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3 - Pmblqm I The cable AH exerts a 32-“ Ihme T{in H1 m I...

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Unformatted text preview: Pmblqm I The cable AH exerts a 32-“: Ihme T {in H1: m I - u]. A. Eaprcsa T in tcn'ns uf wnmonenis. Sulufiun: Tin: Hamlinalu nf poi-I R an“: b' m. T. 4}. Th.- wuur prairiun MB is rm =u+vj+4n Th: raw me [:Iuinl A In fluinl B is. Him by flu] = I'm — rm. Flam mum 1.95. r.“ = 2er + 23:; + 2min 11m ran 2 [I] -— ltfiJl-b- fi' — 1331i + H- — lflTH PI ran F“ -3.fi1l ‘F- 11.61] + 1.331. Th: magnum: is In»! = W = 5.54 n. Th: unit nun: [mintinn from A Lu 5 in u“ E = 44mm + 0.3429] + H.241“: lhusi MMHTlsglmhy Tan = ITuJuau —-— Elm --— — IMI + 3.0] + 'IJ'k :Ib] Problem 2.577 Astronauts: on the space shuttle use radar to determine the magnitudes and direction cmines of the position vectors of two sateliitees A and B. The vector rd from the shuttle to satellite A has magnitude 2 km. and direction engines cos 01 : 0.763,:305 6}, = O.384,co'56P2 = 0.512. The vector 1'3 from the shuttle to sateilite B has magnitude 4 km and direction casinos cost-)1. =O.743, c0561.. = 0.557, cos 6: : -—D.37l_ What is the distance between the satellites '? Solution: The two position vectors. are: rd : 2(0.768i+ [LJde—l— 0.513“: |.53()i + {Hfifij + 1.014;: (km) n r” = 4({}.743i+ 0.557jv— U.371k]=2.‘)72i + 3.223.] ~— l.49.41t (km) The dislunut: is the magnitude of the dilTerencc: in .. rrti : V’nfsseuzmnl +10.768 u 1.223): + ( [.024:t--l.-13-‘1r}t2 z 3.34 [km] I'mblemm The cable BC exerts an B-kN three F on thel bar AB ut B. to} Determine the components of ti unit vecmr that print: from B mwmfl point C. {bl Eat-less F in terms of components. f 13. DJJIIn Up: = —D.It-'Itfl-— (1.351] 141.41% do. I“ = IBM =ueur = —2.:9I-o.ttnj+ 143k it": ...
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