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

2024
Views:

259

Pages:

2
Author:

Nicole Wang

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 71. FBD joint D: By symmetry: FAD = FCD ΣFy = 0: 2FAD sin 25° − 4 kN = 0 FAD = FCD = 4.7324 kN FBD joint A: By symmetry: FAE = FAD ΣFx = 0: FAC − 2 ( 4.7324 kN ) cos 25° = 0 FAC = 8.5780 kN Block and incline at A: 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. COSMOS: Complete Online Solutions Manual Organization System PROBLEM 8.71 CONTINUED θ = tan −1 2 mm = 4.8518° π ( 7.5 mm ) φ s = tan −1 µ s = tan −1 0.15 φ s = 8.5308° φ s − θ = 3.679° Q = ( 8.5780 kN ) tan 3.679° Q = 0.55156 kN Couple at A: M A = Qr  7.5 mm  = ( 0.55156 kN )    2  = 2.0683 N ⋅ m By symmetry: Couple at C : M C = 2.0683 N ⋅ m Total couple M = 2 ( 2.0683 N ⋅ m ) 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. M = 4.14 N ⋅ m

COSMOS Chapter 8 Solution 71

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