Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 8 - Friction - Section 8.8 - Rolling Resistance - Problems - Page 454: 111



Work Step by Step

We know that $M=\frac{2}{3}\mu P(\frac{R_2^3-R_1^3}{R_2^2-R_1^2})$ Now $M_A=\frac{2}{3}(0.20)(12000)[\frac{(0.10)^3-(0.05)^3}{(0.10)^2-(0.05)^2}]=186.67N.m$ and $M_B=\frac{2}{3}(0.20)(4000)[\frac{(0.075)^3-(0.05)^3}{(0.075)^2-(0.05)^2}]=50.67N.m$ We equate the moments about the z-axis equal to zero $\implies \Sigma M_z=0$ $\implies M-M_A-M_B=0$ $\implies M=M_A+M_B$ We plug in the known values to obtain: $M=186.67+50.67=237.34N.m$
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