EN3: Introduction to Engineering. Teach Yourself Vectors. 1. Definition. Problems

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1 EN3: Introducton to Engneerng Tech Yourself Vectors Dvson of Engneerng Brown Unversty. Defnton vector s mthemtcl obect tht hs mgntude nd drecton, nd stsfes the lws of vector ddton. Vectors re used to represent physcl qunttes tht hve mgntude nd drecton ssocted wth them. For exmple, The velocty of n obect s vector. The drecton of the vector specfes the drecton of trvel, nd the mgntude specfes the speed. The force ctng on n obect s vector. The drecton of the vector specfes the lne of cton of the force, nd the mgntude specfes how lrge the force s. Other exmples of vectors nclude poston; ccelerton; electrc feld; electrc current flow; het flow; the norml to surfce. Exmples of qunttes tht re not vectors nclude mss, temperture, electrc potentl, volume, nd energy. These cn ll be descrbed by ther mgntude only (they hve no drecton) nd so re sclrs. vector s often represented pctorlly s n rrow (the rrow s length s ts mgntude, nd t ponts n ts drecton) nd symbolclly by n underlned letter, usng bold type or by n rrow symbol over vrble. The mgntude of vector s denoted, or. There re two specl cses of vectors: the unt vector n hs n ; nd the null vector hs.. Identfy whether the followng physcl qunttes should be descrbed s vectors or sclrs () Your ge (b) The dstnce between the Erth nd the sun (c) Forecst wnd (d) The grdent (slope) of surfce (e) The lner momentum of n obect (f) The speed of lght

2 . frst loo t vector components In prctce, we lmost lwys descrbe 3 dmensonl vectors by specfyng ther components n Crtesn bss. Specfyng the components of vector s lot le sttng the poston of pont on mp. For exmple, suppose we wsh to specfy the poston of Long Islnd Mcrthur rport reltve to JFK rport on the mp below. We mght sy tht Mcrthur s 33 Nutcl mles Est of JFK; Nutcl mles North, nd nd 86 feet (.4 Nutcl Mles) bove JFK. North Est Mcrthur (ISP) 99 ft MSL Kennedy (JFK) 3 ft MSL Nutcl Mles 33 Nutcl Mles The three dstnces (33 NM Est, NM North,.4NM vertclly) re the components of the poston vector of ISP reltve to JFK, n Crtesn bss wth ts xes pontng Est, North nd vertclly. We follow the sme procedure to specfy the components of ny vector. Frst, we choose three convenent, mutully perpendculr, reference drectons s shown n the fgure. The three reference drectons re often gven the symbols, nd or e, e nd e 3, nd re denoted {,,} or le, e, e3q for short. Then, we descrbe vectors by specfyng O how fr you need to trvel long ech of the three reference x drectons to rech the tp of the vector from ts tl, s shown R n the pcture. For exmple, to rech P from O n the fgure, y you need to trvel dstnce x long, dstnce y long nd dstnce z long. In mthemtcl notton ths would be expressed s r x y z The three numbers (x,y,z) re clled the components of the vector r n the bss {,,} P z

3 The reference drectons {,,} re rbtrry, except for two mportnt restrctons. Frst, s we hve lredy stted, the drectons must be mutully perpendculr. Secondly, the drectons must form rght hnded trd, whch mens the rrows must be chosen so tht t s possble to orent your rght hnd so tht your thumb s prllel to, your ndex fnger s prllel to nd your mddle fnger s prllel to. The fgure below shows two bses: the one on the left s correct (t s rght hnded trd) but the one on the rght s not (t s left hnded trd) Note tht the three drectons,, cn (nd wll) be regrded s vectors. Snce they hve drecton but no well defned mgntude, we wll choose them to be unt vectors. Exmple The fgure below shows hevy box suspended from two cbles. The box s subected to vertcl grvttonl force, nd two forces of mgntude T, T ctng prllel to cbles O, nd OB, respectvely. Express ech force s vector components n the bss shown. Soluton pcture s lwys helpful T T 45 O W 3 W T sn(45 ) 3 T 45 T cos(45 ) T T cos(3 ) 45 Ths symbol shows tht ponts towrds you O T sn(3 ) 3 B

