1.2 What is a vector? (Section 2.2) Two properties (attributes) of a vector are and.

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1 Homework 1. Chpters 2. Bsis independent vectors nd their properties Show work except for fill-in-lnks (print.pdf from Textooks Resources). 1.1 Solving prolems wht engineers do. Understnding dynmics results from doing prolems. Mny prolems re guided to help you synthesize processes (imittion). You re encourged to work y yourself or with collegues/instructors nd use the textook s reference theory nd other resources. Confucius 500 B.C. I her nd I forget. I see nd I rememer. I nd I understnd. By three methods we my lern wisdom: First, y reflection, which is nolest; Second, y imittion, which is esiest; Third y experience, which is the itterest. 1.2 Wht is vector? (Section 2.2) Two properties (ttriutes) of vector re nd. 1.3 Wht is zero vector? (Section 2.3) A zero vector 0 hs mgnitude of 0/1/2/. A zero vector 0 hs no direction. True/Flse. 1.4 Unit vectors. (Section 2.4) A unit vector hs mgnitude of 0/1/2/. All unit vectors re equl. True/Flse. 1.5 Drw the following vectors: (Section 2.2) Long, horizontlly-right vector Short, verticlly-upwrd vector Outwrdly-directed unit vector ĉ. 1.6 Vector mgnitude nd direction (orienttion nd sense). (Section 2.2) The figure to the right shows vector v. Drw the following vectors. : Sme mgnitude nd sme direction s v ( = v). : Sme mgnitude nd orienttion s v, ut different sense. c: Sme direction s v, ut different mgnitude. d: Sme mgnitude s v, ut different direction (orienttion). e: Different mgnitude nd different direction (orienttion) s v. v 1.7 Mgnitude of vector. (Section 2.2) Consider rel numer x nd horizontlly-right pointing unit vector î. The mgnitude of the vector x î is (circle one): positive negtive non-negtive non-positive. 1.8 Negting vector. (Section 2.8) Complete the figure to the right y drwing the vector. Negting the vector results in vector with different (circle ll tht pply): mgnitude direction orienttion sense Historicl note: Negtive numers (e.g., 3) were not widely ccepted until 1800 A.D. Copyright c Pul Mitiguy. All rights reserved. 159 Homework 1: Vectors sis independent

2 1.9 Multiplying vector y sclr. (Section 2.7) Complete the figure to the right y drwing the vectors 2 v nd 2 v. The following sttements involve vector v nd rel non-zero sclr s (s 0). If sttement is true, provide numericl vlue for s tht supports your nswer s v cn hve different mgnitude thn v. True/Flse s = s v cn hve different direction thn v. True/Flse s = s v cn hve different sense thn v. True/Flse s = s v cn hve different orienttion thn v. True/Flse s = 1.10 Grphicl vector ddition/sutrction - drw. (Sections 2.6,2.8) Drw + v Drw 1.11 Visul representtion of vector dot-product. (Section 2.10) Write the definition of the dot-product of vector with vector. Include sketch with ech symol in the right-hnd-side of your definition clerly leled. The sketch should include,, fi fi fi fi, fi fi, Visul representtion of vector cross-product. (Section 2.11) Write the definition of the cross-product of vector with vector. Include sketch with ech symol in your definition leled nd descried. (θ) û where û is nd θ is 1.13 Properties of vector dot-products nd cross-products. (Sections nd ) When is prllel to : =0 True/Flse = 0 True/Flse When is perpendiculr to : =0 True/Flse = 0 True/Flse For ritrry vectors nd : = True/Flse = True/Flse Copyright c Pul Mitiguy. All rights reserved. 160 Homework 1: Vectors sis independent

