Question

A lift of mass ${10}^{3}kg$ is tied with thick iron wires. If the maximum acceleration of the lift is $1.2m{s}^{-2}$ and the maximum safe stress is $1.4\times {10}^{8}N{m}^{-2}$, the minimum diameter of wire is ($g=9.8m{s}^{-2}$)

• A. $0.00141m$
• B. $0.00282m$
• C. $0.005m$
• D. $0.001m$

Answer 1

The correct option is D $0.001m$

Consider the tension on the wire

$\begin{array}{c}T=1000\left[9.8+1.2\right]\\ =1000\times 11\\ =11000\end{array}$

so, $N=11000$

Maximum stress $=\frac{T}{\pi r_{{min}^2}}=\frac{{50}^2}{22\times r_{{min}^2}}$

Or

$r_{{min}^2=\frac{11000\times 7}{22\times 1.4\times {10}^8}=\frac{{50}^2}{{\left({10}^4\right)}^2}}$

Or

Minimum radius, $r_{min}=50\times {{10}^{-4}}_m$

Minimum diameter

$\begin{array}{c}=2\times r_{min}=2\times 50\times {{10}^{-4}}_m\\ r_{min}=100\times {{10}^{-4}}_m\\ =0.01\end{array}$