One end of a string is attached to a rigid wall at point O, passes over a smooth pulley and carries a hanger S of mass M at its other end. One end of a string is attached to a rigid wall at point O, passes over a smooth pulley and carries a hanger S of mass M at its other end. Another object P of mass M is suspended from a light ring that can slide without friction, along the string, as is shown in figure. OA is horizontal. Find the additional mass to be attached to the hanger S so as to raise the object P by 10cm.

Solution:

Let us denote the tension in each string as T. $$2Tcos\theta=Mg$$$$2(Mg)cos\theta=Mg$$$$cos\theta=\frac{1}{2}$$$$ \theta=60^\circ$$$$ tan60=\frac{\frac{40\sqrt{3}}{2}}{PQ}$$$$ tan60^\circ=\sqrt{3}$$Hence,$$ PQ=20cm$$Now, when an additional mass m is hung from the pulley, the length of PQ changes to P’Q’.

\(P’Q’=PQ-10=20-10=10\).

$$ Q’S’=\sqrt{P’Q’^2+P’S^2}=\sqrt{1300}$$Now, again considering the force equation$$2Tcos\theta=Mg$$$$2(M+m)g\times\frac{10}{\sqrt{1300}}=Mg$$$$2(M+m)\times\frac{1}{\sqrt{13}}=M$$$$ 2(M+m)=\sqrt{13}M$$$$2m=M(\sqrt{13}-2)$$$$m=\frac{M\times(\sqrt{13}-2)}{2}=0.9M$$

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