3 1.14 Clculting vector dot-products nd cross-products vi definitions. (Sections 2.10 nd 2.11) Drw unit vector k outwrd-norml to the plne of the pper. Knowing vector hs mgnitude 2 nd vector hs mgnitude 4, clculte the following dot-products nd cross-products vi their definitions (2 + significnt digits). = = = = = = 60 o = = 1.15 Property of sclr triple product. (Section 2.12). For ritrry non-zero vectors,, c: ( c) = ( ) c Never/Sometimes/Alwys A property of the sclr triple product is =0. True/Flse Property of vector triple cross-product. (Sections nd 2.12) Complete the following eqution: ( ) ( ) ( ) c = c Circle true or flse (show supporting work): ( c) = ( ) c + ( c) True/Flse 1.17 Optionl: Proof of mgnitude of vector cross product property. (Sections 2.10 nd 2.11) Letting λ λ λ λ λ λ λ λ λ λ λ λ λ e unit vector nd v e ny vector, prove 1 v λ λ λ λ λ λ λ λ λ λ λ λ λ 2 = v v ( v λ λ λ λ λ λ λ λ λ λ λ λ λ) Vector exponentition: v 2 nd v 3. Complete the 3-step proofs. (Section 2.10) Step 1: Complete the definition of v 2 in terms of v. Step 2: Use the definition of the dot-product to show how v v cn e expressed in terms of v. Step 3: Comine these two definitions to provide n lternte wy to clculte v 2 with vector dot-product. v 2 v v v = (2.2) v 2 = Complete the 3-step proof tht reltes v 3 to v v rised to rel numer. v 3 v ( ) = = ( v v) (2.4) 1.19 c â x Clculte vector mgnitude with dot products. (Section 2.10 nd Hw 1.18) Show how the vector dot-product cn e used to show tht the mgnitude of the vector c â x (c is positive or negtive numer nd â x is unit vector) cn e written solely in terms of c (without â x ). c âx = = c 2 = = s(c) 60 o 1.20 Mgnitude of the vector v. Show work. (Section 2.10) Knowing the ngle etween unit vector î nd unit vector ĵ is 110, i clculte numericl vlue for the mgnitude of v = 3î +4ĵ. v Note: The nswer is not 5. j 1 One wy to prove this is to write ( v λ λ) 2 = ( v λ λ) ( v λ λ) = (2.11) v [ λ ( v λ λ)] nd then use the vector triple crossproduct property ( c) = ( c) c( ) from Section Alterntely, it is helpful to write v = v λ + v λ where v λ is the component of v tht is perpendiculr to λ nd v λ is the component of v tht is prllel to λ λ. Copyright c Pul Mitiguy. All rights reserved. 161 Homework 1: Vectors sis independent

4 1.21 Angle etween vectors. (Section 2.10) Referring to the figure to the right, find the numericl vlue for the ngle etween vector nd vector. 30 o (, ) = 30 o 1.22 Visul estimtion of vector dot/cross-products. Show work. (Sections 2.10 nd 2.11) Estimte (e.g., using your pinky) the mgnitude of the vector p shown elow. Note: 1 inch 2.54 cm. Estimte the ngle etween p nd q, p q, nd the mgnitude of p q. Show work. (Provide numericl results with 1 or more significnt digits). q p cm ( p, q ) p p q cm 2 p q cm Form the unit vector û hving the sme direction s c â x. (Section 2.4) û = â x Note: x is unit vector nd c is non-zero rel numer, e.g., 3 or Coefficient of û in cross products definitions nd trig functions. (Section 2.11) The cross product of vectors nd cn e written in terms of rel sclr s s = s û where û is unit vector perpendiculr to oth nd in direction defined y the right-hnd rule. The coefficient s of the unit vector û is inherently non-negtive. True/Flse Orthogonl vectors: Insights vi drwing. (Section 2.11) Consider three unit vectors â,, ndĉ. Vector â is perpendiculr to vector. Vector is perpendiculr to vector ĉ. Vector â is not prllel to vector ĉ. In ll cses, â is perpendiculr to ĉ. True/Flse. Explin your nswer y drwing â,, ĉ nd relevnt ngles Clculting distnce etween point nd line vi cross-products. (Section ) Drw horizontlly-right unit vector â x nd verticlly-upwrd unit vector â y. Drw pointq whose position vector from point P is r Q/P =5â x. Drw line L tht psses through point P nd is prllel to û = 3 5 âx ây. Clculte the distncedistnce/length d etween Q nd L using oth formuls in eqution (2.9). d= (2.9) = 4 d= (2.9) = Rnges of ngles from dot-product nd cross-product clcultions. (Sections 2.10 nd 2.11) Given: Unit vectors â nd nd the numericl vlues of c â nd s â. Complete the following tle with n pproprite rnge of numericl vlues. Quntity Rnge of vlues c c s Angle θ c etween â nd tht cn e uniquely determined solely from c Angle θ s etween â nd tht cn e uniquely determined solely from s Angle θ etween â nd s θc θs θ Note: The rnge of θ s differs from the (correct) rnge for θ. Hence,s nd θ s re insufficient to clculte θ. Copyright c Pul Mitiguy. All rights reserved. 162 Homework 1: Vectors sis independent

