University Physics with Modern Physics with Mastering Physics (11th Edition)

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10.62: At the point of contact, the wall exerts a friction force f directed downward and a normal force n directed to the right. This is a situation where the net force on the roll is zero, but the net torque is not zero , so balancing torques would not be correct. Balancing vertical forces, , cos rod F w f F + + = θ and balacing horizontal forces , With . sin k rod n μ f n θ F = = these equations become , cos k rod w F n μ θ F + + = . sin rod n θ F = (a) Eliminating n and solving for rod F gives N. 266 30 sin ) 25 . 0 ( 30 cos N) 0 . 40 ( ) m/s (9.80 kg) 0 . 16 ( sin
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Unformatted text preview: 2 k rod = °-° + =-+ = θ μ θ F ω F b) With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net torque is n μ f R f F k and , ) ( =-may be found insertion of the value found for rod F into either of the above relations; i.e., N. 2 . 33 sin rod k = = θ F μ f Then, . rad/s 71 . 4 ) m kg (0.260 m) 10 N)(18.0 54 . 31 N . 40 ( 2 2 2 = ⋅ ×-= =-I τ α...
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  • Force, 2 m, 9.80 m/s, 0.260 kg, 4.71 rad, 16.0 kg

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