COSMOS Chapter 8 Solution 21 | Michigan State University - Edubirdie

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

2019
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323

Pages:

1
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 21. FBD ladder: Note: slope of ladder = L = 6.5 m, so AC = 4.5 m 12 13 = , so AC = ( 4.5 m ) = 4.875 1.875 m 5 12 4.875 m 3 = L, 6.5 m 4 and DC = BD = AD = 1 L 2 1 L 4 For impending slip: FA = µ s N A , FC = µ s NC  12  Also θ = tan −1   − 15° = 52.380°  5 FA − W sin15° + FC cosθ − NC sin θ = 0 ΣFx = 0: FA = W sin15° − µ s 10 10 W cosθ + W sin θ 39 39 = ( 0.46192 − 0.15652µ s )W ΣFy = 0: N A − W cos15° + FC sin θ + NC cosθ = 0 N A = W cos15° − µ s 10 10 W sin θ − W cosθ 39 39 = ( 0.80941 − 0.20310µ s )W But FA = µ N A : 0.46192 − 0.15652µ s = 0.80941µ s − 0.20310µ s2 µ s2 − 4.7559µ s + 2.2743 µ s = 0.539, 4.2166 µ s min = 0.539 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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