Question

A small elevator with a mass of 550 kg hangs from a steel cable that is 3.0 m long when not loaded. The wires making up the cable have a total cross-sectional area of 0.20 cm2, and with a 550 kg load, the cable stretches 0.40 cm beyond it unloaded length. Determine the cable's stress and strain. Assuming that the cable is equivalent to a rod with the same cross-sectional area, determine the value of Young's modulus for the cable's steel.

Answer 1

The stress is

$$\sigma=\frac{F}{A}=\frac{mg}{A}=\frac{550\cdot9.8}{0.20\cdot10^{-4}}=270\cdot10^6\text{ Pa}.$$

The strain is change in length compared to the original length:

$$\epsilon=\frac{\Delta l}{l}=\frac{0.40}{300}=1.3\cdot10^{-3}.$$

Assuming that the cable is equivalent to a rod with the same cross-sectional area, calculate the value of Young's modulus for the cable's steel:

$$E=\frac{\sigma}{\epsilon}=2.0\cdot10^{11}\text{ Pa}.$$