A lift is tied with thick iron ropes having m | vedclass

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(B) The tension $T$ in the rope when the lift accelerates upwards with acceleration $a$ is given by $T = M(g + a)$.The stress $\sigma$ is defined as t

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The tension $T$ in the rope when the lift accelerates upwards with acceleration $a$ is given by $T = M(g + a)$.The stress $\sigma$ is defined as the force per unit area,so $\sigma = \frac{T}{A}$,where $A = \pi r^2 = \pi (\frac{d}{2})^2 = \frac{\pi d^2}{4}$.Given the maximum safe stress is $S$,we have $S = \frac{M(g + a)}{\frac{\pi d^2}{4}}$.Rearranging for the diameter $d$,we get $d^2 = \frac{4 M(g + a)}{\pi S}$.Therefore,the minimum diameter is $d = [\frac{4 M(g + a)}{\pi S}]^{\frac{1}{2}}$.

$A$ lift is tied with thick iron ropes having mass $M$. The maximum acceleration of the lift is $a \ m/s^2$ and the maximum safe stress is $S \ N/m^2$. The minimum diameter of the rope is

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