4 Now, remember tht to wrte down the components of vector, you need to specfy the dstnce you trvel n ech of the,, drectons to rech the tp of the vector from ts tl. For the grvty force, we trvel dstnce W n the drecton. Therefore F W For the tenson n O, we trvel dstnce T cos( 45) T / n the drecton, nd T sn( 45) T / n the drecton. Therefore grvty FO T / T / Fnlly, for the tenson n OB, we trvel dstnce T T sn( 3) T 3 / n the drecton. Therefore cos( 3) T / FOB T / T 3 / n the drecton, nd Wrtng down vector components lwys follows ths generl procedure.. Consder the cube shown n the fgure. Identfy whch of the {,,} bses shown n the fgure re rght hnded trds. E H D F B G C. Consder the smple two-dmensonl truss structure shown below. Ech member hs length m. Wrte down the followng poston vectors, expressng your nswer s components n the bss shown, wth physcl dmensons of meters: () The poston vector of reltve to O (.e. the vector pontng from O to ) (b) The poston of B reltve to O (c) The poston of D reltve to O (d) The poston of C reltve to O (e) The poston of C reltve to B (f) The poston of B reltve to C O D C B

5 .3 Consder the VFR eronutcl Sectonl Chrt shown below. Estblsh Crtesn bss wth pontng true Est, pontng true North nd perpendculr to the plne of the pcture. PVD 55ft Newport 7 ft True N True E Bloc Is 9 ft Wrte down the components of the followng vectors n ths bss, expressng your nswer n Nutcl Mles () The poston vector of Newport Stte rport reltve to Provdence T.F. Green rport. (The heght of ech rport n feet bove men se level s shown ner ech rport see fgure) (b) The poston vector of Bloc Islnd Stte rport reltve to Newport Stte rport (c) The poston vector of Bloc Islnd Stte rport reltve to Provdence rport (d) n rcrft t feet on mle fnl pproch to Provdence runwy 3L. (n rcrft on fnl s lgned wth the runwy, nd the number of the runwy (3) ndctes tht the runwy hedng s 3 degrees mgnetc. Mgnetc vrton t PVD s 5 degrees W, so 3 mgnetc s 45 degrees true.)

6 3. How to clculte the mgntude of vector n terms of ts components Let r be vector nd let r x y z where (x,y,z) re three numbers specfyng the Crtesn components of the vector r. Fnd formul for the length (or mgntude) of r n terms of (x,y,z). P O x Elementry geometry, my der Wtson. Consder the fgure shown bove. Observe tht the mgntude of r s equl to the dstnce from O to P. Begn by clcultng the dstnce from O to Q. Observe tht OQR s rght ngled trngle, so Pythgors theorem gves R x+y y z Q OQ x y. Now observe further tht OQP s rght ngled trngle, so pply Pythgors theorem gn to see tht r OP OQ z e x y z r x y z 3. Clculte the mgntudes of ech of the vectors shown below () r 3 6 (b) r 6 6 (c) r For the truss shown below, fnd the mgntude of the poston vector of C wth respect to O. ll members hve length m O 6 D C B 3.3 vector hs mgntude 3, nd nd components of nd, respectvely. Clculte ts component. 3.4 Let {,,} be Crtesn bss. vector hs mgntude 4 nd subtends ngles of 3 degrees nd degrees to the nd drectons, respectvely. Clculte the components of n the bss {,,}

7 4. ddton of vectors Let nd b be vectors. Then (by defnton) c b s lso vector. Vector ddton stsfes b b (gn, by defnton). The vector c my be shown dgrmtclly by plcng rrows representng nd b hed to tl, s shown. b c 4. Formul for the sum of two vectors n Crtesn components Let x y z d d b bx by bz where x, y, z, bx, by, bz re the Crtesn components of vectors, b n bss{,,}. Let d d c cx cy cz d If c=+b clculte cx, c y, cz n terms of x, y, z, bx, by, bz Just Do It! c cx cy cz b x y z bx by bz ( x bx ) ( y by ) (z bz ) nd so comprng coeffcents of, nd cx x bx cy y by, cz z bz 4. Fnd the sum of the vectors lsted below, expressng your nswer s components n the {,,} bss. lso compute the mgntude of ech vector nd the mgntude of ther sum. (), b (b) 3, b 4 6 (c) x y z, b 4. For ech of the vectors lsted n the precedng secton, clculte -b. 4.3 The vectors nd b shown n the fgure hve mgntudes 3, b 5. Clculte the mgntude of the vector c. 5 c b

