error messges: -->A.*B'!--error 9999 inconsistent element-wise opertion -->A'.*B!--error 9999 inconsistent element-wise opertion The dot product of two row or column vectors cn be obtined by combining term-by-term multipliction with the function sum. The function sum, when pplied to vector, produces the sum of the vector components, for exmple: Downlod t InfoCleringhouse.com Gilberto E. Urroz

11 -->sum(a) 7. In the following exmples, the function sum is used to clculte the dot product of two vectors: -->sum(a.*b) 13. The cosine of the ngle between two vectors is obtined from the dot product of the vectors divided by the product of their mgnitudes. For exmple, the cosine of the ngle between vectors A nd B is given by -->costhet = A*B'/(norm(A)*norm(B)) costhet = The corresponding ngle is obtined by using the function cos. -->cos(costhet) Of course, the result is in rdins, the nturl ngulr units. To convert to degrees use: -->thet = 180/%pi*cos(costhet) thet = The projection of vector B over vector A is clculted by dividing the dot product of the two vectors by the mgnitude of A, i.e., -->PB_A = A*B'/norm(A) PB_A = The clcultion of cross products requires the use of determinnts, s described in the following section. Clculting 2 2 nd 3 3 determinnts The clcultion of cross product cn be simplified if the cross product is written s determinnt of mtrix of 3 rows nd 3 columns (lso referred to s 3 3 mtrix). A determinnt is number ssocited with squre mtrix, i.e., mtrix with the sme number of rows nd columns. While there is generl rule to obtin the determinnt of ny squre Downlod t InfoCleringhouse.com Gilberto E. Urroz

12 mtrix, we concentrte our ttention on 2 2 nd 3 3 determinnts. Next, we present simple wy to clculte the determinnt for 2 2 nd 3 3 mtrices. The elements of mtrix re identified with two sub-indices, the first representing the row nd the second the column. Therefore, 2 2 nd 3 3 mtrix will be represented s: , The determinnts corresponding to these mtrices re represented by the sme rrngement of elements enclosed between verticl lines, i.e., , A 2 2 determinnt is clculted by multiplying the elements in its digonl nd dding those products ccompnied by the positive or negtive sign s indicted in the digrm shown below. The 2 2 determinnt is, therefore, = A 3 3 determinnt is clculted by ugmenting the determinnt, n opertion tht consists on copying the first two columns of the determinnt, nd plcing them to the right of column 3, s shown in the digrm below. The digrm lso shows the elements to be multiplied with the corresponding sign to ttch to their product, in similr fshion s done erlier for 2 2 determinnt. After multipliction the results re dded together to obtin the determinnt. Downlod t InfoCleringhouse.com Gilberto E. Urroz

13 Therefore, 3 3 determinnt produces the following result: = ( + + ) Determinnts cn be clculted with SCILAB by using the function det. The following commnds shown exmples of 2x2 nd 3x3 determinnt clcultions in SCILAB. -->A2x2 = [3, -1; 2, 5] A2x2 =! !! 2. 5.! -->det(a2x2) >A3x3 = [2,3, -1 --> 5,5, -2 --> 4, -3, 1] A3x3 =! !! !! ! -->det(a3x3) >det([2,-2; 4, 5]) 18. Downlod t InfoCleringhouse.com Gilberto E. Urroz

14 Cross product s determinnt The cross product A B in Crtesin coordintes cn be expressed s determinnt if the first row of the determinnt consists of the unit vectors i, j, nd k. The components of vector A nd B constitute the second nd third rows of the determinnt, i.e., A B = i A B x x j A B y y k A B z z. Evlution of this determinnt, s indicted bove, will produce the result A B = (B z A y B y A z ) i + (A z B x B z A x ) j + (B y A x A y B x ) k. SCILAB does not provide cross-product function of its own. The following function, CrossProd, clcultes the cross product of two three-dimensionl vectors: function [p] = CrossProd(u,v) //Clcultes the cross-product of two vectors u nd v. //Vectors u nd v cn be columns or row vectors, but //they must hve only three elements ech. [nu,mu] = size(u); [nv,mv] = size(v); if nu*mu <> 3 nv*mv <> 3 then error('vectors must be three-dimensionl only') bort; end A1 = [ u(2), u(3); v(2), v(3)]; A2 = [ u(3), u(1); v(3), v(1)]; A3 = [ u(1), u(2); v(1), v(2)]; px = det(a1); py = det(a2); pz = det(a3); p = [px, py, pz] //end function An ppliction of the function CrossProd follows: -->getf('crossprod') -->u = [2, 3, -1], v = [-3, 1, 4] u =! ! v =! ! -->CrossProd(u,v) Downlod t InfoCleringhouse.com Gilberto E. Urroz

