College Physics (4th Edition)

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

Chapter 10 - Problems - Page 399: 32


(a) $\Delta x = 0.28~cm$ (b) The shear modulus is $2.04\times 10^4~N/m^2$

Work Step by Step

(a) We can find $\Delta x$: $\frac{\Delta x}{L} = tan~\gamma$ $\Delta x = L~tan~\gamma$ $\Delta x = (2.0~cm)~tan~8.0^{\circ}$ $\Delta x = 0.28~cm$ (b) $G = \frac{F/A}{\Delta x/L}$ $G$ is shear modulus $F$ is the force $A$ is the cross-sectional area parallel to the force $\Delta x$ is the distance the surface is displaced $L$ is the perpendicular length We can find the shear modulus: $G = \frac{F/A}{\Delta x/L}$ $G = \frac{F~L}{A~\Delta x}$ $G = \frac{(12~N)(2.0~cm)}{(42\times 10^{-4}~m^2)(0.28~cm)}$ $G = 2.04\times 10^4~N/m^2$ The shear modulus is $2.04\times 10^4~N/m^2$
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