8 4.4 For the structure shown, wrte down the poston vectors of B reltve to O, C reltve to B nd C reltve to O. Verfy your nswer by checng tht D C OC OB BC 6 O The fgure shows hevy box suspended from two cbles. The box s subected to vertcl grvttonl force, nd two forces of mgntude T, T ctng prllel to cbles O, nd OB, respectvely. Express ech force s vector components n the bss shown. Wrte down the vector sum of the forces. 6 6 B B 45 O 3 5. Multplcton of vectors 5. Multplcton by sclr. Let be vector, nd sclr. Then b s vector. The drecton of b s prllel to nd ts mgntude s gven by b. Note tht you cn form unt vector n whch s prllel to by settng n. 5. Formul for the product of sclr nd vector n Crtesn Components Let x y z be vector nd sclr. Fnd n expresson for the components of the vector b Then d b bx by bz x y z x y z nd hence bx x, by y, bz z 5. Fnd the components of unt vector prllel to the vector Let, b 4 6. Fnd 6+b, nd 6-b.

9 5.3 Dot Product (lso clled the sclr product). Let nd b be two vectors. The dot product of nd b s sclr denoted by b, nd s defned by b b cos (, b), where nd b denote the mgntudes of nd b, respectvely, nd (, b) s the ngle subtended by nd b, s shown n the fgure. (,b) b Note tht b b, nd. If nd b then b f nd only f cos (, b) ;.e. nd b re perpendculr. 5.4 Formul for the dot product of two vectors n Crtesn Components Let x y z d d b bx by bz where x, y, z, bx, by, bz re the Crtesn components of vectors, b n bss{,,}. d d Clculte b n terms of x, y, z, bx, by, bz. Ths tme we hve to do some rel wor. Substtute for nd b nd see wht hppens b x y z bx by bz d d xbx xby xbz y bx yby ybz zbx zby zbz Ths s mess. But recll tht I, nd re mutully perpendculr, so the ngle between them s 9 degrees. Recll lso tht cos(9) =. Fnlly, recll the defnton of the dot product. Therefore. Ths leves b xbx yby z bz Fnlly, note tht the vector s lwys prllel to tself, so the ngle between vector nd tself s zero. Recll lso tht I, nd re ll unt vectors. Therefore cos(), nd so on for ll three remnng dot products. So, fnlly b xbx y by zbz

10 5.3 Fnd the dot products of the vectors lsted below (), b (b) 3, b 4 6 (c) x y z, b 5.4 The vectors nd b shown n the fgure below hve mgntudes 3, b 5. Clculte b. 5 b c 5.5 Two vectors nd b re mutully perpendculr. Wht s ther dot product? b 5.6 Clculte 3 3 g 5.7 Clculte the ngle between ech pr of vectors lsted n Problem 5.3.e. fnd the ngle (, b) between nd b n ech cse 5.8 For the structure shown, clculte the ngle between the vectors OB (.e. the vector pontng from O to B) nd b OC. (Use vectors t s possble to do ths by long-wnded trgonometry nd Pythgors theorem but tht s not the pont) D O 6 C B 5.5 Dot Product s Proecton The quntty b / b s sometmes referred to s the component of n drecton prllel to b. The fgure shows why. The vector cn be thought of s the sum of two vectors: one (OX) prllel to b nd nother (X) perpendculr to b. Ths process of dvdng nto two prts s nown s proectng onto components prllel nd perpendculr to b. O Recll tht s the length of O. The length of OX s therefore L cos (, b). But recll tht b b cos (, b), so tht L b / b, s stted. n=b/ b (,b) X n L= cos b

11 5.9 Let nd b be two vectors. Proect onto components prllel nd perpendculr to b s shown n the pcture. F bib () Show tht the vector X G H b JK O n=b/ b (,b) X b n L= cos () Verfy tht the precedng result stsfes X b, s t should (why?) ( ) Show tht the component of n drecton perpendculr to b s X ( b) / b 5.5 Cross Product (lso clled the vector product). Let nd b be two vectors. By defnton, the cross product of nd b s vector, denoted by c b. The drecton of c s perpendculr to both nd b, nd s chosen so tht (,b,c) form rght hnded trd, s shown. The mgntude of c s gven by c b b sn (, b) c b (,b) Note tht b b nd ( b) b ( b). Clcultng the mgntude of the cross product of two vectors s no swet, but fgurng out the drecton s pn. There re vrous de-memors to help you do ths- choose the one you fnd lest confusng, or me up your own. Rght hnd rule To fnd the drecton of b, rrnge your rght hnd so tht your thumb s prllel to, your ndex fnger s prllel to b, nd the ngle between your thumb nd ndex fnger s. Now set your mddle fnger s perpendculr to both nd b. The drecton of b s prllel to your mddle fnger. (Ths rule only relly wors f 9, otherwse you permnently dmge your hnd. Plese don t do ths.) Rght hnd screw rule To fnd the drecton of b, rrnge your rght hnd so tht your thumb s perpendculr to both nd b, nd your fngers curl n the drecton of the lne onng the tp of vector to the tp of vector b. The drecton of b s prllel to your thumb.

