COSMOS Chapter 8 Solution 82 | Michigan State University - Edubirdie

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

2019
Views:

290

Pages:

1
Author:

Madelyn Bradshaw

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 82. Pulley FBDs: Left: rp = 30 mm ( rf = raxle sin φk = raxle sin tan −1 µ k ) * ( = ( 5 mm ) sin tan −1 0.2 ) = 0.98058 mm ( rp + rf ) ( 600 N ) − 2rpTAB = 0 ΣM C = 0: 30 mm + 0.98058 mm ( 600 N ) = 309.81 N 2 ( 30 mm ) TAB = or TAB = 310 N ΣFy = 0: Right: TAB − 600 N + TCD = 0 TCD = 600 N − 309.81 N = 290.19 N or TCD = 290 N ( rp − rf ) TCD − ( rp + rf ) TEF ΣM H = 0: TEF = or =0 30 mm − 0.98058 mm ( 290.19 N ) 30 mm + 0.98058 mm TEF = 272 N 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.

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