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

2021
Views:

300

Pages:

1
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 23. FBD rod: (a) Geometry: BE = L cosθ 2 EF = L sin θ So or Also, or L  DE =  cosθ  tan β 2  DF = L cosθ 2 tan φ s 1  L cosθ L  cosθ tan β + sin θ  = 2  2 tan φs tan β + 2 tan θ = 1 1 1 = = = 2.5 tan φ s µ s 0.4 (1) L sin θ + L sin β = L sin θ + sin β = 1 Solving Eqs. (1) and (2) numerically θ1 = 4.62° (2) β1 = 66.85° θ 2 = 48.20° β 2 = 14.75° θ = 4.62° and θ = 48.2° Therefore, (b) Now and or φ s = tan −1 µ s = tan −1 0.4 = 21.801° T W = sin φs sin ( 90 + β − φ s ) T =W sin φs sin ( 90 + β − φ s ) For 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. θ = 4.62° T = 0.526W θ = 48.2° T = 0.374W

COSMOS Chapter 8 Solution 23

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