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Unformatted text preview: 10.1: Equation (10.2) or Eq. (10.3) is used for all parts. a) m, N 00 . 40 90 sin N) m)(10.0 00 . 4 ( ⋅ = ° out of the page. b) m, N 6 . 34 120 sin N) m)(10.0 00 . 4 ( ⋅ = ° out of the page. c) m, N . 20 30 sin N) m)(10.0 00 . 4 ( ⋅ = ° out of the page. d) m, N 3 . 17 60 sin N) m)(10.00 00 . 2 ( ⋅ = ° into the page. e) The force is applied at the origin, so 0. = τ f) . 180 sin N) m)(10.0 00 . 4 ( = ° 10.2: m, N 40.0 m) N)(5.00 00 . 8 ( 1 ⋅ = = τ m, N . 12 30 sin m) N)(2.00 (12.0 2 ⋅ = ° = τ where positive torques are taken counterclockwise, so the net torque is m, N . 28 ⋅ with the minus sign indicating a clockwise torque, or a torque into the page. 10.3: Taking positive torques to be counterclockwise (out of the page), m, N 2.34 N) m)(26.0 (0.09 m, N 1.62 N) (180.0 m) 090 . ( 2 1 ⋅ = = ⋅ = × = τ τ ( 29 m, N 78 . 1 N) (14.0 m) 090 . ( 2 3 ⋅ = = τ so the net torque is m, N 50 . 2 ⋅ with the direction counterclockwise (out of the page). Note that for 3 τ the applied force is perpendicular to the lever arm. 10.4: R F F R F R F τ τ ) ( 1 2 2 1 2 1 = + = + m. N 726 . m) N)(0.330 7.50 N 30 . 5 ( ⋅ = = 10.5: a) b) Into the plane of the page. c) ] ˆ N) 00 . 4 ( ˆ n) 00 . 5 [( ] ˆ m) 150 . ( ˆ m) 450 . [( i i j i F r + × + = × ˆ m) N 1.05 ( ˆ N) 00 . 5 m)( 150 . ( N) 00 . 4 ( m) 450 . ( k k ⋅ = = 10.6: (a) CCW m, N 8.7 m) 2 . )( 60 N)(sin 50 ( A ⋅ = ° = τ CW m, N 10 m) 2 . ( N) 50 ( CW m, N 5 m) 2 . )( 30 N)(sin 50 ( D C B ⋅ = = ⋅ = ° = = τ τ τ (b) m N 10 m N 5 m N 8.7 ⋅ ⋅ ⋅ = ∑ τ CW m, N 3 . 6 ⋅ = 10.7: kg 00 . 2 kg, 40 . 8 where , 2 2 2 3 2 = = + = m M mR MR I 2 m kg 600 . ⋅ = I m N 0524 . , ; s rad 0.08726 gives ? s, . 30 ; s rad 5.236 rpm . 50 ; s rad 7.854 rpm 75.0 2 ⋅ = = = Σ = + = = = = = = = Iα τ Iα τ α αt ω ω α t ω ω f 10.8: ( 29 ( 29 ( 29 m. N 1 . 13 s 00 . 8 min rev 400 m kg 50 . 2 a) min rev s rad 60 2 2 ⋅ = × ⋅ = ∆ ϖ ∆ = = τ π t I Iα b) . J 10 2.19 min rev s rad 60 2 min rev 400 ) m kg 50 . 2 ( 2 1 2 1 3 2 2 2 × = × ⋅ = ϖ π I 10.9: ( 29 ( 29 in found that as same the , s m 2 . 1 m . 2 s m 36 . 2 2 2 = = = as v Example 98. 10.10: ( 29 ( 29 ( 29 . s rad 00 . 2 m kg . 5 m 250 . N . 40 2 2 = ⋅ = = = I FR I τ α 10.11: a) + + = + + = + = M m m M g M m m M g T Mg n 2 1 3 2 1 b) This is less than the total weight; the suspended mass is accelerating down, so the tension is less than mg. c) As long as the cable remains taut, the velocity of the mass does not affect the acceleration, and the tension and normal force are unchanged. 10.12: a) The cylinder does not move, so the net force must be zero. The cable exerts a horizontal force to the right, and gravity exerts a downward force, so the normal force must exert a force up and to the left, as shown in Fig. (10.9)....
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This note was uploaded on 08/23/2008 for the course PHYS 2048 taught by Professor Hiebig during the Spring '08 term at Nova Southeastern University.
 Spring '08
 HIEBIG
 Physics

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