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5 - Solution Assume that lmlh I1 and V Iiu in ill.T-J Mama...

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Unformatted text preview: Solution: Assume that lmlh I1 and V Iiu: in ill: .T-J' Mama. Thu: Slmlcgy ii In use the deliniliun 01' the crass product lEq. 2.28} and [he Eu; (2.34]. and equal: the two. me Eq. (2.28] I] x V :_- {UHVI sinﬂil -- 3311:. Since [he p-nsitive 2-ax15 is But of ”It: paper, and L' paints inlu mu {Japan then u : «It. Talk: 11:: um prnducl al‘bmh sldus with c. and not: that lick =1. Thus IUxVJ-k) smith — 191] 2 —( WINE The vectors are; U : |U|Elcusﬂ| +.|sin93}. and V = |V|ticusﬂl +jsin91). The crass prudutl is i j k “XV: llilctlsfh IUISinHl D |\'|r_'usbh EVlsim‘LI [I- aim] wjlm—f- ILHUIIW Nuns b" sinﬂg — uusflg sin 0.] Summit: imn Inc deﬁnition to uhmin: sinm. “—331 2 mm emu; -- LUSH] sin 61. (LED. Problem 2.138 Th: rope AH exerts tl SCI-H farce T _v m the calla: ut A. but rm be the pmilinn vector from point C m puinl A. Determine the cross pmduct rm x T. Solution: We delim- ll“: [lpfll'DIWiiltE vmurs. n1; : {—0.21—{131' +{115kl n1 l‘ [Ta = [(1.2 m1~£“— = [~E}.Illlll — 0.1313 + Ill mm In Irml rm: :- (Uﬁj -I- [LISHJ m l‘ut.‘ = Mall «1- ﬂjj] In my = rut: - trut: +113.) 2 Illﬁli— LIZ; ~— ﬂ.3ﬂ5k] In T = :50 N13 = {—33.1i+ 35.11 +3.51%]: N - lust Now take the cross pruduct i ' i I Pm 241': —ﬂ.L'IBI -u'.i31 mm 43.1 343.1 3.93 : tw—LTZI — 3.43] + —7.‘J{Ilt3 N-In rm :n' T = (“4-7.3 - 3.48] -I- —7.'J'l'}lt} N-rn Problem 2.141'" Dclcrmini: the minimum distancc .1' frnm point 'P to the plane deﬁned by the three points A. B, and C. (0.10m {J Solutiu: _--.- '- ' . ' . . . H The Hll'lttgy Ih 10 Iiml ilii. unit chlUi‘ le'andlLuldI' hi E9516. 5} m _ the plane. The pmjcctiun of Ibis unit vcclur on Ilac vector 0.”: 1'0}? - e is 5: ﬁle dislanuu from the origin in P Mung the penicndicuiar In the plume. The pmjcction an e or any vecmr into the plum: [rug - 1:. run -c. or :jj' l'nr- I1} is the dimmer fmm the origin In the plane along this same :5 papeudiuular. Thus UIL' distance of P [mm the plum: is . '.' .‘J' [3. U. ii} In .I.‘ *r01'-n~raA-ui 11: pusiﬁun vccmrs are; rm 2 3], rm.- 2 Sj, rm: 2 4]: and rm: 2 5+ [r] + 5h. The unit vector purpcndlcular lo the plain: is found mm the cmss pruduct 0! any two veclurs lying in llu: plane. Noting: m: ': I‘m “4 1'03 = *5] -E- 41;. ile FHA = I'm. - ray 2 3i - 5]. Tilt: FIRES-I .953 prod ncl: ‘ 0 [\$10.51)] 5 j ﬁfm= 0 —-5 4 =Zﬂi+izj+lﬁk. 3 s 0 1mflglliludc is Il'nc .5: l‘ml : E7173, thus Illc unit VECLDI‘ is 1:: ~12: + 0.4327; + 0.540‘Jk‘ The distance or point P from “It plum: rim" ' I: "' I‘m - E a I 1.792 -- 116-1- 2 9.113 m. The sucniid term iii {Eiﬂfeluf Elie Flam: from the origin; the venom rag, or rm- m1 used instead of rm. AllOJJI :z/ Cllmd] ...
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5 - Solution Assume that lmlh I1 and V Iiu in ill.T-J Mama...

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