University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 11 - Equilibrium and Elasticity - Problems - Exercises - Page 359: 11.25


The minimum diameter of the wire is 1.89 mm.

Work Step by Step

The Young's Modulus for steel is $20\times 10^{10}~N/m^2$. We can find the required area of the wire. $Y = \frac{F/A}{\Delta L/L_0}$ $A = \frac{F~L_0}{Y~\Delta L}$ $A = \frac{(700~N)(2.00~m)}{(20\times 10^{10}~N/m^2)(0.0025~m)}$ $A = 2.80\times 10^{-6}~m^2$ We can find the radius of the wire with this area. $A = 2.80\times 10^{-6}~m^2$ $\pi~R^2 = 2.80\times 10^{-6}~m^2$ $R^2 = \frac{2.80\times 10^{-6}~m^2}{\pi}$ $R = \sqrt{\frac{2.80\times 10^{-6}~m^2}{\pi}}$ $R = 0.944\times 10^{-3}~m$ $R = 0.944~mm$ Since the diameter is twice the radius, the minimum diameter of the wire is 1.89 mm.
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