5 1.28 Vector opertions nd units. (Chpter 2) Circle the vector opertions elow (sclr multipliction, ddition, dot-product, etc.) tht re defined for position vector (with units of m) nd velocity vector (with units of m s ). 5 / Using vector identities to simplify expressions (refer to Homework 1.13). One reson to tret vectors s sis-independent quntities is to simplify vector expressions without resolving the vectors into orthogonl x, y, z or i, j, k components. Simplify the following vector expressions using vrious properties of dot-products nd cross-products. Express your results in terms of dot-products nd crossproducts of the ritrry vectors u, v, w (i.e., u, v, w re not orthogonl). Vector expression (3 u 2 v) ( u + v) Simplified vector expression u v (3 u 2 v) ( u + v) u 2 v 2 + u v ( u v) ( u + v) (3 u 2 v) ( u + v) (2 u 7 v) ( u + v) ( v +2 w) ( w +2 u) u v w 1.30 Vector concepts: Solving vector eqution. (Section ) Consider the vector eqution to the right nd the process tht follows tht solves for θ (â x is unit vector nd v x, θ, R re sclrs). This process is vlid wy to solve for θ. True/Flse. Explin: v x â x = θrâ x θ = v x â x R â x = v x R 1.31 Chnging vector eqution to sclr equtions. Show work. (Section ) () Drw three mutully orthogonl unit vectors p, q, r. Use vector opertion (e.g., +,,,, ) to trnsform the following vector eqution into one sclr eqution nd susequently solve the sclr eqution.?? (2x 4) p = 0 x =2 () Show every vector opertion (e.g., +,,,, or ) tht trnsforms the following vector eqution into three sclr equtions nd susequently solve the sclr equtions for x, y, z. (2 x 4) p + (3y 9) q + (4z 16) r = 0 x =2 y =3 z = k (c). Optionl: The figure to the right shows three non-zero, non-orthogonl, non-coplnr vectors i, j, k. Show every vector opertion tht trnsforms the following vector eqution into three uncoupled sclr equtions nd susequently solve those sclr equtions for x, y, z (think, not mtrix lger). j (2 x 4) i + (3y 9) j + (4z 16) k = 0 x =2 y =3 z = i Copyright c Pul Mitiguy. All rights reserved. 163 Homework 1: Vectors sis independent

6 1.32 Numer of independent sclr equtions from one vector eqution. (Section ) Consider the vector eqution shown to the right tht cn e useful for sttic nlyses of ny system S. Complete the lnks in the tle to the right with ll integers tht could e equl to the numer of independent sclr equtions produced y the previous vector eqution for ny system S. Hint: See Homework 1.31 for ides. S F = 0 System type Integer(s) 1D (line) 0, 2D (plnr) 0, 3D (sptil) 0, Note: Regrd 1D/liner s mening F S cneexpressedintermsofsingleunitvector i wheres 2D/plnr mens F S cn e expressed in terms of two non-prllel unit vectors i nd j, nd 3D/sptil mens F S cn e expressed in terms of three non-coplnr unit vectors i, j, k Vector concepts: Solving vector eqution (just circle true or flse nd fill-in the lnk). Consider the following vector eqution written in terms of the sclrs x, y, z nd three unique non-orthogonl coplnr unit vectors â 1, â 2, â 3. (2 x 4) â 1 + (3y 9) â 2 + (4z 16) â 3 = The unique solution to this vector eqution is x =2, y =3, z =4. True/Flse. Explin: â 2 cn e expressed in terms of â 1 nd â 3 (i.e., 2 is liner comintion of 1 nd 3). Hence the vector eqution produces linerly independent sclr equtions A vector revolution in geometry. (Chpter 2) The reltively new invention of vectors (Gis 1900 AD) hs revolutionized Eucliden geometry (Euclid 300 BC). For ech geometricl quntity elow, circle the vector opertion(s) (either the dot-product, cross-product, or oth) tht is most useful for their clcultion. Length: Angle: Are: Volume: 1.35 Microphone cle lengths (non-orthogonl wlls) It s just geometry. Show work. A microphone Q is ttched to three pegs A, B, C y three cles. Knowing the peg loctions, microphone loction, nd the ngle θ etween the verticl wlls, express L A, L B, L C solely in terms of numers nd θ. Next, complete the tle y clculting L B when θ = 120. Hint: Todothisefficiently, use only unit vectors u, v, w, nd(donot introduce n orthogonl set of unit vectors). Hint: Use the distriutive property of the vector dot-product s shown in Section nd Homework 2.4. Note: Synthesis prolems re difficult. Think, tlk, drw, sleep, wlk, get help,... (if necessry, red Section 3.3). A w 20 8 v B N o θ 7 5 Q 15 Note: The floor is horizontl, the wlls re verticl. u 8 C Distnce etween A nd B Distnce etween B nd C Distnce etween N o nd B Distnce long ck wll (see picture) Q s height ove N o Distnce long side wll (see picture) L A : Length of cle joining A nd Q L B : Length of cle joining B nd Q L C : Length of cle joining C nd Q r Q/No = 7û + 5 v + 8ŵ L A = cos(θ) L B = L C = 20 m 15 m 8m 7m 5m 8m 16.9 m 8.1 m 14.2 m 137 cos(θ) Copyright c Pul Mitiguy. All rights reserved. 164 Homework 1: Vectors sis independent

