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: 1

Answer

(a) $160N$ (b) $146.1N$

Work Step by Step

(a) We can determine the required friction as follows: As the sum of forces in the y- direction is zero $\Sigma F_y=0$ $\implies N-P_y-W=0$ $N-200\times 0.6-50\times 9.81=0$ $\implies N=610.5N$ As $F_{max}=\mu_s N$ $\implies F_{max}=0.30\times 610.5$ $\implies F_{max}=183.15N$ Now $P=\frac{4}{5}(200)=160N$ As the magnitude of the force $P$ is less than the maximum static friction $F_{max}$, thus, the crate will not move and hence the crate is in equilibrium. The required friction is $F=160N$ (b) We can find the required friction as follows: $\Sigma F_y=0$ $\implies N-P_y-W=0$ $\implies N-400(0.6)-50(9.81)=0$ $\implies N=730.5N$ $F_{max}=\mu_s N$ $F_{max}=0.30(730.5)=219.15N$ As $P=\frac{4}{5}(400)=320N$ The magnitude of the force $P$ is greater than the value of maximum static friction, which shows that the crate is in motion and is not in equilibrium. Now the expression for kinetic friction is given as $F=\mu_k N$ $F=(0.20)(730.5)$ $\implies F=146.1N$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.