Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 8 - Friction - Section 8.2 - Problems Involving Dry Friction - Fundamental Problems - Page 416: 5

Answer

$83.33lb$

Work Step by Step

We can find the required forced $P$ as follows: $P-\mu mg=0$ We plug in the known values to obtain: $P-(0.4)(250)=0$ $\implies P=100$ We know that $\Sigma M_A=0$ $\implies P\times 4.5-250\times 1.5=0$ $P=83.33lb$
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