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Q.
A mass $M$ is hung with a light inextensible string as shown in the figure. Find the tension of the horizontal string.
Laws of Motion
Solution:
As here is a load at $P$, so tension in $A P$ and $P B$ will be different as shown in figure. Let these tensions be $T_{1}$ and $T_{2}$, respectively.
For vertical equilibrium of $P$,
$T_{2} \cos 60^{\circ}=M g$ ...(i)
i.e. $T_{2}=2 M g$
and for horizontal equilibrium of $P$,
$T_{1}=T_{2} \sin 60^{\circ}=T_{2}(\sqrt{3} / 2)$ ...(ii)
Substituting the value of $T_{2}$ from Eq. (i), we get
$T_{1}=(2 M g) \times(\sqrt{3} / 2)=\sqrt{3} M g$