15 ! ! These opertions re interpreted s u = 2i+3j-k, v = -3i+j+4k, nd u v = 13i+j+4k. Erlier in this chpter we indicted tht the volume of the prllelepiped defined by vectors A, B, nd C, is given by the mgnitude of the vector triple product A (B C). As n exmple in SCILAB, try the following clcultion: -->A = [2, 2, -1], B = [5, 3, 1], C = [1, 2, 3] A =! ! B =! ! C =! ! -->bs(a*crossprod(b,c)') 21. Polynomils s vector components Although SCILAB is not symbolic environment, some pseudo-symbolic clcultions re possible using polynomils. Polynomils re specil type of SCILAB objects useful in liner system nlysis. They re presented in more detil elsewhere in this book. In this section we present simple pplictions of polynomils s vector components. Simple symbolic vribles cn be defined using the following cll to the SCILAB function poly. For exmple, the following sttement cretes the symbolic vrible c: -->c=poly(0,'c') c = c With the symbolic vrible c defined bove, you cn produce the following vector opertions: -->c*a, c*b, c*(a+b), c*(a-b)! 3c 5c - c!! 2c 2c 3c!! 5c 7c 2c! Downlod t InfoCleringhouse.com Gilberto E. Urroz

16 ! c 3c - 4c! It should be kept in mind tht these re not symbolic results in the most generl sense of the word (i.e., s those obtined in symbolic mthemticl environment such s Mple or Mthemtic). The results obtined bove, in terms of the polynomil c, re threedimensionl vectors whose components re polynomils. Other exmples of vectors with polynomil components would be: -->u = [ 1+c, (c+1)^2, c^3+2*c^2 ] u =! 2 2 3!! 1 + c 1 + 2c + c 2c + c! -->u+c*a+3*c*b! 2 2 3!! c c + c 8c + 2c + c! Opertions such s term-by-term multipliction, or liner combintions, re permitted with vectors whose components re polynomils. For exmple, -->A.*u! 2 2 3!! 3 + 3c c + 5c - 2c - c! -->u.*b! 2 2 3!! 2 + 2c 2 + 4c + 2c 6c + 3c! -->A-B.*u! 2 2 3!! 1-2c 3-4c - 2c - 1-6c - 3c! -->u +3*A.*u! 2 2 3!! c c + 16c - 4c - 2c! Some opertions, such s norm, re not defined when vector hs polynomil components: -->norm(u)!--error 4 undefined vrible : %p_norm -->norm(c*a)!--error 4 undefined vrible : %p_norm Downlod t InfoCleringhouse.com Gilberto E. Urroz

17 Also, in keeping with SCILAB s numericl nture, opertions tht ttempt to combine two polynomil vribles (i.e., symbolic opertions) re not llowed. For exmple, the following commnd defines new polynomil symbolic vrible r: -->r=poly(0,'r') r = r The following liner combintion of the polynomil vribles c nd r fils to produce result: -->c*a+r*b!--error 4 undefined vrible : %p p Assignment sttements cn replce the vlues of polynomil vribles. In the next exmple, the vlue of c, which so fr hs been used s polynomil vrible, gets redefined s constnt: -->c=2 c = 2. With the new vlue of c the following opertion produces constnt vector: -->c*a! ! As n ppliction of polynomils s components of vectors, suppose tht you wnt to determine the vlue of c such tht vector A = 2i+cj-k is norml to vector B = -5i+3j-2k. We will use the fct tht if vectors A nd B re perpendiculr then their dot product is zero, i.e., A B = 0. Using SCILAB: -->c = poly(0,'c') c = c -->A = [2, c, -1], B = [-5, 3, -2] A =! 2 c - 1! B =! ! -->A*B' c This trnsltes into the eqution c = 0. From which we cn find c = 8/3. Downlod t InfoCleringhouse.com Gilberto E. Urroz

18 Applictions of vector lgebr using SCILAB In this section we present exmples of opertions with 2- nd 3-dimensionl vectors using SCILAB functions. The exmples re tken from pplictions in different physicl sciences. Exmple 1 - Position vector The position vector of prticle is vector tht strts t the origin of system of coordintes nd ends t the prticle s position. If the current position of the prticle is P(x,y,z), then the position vector is r = x i+y j+z k, s illustrted in the figure below. If prticle is locted t point P(3, -2, 5), the position vector for this prticle is r = 3i -2j+5k. In SCILAB this position vector is written s [3 2 5]. -->r = [3,-2,5] r =! ! To determine the mgnitude of the position vector the function norm. -->bs_r = norm(r) bs_r = In pper, you cn write r = To determine the unit vector, use: Downlod t InfoCleringhouse.com Gilberto E. Urroz

19 -->e_r = r/bs_r e_r =! ! In pper, we cn write: e r = r/ r = [ ]. Exmple 2 - Center of mss of system of discrete prticles Consider system of discrete prticles of mss m i, locted t position P i (x i, y i, z i ), with i = 1, 2, 3,, n. We cn write position vectors for ech prticle s r i = x i i+y i j+z i k. The center of mss of the system of prticles will be locted t position r cm defined by r cm n i= 1 = n mi ri. m i= 1 i A system of four prticles is illustrted in the figure below. The following tble shows the coordintes nd msses of 5 prticles. Determine their center of mss. i x i y i z i m i Downlod t InfoCleringhouse.com Gilberto E. Urroz

