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

Draw a diagram of given problem


Calculate maximum upward acceleration of steel cable

Formula used: stress $=\frac{T}{A}$
Given,
Elastic limit of cable $=2.40\times {10}^8$$Pa$
Stress, $\sigma =\frac{1}{3}$ (elastic limit)
$=\frac{1}{3}(2.40\times {10}^{8}Pa$)
Cross sectional area $A=4.00c{m}^2$
Mass $m=800kg$)
From FBD,
$T-mg=ma$
$T=m(g+a)$
Stress developed in the cable, $\sigma =T/A$
$\frac{1}{3}(2.40\times {10}^8)=\frac{m(g+a)}{A}=\frac{800(g+a)}{4\times {10}^{-4}}$
$(g+a)=\frac{1}{3}\frac{(2.4\times {10}^8)}{2\times {10}^6}$
$(g+a)=40$
$\Rightarrow a=30m/s^2$