Question

A $900kg$ elevator hangs by a steel cable for which the allowable stress is $1.15\times {10}^{8}N/m^2$. What is the minimum diameter required if the elevator accelerates upward at $1.5m/s^2$. Take $g=10m/s^2$

• A.
  $\frac{6\times {10}^{-2}}{\sqrt{10\pi }}M$
  
• B.
  $\frac{6\times {10}^{-2}}{\sqrt{5\pi }}M$
  
• C.
  $\frac{3\times {10}^{-2}}{\sqrt{5\pi }}M$
  
• D.
  $\frac{6\times {10}^{-2}}{\sqrt{10\pi }}M$
  

Answer 1

The correct option is A

$\frac{6\times {10}^{-2}}{\sqrt{10\pi }}M$

Calculate tension for elevator

Given, mass of the lift, $M=900kg$
Acceleration of the lift, $a=1.5m/s^2$
$T=M(g+a)$
$T=9000(10+1.2)$
$=900\left(11.2\right)N$

Calculate diameter required by elevator

Formula used: stress $=\frac{T}{A}$
Given, stress in the cable $=1.15\times {10}^{8}N/M^2$
Let the diameter of the cable be $d$. Therefore,
stress $=\frac{T}{A}$
$1.15\times {10}^8=\frac{900\left(11.2\right)}{\left(\frac{\pi d^2}{4}\right)}$
$\Rightarrow d=\frac{6\times {10}^{-2}}{\sqrt{10\pi }}m$