COSMOS Chapter 8 Solution 27 - 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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2024
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Quintin Ware

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 27. FBD pin A: From FBD Whole the force at A = 750 lb 4 ( FAB′ − FAB ) = 0, FAB′ = FAB 5 ΣFx = 0: 3 750 lb − 2 FAB = 0, FAB = 625 lb 5 ΣFy = 0: FBD Casting: ΣFx = 0: N D′ − N D = 0, N D′ = N D = N FD = FD′ = µ N D , or N D = Impending slip ΣFy = 0: FD µs 2FD − 750 lb = 0, FD = 375 lb ND = 375 lb µs FBD ABCD: ΣM C = 0: (12 in.) 375 lb µs 4 5 (12 in.) N − ( 6 in.) F − ( 42.75 in.) ( 625 lb ) = 0 = ( 6 in.)( 375 lb ) + ( 42.75 in.) 4 ( 625 lb ) = 0 5 µ s = 0.1900 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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