College Physics (4th Edition)

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

Chapter 10 - Problems - Page 398: 22


(a) The stress is $2.8\times 10^7~N/m^2$ (b) The strain is $4.67\times 10^{-4}$ (c) The column is compressed by $9.34\times 10^{-4}~m$ (d) The maximum load is $5.0\times 10^5~N$

Work Step by Step

(a) We can find the stress: $\frac{F}{A} = \frac{7.0\times 10^4~N}{25\times 10^{-4}~m^2}$ $\frac{F}{A} = 2.8\times 10^7~N/m^2$ The stress is $2.8\times 10^7~N/m^2$ (b) We can find the strain: $\frac{\Delta L}{L} =\frac{F/A}{Y}$ $\frac{\Delta L}{L} = \frac{2.8\times 10^7~N/m^2}{6.0\times 10^{10}~Pa}$ $\frac{\Delta L}{L} = 4.67\times 10^{-4}$ The strain is $4.67\times 10^{-4}$ (c) We can find the change in length: $\frac{\Delta L}{L} = 4.67\times 10^{-4}$ $\Delta L = (4.67\times 10^{-4})(L)$ $\Delta L = (4.67\times 10^{-4})(2.0~m)$ $\Delta L = 9.34\times 10^{-4}~m$ The column is compressed by $9.34\times 10^{-4}~m$ (d) We can find the maximum load: $\frac{F}{A} = 2.0\times 10^8~Pa$ $F = (2.0\times 10^8~Pa)~(A)$ $F = (2.0\times 10^8~Pa)~(25\times 10^{-4}~m^2)$ $F = 5.0\times 10^5~N$ The maximum load is $5.0\times 10^5~N$
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