Question

The elastic limit of a steel cable is $2.40\times 10^{8}Pa$ and the cross-sectional area is $4.00c{m}^2$. Find the maximum upward acceleration (in $m/s^2$ ) that can be given to a $800kg$ elevator supported by the cable, if the stress is to not exceed one third of the elastic limit of the cable.

Answer 1

Answer: 30

$T-mg=ma$
$T=mg+ma$
$\frac{800(g+a)}{4\times 10^{-4}}=\frac{T}{A}=\frac{1}{3}\times 2.4\times 10^8=8\times 10^7$
$\Rightarrow (g+a)=\frac{8\times 10^7}{2\times 10^6}=40$
$\Rightarrow a=30m/s^2$