131_Physics ProblemsTechnical Physics

131_Physics ProblemsTechnical Physics - Chapter 5 Section...

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Chapter 5 133 Section 5.8 Forces of Friction *P5.35 22.0° n F ground g /2 = = 85.0 lb F 1 F 2 F g = 170 lb 22.0° + x + y n tip f F = 45.8 lb 22.0° + x + y Free-Body Diagram of Person Free-Body Diagram of Crutch Tip FIG. P5.35 From the free-body diagram of the person, FF F x °= 12 22 0 22 0 0 sin . sin . af , which gives FFF == . Then, y + = 22 2 0 8 5 0 1 7 0 0 cos . . lbs lbs yields F = 45 8 . lb . (a) Now consider the free-body diagram of a crutch tip. Ff x = () ° = 45 8 22 0 0 .s i n . lb , or f = 17 2 l b . Fn y = °
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