7 Homework 2. Chpters 1, 2, 3, 4. Vector ddition, dot products, nd cross products Show work except for fill-in-lnks (print.pdf from Textooks Resources). 2.1 Right-hnded, orthogonl, unitry, vector sis. (Section 4.1) Drw right-hnded orthogonl (mutully perpendiculr) vector sis consisting of the unit vectors â x, â y, â z. 2.2 Adding nd sutrcting vectors with ses. (Sections 2.6 nd 2.8) Shown right re right-hnded orthogonl unit vectors î, ĵ, k nd vectors u, v, nd w. Form the vector sums nd differences elow. u = 2î + 3ĵ + 4 k v = x î + y ĵ + z k j w = 5î 6 ĵ + 7 k k i u + v = (2+x) î + ĵ + k u v = (2 x) î + ĵ + k 2.3 Column mtrices nd vectors (Hint: Wht is vector Hw 1.2). (Section 2.9) 1 The vector â x +2â y +3â z is equl to the column mtrix 2. True/Flse. Note: x, y, z re orthogonl unit vectors shown right. 3 y z Adding the following vectors nd column mtrices produce equivlent results. True/Flse. Note: x, y, z nd x, y, z re the sets of orthogonl unit vectors shown elow y y x â x +2â y +3â z + 4 x +5 y +6 z = = 7 x z z x 2.4 Clculting vector dot products with ses. (Sections 2.10 nd ) Given: Right-hnded orthogonl unit vectors, n y, n z nd: u = n y + 4 n z n v = x + y n y + z n y z w = 5 6 n y + 7 n z n z () Use the distriutive lw for dot products to write u v in terms of, n y, etc. u v = 2x + 2y n y + 2z n z + 3x n y + 3y () Use the definition of the dot product in eqution (2.2) to clculte, n y, etc. = n y = n z = n y = n y n y = n y n z = 0 n z = n z n y = n z n z = 1 (c) In view of your previous two results, clculte u v. u v = (d) As shown elow for the ritrry vectors nd (lso shown in Section ), the dot product Copyright c Pul Mitiguy. All rights reserved. 165 Homework 2: Vectors with sis

8 is reltively esy to clculte when nx, n y, n z re orthogonl unit vectors. = x + y n y + z n z } = x + y n y + z n z = x x + y y + z z Use this orthogonl-short-cut to clculte: u v =2x +3y + z u w = v w = 2.5 Perpendiculr vectors. (Note: i, j, k re orthogonl unit vectors). (Section 2.10) When vector v = x î +2ĵ +3 k is perpendiculr to vector w =4î +5ĵ +6 k, x =. 2.6 Dot product for clculting ngles. (Sections 2.10 nd 3.3) n z The figure to the right shows rectngulr prllelepiped (lock) of sides 2, 3, nd 4 with points A, B, C locted t corners., n y, n z re right-hnded orthogonl unit vectors with directed from B to C nd n y from B to A. n y () Express r C/A (C s position vector from A) in terms of, n y, n z. Find numericl vlue for r C/A 2 = r C/A r C/A. Next, use eqution (2.4) to clculte the mgnitude of r C/A (the distnce from A to C). r C/A = n y r C/A 2 = r C/A r C/A = r C/A = () Using eqution (2.1), clculte the unit vector û directed from A to C in terms of, n y, n z. Next, find the unit vector v directed from A to D in terms of, n y, n z. û = 3 2 n y v = 13 (c) Clculte BAC, the ngle etween line AB nd line AC. Next, clculte CAD, the ngle etween line AC nd line AD. BAC = CAD = Vector components: Sine nd cosine. (Section 1.4) j? Trigonometry plys centrl role in rottion mtrices. Replce ech? in the figure to the right with sin(θ) orcos(θ). Express unit vectors â nd in terms of sin(θ), cos(θ), î, ĵ. 1? â = î + ĵ θ 1 SohChTo = î + cos(θ) ĵ θ?? i Copyright c Pul Mitiguy. All rights reserved. 166 Homework 2: Vectors with sis

