Question

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

Answer 1

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