Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 9 - Static Equilibrium; Elasticity and Fracture - General Problems - Page 258: 72

Answer

a) $(4.506\times 10^{-6})$% b) $9.012\times 10^{-18}m$

Work Step by Step

a) The fractional decrease in the rod's length is the strain. Use Eq. 9-4. The force applied is the weight of the man. $$\frac{\Delta l}{l_0}=\frac{F}{AE}=\frac{mg}{\pi r^2E}=\frac{(65kg)(9.80m/s^2)}{\pi(0.15)^2(200\times 10^9)N/m^2}=4.506\times 10^{-8}$$ $=(4.506\times 10^{-6})$% b) The fractional change is the same for the atoms as for the macroscopic material. Let $d$ represent the interatomic spacing. $$\frac{\Delta d}{d_0}=\frac{\Delta l}{l_0}=4.506\times10^{-8}$$ $$\Delta d=(4.506\times10^{-8})d_0=(4.506\times10^{-8})(2.0\times10^{-10}m)$$ $$=9.012\times 10^{-18}m$$
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