University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2024
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221

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1
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strofamay56

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 65. ( ) ( ) WA = (10 kg ) 9.81 m/s 2 = 98.1 N, WB = ( 50 kg ) 9.81 m/s 2 = 490.5 N Slip impends at all surfaces simultaneously FBD I: A + B ΣFx = 0: ΣFy = 0: N A − FB = 0, N A = FB = µ s N B (1) FA − (150 N + 98.1 N + 490.5 N ) + N B = 0 µ s N A + N B = 738.6 N Solving (1) and (2) N B = FBD II: B ΣFx′ = 0: FB = 738.6 µ s N 1 + µ s2 N AB + ( 490.5 N ) cos 70° − N B cos 70° − FB sin 70° = 0 N AB = ΣFy′ = 0: 738.6 N , 1 + µ s2 (2) 738.6 N ( cos 70° + µ s sin 70° ) − ( 490.5 N ) cos 70° (1) 1 + µ s2 −FAB − ( 490.5 N ) sin 70° + N B sin 70° − FB cos 70° = 0 FAB = 738.6 N ( sin 70° − µs cos 70° ) − ( 490.5 N ) sin 70° = 0 1 + µ s2 Setting FAB = µ s N AB , µ s3 − 6.8847µ s2 − 2.0116µ s + 1.38970 = 0 Solving numerically, µ s = − 0.586, 0.332, 7.14 Physically meaningful solution: 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. µ s = 0.332

COSMOS Chapter 8 Solution 65

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