COSMOS Chapter 8 Solution 102 | Michigan State University - Edubirdie

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COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 102. FBD A: ( ) WA = (10 kg ) 9.81 m/s 2 = 98.1 N ΣFx = 0: TA − WA sin 30° = 0, TA = WA 2 FBD B: ΣFx′ = 0: WB sin 30° − TB = 0, TB = WB 2 (a) Motion of B impends up incline and mB = 8 kg TA = eµs β , TB µs = = 1 β 1 β ln ln TA 1 W = ln A TB β WB mA 3  10 kg  = ln   π  8 kg  mB From hint, β is not dependent on shape of support µ s = 0.21309 µ s = 0.213 (b) For maximum mB , motion of B impend down incline TB = eµs β , TA TB = TAe 0.21309 π 3 = 1.250TA ∴ WB = 1.25WA and mB = 1.25 mA = 1.25 (10 kg ) mB max = 12.50 kg 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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