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.

$$2Tcos\theta=Mg$$

$$\Rightarrow2(Mg)cos\theta=Mg$$

$$\Rightarrow cos\theta=\frac{1}{2}$$

$$ \Rightarrow\theta=60^\circ$$

$$ \Rightarrow tan60=\frac{\frac{40\sqrt{3}}{2}}{PQ}$$

$$ \Rightarrow tan60^\circ=\sqrt{3}$$

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

Hence, P’Q’=1cm.

$$ Q’S’=\sqrt{P’Q’^2+P’S^2}=\sqrt{1300}$$

Now, again considering the force equation

$$\Rightarrow 2Tcos\theta=Mg$$

$$\Rightarrow 2(M+m)g\times\frac{10}{\sqrt{1300}}=Mg$$

$$\Rightarrow 2(M+m)\times\frac{1}{\sqrt{13}}=M$$

$$ \Rightarrow (M+m)=\sqrt{13}M$$

$$\Rightarrow 2m=M(\sqrt{13}-2)$$

$$\Rightarrow m=\frac{M\times(\sqrt{13}-2)}{2}=0.9M$$