20 First, we enter vectors contining the coordintes x i, y i, nd z i, s well s the msses m i : -->x = [2,-1,3,5,5] x =! ! -->y = [3,6,-1,4,3] y =! ! -->z = [5,4,2,-7,2] z =! ! -->m = [12,15,25,10,30] m =! ! Then, we proceed to clculte the coordintes of the center of mss, (x cm, y cm, z cm ) using the formuls: x cm n x i mi y i mi z i mi = i= 1 i= 1 i= 1, ycm =, xcm = n n n. m m m i= 1 i n i= 1 i n i= 1 i Using SCILAB, these formuls re clculted using the function sum: -->x_cm = sum(x.*m)/sum(m) x_cm = >y_cm = sum(y.*m)/sum(m) y_cm = >z_cm = sum(z.*m)/sum(m) z_cm = Exmple 3 - Resultnt of forces Downlod t InfoCleringhouse.com Gilberto E. Urroz

21 The figure below shows verticl pole buried in the ground nd supported by four cbles EA, EB, EC, nd ED. The mgnitude of the tensions in ech of those cbles, s shown in the figure, re T 1 = 150 N, T 2 = 300 N, T 3 = 200 N, nd T 4 = 150 N. To find the vector resultnt of ll those forces you need to dd the four tensions written out s vectors. The tension in cble i will be given by T i = T i e i, where e i is unit vector in the direction of the cble where the tension cts. (The tension vectors ct so tht the cbles pull wy from point E, where the cbles re ttched to the pole.) Writing out tension T 1 Tension T 1 cts long cble EA. To determine unit vector long cble EA, you need to write the vector r EA = r A r E, where r A is the position vector of point A (the tip of the vector r EA ), nd r A is the position vector of point A (the origin of the vector r EA ). The coordintes of these points, obtined from the figure, re A(-1.0 m, -1.1 m, 0) nd E(0, 0, 2.5 m). Therefore, we cn write r E = (2.5k) m, nd r A = (-i 1.1j) m. Using SCILAB we would clculte the vector r EA s follows. First, enter r E nd r A : --> re = [ ]; ra = [ ]; To obtin, r EA = r A r E, we will use --> rea = ra - re; Downlod t InfoCleringhouse.com Gilberto E. Urroz

22 rea =! ! i.e., r EA = -i 1.1j-2.5k. To find the unit vector long which T 1, first we find the mgnitude, --> rea_bs = norm(rea) i.e., ( r EA = ). Finlly, the unit vector, is clculted s --> eea = rea/rea_bs eea =! ! i.e., e EA = r EA / r EA = [ ] = i j k. (To verify tht this is indeed unit vector, use the commnd --> norm(eea). You should get s result 1.000). The tension vector is clculted by multiplying the mgnitude of the tension T 1 = 150, times the unit vector e EA, i.e., --> T1 = 150*eEA T1 =! ! The result is the vector [ ], or T 1 = (-51.57i j k) N. Writing the vector tht joins two points in spce The vector tht joints points A nd B, where A(x A, y A, z A ) is the origin nd point B(x B, y B, z B ) the tip of the vector, is written s r AB = r B - r A.= (x B - x A ) i+(y B - y A ) j+(z B - z A ) k, where r A = x A i+y A j+z A k is the position vector of point A, nd r B = x B i+y B j+z B k is the position vector of point B. This opertion is illustrted in the figure below. Downlod t InfoCleringhouse.com Gilberto E. Urroz

23 Reltive position vector The previous opertion cn lso be interpreted s obtining the reltive position vector of point B(x B, y B, z B ) with respect to point A(x A, y A, z A ), nd written s r B/A = r B - r A.= (x B - x A ) i+(y B - y A ) j+(z B - z A ) k. Writing out tensions T 2, T 3, nd T 4 Hving developed procedure to determine the unit vector long ny of the cbles, we cn proceed to write out the vectors representing the tensions T 2, T 3, nd T 4. First, we determine the coordintes of relevnt points from the figure describing the problem. These points re: A(-1.0 m, -1.1m, 0) B(0.8 m, -1.0 m, 0.0), C(-1.5 m, 0.0, 0.5 m), D(1.2 m, 1.3m, 0.0), nd E(0, 0, 2.5 m). The tensions, s vectors, will be written s T 2 = T 2 e EB, T 3 = T 3 e EC, nd T 4 = T 4 e ED. (Recll tht T 2 = 300 N, T 3 = 200 N, nd T 4 = 150 N.). Thus, we need to write out the vectors r EB, r EC, nd r ED, find their mgnitudes, r EB, r EC, nd r ED, nd clculte the unit vectors e EB = r EB / r EB, e EC = r EC / r EC, nd e ED = r ED / r ED, in order to obtin T 2, T 3, nd T 4. The following SCILAB commnds perform the required clcultions. First, we enter the position vectors for points A, B, C, D, nd E: -->ra =[-1, -1.1, 0], rb = [0.8, -1, 0], rc = [-1.5, 0,0.5] ra =! ! rb =! ! rc =! ! -->rd = [1.2, 1.3, 0], re = [0, 0, 2.5] rd = Downlod t InfoCleringhouse.com Gilberto E. Urroz

