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 - Problems: 43

Answer

a. $1.4 \times 10^6 N/m^2$ b. $6.9 \times 10^{-6}$ c. $6.6 \times 10^{-5}m$

Work Step by Step

a. The stress is $\frac{F}{A}=\frac{(1700 kg)(9.80 m/s^2)}{0.012m^2}\approx1.4 \times 10^6 N/m^2$ b. The strain is $\frac{stress}{E}=\frac{1388333N/m^2}{200\times 10^9 N/m^2}=6.9 \times 10^{-6}$ c. Use the definition of strain to find the amount by which it is lengthened. $$\Delta \mathcal{l}=\mathcal{l}_o(strain)$$ $$=(9.50m)(6.9 \times 10^{-6})=6.6 \times 10^{-5}m$$
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