COSMOS Chapter 8 Solution 59 - Edubirdie

University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2019
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1
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Lamghari6kx

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 59. FBD Wedge: φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900° by symmetry R1 = R2 ΣFy = 0: 2 R1 sin 22.29° − 60 lb = 0 R2 = 79.094 lb When P is removed, the vertical component of R1 and R2 will vanish, leaving the horizontal components H1 = H 2 = ( 79.094 lb ) cos 22.29° = 73.184 lb Final forces H1 = H 2 = 73.2 lb Since these are at 3° ( < φs ) from the normal, the wedge is self-locking and will remain in place. 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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