12 Bottle-cp rule. Obtn twst-top bottle of your fvorte beverge. Drw n rrow on the cp. rrnge the bottle so tht, by twstng the cp through the ngle, you cn rotte the rrow from prllel to to prllel to b. The drecton of b s prllel to the drecton of moton of the bottle-cp s t s turned. (Full beverge contners re not be permtted n EN3 exmntons) If none of these trcs help you Extend your mddle fnger nto the r. Shout your fvorte expletve. Ths wll not help, but t my me you feel better. Problem 5. Let {,,} be Crtesn bss. Use the defnton of the cross product gven bove to clculte ll possble cross products of the bss vectors.e., clculte, You wll fnd tht the results re ll very smple. For exmple, sn (, ), snce I s prllel to tself. Hence. Smlrly sn (, ), snce I nd re both unt vectors nd the ngle between them s 9 degrees. The rules governng the drecton of cross product lso show tht s prllel to. Therefore. See f you cn wor out the rest on your own. 5.6 Formul for the cross product of two vectors n Crtesn Components. Let x y z d b bx by bz d Clculte b n terms of d,,, db, b, b. where x, y, z, bx, by, bz re the Crtesn components of vectors, b n bss{,,}. x y z x y z More wor for the wced. Substtute for nd b b x y z bx by bz d d xbx xby xbz ybx yby ybz zbx zby zbz

13 Ths s nother mess. Ths tme, note tht sn (, ) (nd smlrly for nd ), note tht the ngle (, ) between I nd tself s zero, nd recll tht sn()=. Therefore. The remnng cross products between,, hve to be clculted lborously one t tme, usng the defnton gven n the precedng secton. The fgure shows the drecton of ll sx possble cross products between the bss vectors (mgntudes re not shown to scle, for clrty). Thus, we conclude tht x x x x x x THESE FORMULS RE IMPORTNT! You need to remember them. There s nce lttle trc to help you. Wrte down the 3 vectors,, n crcle, gong clocwse, s shown below. + - Now, to fnd the cross product of ny pr of bss vectors, you trvel round the crcle. Thus, to get, you strt t, move to nd then on to. If you go round the crcle clocwse, the nswer s postve, f you go counter-clocwse, t s negtve. Thus,, nd so on, whle, etc. If we substtute these results nto our expresson for b we determne tht d d b x y z bx by bz x by x bz y bx y bz zbx zby ( y bz zby ) (zbx xbz ) ( x by y bx ) Hence cx ( y bz zby ), cy ( z bx x bz ), cz ( x by y bx ) Ths s not n esy formul to remember, but t s so mportnt tht you must memorze t. The followng trc s sometmes used to help remember the formul f you now how to clculte the determnnt of mtrx, then you wll note tht

14 LM b det M MNb x x y by OP P ( b b ) b PQ z y z z y (zbx xbz ) ( x by y bx ) z It s frly esy to remember the cover-up rule for computng the determnnt of mtrx, so ths s populr trc. nother wy to remember the formul s to notce the pttern n the ndces. The ndces re wrtten out below to show the pttern more clerly cx ( y bz zby ), ndces x, y, z x, z, y cy (zbx x bz ), ndces y, z, x y, x, z cz ( xby y bx ), ndces z, x, y z, y, x There re two thngs to notce bout ths pttern. Frst, note tht the expresson for cx nvolves only y, z nd by, bz, smlrly, the expresson for cy nvolves only x, z nd bx, bz, nd the thrd expresson hs the sme feture. Secondly, notce tht the ndces lwys pper both forwrds (x,y,z or y,z,x or z,x,y) nd bcwrds (x,z,y or y,x,z or z,y,x) n ech expresson. The forwrd terms (x,y,z or y,z,x or z,x,y) re ll postve, whle the bcwrd terms (x,z,y or y,x,z or z,y,x) re ll negtve. 5. Fnd the cross products of the vectors lsted below (), b (b) 3, b 4 6 (c) x y z, b 5. The vectors nd b shown n the fgure hve mgntudes 3, b 5. Clculte b. Wht s the drecton of b? 5 b c 5.3 force F cts t some pont P on sold obect, s shown n the fgure. By defnton, the moment of the force bout n rbtrry pont O s vector M, defned s M rop F where rop s the poston vector of pont P reltve to O. F P rop O For ech fgure shown below, wrte down the force F nd the poston vector rop s components n the bss shown, nd hence clculte the vector moment M of the force bout the pont O.