9 2.8 Vector components for crne-oom. (Section 1.4) Shown right is crne whose c A supports oom B tht swings wrecking ll C o. î, ĵ, k re right-hnded orthogonl unit vectors with î horizontlly-right, ĵ verticllyupwrd, nd k perpendiculr to the plne contining points N o, A B, B C, C o. Drw ech position vector listed elow nd then use your knowledge of sine/cosine to resolve these vectors into î nd ĵ components. N o N x Drw position vectors A B s position vector from N o r A B/N o = î + ĵ B C s position vector from A B r B C /A B = î + ĵ C o s position vector from B C r Co/B C = î + ĵ B C s position vector from N o r B C /N o = î + ĵ N o s position vector from C o r No/Co = î +[ L B sin(θ B )+L C cos(θ C )] ĵ A A B L B θ B B B C θ C L C C o 2.9 Dot products nd distnce clcultions. Show work. (Section 2.10) Shown right is crne whose c A supports oom B tht swings wrecking ll C o. To prevent the wrecking ll from hitting cr t point N o, the distnce etween N o ndthetipoftheoom (point B C) must e controlled. N To strt this prolem, express B C s position vector from N o in terms of x, L B,nd the unit vectors nd x. N o x A A B L B θ B B B C θ C L C C o r B C/N o = + x () Without resolving r B C/N o into nd n y components (done in the next step), use eqution (2.4) nd the distriutive property to clculte the distnce etween N o nd B C in terms of x, L B, θ B. Then clculte its numericl vlue whe =20m, L B =10m, θ B =30. (If necessry, complete the footnote hint elow). 1 Distnce etween N o nd B C : r B C/N o = + + =29.1 m () Two collegues re confused y your use of mixed-ses vectors (i.e., x + L B x ),ndsk you to verify B s position vector from N o cneexpressedintheuniform-sis s shown elow. Use this uniform-sis expression to verify your previous result for r B C/N o. Note: This inefficient uniform-sis pproch requires the simplifying trigonometric identity sin 2 (θ B)+cos 2 (θ B)=1. r B C/N o =[x + L B cos(θ B )] + L B sin(θ B ) n y r B C /N o simplifies to previous result. (c) Optionl: Clculte the distnce etween N o nd C o in terms of x, L B, L C, θ B,ndθ C. r Co/No = x 2 + L 2 B + L 2 C +2xL B cos(θ B )+2xL C sin(θ C ) 2 L B L C sin(θ B θ C ) 1 Hint: Use the distriutive property to express r B C /No r B C /N o in terms of x, L B,nd x. Next, use the dot-product definition in eqution (2.2) to clculte x =, nd then rewrite r B C /No r B C /N o = xLB ( ) = 2 + Note: The distriutive property for vector dot-multipliction is xLB cos( ). c + d = c + d + c + d. Copyright c Pul Mitiguy. All rights reserved. 167 Homework 2: Vectors with sis

