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10_InstSolManual_PDF_Part4 - n is the normal force applied...

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10.12. Set Up: Apply to the suitcase. Let be downward. Apply to the wheel. Let the counterclockwise sense of rotation be positive. The angular velocity and angular acceleration of the wheel are related to the linear velocity and linear acceleration a of the suitcase by and Solve: (a) (b) For the suitcase, gives The free-body diagram for the suitcase is given in Figure 10.12a. Figure 10.12 gives and The free- body diagram for the wheel is given in Figure 10.12b. gives and 10.13. Set Up: For the pulley The elevator has The free-body diagrams for the elevator, the pulley and the counterweight are given in Figure 10.13. Apply to the elevator and to the counterweight. For the elevator take upward and for the counterweight take downward, in each case in the direction of the acceleration of the object. Apply to the pulley, with clockwise as the positive sense of rotation.
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Unformatted text preview: n is the normal force applied to the pulley by the axle. The elevator and counterweight each have acceleration a . Figure 10.13 x y m 1 g T 1 a Elevator x m 2 g T 2 y a Counterweight Mg n T 1 T 2 1 Pulley a 5 R a . g t 5 I a 1 y 1 y g F S 5 m a S m 1 5 22,500 N 9.80 m / s 2 5 2300 kg. I 5 1 2 MR 2 . I 5 TR a 5 1 124 N 21 0.400 m 2 3.82 m / s 2 5 13.0 kg # m 2 TR 5 I a g t 5 I a T 5 m 1 g 2 a 2 5 1 15.0 kg 21 9.80 m / s 2 2 1.53 m / s 2 2 5 124 N. mg 2 T 5 ma g F y 5 ma y y mg Mg x a T T n (a) (b) a 5 a y R 5 1.53 m / s 2 0.400 m 5 3.82 rad / s 2 . a y 5 v y 2 2 v y 2 2 1 y 2 y 2 5 1 3.50 m / s 2 2 2 2 1 4.00 m 2 5 1 1.53 m / s 2 . v y 2 5 v y 2 1 2 a y 1 y 2 y 2 v y 5 3.50 m / s. v y 5 0, y 2 y 5 4.00 m, v 5 v R 5 3.50 m / s 0.400 m 5 8.75 rad / s a 5 R a . v 5 R v v a v g t 5 I a 1 y g F y 5 ma y 10-4 Chapter 10...
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