A man slides down a light rope whose breaking | vedclass

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(A) Let the mass of the man be $m$ and his weight be $w = mg$.The breaking strength of the rope is given as $T_{\max} = \eta w = \eta mg$.When the man

Vedclass · Accepted Answer

$g(1-\eta)$

Vedclass · Answer

(A) Let the mass of the man be $m$ and his weight be $w = mg$.The breaking strength of the rope is given as $T_{\max} = \eta w = \eta mg$.When the man slides down with an acceleration $a$,the equation of motion is given by Newton's Second Law:$mg - T = ma$To prevent the rope from breaking,the tension $T$ must be less than or equal to $T_{\max}$. For the limiting case where the rope is just about to break,we set $T = T_{\max} = \eta mg$.Substituting this into the equation of motion:$mg - \eta mg = ma$Dividing both sides by $m$:$g(1 - \eta) = a$Thus,the maximum acceleration the man can have without breaking the rope is $a = g(1 - \eta)$.

$A$ man slides down a light rope whose breaking strength is $\eta$ times the weight of the man $(\eta < 1)$. What is the maximum acceleration of the man so that the rope does not break?

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