Question

A steel wire 2.02 m long with a circular cross-section must stretch no more than 0.250 cm

when a 400.8 N weight is hung from one of its ends. What is the minimum diameter in millimeters? (thank you in advance, I've spent so long on this with no luck)

Answer 1

Given:

$l_0=2.02\:\rm m$

$\Delta l=0.250*10^{-2}\:\rm m$

$F=400.8\:\rm N$

$E=200*10^9\:\rm Pa$

The Hooke's law says

$$F=k\Delta l$$

The stiffness of a wire

$$k=\frac{EA}{l_0}$$

Hence, the area of a wire

$$A=\frac{Fl_0}{E\Delta l}=\frac{400.8*2.02}{200*10^9*0.250*10^{-2}}=1.62*10^{-6}\:\rm m^2$$

The diameter of a wire

$$d=\sqrt{4A/\pi}=\sqrt{4*1.62*10^{-6}\:\rm m^2/3.14}\\=1.44*10^{-3}\:\rm m=1.44\: mm$$