15 N P ll members hve length m ll members hve length m P N O O There s more drect wy to clculte the moment of force, whch vods hvng to wrte out components of the poston vector nd force nd then tng the cross product. Insted, recllng the defnton of cross product, we note tht the mgntude of the moment s M r F F r sn (r, F) Observe tht r sn (r, F) s the perpendculr dstnce from O to lne drwn prllel to F through ts pont of cton, s shown n the pcture. Thus, to clculte the mgntude of moment, you only need to fnd ths perpendculr dstnce, nd multply t by the mgntude of the force. The drecton of M cn be deduced usng the usul rules. 6 6 F d P rop O dsn Clculte the requred perpendculr dstnce n ech problem below, nd hence deduce M for ech fgure. ll members hve length m N P ll members hve length m P N O O

16 6. New Loo t Vector Components Fnlly, we te new loo t wht we re dong when we express vectors s components n bss. Frst, two theorems. 6. THEOREM. Let nd b be two non-collner vectors. Then ny vector r whch s coplnr wth nd b cn be expressed s lner combnton of nd b, tht s to sy, there exst two sclr numbers nd such tht r b. b r b It s esest to see ths grphclly. Recll tht vector r cn be regrded s connectng two ponts n plne. If nd b le n the sme plne, t s lwys possble to get from one end of the vector to the other by trvelng long pth prllel to nd b. In fct, we cn even fnd formul for the two numbers nd. Recll tht r b. We cn turn ths nto two sclr equtons by tng dot products of both sdes wth nd b n turn r b r b b b b Solve for nd. br gbb bg br bgb bg b gbb bg b bg br bgb g br gb bg b gbb bg b bg Ths s messy, but t loos bt better f we choose nd b to be unt vectors, n whch cse b b r r b b b b g d d d dr b br gd b d b (We put lttle hts on the vectors to show tht you cn only use the formul for unt vectors) We relly blow our mnds f we lso choose nd b to be mutully perpendculr so b r r b b b g d

17 6. THEOREM II The sme sort of thng wors n three dmensons. In ths cse, let, b nd c be three non-coplnr, noncollner vectors. Then ny vector r cn be represented s lner combnton of, b, nd c,.e. there exst three sclrs, nd such tht r b c We could, f we relly wnted to, fnd generl expresson for, nd, but the results re so complcted t s not relly worth the effort. However, f we choose, b, nd c to be mutully perpendculr, unt vectors, we fnd tht r r cos (r, ) r b r cos (r, b ) b c b c r c r cos (r, c ) These results gve us new nsght nto wht t mens to express vector s components n bss. Here s the scoop. Truth be told, we don t ctully le vectors very much. (Ths my be the frst sttement n ths tutorl you relly pprecte). Clcultng sums nd products of rbtrry vectors s pn. So, n ny problem we solve, we use s few vectors s possble. In two dmensons, we pc two convenent vectors {,}nd then express ll vectors s sum of these two, r x y. In three dmensons, we need to pc three reference vectors, {,,}, then we cn me ll other vectors sum of these r x y z. We now from our Theorems tht t s very helpful f we pc our reference vectors to be mutully perpendculr unt vectors. In ths cse, we get very convenent formul for the three numbers x, y nd z. x r r cos (r, ) y r r cos (r, ) z r r cos (r, ) The fgure shows the three ngles (r, ), (r, ), ( r, ). We see tht x, y nd z correspond to the proectons of r on the three bss vectors, precsely s we ssumed when frst wrtng down vectors s components n bss. z O r,) r P r,) r,) y Ths dscusson hs gven us new nsght nto wht we re dong n expressng vectors s components n bss. It shows why our three x reference drectons cn be regrded s vectors; t shows why they should be unt vectors, nd why the vectors should be mutully perpendculr. It does not expln why the three bss vectors must form rght hnded trd -- Ths s done so tht we get the correct expresson for the component form for cross product (Sect 5.6)

