05_InstSolManual_PDF_Part21 - is increased T decreases...

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5.46. Set Up: The free-body diagram is given in Figure 5.46a for going up the hill and in Figure 5.46b for going down the hill. In each case the kinetic friction force is directed opposite to the velocity. Use coordinates where the axes are parallel and perpendicular to the ground. Let m be the mass of the toboggan. Figure 5.46 Solve: In either case gives and (a) gives and The acceleration is directed down the hill. (b) gives and The acceleration is directed down the hill. 5.47. Set Up: The free-body diagram for the load is given in Figure 5.47. The tension T in the rope has been resolved into its x and y components. Figure 5.47 Solve: (a) gives and gives so (b) (i) No friction means The result in (a) becomes (ii) Pulls horizontally means The result in (a) becomes Reflect: Pulling above the horizontal reduces the normal force and therefore reduces the friction force. But it also reduces the component of T in the direction of the acceleration. As
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Unformatted text preview: is increased, T decreases, reaches a minimum value at an optimum value of and then increases. The optimum angle depends on the value of m . f f T 5 M 1 a 1 m g 2 . f 5 0°. T cos f 5 Ma . m 5 0. T 5 M 1 a 1 m g 2 cos f 1 m sin f T cos f 2 m Mg 1 m T sin f 5 Ma . T cos f 2 f 5 Ma g F x 5 ma x f 5 m n 5 m Mg 2 m T sin f n 5 Mg 2 T sin f . T sin f 1 n 2 Mg 5 g F y 5 ma y T T cos f Tsin f x y f k n Mg f a 4.05 m / s 2 , a x 5 2 g 1 sin f 2 m k cos f 2 5 2 4.05 m / s 2 2 mg sin f 1 f k 5 ma x g F x 5 ma x 8.55 m / s 2 , a x 5 2 g 1 m k cos f 1 sin f 2 5 2 1 9.80 m / s 2 21 0.30 cos 40.0° 1 sin 40.0° 2 5 2 8.55 m / s 2 2m k mg cos f 2 mg sin f 5 ma x 2 f k 2 mg sin f 5 ma x g F x 5 ma x f k 5 m k mg cos f . n 5 mg cos f g F y 5 ma y ( b ) ( a ) mg cos f sin y n f k f x n f k f x v Applications of Newton’s Laws 5-21...
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