Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

Chapter 8 - Friction - Section 8.8 - Rolling Resistance - Problems - Page 454: 107

Answer

$132.14N.m$

Work Step by Step

We know that $M=\frac{2}{3}\mu P(\frac{R_2^3-R_1^3}{R_2^2-R_1^2})$ We plug in the known values to obtain: $M=\frac{2}{3}(0.30)(5000)[\frac{(0.100)^3-(0.075)^3}{(0.100)^2-(0.075)^2}]$ This simplifies to: $M=132.14N.m$

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