The system shown in figure is in equilibrium. Find the magnitude of tension in each string; $T_{1}, T_{2}, T_{3}$ , and $T_{4}, (g = 10 m / s^{- 2})$

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Solution

Using sine law at point A,
$\frac{T_{1}}{s i n (150^{o})} = \frac{T_{2}}{s i n (90^{o})} = \frac{10 g}{s i n (120^{o})}$
$\Rightarrow T_{1} = \frac{10 \times 10 s i n (150^{o})}{s i n (120^{o})}$ = $\frac{100 \sqrt{3}}{3}$ ,
$\Rightarrow T_{2} = \frac{10 \times 10 s i n (90^{o})}{s i n (120^{o})}$ = $\frac{200 \sqrt{3}}{3}$

Using Sine law at point B,
$\frac{T_{2}}{s i n (150^{o})} = \frac{T_{3}}{s i n (150^{o})} = \frac{T_{4}}{s i n (60^{o})}$
$\Rightarrow T_{3} = \frac{T_{2} \times s i n (150^{o})}{s i n (150^{o})}$ = $\frac{200 \sqrt{3}}{3}$
$\Rightarrow T_{4} = \frac{T_{2} \times s i n (60^{o})}{s i n (150^{o})} = 200$

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