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

2023
Views:

498

Pages:

2
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 70. FBD joint D: FAD = FCD By symmetry: ΣFy = 0: 2FAD sin 25° − 4 kN = 0 FAD = FCD = 4.7324 kN FBD joint A: FAE = FAD By symmetry: ΣFx = 0: FAC − 2 ( 4.7324 kN ) cos 25° = 0 FAC = 8.5780 kN Block and incline A: θ = tan −1 2 mm = 4.8518° π ( 7.5 mm ) φ s = tan −1 µ s = tan −1 0.15 = 8.5308° 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 8.70 CONTINUED Q = ( 8.578 kN ) tan (13.3826° ) = 2.0408 kN Couple at A: M A = rQ  7.5  = mm  ( 2.0408 kN ) 2   = 7.653 N ⋅ m By symmetry: Couple at C: M C = 7.653 N ⋅ m Total couple M = 2 ( 7.653 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 = 15.31 N ⋅ m

COSMOS Chapter 8 Solution 70

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