College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 399: 31


The magnitude of the tangential force is $0.301~N$

Work Step by Step

$G = \frac{F/A}{\Delta L/L}$ $G$ is shear modulus $F$ is the force $A$ is the cross-sectional area parallel to the force $\Delta L$ is the distance the surface is displaced $L$ is the perpendicular length We can find the required force: $F = \frac{G~A~\Delta L}{L}$ $F = \frac{(940~N/m^2)(5.0\times 10^{-2}~m)^2(0.64\times 10^{-2}~m)}{5.0\times 10^{-2}~m}$ $F = 0.301~N$ The magnitude of the tangential force is $0.301~N$.
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