Question

The maximum stress that can be applied to the material of a wire used to suspend an elevator is ${10}^8{\text{Nm}}^{-2}$. If the mass of the elevator is $1000\text{kg}$ and it moves up with an acceleration of $0.2{\text{ms}}^{-2}$, what is the minimum diameter of the wire required? (Take $g=9.8\text{m}/{\text{s}}^2$)

• A. $2\times \frac{{10}^{-2}}{\sqrt{\pi }}\text{m}$
• B. $\frac{{10}^{-2}}{\sqrt{\pi }}\text{m}$
• C. $2\times \frac{{10}^{-2}}{\sqrt{2\pi }}\text{m}$
• D. $4\times {10}^{-2}\text{m}$

Answer 1

The correct option is A $2\times \frac{{10}^{-2}}{\sqrt{\pi }}\text{m}$

As the elevator moves up, the tension in the wire is
$F=mg+ma=m(g+a)$
[Here, $ma=$ pseudo force acting on the elevator]
$\Rightarrow F=1000\times (9.8+0.2)=10,000\text{N}$
Stress in the wire $=\frac{F}{A}=\frac{F}{\pi r^2}$
Clearly, when the stress is maximum, $r$ is minimum.
$\therefore$ Maximum stress $=\frac{F}{\pi r_{\text{min}}^2}$
or $r_{\text{min}}^2=\frac{F}{\pi \times \text{Maximum stress}}$
$r_{\text{min}}^2=\frac{10000}{\pi \times {10}^8}=\frac{{10}^{-4}}{\pi }$
$\therefore r_{\text{min}}=\frac{{10}^{-2}}{\sqrt{\pi }}\text{m}$
Hence, minimum diameter
$d_{min}=2r_{\text{min}}=2\times \frac{{10}^{-2}}{\sqrt{\pi }}$
$d_{min}=2\times \frac{{10}^{-2}}{\sqrt{\pi }}\text{m}$