Question

A mass of 2kg is supported by a copper wire of length 4.0 m and diameter of 4.0 mm. Modulus of elasticity of copper is 1.10 x 10 raised to eleven Pa. Calculate the stress and elongation of wire.

Answer 1

(a)

$$\sigma=\dfrac{F}{A}=\dfrac{mg}{\pi r^2},$$ $$\sigma=\dfrac{2\ kg\times9.8\ \dfrac{m}{s^2}}{\pi\times(2\times10^{-3}\ m)^2}=1.56\times10^6\ Pa.$$

(b)

$$E=\dfrac{\sigma}{\epsilon}=\dfrac{FL}{A\Delta L},$$ $$\Delta L=\dfrac{FL}{AE}=\dfrac{mgL}{\pi r^2E},$$ $$\Delta L=\dfrac{2\ kg\times9.8\ \dfrac{m}{s^2}\times4\ m}{\pi\times(2\times10^{-3}\ m)^2\times1.1\times10^{11}\ Pa}=5.7\times10^{-5}\ m.$$