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

2024
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452

Pages:

1
Author:

hacedoras1kxy

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 89. FBD Each wheel: For equilibrium (constant speed) the two forces R and W must be equal 2 and opposite, tangent to the friction circle, so rf sin θ = where θ = tan −1 ( slope ) rw ( ) sin tan −1 0.03 = rw = (12.5 mm ) ( rB sin tan −1 µ k ( sin ( tan rw ) ) = 49.666 mm 0.03) sin tan −1 0.12 −1 d w = 99.3 mm NOTE FOR PROBLEMS 8.75–8.89 ( ) Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal bearing and axle friction problems. While this approximation may be valid for very small values of µ , there is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79, µ s = 0.50, and the error made by using the approximation is about 11.8%. 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 Chapter 8 Solution 89

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