Question

A load of 50 kg is applied to the lower end of a steel rod 80 cm long and 0.6 cm in radius.

How much will the rod stretch? Y = 180 GPa for steel?

A force of 4.8 N acts through a distance of 18 m in the direction of the force. Find the work done.

Answer 1

1) The stress is

$$\sigma=\frac{F}{A}=\frac{F}{\pi r^2}.$$

The strain is

$$\epsilon=\frac{\sigma}{Y}=\frac{F}{\pi r^2Y}.$$

The rod will stretch for

$$\Delta l=\epsilon l=\frac{Fl}{\pi r^2Y},\\\space\\ \Delta l=\frac{(50\cdot9.8)0.8}{3.14 \cdot0.006^2\cdot180\cdot10^9}=1.93\cdot10^{-5}\text{ m}.$$

Less than 0.02 mm.

2) The work done is

$$W=F\cdot d=4.8\cdot18=86.4\text{ J}.$$