24 ! ! re =! ! Next, we clculte the reltive position vectors, r EB, r EC, nd r ED : -->reb = rb - re, rec = rc - re, red = rd - re reb =! ! rec =! ! red =! ! The mgnitudes of the vectors,, r EB, r EC, nd r ED, re clculted next: -->bs_reb = norm(reb), bs_rec = norm(rec), bs_red= norm(red) bs_reb = bs_rec = 2.5 bs_red = The unit vectors long the cbles, e EB = r EB / r EB, e EC = r EC / r EC, nd e ED = r ED / r ED, re clculted s: -->eeb = reb/bs_reb, eec = rec/bs_rec, eed = red/bs_red eeb =! ! eec =! ! eed =! ! Next, we enter the mgnitudes of the tensions, T 2 = 300 N, T 3 = 200 N, nd T 4 = 150 N. These re identified in SCILAB s T1m, T2m, nd T3m, respectively: -->T2m = 300, T3m = 200, T4m = 150 T2m = 300. T3m = 200. T4m = 150. Downlod t InfoCleringhouse.com Gilberto E. Urroz

25 The tension vectors, themselves, re clculted by using T 2 = T 2 e EB, T 3 = T 3 e EC, nd T 4 = T 4 e ED. In SCILAB we use: -->T2 = T2m*eEB, T3 = T3m*eEC, T4 = T4m*eED T2 =! ! T3 =! ! T4 =! ! The results re: T 2 = (85.44i j k) N, T 3 = (-120i 160k) N, nd for cble ED, T 4 = (58.77i j k) N. The resultnt of these four forces, nmely, R = T 1 + T 2 + T 3 + T 4, turns out to be R = (-27.36i 99.86j k) N. Exmple 4 Eqution of plne in spce The eqution of plne in spce, in Crtesin coordintes, cn be obtined given point on the plne A(x A,y A,z A ), nd vector norml to the plne n = n x i+n y j+n z k. Let P(x,y,z) be generic point on the plne of interest. We cn form reltive position vector r P/A = r P r A = (x-x A ) i+(y-y A ) j+(z-z A ) k, which is contined in the plne, s shown in the figure below. Becuse the vectors n nd r P/A re perpendiculr to ech other, then, we cn write n r P/A = 0, or (n x i+n y j+n z k) ((x-x A ) i+(y-y A ) j+(z-z A ) k) = 0, which results in the eqution n x (x-x A ) + n y (y-y A ) + n z (z-z A ) = 0. To implement SCILAB function to produce the eqution of plne in spce we use the fct tht the eqution cn be re-written s n x x + n y y + n z z = n r A. The function PlneEqution, shown below, clcultes the right-hnd side of the eqution, n r A, nd uses the function string Downlod t InfoCleringhouse.com Gilberto E. Urroz

26 to convert numericl results to strings. The finl result of the function is string representing the eqution. function [eq] = PlneEqution(n,p) //This function produces the eqution of plne //in spce given the norml vector n nd the point p. [nn,mn] = size(n); [np,mp] = size(p); if nn*mn <> 3 np*mp <> 3 then error('vector n or point p in PlneEqution must hve 3 elements') bort; end c = n*p'; eq = string(n(1))+'*x+'+string(n(2))+'*y+'+string(n(3))+'*z='+string(c) //end function The function requires s input the norml vector n nd the point A (referred to s p in the function), both entered s SCILAB vectors. For exmple, to find the eqution of the plne with norml vector n = 3i-5j+6k pssing through point A(3, 6, -2) use: -->getf('plneeqution') -->n = [3, -5, 6], A = [3, 6, -2] -->myeqution = PlneEqution(n,A) myeqution = 3*x+-5*y+6*z=-33 To verify tht the eqution is stisfied by point A use: -->x = 3, y = 6, z = -2 x = 3. y = 6. z = >3*x-5*y+6*z Exmple 5 Moment of force The figure below shows force F cting on point P in rigid body. Suppose tht the body is llowed to rotte bout point O. Let r be the position vector of point P with respect to the Downlod t InfoCleringhouse.com Gilberto E. Urroz

