05_InstSolManual_PDF_Part15 - Reflect If some bricks are...

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5.32. Set Up: Assume that the normal force n at each hip is vertical. Solve: (a) The free-body diagram for the upper body of the person is shown in Figure 5.32. gives and Figure 5.32 (b) (c) 5.33. Set Up: The free-body diagram for the two crates treated as a single object, weight is shown in Figure 5.33a. The system doesn’t move so the friction force exerted by the roof is static friction. For the heaviest pallet of bricks this force has its maximum possible, The free-body diagram for the pallet of bricks is given in Figure 5.33b. Figure 5.33 Solve: (a) For the crates, gives and Then gives and For the bricks, gives and (b) For the upper crate the only horizontal force on the crate would be friction. This crate has so and the friction force is zero.
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Unformatted text preview: Reflect: If some bricks are removed, so the weight of the pallet is reduced, the system remains at rest. The friction force on the crate is equal to the new weight of the pallet and is less than m s n . g F x 5 a x 5 w b 5 T 5 266 lb T 2 w b 5 g F y 5 ma y T 5 f s 5 266 lb T 2 f s 5 g F x 5 ma x f s 5 m s n 5 1 0.666 21 400 lb 2 5 266 lb. n 5 400 lb. n 2 w C 5 g F y 5 ma y ( a ) ( b ) x y w b T a 5 x y w c 5 400 lb n f s T f s 5 m s n . w C , f k 5 m k n 5 1 0.30 21 2.5 21 0.65 2 mg 5 1 0.30 / 0.0050 21 5.2 N 2 5 310 N f k 5 m k n 5 1 0.0050 21 2.5 21 0.65 2 mg 5 5.2 N x y w u 5 0.65 w n n a 5 n 5 w u / 2. 2 n 2 w u 5 g F y 5 ma y Applications of Newton’s Laws 5-15...
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