COSMOS Chapter 8 Solution 62 | Michigan State University - Edubirdie

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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2024
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COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 62. FBD plank + wedge: (8 ft ) N B − (1.5 ft )( 48 lb/ft )( 3 ft ) ΣM A = 0: − ( 2 ft ) 1 ( 48 lb/ft )( 3 ft ) 2  5  1 −  3 +  ft  ( 96 lb/ft )( 5 ft ) = 0 3  2  N B = 185 lb ΣFy = 0:  48 + 96  NW + 185 lb −  lb/ft  ( 3ft ) 2   + 1 ( 96 lb/ft )( 5 ft ) = 0 2 NW = 271 lb Since NW > N B , and all µ s are equal, assume slip impends at B and between wedge and floor, and not at A. Then FW = µ s NW = 0.45 ( 271 lb ) = 121.95 lb FB = µ s N B = 0.45 (185 lb ) = 83.25 lb ΣFx = 0: P − 121.95 lb − 83.25 lb = 0, P = 205.20 lb Check Wedge for assumption 271 lb − RA cosθ = 0 ΣFy = 0: ΣFx = 0: so tan θ = 205.2 lb − 121.95 lb − RA sin θ = 0 83.25 = 0.3072 < µ s + tan 9° 271 so no slip here ∴ (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 = 205 lb impending slip at B

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