COSMOS Chapter 8 Solution 111 | 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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2020
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COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 111. FBD Flywheel: Slip of belt: TB = TA e µk β = TA e0.20π Also, since the belt doesn’t change length, the increase in stretch of spring B equals the decrease in stretch of spring A. Therefore the increase in TB equals the decrease in TA , and the sum is unchanged, so TA + TB = 80 N + 80 N = 160 N ( ) ∴ TA 1 + e0.20π = 160 N, so TA = 55.663 N TB = 104.337 N (a) TA = 55.7 N TB = 104.3 N ΣM C = 0: ( 0.225 m )(TB − TA ) − M C = 0 M C = ( 0.225 m )(104.337 N − 55.663 N ) (b) M C = 10.95 N ⋅ m 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.

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