10 2.10 Vector components, free-ody digrm (FBD), nd motion grphs for rocket-sled. The following figure shows rocket-sled moving long smooth (frictionless) inclined rils. Description Symol Type Mss of rocket-sled nd rider m Constnt Erth s grvittionl ccelertion g Constnt Angle etween horizontl nd inclined-rils θ Constnt i mesure of thrust force on sled F T Specified j mesure of norml force on sled F N Vrile i mesure of rocket-sled positio Vrile Drw unit vector î upwrd-right nd prllel to the rils. Drw unit vector ĵ outwrd-norml to the rils (perpendiculr to i ) nd in the plne of the pper. Free-ody digrm Drw prticle representing the rocket-sled. Drw the thrust, norml, nd grvity forces on the rocket-sled. Express the net force on the rocket-sled in terms of î nd ĵ. FNet = F Thrust + F Norml + F grvity = î + ĵ F = m The rocket-sled s ccelertion cn e clculted s = d 2 x dt 2 î = ẍ î. Sustitute the right-hnd sides of F Net nd into F Net = m nd solve for ẍ nd F N. î + ĵ = m ẍ î î : ẍ = m ĵ : F N = Solve for ccelertion : Knowing m = 100 kg, g =10 m s 2, θ =30, F T = 700 N, grph ẍ. Solve for velocity nd position : Knowing the rocket-sled strts t x = 8 m, nd is initilly moving downwrd-left long the ril t 4 m s, solve nd sketch ẋ(t) nd x(t) for 0 t 8. ẍ(t) = 2 m/s 2 ẋ(t) = m/s x(t) = meters θ j i = x (m/s 2 ).. x ẋ = v x (m/s) x(t) (meters) time (seconds) time (seconds) time (seconds) Include friction etween the rils nd rocket-sled, modeled vi coefficient of kinetic friction µ k. Express ẍ nd F N in terms of some/ll of µ k nd symols in the tle. Knowing µ k x(0) = 8 m, nd the rocket-sled initilly moves upwrd-right t 4 m s, find ẋ(t) nd x(t). ẋ(t) ẍ(t) = 1 m F s m 2 N = x(t) Clculte the minimum thrust (redrw FBDs) to: Result. Keep the rocket-sled moving uphill t constnt speed F T N. Keep the rocket-sled moving downhill t constnt speed F T N For prt, ssume the rocket is initilly moving downwrd-left t 4 m/s. Solution t Get Strted Rocket sled. F T m g j i F N Copyright c Pul Mitiguy. All rights reserved. 168 Homework 2: Vectors with sis

11 2.11 Construct unit vector û in the direction of ech vector given elow. (Section ) Vector Unit vector û 3 3 nx 3 4 n y 3 4 n y +12 n z n y n z c or Note:, n y, n z re orthogonl unit vectors. c is rel non-zero numer Note: Ensure your lst nswer grees with your first two nswers, e.g., if c =3 or c = Clculting vector cross products with ses. (Section 2.11) Given: Right-hnded orthogonl unit vectors, n y, n z nd: u = n y + 4 n z n y v = x + y n y + z n z w = 5 6 n y + 7 n z n z () Use the distriutive lw for cross products to write u v in terms of, n y, etc. u v = 2x + 2y n y + 2z n z () Use the definition of the cross product to clculte, n y, etc. = 0 n y = n z n z = n y n y = n y n y = n y n z = n z = n z n y = n z n z = (c) In view of your previous two results, clculte u v. u v = + n y + n z 2.13 Cross products nd determinnts. (Section ) Given right-hnded orthogonl unit vectors, n y, n z nd two ritrry vectors nd expressed s shown to the right, show tht clculting with the distriutive property of the cross product hppens to e equl to the determinnt of the mtrix shown to the right. = x + y n y + z n z = x + y n y + z n z n y n z = det x y z x y z Next use this determinnt method to clculte the following cross products (refer to Hw 2.12). u v = (3z 4 y) + (4x 2 z) n y + (2y 3 x) n z Optionl: u w = n y 27 n z Optionl: v w = ( ) + ( ) n y ( ) n z Copyright c Pul Mitiguy. All rights reserved. 169 Homework 2: Vectors with sis

