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

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(C) The total force acting on the rope when the lift accelerates upwards is $F = M(g+a)$.Stress is defined as force per unit area,so $s = \frac{F}{A}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The total force acting on the rope when the lift accelerates upwards is $F = M(g+a)$.Stress is defined as force per unit area,so $s = \frac{F}{A} = \frac{M(g+a)}{\pi r^2}$,where $r$ is the radius of the rope.Rearranging for $r^2$,we get $r^2 = \frac{M(g+a)}{\pi s}$.Since the diameter $D = 2r$,we have $r = \frac{D}{2}$.Substituting this into the equation: $(\frac{D}{2})^2 = \frac{M(g+a)}{\pi s} \implies \frac{D^2}{4} = \frac{M(g+a)}{\pi s}$.Solving for $D$,we get $D^2 = \frac{4 M(g+a)}{\pi s}$,which gives $D = \left[\frac{4 M(g+a)}{\pi s}\right]^{\frac{1}{2}}$.

$A$ 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 $(g = \text{acceleration due to gravity})$

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