297_Physics ProblemsTechnical Physics

297_Physics ProblemsTechnical Physics - Chapter 10 P10.37...

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Chapter 10 299 P10.37 For m 1 , Fm a yy = : +− = nmg 1 0 nm g 11 19 6 == . N fn kk 706 µ . N a xx = : −+ = 200 1 .. N k g Ta bg (1) For the pulley, τα = I : = F H G I K J TR TR MR a R 12 2 1 2 −+ = TT a 1 2 10 0 . kg a 500 k g (2) For m 2 , = g 22 0 cos θ n 2 6 00 9 80 30 0 50 9 = c o s . . kg m s N 2 ej a f = = 18 3 : −− + = 18 3 2 .s i n N Tm m a + = 18 3 29 4 6 00 2 . N k g (3) FIG. P10.37 (a) Add equations (1), (2), and (3): −− + = 183 294 130 401 0309 . . N k g N 13.0 kg ms 2 a a (b) T 1 2 00 0 309 7 06 7 67 =+ = . . kg m s N N 2 T 2 767 922 = . . k g m s N 2 P10.38 Im R = 1 2 1 2 100 0 500 12 5 2 2 kg m kg m 2 af ω α ωω i fi t = = =− 50 0 5 24 05 2 4 600 0873 . . . rev min rad s rad s s rad s 2 = I 12 5 0 873 10 9 . kg m rad s N m The magnitude of the torque is given by
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