Problem_Set_7_Solutions

Problem_Set_7_Solutions - 22 If we write r = x i y j z k...

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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 is () 32i 24k N m. F = GG (d) Now G G G =− rr r o where o ˆ 3.0i 2.0j 4.0k. r =++ G Therefore, in the above expression, we set 0, 4.0, 0, xy z ′′ == = and 3.0 3.0 0 4.0 4.0 8.0 5.0 5.0 0. x y z F F F =−= We get G G G G = ×+ = F 0 . 22. If we write G ′ = + + ′ rx y z ### i j k, then (using Eq. 3-30) we find G G ′ × r F is equal to
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30. (a) The acceleration vector is obtained by dividing the force vector by the (scalar) mass: a = F / m = (3.00 m/s 2 )i ^ – (4.00 m/s 2 )j ^ + (2.00 m/s 2 )k ^ . (b) Use of Eq. 11-18 leads directly to L = (42.0 kg . m 2 /s)i ^ + (24.0 kg . m 2 /s)j ^ + (60.0 kg . m 2 /s)k ^ . (c) Similarly, the torque is rF τ G GG = (–8.00 N . m)i ^ – (26.0 N . m)j ^ – (40.0 N . m)k ^ . (d) We note (using the Pythagorean theorem) that the magnitude of the velocity vector is 7.35 m/s and that of the force is 10.8 N. The dot product of these two vectors is v .
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Problem_Set_7_Solutions - 22 If we write r = x i y j z k...

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