COSMOS Chapter 8 Solution 110 | Michigan State University - 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 110. FBD Flywheel: ( 0.225 m )(TB − TA ) − 12.60 N ⋅ m = 0 ΣM C = 0: TB − TA = 56 N, TB = TA + 56 N Also, since the belt doesn’t change length, the additional stretch in spring B equals the decrease in stretch of spring A. Thus the increase in TB equals the decrease in TA. Thus TB + TA = ( 70 N + ∆T ) + ( 70 N − ∆T ) = 140 N (TA + 56 N ) + TA = 140 N, TA = 42 N TB = 42 N + 56 N = 98 N (a) TA = 42.0 N TB = 98.0 N For slip TB = TAe µk β , or µ k = µk = 1 π ln 1 β ln TB TA 98 = 0.2697 42 (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 k = 0.230

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