# Chapter 10 - Problems - Page 402: 81

Each of the cables is stretched by $\frac{\Delta L}{9}$

#### Work Step by Step

$Y = \frac{F/A}{\Delta L/L}$ $Y$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L$ is the original length We can find an expression for $\Delta L$ before the cable is cut: $\Delta L = \frac{W~L}{A~Y}$ After the able is cut into three pieces, each cable supports a weight of $\frac{W}{3}$. We can find an expression for $\Delta L'$: $\Delta L' = \frac{(W/3)~(L/3)}{A~Y}$ $\Delta L' = \frac{1}{9}\times \frac{W~L}{A~Y}$ $\Delta L' = \frac{1}{9}\times \Delta L$ $\Delta L' = \frac{\Delta L}{9}$ Each of the cables is stretched by $\frac{\Delta L}{9}$.

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