27 point of rottion O, which we mke coincide with the origin of our Crtesin coordinte system. The moment of the force F bout point O is defined s M = r F. Referring to the figure of Exmple 3, if we wnt to clculte the moment of tension T 1 = ( i j k) N, bout point O, we will use s the position vector r of the force s line of ction through the pole, the vector r OE = (2.5k) m. The moment M 1 = r OE T 1, cn be clculted s follows: -->roe = [ ] Enter r OE -->T1 = [ ] Enter T 1 -->M1 = CrossProd(r0E,T1) Clculte M 1 = r OE T 1 The result is [ ], or M 1 = ( i j ) m N. To find the mgnitude of the moment, use [!][ABS], thus, M 1 = M 1 = m N. Note: Moments re vector quntities nd obey ll rules of vectors, i.e., they cn be dded, subtrcted, multiplied by sclr, undergo internl nd externl vector products. As n exercise, the reder my wnt to clculte the moments corresponding to the other tensions in Exmple 3, s well s the resultnt moment from ll four tensions. Exmple 6 Crtesin nd polr representtions of vectors in the x- y plne A position vector in the x-y plne cn be written simply s r = x i+y j. Let its mgnitude be r = r. A unit vector long the direction of r is given by e r = r/r = (x/r) i+(y/r) j. If we use polr coordintes, we recognize thus, we cn write the unit vector s x/r = cos θ, nd y/r = sin θ, e r = r/r = cos θ i+ sin θ j. Downlod t InfoCleringhouse.com Gilberto E. Urroz

28 Thus, if we re given the mgnitude, r, nd the direction, θ, of vector (i.e., its polr coordintes (r,θ)), we cn esily put together the vector s This result is illustrted in the figure below. r = r e r = r (cos θ i+ sin θ j). To enter vector in SCILAB, given the mgnitude r nd the ngle θ, you would use, for exmple: -->r = 2.5*[cos(0.75), sin(0.75)] r =! ! In this cse the ngle represents rdins. If you wnt to use n ngle in degrees, you need to trnsform it to rdins in the trigonometric functions, for exmple: -->r = 5.2*[cos(32*%pi/180), sin(32*%pi/180)] r =! ! Exmple 7 Plnr motion of rigid body In the study of the plnr motion of rigid body the following equtions re used to determine the velocity nd ccelertion of point B (v B, B ) given the velocity nd ccelertion of reference point A (v A, A ): v B = v A + ω AB r B/A, B = A + α AB r B/A - ω AB2 r B/A. Downlod t InfoCleringhouse.com Gilberto E. Urroz

29 The equtions lso use the ngulr velocity of the body connecting A nd B, ω AB, whose mgnitude is ω AB ; the ngulr ccelertion of the body connecting A nd B, α AB, nd the reltive position vector of point B with respect to point A, r B/A. The following figure illustrtes the clcultion of reltive velocity nd ccelertion in plnr motion of rigid body. To present pplictions of this eqution using the HP 49 G clcultor, we use the dt from the mechnism shown in the figure below. In this mechnism there re two brs AB nd AC pin-connected t B. Br AB is pin supported t A, nd br BC is ttched through pin to piston C. Piston C is llowed to move in the verticl direction only. At the instnt shown the ngulr velocity nd ccelertion of br AB re 10 rd/s nd 20 rd/s 2 clockwise. You re sked to determine the ngulr velocity nd ccelertion of br BC nd the liner velocity nd ccelertion of piston C. Downlod t InfoCleringhouse.com Gilberto E. Urroz

30 Angulr velocity nd ccelertion Angulr velocities nd ccelertions in the x-y plne re represented s vectors in the z- direction, i.e., norml to the x-y plne. These vectors re positive if the ngulr velocity or ccelertion is counterclockwise. In generl, thus we cn write, ω AB = ±ω AB k, nd α AB = ±α AB k. For the dt in this problem we cn write, therefore, ω AB = (-10k) rd/s, α AB = (-20k)rd/s 2, ω BC = (ω BC k) rd/s, nd α BC = (α BC k). Using SCILAB, we will write: -->wab = [0, 0, -10], lphab = [0, 0, -20] wab =! ! lphab =! ! The next sttements define the polynomil vribles wbcm (stnding for ω BC ) nd lbcm (stnding for α BC ). The m in the polynomil vrible nmes stnds for mgnitude : -->wbcm = poly(0,'wbcm'), lbcm = poly(0,'lbc') wbcm = wbcm lbcm = lbc With the vribles wbcm nd lbcm we cn define the vectors ω BC nd α BC s: -->wbc = [0, 0, wbcm], lbc = [0, 0, lbcm] wbc =! 0 0 wbcm! lbc =! 0 0 lbc! Reltive position vector The reltive position vectors of interest in this problem re r B/A nd r C/B, which re obtined s follows: To obtin r B/A, we use the fct tht for the xy coordinte system shown, the ngle corresponding to vector r B/A is θ B/A = 90 o + 45 o = 135 o. Thus, we cn write -->rb_a = 0.25*[cos(135*%pi/180), sin(135*%pi/180), 0] rb_a =! ! In pper, we would write, therefore, r B/A = (-0.177i j)ft. For r C/B, the ngle θ C/B = 90 o 13.6 o = 76.4 o. Thus, in SCILAB, Downlod t InfoCleringhouse.com Gilberto E. Urroz