12 2.14 Cross products: Commercil lgorithm for re clcultions (surveying). (Section ) One reson tringles re importnt is tht complex plnr ojects such s the polygon B elow cn e decomposed into tringles. Plnr mesurements re importnt in vrious professions, including frming crege, uilding costs, nd mss nd re properties of 2D ojects. B 7 B 5 B 4 r B 1/B 0 = 2.0 x y r B 2/B 0 = 0.5 x y B 8 B 6 B 3 r B 3/B 0 = 3.0 x y r B 4/B 0 = 0.2 x y B B 2 B cm B 1 r B 5/B 0 = 0.5 x y r B 6/B 0 = 1.0 x y B 9 B 0 z y x r B 7/B 0 = 2.0 x y r B 8/B 0 = 4.0 x y r B 9/B 0 = 2.0 x y A commercil lgorithm for clculting the re of the polygon B shown ove is to: Lel vertex B 0 nd numer the remining vertices sequentilly in counter-clockwise fshion. Form r B i/b 0, the position vector of vertex B i (i =1, 2,...) from vertex B 0 Clculte A 2 nd A 4, the vector-res of the tringles defined y vertices B 0 B 2 B 3, nd B 0 B 4 B 5, respectively. Formuls for the vector-res of ech tringle re given elow long with the vector sum of these res A nd the polygon s re (the mgnitude of A). (Just fill in the lnks. You only need to clculte A 2, A 4,ndA) A 1 = 1 2 r B 1/B 0 r B 2/B 0 = 2 z A 2 = 1 2 r B 2/B 0 r B 3/B 0 = Accounting for overlpped res is done with positive nd negtive signs on vectors. A 3 =... = 8.6 z A 4 =... = A 5 =... = 2.25 z A 6 =... = 1.5 z A 7 =... = 9 z A 8 = 1 2 r B 8/B 0 r B 9/B 0 = 5 z A = Are = 8 i =1 A i = A = 27.8 Note: Compute cross products with the distriutive property ( + ) ( c+ d)= c + d + c + d nd the cross-product-definition with the right-hnd rule (not determinnts or specil formuls in ook). Also, use the fct tht x, y, z re orthogonl unit vectors. Plnr ojects, courtesy of Working Model nd Design-Simultion Technologies Copyright c Pul Mitiguy. All rights reserved. 170 Homework 2: Vectors with sis

13 2.15 Sclr triple product with ses (Section 2.12). The figure shows right-hnded orthogonl unit vectors, n y, n z. Given Clculte u =2 + 3 n y + 4 n z v = x + y n y + z n z w =5 6 n y + 7 n z u v u = u v w = z x 6 y u v w = z 45 x y n y n z Note: Although the order of opertions in u v u is unmiguous, prentheses my clrify your work. u v w = u v w nd it is OK to switch nd in sclr triple products. True/Flse Locting microphone (2D). Show work. (Section 1.4) A microphone Q is ttched to two pegs B nd C y two cles. Knowing the peg loctions, cle lengths, nd points B, C, Q, N o ll lie in the sme plne, determine the distnce etween Q nd N o. Try to do the prolem first using Eucliden geometry - nd then try vectors. 2 B 9 n y? N o x 15 Q y 8 8 C Quntity Distnce from B to C Distnce from N o to B Length of cle joining B nd Q Length of cle joining C nd Q Distnce etween N o nd Q If Q is ove ceiling, distnce 12 m Vlue 15 m 8m 9m 8m 9.01 m Note: Although there re two mthemticl nswers to this prolem, one is ove the ceiling nd requires the cles to e in compression Locting microphone (3D). A microphone Q is ttched to three pegs A, B, ndc y three cles. Knowing the peg loctions nd cle lengths, determine the distnce etween Q nd point N o. Show work. (If needed, hint elow). 3 A B n y N o 13 Q C Quntity Distnce from A to B Distnce from B to C Distnce from N o to B Length of cle joining A nd Q Length of cle joining B nd Q Length of cle joining C nd Q Distnce etween N o nd Q If Q is ove ceiling, distnce 17 m Vlue 20 m 15 m 8m 15 m 13 m 11 m 13.3 m n z Note: The floor is horizontl, the wlls re verticl. Note: This is prt of the process of cmer trgeting footll/sell in stdium or lser trgeting cncer or Optionl: FBD nd sttics review prolems from Homework 12. The following prolems require n understnding of dot-products, force, nd free-ody digrms. Hw 12.1 Hw Hw Hw Hw 12.16() 2 Hint: For Eucliden geometry, use the lw of cosines (t lest once). For vectors, see Homework Hint: See Homework 1.35 or Section 3.3. Introduce unknowns x, y, z so Q s position vector from N o is x + y n y + z n z. Although nonliner lgeric equtions re usully solved with computer, these cn lso e solved y-hnd. Solution t Alterntely, solves sets of nonliner equtions, e.g., type Solve x^2 + (-20+z)^2 + (-8+y)^2 = 225, x^2 + z^2 + (-8+y)^2 = 169, z^2 + (-15+x)^2 + (-8+y)^2 = 121 Copyright c Pul Mitiguy. All rights reserved. 171 Homework 2: Vectors with sis