05_InstSolManual_PDF_Part13

05_InstSolManual_PDF_Part13 - same magnitude. The box has...

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(b) Since the speed is decreasing, the acceleration is opposite to the direction of the motion and is to the right, as in part (a). The same equation applies as in part (a) and The accelerometer deflects to the left of vertical. (c) Constant speed means The hanging mass has no acceleration and hence no horizontal force on it. The accelerometer hangs vertically. 5.27. Set Up: Use coordinates where is parallel to the slope and directed up the slope. Let the skier have mass m . The normal force n exerted by the ground is perpendicular to the slope. Solve: (a) The free-body diagram for the skier is given in Figure 5.27. The weight mg of the skier has been replaced by its x and y components. gives and Her accelera- tion is directed down the slope. Figure 5.27 (b) and (c) Nothing changes in the free-body diagram and her acceleration is the same as in (a). 5.28. Set Up: In each free-body diagram take a positive coordinate direction to be the direction of the acceleration of the object. The box accelerates to the right and the bucket accelerates downward. Their accelerations have the
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Unformatted text preview: same magnitude. The box has mass Solve: (a) The free-body diagram for the box and bucket are shown in Figure 5.28. Figure 5.28 (b) applied to the box gives applied to the bucket gives Combining these two equations to eliminate T gives (c) gives v y 5 " 2 a y 1 y 2 y 2 5 " 2 1 3.58 m / s 2 21 1.50 m 2 5 3.28 m / s v y 2 5 v y 2 1 2 a y 1 y 2 y 2 y 2 y 5 1.50 m. a y 5 3.58 m / s 2 , v y 5 0, a 5 1 m bucket m box 1 m bucket 2 g 5 3.58 m / s 2 m bucket g 2 T 5 m bucket a . g F y 5 ma y T 5 m box a . g F x 5 ma x x y w box n T box x y m bucket g a T bucket a m box 5 1 375 N 2 / 1 9.80 m / s 2 2 5 38.3 kg. y n x mg cos24 mg sin24 24 4.0 m / s 2 , a x 5 2 g sin 24 5 2 4.0 m / s 2 . 2 mg sin 24 5 ma x g F x 5 ma x 1 x a 5 0. 25 b 5 25. tan b 5 a g 5 4.5 m / s 2 9.80 m / s 2 . Applications of Newtons Laws 5-13...
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