31 -->rc_b = 0.75*[ cos(76.4*%pi/180), sin(76.4*%pi/180), 0] rc_b =! ! In pper, we write, r C/B = (0.175i j)ft. Velocity Since point A is fixed point, v A = 0, nd we cn write v B = ω AB r B/A = (-10k) rd/s (-0.177i j)ft. Using SCILAB, we would write: -->getf('crossprod') -->va = 0 va = 0. -->vb = va + CrossProd(wAB,rB_A) vb =! ! Thus, v B = (1.77i+1.77j) ft/s. Note: Rdins re bsiclly dimensionless units, thus rd ft = ft, rd m = m. To find the velocity of point C, we use v C = v B + ω BC r C/B = (-1.77i-1.77j) ft/s + (ω BC k) rd/s (0.175i j)ft. With SCILAB, this opertion is written s: -->vc = vb + CrossProd(wBC,rC_B) vc =! wBCm wBCm 0! The result cn be written in pper s: v C = (( ω BC ) i + ( ω BC ) j) ft/s. The figure bove shows tht the piston C is forced to move in the verticl direction, thus, the velocity of point C cn be written s v C = (v C j) ft/s. Equting the two results presented immeditely bove for v C we get: ( ω BC ) i + ( ω BC ) j = 0 i + v C j. Since the x- nd y-components of the two vectors in ech side of the equl sign must be the sme, we cn write the system of equtions: Downlod t InfoCleringhouse.com Gilberto E. Urroz

32 ω BC = ω BC = vc Solution of equtions one t time Using SCILAB, we cn solve for the two unknowns (ω BC nd v C ) s follows: 1. From the first eqution, ω BC = 1.77/0.729, i.e., -->wbcm = 1.77/0.729 wbcm = From the second eqution, vc = 1.77+(0.176) (2.43), i.e., -->vcm = *2.43 vcm = Thus, the solution of the system of equtions is: ω BC = 2.43 rd/s, nd v C = 2.20 ft/s. The positive sign in ω BC mens tht the ngulr velocity is counterclockwise. The positive sign in v C mens tht point C is moving upwrds in the verticl direction. Accelertion Agin, becuse A is fixed point, A = 0. Thus, the ccelertion of point B is given by B = α AB r B/A - ω AB2 r B/A = (-20k)rd/s 2 (-0.177i j)ft (10 rd/s) 2 (-0.177i j)ft Using SCILAB we cn obtin B s follows: -->A = 0 A = 0. -->wabm = -10 wabm = >B = A + CrossProd(lphAB,rB_A)-wABm^2*rB_A B =! ! The result is [ ], or B = (21.24i j)ft/s 2. To clculte the ccelertion of point C we use: Downlod t InfoCleringhouse.com Gilberto E. Urroz

33 C = B + α BC r C/B - ω BC2 r C/B = (21.24i j)ft/s 2 +(α BC k) (0.175i j)ft - (-2.42 rd/s) 2 (0.175i j)ft In Scilb: -->C = B + CrossProd(lphBC,rC_B) - wbcm^2*rc_b C =! lBC lBC 0! The result is: C = (( α BC ) i+( α BC )) j) ft/s 2. Also, becuse the motion of point C is in the verticl direction, we cn write C = C j. Equting the two expressions for the vector C shown bove, we get the following equtions: α BC = 0, α BC = C. Solution of system of liner equtions using mtrices The system of liner equtions in two unknowns (α BC nd C ) obtined bove cn be re-written in mtricil form s follows: 1. The equtions re first re-written s c α BC = α BC = or, s mtrix eqution: 1 0 C α BC = This mtrix eqution cn be solved by using the bckwrd-slsh opertor (\), i.e., x = A\b. This result follows from the originl mtrix eqution, A x = b, by dividing both sides of the eqution by A. However, since A is mtrix, this division is not regulr rithmetic division. Typiclly, this opertion would be represented s x = A -1 b, where A -1 is the inverse of mtrix A. Modern mtrix-bsed numericl environments, such s SCILAB, provide the user with the bckwrd-slsh opertor insted of using the inverse mtrix. The result, however, is exctly the sme. Using SCILAB, we would enter: -->A = [1, ; 0, 0.729], b = [-18.44;20.17] A =! !! ! Downlod t InfoCleringhouse.com Gilberto E. Urroz

