## College Physics (4th Edition)

(a) $2F$ (b) $\frac{F}{4}$
$E = \frac{F/A}{\Delta L/L_0}$ $E$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L_0$ is the original length (a) Since a force of $F$ is applied at each end, the total applied force is $2F$: $2F = \frac{\Delta L ~A_0~E}{L_0}$ $F = \frac{\Delta L ~A_0~E}{2L_0}$ We can find the required force at each end when the length is $\frac{L_0}{2}$: $2F' = \frac{\Delta L ~A_0~E}{\frac{L_0}{2}}$ $F' = 2\times \frac{\Delta L ~A_0~E}{2L_0}$ $F' = 2~F$ (b) Since the bar has half the radius, the bar has one-fourth of the area. We can find the required force at each end when the area is $\frac{A_0}{4}$: $2F' = \frac{\Delta L ~(\frac{A_0}{4})~E}{L_0}$ $F' = \frac{1}{4}\times \frac{\Delta L ~A_0~E}{2L_0}$ $F' = \frac{F}{4}$