Discussion Forum



1)

 A lift is tied with thick iron ropes having mass 'M'. The maximum acceleration of the lift is 'a' m/s2 and the maximum safe stress is 'S' N/m2. The minimum  diameter  of the rope is 


A) $\left[\frac{6M(g+a)}{\pi S}\right]^{1/2}$

B) $\left[\frac{4M(g+a)}{\pi S}\right]^{1/2}$

C) $\left[\frac{M(g+a)}{\pi S}\right]^{1/2}$

D) $\left[\frac{M(g-a)}{\pi S}\right]^{1/2}$

Answer:

Option B

Explanation:

The maximum stress produced in a rope is given by 

$\sigma_{max}=\frac{Force}{Area}=\frac{Mg}{\pi r^{2}}$

As the lift is accelerationg with acceleration a, then

$\sigma_{max}=\frac{M(g\pm a)}{\pi r^{2}}$

$\Rightarrow $   $r^{2}=\frac{M(g\pm a)}{\pi S}$    [ Given , $\sigma_{max}$ = S]

$ \frac{d^{2}}{4}=\frac{M(g\pm a)}{\pi S}$           $\left[\because r=\frac{d}{2}\right]$

$d= \sqrt{\frac{4M(g\pm a)}{\pi S}}$

 As acceleration is maximum i.e, g'= g+a, so

$d= \sqrt{\frac{4M(g\pm a)}{\pi S}}$