ch11-p022 - 22. If we write r = x i + y j + z k, then...

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− ′ + + yF zF zF xF yF zy xz y x di bg ## # . ijk (a) Here, G G ′ = r r where ˆˆ ˆ 3.0i 2.0j 4.0k, r =−+ G and G G FF = 1 . Thus, dropping the prime in the above expression, we set (with SI units understood) x = 3.0, y = –2.0, z = 4.0, F x = 3.0, F y = –4.0 and F z = 5.0. Then we obtain G G G τ =× = rF 1 60 30 . # . # . # ij k N m . e j (b) This is like part (a) but with G G = 2 . We plug in F x = –3.0, F y = –4.0 and F z = –5.0 and obtain G G G + 2 26 18 # . k N m . e j (c) We can proceed in either of two ways. We can add (vectorially) the answers from parts (a) and (b), or we can first add the two force vectors and then compute G G G G =× + F 12 (these total force components are computed in the next part). The result
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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