18 Note, however, tht mthemtclly speng we dd everythng bcwrds n ths tutorl. Strctly speng, we should hve strted wth the defnton of vector sum (f nd b re vectors then c=+b s vector), defne the dot product ( b b cos (, b) ), nd then deduce the exstence of bss s we dd n ths secton. Then, fnlly, we cn deduce expressons for vector opertons n component form. 6. Let =5-6; b=4+ Let r=5. Express r s components prllel to nd b,.e. fnd two sclrs nd such tht r b 6.3 Drecton Cosnes of vector We see from the precedng secton tht ll vectors cn be represented n Crtesn bss s r r (cos (r, ) cos (r, ) cos (r, ) ) The three numbers cos (r, ), cos (r, ), cos (r, ) re nown s the drecton cosnes of vector. Ths s becuse they re cosnes, nd specfy the drecton of the vector. Duh. It s strghtforwrd to clculte the drecton cosnes of vector f you now ts components. For exmple, f r x y z the three drecton cosnes re x y z,, x y z x y z x y z See f you cn show ths for yourself. 6. Fnd the drecton cosnes of the followng vectors () (b) Chnge of bss Next, we s n obvous queston. Supposng we re gven ll our vectors s components n some bss {,,}, but for some reson we don t le ths bss, nd would prefer to now our vector s components n nother bss {e, e, e 3} (Here, the three es represent mutully perpendculr unt vectors, ust le,,). How do we convert from one to the other? For two dmensonl problems, the esest procedure s to fnd how to construct ech of the,, vectors by ddng up {e, e, e 3}, nd then substtute. We wll llustrte ths usng n exmple.

19 In the fgure shown below, the slbot trvels n Northesterly drecton. The wnd s 5 nots. Fnd the components of the wnd vector n bss {e, e, e 3} lgned wth the slbot.. e e -sn(45) e sn(45) e 45 o e3 cos(45) e North 45o sn(45) e To proceed, we wll express nd n terms of {e, e, e 3}. Recll tht nd hve unt length, therefore we cn combne the e bss vectors s shown n the fgure to me up the nd vectors. From the fgure cos(45)e sn(45)e e e sn(45)e cos(45)e e e so plug nto the expresson for the wnd LM e N w 5 5 OP LM Q N e e e OP Q 5 5 e e nots For three dmensonl problems, t pys to be more systemtc. Suppose we now the components of vector r n {,,} re (x,y,z), nd we wsh to clculte the components (,, ) n {e, e, e 3}. To proceed, we go bc to the fundmentl defnton of the bss vectors, nd note tht r cn be wrtten s r x y z e e e 3 clever trc llows us to solve for (,, ). If we te the dot product of both sdes wth e, then x e y e z e e e e e e3 e But recll tht e e, e e e e 3, so tht x e y e z e

20 Smlrly, tng dot products wth the other two e vectors gves x e y e z e x e y e z e x e 3 y e 3 z e 3 Fnlly, we hve to clculte e nd ll the rest. Ths s done ether by fndng the ngles between the pproprte vectors nd usng the defnton of dot product, or, f we re lucy, we now the components of ech e vector n {,,} n whch cse we cn evlute the dot product drectly. Let s try ths out on our slbot exmple. In ths cse x=-5, y=, z=, nd e cos(45) e sn(45) e3 e sn(45) e cos(45) e3 So, plug everythng nto the mgc formul to see tht 5 5 nots, gvng the sme nswer s before (Phew!) 6.3 Repet the slbot problem gn, but wth the bot trvelng t 3 degrees to the drecton. 6.4 Let 5 6 3, b 3 6, c () Verfy tht, b nd c re mutully perpendculr nd tht c b (b) In vew of (), three unt vectors {e, e, e 3} prllel to b nd c cn form bss. Clculte the components of {e, e, e 3} n the {,,} bss. (c) Let r=4+6. Clculte the components of r n {e, e, e 3}. (Use your nswer to (b) to clculte the requred dot products n the formul) 7. Fun ctvtes wth Vectors You wll use vectors n morty of your engneerng courses (f you pln to leve engneerng fter EN3, you wll use vectors n ll your engneerng courses!). They re used n dynmcs, flud flow problems, mechncs of deformble solds, electrc nd mgnetc felds, het flow problems, mong others. Here, we ust llustrte few pplctons.