34 b =! !! 20.17! -->A\b! !! ! The results re interpreted s C = ft/s 2, α BC = rd/s 2. The negtive sign in c indictes tht point C is decelerting. The positive sign in α BC indictes tht br BC is ccelerting ngulrly in the counterclockwise direction. We hve presented here two methods for solution of systems of liner equtions. These methods, nd others, re presented in more detil in different chpter. Exercises For the following exercises use the Crtesin vectors: u = 3i+2j-5k, v = -i-3k, w = 5i-10j-3k, r = -8i+10j-2.5k, s = -3i-2j-5k, t = 6i-2j+15k, [1]. Determine the result of the following opertions: () u (b) w (c) v r (d) s / w (e) = u + v (f) b = r-t (g) c = 3r-2v (h) d = -t + 2s (i) unit vectors: e u, e v, e w (j) ngles between vectors u,v; r,s; nd v,w. (k) u v (l) w r (m) s t (n) u v (o) r w (p) u (v w) (q) (s t) r (r) w (r u) (s) (u v) t (t) (u v) (s t) [2]. Four different cbles re ttched to point E(0,0,0) on structure. The four cbles re nchored to points A(-1,-1,-1), B(2,3,-5), C(-2,2,4), nd D(2,3,-1). The tensions in the four cbles re: AE = 150 lb, BE = 250 lb, CE = 100 lb, DE = 50 lb. Determine the resultnt force from the four cbles. [3]. Determine the torque of the following forces F given the rm r, i.e., M = r F: () F = (3i+2j-5k) N, r = (-2i+5j-3k) m (b) F = (i-4j-3k) lb, r = (i+8j-13k) ft () F = 100(i+j-k) lb, r = (-2i+5j-3k)/2 ft (b) F = 200(i-4j-k) N, r = 3(i+j-10k) m [4]. Determine the eqution of the plne through point A with norml vector n. Sketch the plne: () A(-2,3,5), n = [2,-2,3] (b) A(0, -1, 2), n = [5,5,-1] (c) A(2,5,-1), n = [1, 1, 1] (d) A(-1,5,-2), n = [3,3,-1] Downlod t InfoCleringhouse.com Gilberto E. Urroz

35 [5]. Two vectors n nd m re sid to be orthogonl if n m = 0. Determine the missing components in the following vectors m nd n so tht they re orthogonl: () n = [2, y,-2], m = [5,5,-1] (b) n = [x,5,4], m = [-1,0,2] (c) n = [4,2,y], m = [3,3,2] (d) n = [5,5,2], m = [x,-2,3] Note: define the unknown vrible s SCILAB polynomil vrible, e.g., for (): -->y = poly(0, y ) Then, define the vectors s follows: --> n = [2,y,-2]; m = [5,5,1]; Clculte the dot product s polynomil: --> p = n*m nd solve for y using function roots: --> roots(p) [6]. For the mechnism presented in Exmple 7, if the velocity of point C is v c = 1.2 ft/s nd its ccelertion is c = -0.2 ft/s 2, determine the ngulr velocity nd ccelertion of brs AB nd BC. Downlod t InfoCleringhouse.com Gilberto E. Urroz

36 REFERENCES (for ll SCILAB documents t InfoCleringhouse.com) Abrmowitz, M. nd I.A. Stegun (editors), 1965,"Hndbook of Mthemticl Functions with Formuls, Grphs, nd Mthemticl Tbles," Dover Publictions, Inc., New York. Aror, J.S., 1985, "Introduction to Optimum Design," Clss notes, The University of Iow, Iow City, Iow. Asin Institute of Technology, 1969, "Hydrulic Lbortory Mnul," AIT - Bngkok, Thilnd. Berge, P., Y. Pomeu, nd C. Vidl, 1984,"Order within chos - Towrds deterministic pproch to turbulence," John Wiley & Sons, New York. Brs, R.L. nd I. Rodriguez-Iturbe, 1985,"Rndom Functions nd Hydrology," Addison-Wesley Publishing Compny, Reding, Msschussetts. Brogn, W.L., 1974,"Modern Control Theory," QPI series, Quntum Publisher Incorported, New York. Browne, M., 1999, "Schum's Outline of Theory nd Problems of Physics for Engineering nd Science," Schum's outlines, McGrw-Hill, New York. Frlow, Stnley J., 1982, "Prtil Differentil Equtions for Scientists nd Engineers," Dover Publictions Inc., New York. Friedmn, B., 1956 (reissued 1990), "Principles nd Techniques of Applied Mthemtics," Dover Publictions Inc., New York. Gomez, C. (editor), 1999, Engineering nd Scientific Computing with Scilb, Birkhäuser, Boston. Gullberg, J., 1997, "Mthemtics - From the Birth of Numbers," W. W. Norton & Compny, New York. Hrmn, T.L., J. Dbney, nd N. Richert, 2000, "Advnced Engineering Mthemtics with MATLAB - Second edition," Brooks/Cole - Thompson Lerning, Austrli. Hrris, J.W., nd H. Stocker, 1998, "Hndbook of Mthemtics nd Computtionl Science," Springer, New York. Hsu, H.P., 1984, "Applied Fourier Anlysis," Hrcourt Brce Jovnovich College Outline Series, Hrcourt Brce Jovnovich, Publishers, Sn Diego. Journel, A.G., 1989, "Fundmentls of Geosttistics in Five Lessons," Short Course Presented t the 28th Interntionl Geologicl Congress, Wshington, D.C., Americn Geophysicl Union, Wshington, D.C. Julien, P.Y., 1998, Erosion nd Sedimenttion, Cmbridge University Press, Cmbridge CB2 2RU, U.K. Keener, J.P., 1988, "Principles of Applied Mthemtics - Trnsformtion nd Approximtion," Addison-Wesley Publishing Compny, Redwood City, Cliforni. Kitnidis, P.K., 1997, Introduction to Geosttistics - Applictions in Hydogeology, Cmbridge University Press, Cmbridge CB2 2RU, U.K. Koch, G.S., Jr., nd R. F. Link, 1971, "Sttisticl Anlysis of Geologicl Dt - Volumes I nd II," Dover Publictions, Inc., New York. Korn, G.A. nd T.M. Korn, 1968, "Mthemticl Hndbook for Scientists nd Engineers," Dover Publictions, Inc., New York. Kottegod, N. T., nd R. Rosso, 1997, "Probbility, Sttistics, nd Relibility for Civil nd Environmentl Engineers," The Mc-Grw Hill Compnies, Inc., New York. Kreysig, E., 1983, "Advnced Engineering Mthemtics - Fifth Edition," John Wiley & Sons, New York. Lindfield, G. nd J. Penny, 2000, "Numericl Methods Using Mtlb," Prentice Hll, Upper Sddle River, New Jersey. Mgrb, E.B., S. Azrm, B. Blchndrn, J. Duncn, K. Herold, nd G. Wlsh, 2000, "An Engineer's Guide to MATLAB ", Prentice Hll, Upper Sddle River, N.J., U.S.A. McCuen, R.H., 1989, Hydrologic Anlysis nd Design - second edition, Prentice Hll, Upper Sddle River, New Jersey. Middleton, G.V., 2000, "Dt Anlysis in the Erth Sciences Using Mtlb," Prentice Hll, Upper Sddle River, New Jersey. Downlod t InfoCleringhouse.com Gilberto E. Urroz

