COSMOS Chapter 8 Solution 38 | Michigan State University - Edubirdie

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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2023
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COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 38. FBD A: Note: Rod is a two force member. For impending slip the reactions are at angle φ s = tan −1 µ s = tan −1 ( 0.40 ) = 21.801° FBD B: Consider first impending slip to right 9 lb FAB = = 3.8572 lb tan 66.801 ΣFy = 0: N B − ( 3.8522 lb ) sin 30° − ( 6 lb ) cos 30° = 0 N B = 7.1223 lb, FB = µ s N B = 0.40 ( 7.1223 lb ) FB = 2.8489 lb ΣFx = 0: − 2.8489 lb + ( 3.8572 lb ) cos 30° − ( 6 lb ) sin 30° − P = 0 Pmin = − 2.508 lb FBD A: Next consider impending slip to left FAB = ( 9 lb ) tan 66.801° = 21.000 lb FBD B: ΣFy = 0: N B − ( 21 lb ) sin 30° − ( 6 lb ) cos 30° = 0, N B = 15.6959 lb FB = µ s N B = 0.4 (15.6959 lb ) = 6.2784 lb ΣFx = 0: 6.2784 lb + ( 21 lb ) cos 30° − ( 6 lb ) sin 30° − P = 0 Pmax = 21.465 lb equilibrium for − 2.51 lb ≤ P ≤ 21.5 lb W 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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