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: 109

Answer

$10.7lb$

Work Step by Step

We know that $M=\frac{2}{3}\mu_s PR$ We plug in the known values to obtain: $F(1.5)=\frac{2}{3}(0.3)(80)$ This simplifies to: $F=10.7lb$
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