05_InstSolManual_PDF_Part26

# 05_InstSolManual_PDF_Part26 - 5-26 Chapter 5 5.63 Set Up...

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5.63. Set Up: Take downward. (a) Assume the hip is in free-fall. (b) The free-body diagram for the person is given in Figure 5.63. It is assumed that the whole mass of the person has the same acceleration as her hip. Figure 5.63 Solve: (a) gives (b) gives The acceleration is upward. gives and (c) gives Reflect: When the velocity change occurs over a small distance the acceleration is large. 5.64. Set Up: Hooke’s law says the magnitude F of the force from the pad is given by where x is the distance the pad is compressed. From Problem 5.63, set when Solve: 5.65. Set Up: For each spring so and Solve: (a) The end of each spring moves the same distance, so (b) The free-body diagram for the point where F is applied is given in Figure 5.65. Figure 5.65 (c) For the single effective spring Comparing this to the result in (b) gives (d) Reflect: For two springs in parallel the effective force constant for the combination is greater than the force constant
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Unformatted text preview: of either spring. k eff 5 k 1 1 k 2 5 25 N / cm 1 45 N / cm 5 70 N / cm k eff 5 k 1 1 k 2 . F 5 k eff x . F 5 F 1 1 F 2 5 k 1 x 1 k 2 x 5 1 k 1 1 k 2 2 x x y a 5 F 1 F 2 F x 1 5 x 2 5 x . F 2 5 k 2 x . F 1 5 k 1 x F 5 kx , k 5 F x 5 25,000 N 0.020 m 5 1.3 3 10 6 N / m x 5 0.020 m. F 5 25,000 N F 5 kx , t 5 v y 2 v y a y 5 1.3 m / s 2 4.4 m / s 2 440 m / s 2 5 7.0 ms v y 5 v y 1 a y t n 5 w 1 ma 5 m 1 a 1 g 2 5 1 55 kg 21 440 m / s 2 1 9.80 m / s 2 2 5 25,000 N w 2 n 5 2 ma g F y 5 ma y 440 m / s 2 , a y 5 v y 2 2 v y 2 2 1 y 2 y 2 5 1 1.3 m / s 2 2 2 1 4.4 m / s 2 2 2 1 0.020 m 2 5 2 440 m / s 2 v y 2 5 v y 2 1 2 a y 1 y 2 y 2 v y 5 1.3 m / s. y 2 y 5 0.020 m, v y 5 4.4 m / s, v y 5 " 2 a y 1 y 2 y 2 5 " 2 1 9.80 m / s 2 21 1.0 m 2 5 4.4 m / s v y 2 5 v y 2 1 2 a y 1 y 2 y 2 a y 5 1 9.80 m / s 2 . y 2 y 5 1.0 m, v y 5 0, x y w n a 1 y 5-26 Chapter 5...
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