05_InstSolManual_PDF_Part20

05_InstSolManual_PDF_Part20 - 7.39 m / s 2 . 1 283 N 2 / 1...

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5.44. Set Up: (a) The free-body diagram for the person is given in Figure 5.44a. F is the traction force along the spinal column and is the person’s weight. (b) The free-body diagram for the collar where the cables are attached is given in Figure 5.44b. The tension in each cable has been resolved into its x and y components. Figure 5.44 Solve: (a) and (b) so 5.45. Set Up: The free-body diagram for the crate is given in Figure 5.45. Use coordinates with axes parallel and perpendicular to the ramp. Constant speed means The mass of the crate is The weight mg has been resolved into its x and y components. Figure 5.45 Solve: (a) gives with gives so (b) so and The acceleration would be directed down the ramp. Reflect: In part (a) the force up the incline is 283 N and the force down the incline is 283 N. When the rope breaks and the force down the incline remains 283 N. The acceleration down the incline is
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Unformatted text preview: 7.39 m / s 2 . 1 283 N 2 / 1 38.3 kg 2 5 T S 0.754 g 5 7.39 m / s 2 , a x 5 2 g 1 m k cos f 1 sin f 2 5 2 0.754 g 2m k mg cos f 2 mg sin f 5 ma x 2 f k 2 mg sin f 5 ma x T 5 T 5 mg sin f 1 f k 5 1 375 N 2 sin 33 1 1 0.250 21 375 N 2 cos 33 5 283 N f k 5 0, T 2 mg sin f 2 a x 5 g F x 5 ma x f k 5 m k n 5 m k mg cos f . n 5 mg cos f . g F y 5 ma y mg y x n T f k mg sin f mg cos f f m 5 w / g 5 38.3 kg. a 5 0. T 5 F 2 sin 65 5 0.75 w 2 sin 65 5 0.41 w 5 1 0.41 21 9.80 m / s 2 21 78.5 kg 2 5 315 N 2 T sin 65 2 F 5 F 5 f s 5 m s n 5 0.75 w 5 0.75 1 9.80 m / s 2 21 78.5 kg 2 5 577 N n 5 w F x y T cos65 T cos65 T sin65 T sin65 65 65 T T a 5 ( b ) ( a ) y x w n f s F a 5 w 5 mg 5-20 Chapter 5...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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