**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 Newton’s Laws 5-13...

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- Spring '07
- Shoberg
- Acceleration, Force, Velocity