COSMOS Chapter 8 Solution 40 - Edubirdie

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Michigan State University
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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2019
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COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 40. FBD yoke: ΣFx = 0: P − N = 0, N = P = 8 lb F = µ s N = 125 ( 8 lb ) For impending slip, F = 2 lb For M max , F on yoke is down as shown FBD wheel and slider: For M min , F on yoke is up. (a) For M max the 2 lb force is up as shown. ΣM B = 0: M B − ( 3 in.) sin 65° ( 8 lb ) − ( 3 in.) cos 65° ( 2 lb ) = 0 M B max = 24.3 lb ⋅ in. W (b) For M min the 2 lb force is reversed, and ΣM B = 0: M B − ( 3in .) sin 65°  ( 8 lb ) + ( 3 in.) cos 65°  ( 2 lb ) = 0 M B min = 19.22 lb ⋅ in. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. W

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