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 - Discussion Questions - Page 356: Q11.21

Answer

The diameter would have to be $1.73~D$.

Work Step by Step

We can write an expression for the Young's Modulus of the wire for the first condition. $Y = \frac{F_1/A_1}{\Delta~L/L_0}$ We can write an expression for the Young's Modulus of the wire for the second condition. $Y = \frac{F_2/A_2}{\Delta~L/L_0}$ The Young's Modulus is the same in both cases since the material is the same, so we can equate the two expressions. $\frac{F_1/A_1}{\Delta~L/L_0}=\frac{F_2/A_2}{\Delta~L/L_0}$ $A_2 = \frac{F_2~A_1}{F_1}$ $A_2 = \frac{(3~F_1)~A_1}{F_1}$ $A_2 = 3~A_1$ The area of the wire in the second condition must be three times larger. Therefore, the radius (and thus the diameter) needs to increase by a factor of $\sqrt{3}$, (since $Area = \pi~r^2$). Note that $\sqrt{3} = 1.73$ Therefore, the diameter would have to be $1.73~D$.
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