College Physics (4th Edition)

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

Chapter 10 - Problems - Page 397: 8


The abductin ligament is compressed by $0.48~mm$

Work Step by Step

$Y = \frac{F/A}{\Delta L/L}$ $Y$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L$ is the original length We can find the change in length: $\Delta L = \frac{F~L}{Y~A}$ $\Delta L = \frac{(1.5~N)(1.0\times 10^{-3}~m)}{(4.0\times 10^6~N/m^2)(0.78\times 10^{-6}~m^2)}$ $\Delta L = 4.8\times 10^{-4}~m = 0.48~mm$ The abductin ligament is compressed by $0.48~mm$.
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