COSMOS Chapter 8 Solution 63 | 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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2019
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COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 63. FBD plank + wedge: ΣM A = 0: (8 ft ) N B − (1.5 ft )( 48 lb/ft )( 3 ft ) 1 ( 48 lb/ft )( 3 ft ) 2  5  −  3 +  ft  ( 96 lb/ft )( 5 ft ) = 0 3   − ( 2 ft ) NW = 185 lb ΣFy = 0:  48 + 96  N A + 185 lb −  lb/ft  ( 3ft ) 2   1 − ( 96 lb/ft )( 5 ft ) = 0 2 N A = 271 lb Since N A > NW , and all µ s are equal, assume impending slip at top and bottom of wedge and not at A. Then FW = µ s NW = 0.45 (185 N ) FW = 83.25 lb FBD Wedge: φ s = tan −1 µ s = tan −1 ( 0.45 ) = 24.228° ΣFy = 0: 185 lb − RB cos ( 24.228° + 9° ) = 0 RB = 221.16 lb ΣFx = 0: ( 221.16 lb ) sin 33.228° + 83.25 lb − P = 0 P = 204.44 lb Check assumption using plank/wedge FBD ΣFx = 0: FA + FW − P = 0, FA = 204.44 lb − 83.25 lb = 121.19 lb FA max = µ s N A = 0.45 ( 271 lb ) = 121.95 lb FA < FA max , OK ∴ (a) (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. P = 204 lb no impending slip at A

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