Question

The elastic limit of steel cable is $3.0\times {10}^{8}N/m^2$ and the cross-section area is $4c{m}^2$. Find the maximum upward acceleration that can be given to a $900kg$ elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.If your answer is $x$ ,then mark the value of $\frac{x}{5}$

Answer 1

Elastic limit means breaking stress $\sigma =\frac{mg}{A}$
If $a$ be the maximum upward acceleration, then
$\frac{m(g+a)}{A}=\frac{1}{3}\sigma$
$\therefore a=\frac{A\sigma }{3m}-g=\frac{4\times {10}^{-4}\times 3\times {10}^8}{3\times 900}-9.8=34.64m/s^2\approx 35m/s^2$
$\therefore$The values of $\frac{x}{5}=35/5=7m/s^2$