COSMOS Chapter 8 Solution 92 | 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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2020
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COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 92. Let the normal force on ∆A be ∆N , and ∆N k = r ∆A ∆F = µ∆N , ∆M = r ∆F As in the text The total normal force 2π  R k  P = lim Σ∆N = ∫ 0  ∫ 0 rdr  dθ ∆A → 0  r  ( R ) P = 2π ∫ 0 kdr = 2π kR The total couple or P 2π R k 2π  R  M worn = lim Σ∆M = ∫ 0  ∫ 0 r µ rdr  dθ ∆A → 0 r   R M worn = 2πµ k ∫ 0 rdr = 2πµ k R2 P R2 = 2πµ 2 2π R 2 or M worn = 1 µ PR 2 Now M new = 2 µ PR 3 Thus k = M worn = M new 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. [Eq. (8.9)] 1 2 2 3 µ PR 3 = = 75% µ PR 4

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