A copper wire of length $2.2 ext{ m}$ and a steel wire of length $1.6 ext{ m}$, both of diameter $

Question

A copper wire of length $2.2	ext{ m}$ and a steel wire of length $1.6	ext{ m}$, both of diameter $3.0	ext{ mm}$, are connected end to end. When stretched by a load, the net elongation is found to be $0.70	ext{ mm}$. Given $Y_{	ext{copper}} = 1.1 	imes 10^{11}	ext{ N/m}^2$ and $Y_{	ext{steel}} = 2.0 	imes 10^{11}	ext{ N/m}^2$, what is the approximate load applied?

Professor PC Thomas Classes · Answer

We know $\Delta L_c / \Delta L_s = 2.5$ and $\Delta L_c + \Delta L_s = 0.70	ext{ mm}$. This gives $\Delta L_s = 0.20	ext{ mm}$. Load $W = \frac{A \cdot Y_s \cdot \Delta L_s}{L_s} = \frac{\pi (1.5 	imes 10^{-3})^2 	imes 2.0 	imes 10^{11} 	imes 0.20 	imes 10^{-3}}{1.6} \approx 177	ext{ N}$.