A loop of light inextensible string passes ov | vedclass

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Question

(A) Let the horizontal distance between $A$ and $B$ be $2l$. Let the vertical distance of $C$ below the line $AB$ be $y$. The length of the string seg

Vedclass · Accepted Answer

$\frac{m}{M} > 2\sqrt {\frac{{\eta + 1}}{{\eta + 3}}} - 1$

Vedclass · Answer

(A) Let the horizontal distance between $A$ and $B$ be $2l$. Let the vertical distance of $C$ below the line $AB$ be $y$. The length of the string segment $AC$ is $\eta l$. By Pythagoras theorem,the horizontal distance of $C$ from the midpoint $O$ is $x = \sqrt{(\eta l)^2 - y^2}$.For the masses to cross,the mass $m$ must be able to move downwards past the level of $C$. The system is in equilibrium when the tension $T$ in the string balances the weights. At point $C$,$2T \cos 	heta = Mg$,where $\cos 	heta = y/(\eta l)$. Thus $T = Mg \eta l / (2y)$.At point $O$,the mass $m$ is supported by the tension in the vertical string. For $m$ to move down,its weight $mg$ must be greater than the upward force exerted by the string at $O$. As $m$ moves down,the angle $	heta$ changes. The condition for $m$ to reach $C$ is derived from the potential energy or force balance at the limit where $y 	o 0$. The correct condition is $\frac{m}{M} > 2\sqrt{\frac{\eta+1}{\eta+3}} - 1$.

$A$ loop of light inextensible string passes over smooth small pulleys $A$ and $B$. Two masses $m$ and $M$ are attached to the points $O$ and $C$ respectively. The condition that $m$ and $M$ will cross each other (given $AB = 2l$ and $AC = BC = \eta l$) is:

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