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A mass M is hung with a light inextensible string as shown in the figure. Find the tension of the horizontal string.
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Verified Answer
The correct answer is:
$\sqrt{3} M g$
As there is a load at P , so tension in AP and PB will be different. Let these be $\mathrm{T}_1$ and $\mathrm{T}_2$ respectively. For vertical equilibrium of P ,
$\mathrm{T}_2 \cos 60^{\circ}=\mathrm{Mg}$
i.e., $\quad \mathrm{T}_2=2 \mathrm{Mg}$ ...(i)
And for horizontal equilibrium of P ,
$\mathrm{T}_1=\mathrm{T}_2 \sin 60^{\circ}=\mathrm{T}_2(\sqrt{3} / 2)$ ...(i)
Substituting the value of $\mathrm{T}_2$ from Eq. (i),
$\mathrm{T}_1=(2 \mathrm{Mg}) \times(\sqrt{3} / 2)=\sqrt{3} \mathrm{Mg}$
$\mathrm{T}_2 \cos 60^{\circ}=\mathrm{Mg}$
i.e., $\quad \mathrm{T}_2=2 \mathrm{Mg}$ ...(i)
And for horizontal equilibrium of P ,
$\mathrm{T}_1=\mathrm{T}_2 \sin 60^{\circ}=\mathrm{T}_2(\sqrt{3} / 2)$ ...(i)
Substituting the value of $\mathrm{T}_2$ from Eq. (i),
$\mathrm{T}_1=(2 \mathrm{Mg}) \times(\sqrt{3} / 2)=\sqrt{3} \mathrm{Mg}$
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