Question

Tension in the string $AB$ (in $\text{N}$) is -

Answer 1

Along Y-direction applying equilibrium condition,
$\sum F_y=0$
$T_2\sin {30}^{\circ }+T_1\sin {60}^{\circ }-10=0$
[$\because a_y=0$ vertical equilibrium]
$\Rightarrow \frac{T_2}{2}+\frac{\sqrt{3}{T}_1}{2}=10$
$\Rightarrow T_2+\sqrt{3}{T}_1=20...\left(1\right)$
Along X-direction applying equilibrium condition,
$\sum F_x=0$
$T_1\cos {60}^{\circ }-T_2\cos {30}^{\circ }=0$
[$\because a_x=0$ horizontal equilibrium]
$\Rightarrow \frac{T_1}{2}-\frac{\sqrt{3}}{2}{T}_2=0$
$\Rightarrow T_1=\sqrt{3}{T}_2...\left(2\right)$
From Eq.$\left(1\right)\&\left(2\right)$,
$T_2+\sqrt{3}\left(\sqrt{3}{T}_2\right)=20$
$\Rightarrow 4T_2=20$
$\therefore T_2=5\text{N}$
Tension in string $AB$ is $5\text{N}$.