COSMOS Chapter 8 Solution 109 | Answer Key - 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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2021
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COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 109. FBD lower portion of belt: ΣFy = 0: 48 N − N D = 0, N D = 48 N Slip on both platen and wood FD = µkD N D = 0.10 ( 48 N ) = 4.8 N FE = µkE N E = ( 48 N ) µ kE FBD Drum A (assume free to rotate) ΣFx = 0: TA − TB − 4.8 N − µ kE ( 48 N ) = 0 TB = TA + 4.8 N + µ kE ( 48 N ) (1) ΣM A = 0: rA (TA − TT ) = 0, (2) ΣM B = 0: M B + r (TT − TB ) = 0 TT = TA FBD Drive drum B TB = TT + 2.4 N ⋅ m = TT + 96 N 0.025 N Impending slip on drum, TB = TT e µs β = TT e0.35π so TT + 96 N = TT e0.35π , TT = 47.932 N TB = 143.932 N From (2) above, TA = TT , so (a) Tmin lower = 47.9 N From (1) above, 143.932 N = 47.932 N + 4.8 N + µkE ( 48 N ) So (b) 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. µ kE = 1.900

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