# A lift is tied with four iron wires and its mass is 1500Kg.If maximum acceleration of lift is $2m/s^2$ and maximum safe stress is $2\times 10^8N/m^2$ find the minimum diameter of the wire $g=10m/s^2$

$\begin{array}{1 1}(A)\;.01m\\(B)\;.005m\\(C)\;.5m\\(D)\;\text{None of above}\end{array}$

If the maximum acceleration of lift is $2m/s^2$ and then maximum force is $m(g+a)$
$\Rightarrow 1500\times 12$
$\Rightarrow 18000N$
Force on 1wire=$\large\frac{18000}{4}$N
$\Rightarrow 4500N$
Maximum stress =$\large\frac{4500}{\pi r^2}$
This should be less than given safe stress
$\large\frac{4500}{\pi r^2}$$< 2\times 10^8$
$716.2\times 10^{-8} < r^2$
$26.76\times 10^{-4} < r$
$.00535 < D$
Hence (B) is the correct answer.
answered May 8, 2014