05_InstSolManual_PDF_Part27

# 05_InstSolManual_PDF_Part27 - Let be upward. Solve: The...

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5.66. Set Up: If F is the magnitude of the force applied to each end of a spring and the spring stretches a distance x , then Hooke’s law says Let the springs have unstretched lengths and Solve: (a) The unstretched and stretched springs are shown in Figure 5.66. Call x the amount the combination has increased in length (stretched). Figure 5.66 (b) The force at each end of spring 2 is F . The force at the left-hand end of spring 2 is applied by spring 1. By Newton’s 3 rd law, spring 2 applies a force of the same magnitude to the right-hand end of spring 1. So, both springs are stretched by the same force F . (c) Hooke’s law applied to springs 1 and 2 and to the single effective spring gives and (d) From (a), and so F divides out and (e) By finding a common denominator the result in (d) can be written as 5.67. Set Up: For the climber, The rock face exerts a horizontal force n on the climber and no friction force.
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Unformatted text preview: Let be upward. Solve: The free-body diagram for the climber is given in Figure 5.67. Figure 5.67 gives (b) gives Reflect: The tension in the rope is greater than his weight because only the vertical component of the tension holds him up. n 5 T sin 15 5 161 N. n 2 T sin 15 5 0. g F x 5 ma x T 5 mg cos 15 5 600.0 N cos 15 5 621 N. T cos 15 2 mg 5 0. g F y 5 ma y y x T T cos15 n T sin15 mg 15 a 5 1 y a 5 0. k eff 5 1 25.0 N / cm 21 50.0 N / cm 2 25.0 N / cm 1 50.0 N / cm 5 16.7 N / cm k eff 5 k 1 k 2 k 1 1 k 2 . 1 k eff 5 1 k 1 1 1 k 2 . F k eff 5 F k 1 1 F k 2 . x 2 5 F k 2 , x 1 5 F k 1 , x 5 F k eff , x 5 x 1 1 x 2 . F 5 k eff x . F 5 k 2 x 2 F 5 k 1 x 1 , x 5 1 l 1 1 x 1 1 l 2 1 x 2 2 2 1 l 1 1 l 2 2 5 x 1 1 x 2 l 1 l 2 l 1 1 x 1 l 2 1 x 2 l 2 . l 1 F 5 kx . Applications of Newtons Laws 5-27...
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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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