37 Montgomery, D.C., G.C. Runger, nd N.F. Hubele, 1998, "Engineering Sttistics," John Wiley & Sons, Inc. Newlnd, D.E., 1993, "An Introduction to Rndom Vibrtions, Spectrl & Wvelet Anlysis - Third Edition," Longmn Scientific nd Technicl, New York. Nicols, G., 1995, Introduction to Nonliner Science, Cmbridge University Press, Cmbridge CB2 2RU, U.K. Prker, T.S. nd L.O. Chu,, "Prcticl Numericl Algorithms for Chotic Systems, 1989, Springer-Verlg, New York. Peitgen, H-O. nd D. Supe (editors), 1988, "The Science of Frctl Imges," Springer-Verlg, New York. Peitgen, H-O., H. Jürgens, nd D. Supe, 1992, "Chos nd Frctls - New Frontiers of Science," Springer-Verlg, New York. Press, W.H., B.P. Flnnery, S.A. Teukolsky, nd W.T. Vetterling, 1989, Numericl Recipes - The Art of Scientific Computing (FORTRAN version), Cmbridge University Press, Cmbridge CB2 2RU, U.K. Rghunth, H.M., 1985, "Hydrology - Principles, Anlysis nd Design," Wiley Estern Limited, New Delhi, Indi. Recktenwld, G., 2000, "Numericl Methods with Mtlb - Implementtion nd Appliction," Prentice Hll, Upper Sddle River, N.J., U.S.A. Rothenberg, R.I., 1991, "Probbility nd Sttistics," Hrcourt Brce Jovnovich College Outline Series, Hrcourt Brce Jovnovich, Publishers, Sn Diego, CA. Sgn, H., 1961,"Boundry nd Eigenvlue Problems in Mthemticl Physics," Dover Publictions, Inc., New York. Spnos, A., 1999,"Probbility Theory nd Sttisticl Inference - Econometric Modeling with Observtionl Dt," Cmbridge University Press, Cmbridge CB2 2RU, U.K. Spiegel, M. R., 1971 (second printing, 1999), "Schum's Outline of Theory nd Problems of Advnced Mthemtics for Engineers nd Scientists," Schum's Outline Series, McGrw-Hill, New York. Tnis, E.A., 1987, "Sttistics II - Estimtion nd Tests of Hypotheses," Hrcourt Brce Jovnovich College Outline Series, Hrcourt Brce Jovnovich, Publishers, Fort Worth, TX. Tinker, M. nd R. Lmbourne, 2000, "Further Mthemtics for the Physicl Sciences," John Wiley & Sons, LTD., Chichester, U.K. Tolstov, G.P., 1962, "Fourier Series," (Trnslted from the Russin by R. A. Silvermn), Dover Publictions, New York. Tveito, A. nd R. Winther, 1998, "Introduction to Prtil Differentil Equtions - A Computtionl Approch," Texts in Applied Mthemtics 29, Springer, New York. Urroz, G., 2000, "Science nd Engineering Mthemtics with the HP 49 G - Volumes I & II", Chrleston, S.C. Urroz, G., 2001, "Applied Engineering Mthemtics with Mple", Chrleston, S.C. Winnick, J.,, "Chemicl Engineering Thermodynmics - An Introduction to Thermodynmics for Undergrdute Engineering Students," John Wiley & Sons, Inc., New York. Downlod t InfoCleringhouse.com Gilberto E. Urroz