Discussion Forum

A lift of mass 'm' is connected to a rope that is moving upward with maximum acceleration 'a' . For maximum safe stress, the elastic limit of the rope is 'T' . The minimum  diameter of the rope is (g= gravitational acceleration)

Answer:

Explanation:

 The maximum tension in the rope =m(g+a)

Stress in the rope ,  $T=\frac{m(g+a)}{\pi r^{2}}$

$\therefore$    $T=\frac{m(g+a)}{\pi r^{2}}=\frac{m(g+a)}{\pi (\frac{d}{2})^{2}}$

 $\Rightarrow$    $T=\frac{4m(g+a)}{\pi d^{2}}$

$\Rightarrow$    $d^{2}=\frac{4m(g+a)}{\pi T}$

$\therefore$     d=$\left[\frac{4m(g+a)}{\pi T}\right]^{1/2}$