21 7. Clcultng res. g b rc Recll tht the re of the trngle s B C sn. Ths quntty cn be found qucly usng cross product B BC rb r rc rb If you relly wnt to loo cool, you cn smplfy ths formul to r rb rb rc rc r b C Mny geometry problems re ncely solved usng vectors. We ll show one exmple. trngle hs corners t ponts r, rb, rc s shown n the pcture. Clculte the re of the trngle. rb B r g 7. Solvng Vector Equtons In EN3, by fr the most mportnt pplcton of vectors wll be n solvng the equtons of equlbrum for structures nd mchnes. Newton s lws (wth n extenson by Euler), sys tht f body s subected to forces F (= N) ctng t postons r, together wth moments M (=.M), then the resultnt force nd moment on the sold must vnsh f the body s t equlbrum. The equlbrum equtons re N F N M r F M In typcl sttcs problem, our ts wll be to dentfy ll the forces ctng on our system of nterest, express them ll s vectors, nd then use the equtons of equlbrum to solve for ny unnown forces. Exmple typcl force equlbrum equton loos le ths R T (4 3) W ( ) 5 Here, T R nd W re three forces whose drecton s nown, but whose mgntude s not nown. We need to solve the equton for T, R nd W. To do so, we collect together ll the, nd components 4R 3R T (W ) 5 5 then, we note tht f the vector s zero, ll ts components must be zero, so tht FG H IJ FG K H IJ K

22 FG T 4 R IJ H 5 K FG 3R IJ H5 K gvng us three equtons to solve (the nwer s R 5 3 T (W ) 4 3 W Newtons) Exmple Here s typcl force-nd-moment equlbrum problem, n two dmensons R (Tx Ty ) b4 6g R ( ) (T T ) x y The frst equton s force equlbrum, the second s moment equlbrum. R s force wth nown drecton but unnown mgntude, Tx Ty s force wth unnown mgntude nd drecton. To proceed, we hve to multply out the cross product n the second equton (remember the rules for cross products of the bss vectors?) 4R Tx Ty Then, collect ll the vector components from both equtons to see tht Tx R Ty 4 R Tx Ty whch re esly solved to get Tx, R Ty 8 Newtons. The sme procedure wors for three dmensonl problems. Usully, n 3D problem we end up wth 6 equtons for 6 unnowns nsted of ust 3, so thngs cn get relly messy. But tht s wht computers (or entry level engneers n your frm) re for. 7. Solve the followng vector equtons () x 7 y (b) x 7 y FG H IJ K 7. The fgure shows hevy box suspended from two cbles. The box s subected to vertcl grvttonl force W, nd two forces of mgntude T, T ctng prllel to cbles O, nd OB, respectvely. Express ech force s vector components n the bss shown. Wrte down the vector sum of the forces. Use the fct tht the vector sum of the forces s zero to clculte T, nd T n terms of W B 45 O 3

23 Summry Checlst Before tng the EN3 Vector Profcency Exm, you should me sure you cn ccomplsh the followng tss: () Identfy physcl qunttes s sclrs or vectors () Set up Crtesn bss () Identfy whether trd of vectors s rght hnded (v) Usng geometry, wrte down the components of vectors such s poston, force, etc s components n Crtesn bss (v) Clculte the mgntude of vector whose Crtesn components re gven (v) dd nd subtrct vectors, both grphclly nd usng components (v) Multply vector by sclr (v) Clculte the dot product of two vectors; (x) Use dot products to clculte ngles between vectors (x) Clculte the component of vector n drecton prllel to nother vector (x) Clculte the cross product of two vectors (x) Clculte moments of forces bout gven pont, both usng the cross product method nd usng the perpendculr dstnce method. (x) Clculte the drecton cosnes of vector (xv) Gven vector components n one bss, compute new components n second bss (xv) Clculte the re of trngle gven vector expressons for the poston of ts corners (xv) Solve vector equtons for unnown components or mgntudes of vectors (xv) Derve ny formul or expresson whch s derved for you n ths tutorl (xv) Compose n epc poem extollng the oys of vectors nd recte t whle wlng on wter.