An elevator of mass m is connected to a rope | vedclass

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(B) The force equation for the elevator moving upward with acceleration $a$ is given by $T_{actual} = m(g+a)$.For a safe journey,the maximum tension $

Vedclass · Accepted Answer

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

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(B) The force equation for the elevator moving upward with acceleration $a$ is given by $T_{actual} = m(g+a)$.For a safe journey,the maximum tension $T$ that the rope can bear must be at least equal to the actual tension,so $T = m(g+a)$.The stress in the rope is defined as $\sigma = \frac{T}{A}$,where $A$ is the cross-sectional area of the rope.Assuming the rope is cylindrical with diameter $d$,the area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.Substituting this into the tension equation: $T = \frac{m(g+a)}{\pi d^2 / 4} = \frac{4 m(g+a)}{\pi d^2}$.Rearranging for $d^2$,we get $d^2 = \frac{4 m(g+a)}{\pi T}$.Taking the square root of both sides,the minimum diameter is $d = [\frac{4 m(g+a)}{\pi T}]^{1/2}$.

An elevator of mass $m$ is connected to a rope which pulls the elevator up with a maximum acceleration equal to $a$. If the maximum bearable tension in the rope is $T$,then for a safe journey the minimum diameter of the rope is $[g=$ acceleration due to gravity$]$

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