06_InstSolManual_PDF_Part6

06_InstSolManual_PDF_Part6 - In(c use coordinates where is...

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Figure 6.11 The free-body diagram for the person at the bottom of his path is given in Figure 6.11b. The acceleration is upward, so take upward. gives and (c) Apparent means and and The time for one revolution would be Note that (d) twice his true weight. Reflect: At the top of his path his apparent weight is less than his true weight and at the bottom of his path his appar- ent weight is greater than his true weight. 6.12. Set Up: directed toward the center of the circular path. At the bottom of the dive, is upward. The apparent weight of the pilot is the normal force exerted on her by the seat on which she is sitting. Solve: (a) gives (b) The free-body diagram for the pilot is given in Figure 6.12. gives Figure 6.12 6.13. Set Up: A angle is of a full rotation, so in a hand travels through a distance of
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Unformatted text preview: In (c) use coordinates where is upward, in the direction of at the bottom of the swing. a S rad 1 y 1 8 1 2 p R 2 . 1 2 s 1 8 45° R 5 0.700 m. n 5 m 1 g 1 a rad 2 5 m 1 g 1 4.00 g 2 5 5.00 mg 5 1 5.00 21 50.0 kg 21 9.80 m / s 2 2 5 2450 N mg n y x a rad n 2 mg 5 ma rad . g F y 5 ma y R 5 v 2 a rad 5 1 95.0 m / s 2 2 4.00 1 9.80 m / s 2 2 5 230 m. a rad 5 v 2 R a S rad a rad 5 v 2 / R , n 5 m 1 g 1 a rad 2 5 2 mg 5 2 1 882 N 2 5 1760 N, a rad 5 g . T 5 2 p R v 5 2 p 1 50.0 m 2 22.1 m / s 5 14.2 s. v 5 " gR 5 22.1 m / s. g 5 v 2 R mg 5 ma rad . n 5 weight 5 n 5 m 1 g 1 a rad 2 5 931 N. n 2 mg 5 ma rad g F y 5 ma y 1 y n 5 m 1 g 2 a rad 2 5 1 90.0 kg 21 9.80 m / s 2 2 0.549 m / s 2 2 5 833 N. y x n y n mg mg x ( a ) ( b ) a rad a rad 6-6 Chapter 6...
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