Question

A weight of 200 N is suspended with the help of the string as shown in the figure. Find the tension $t_1$ and $t_2$

Answer 1

The resolution of forces along the horizontal direction is given as,

$T_1\cos {30}^{\circ }=T_2\cos {45}^{\circ }$

$T_2=\sqrt{\frac{3}{2}}{T}_1$

The resolution of forces along the vertical direction is given as,

$T_1\sin {30}^{\circ }+T_2\sin {45}^{\circ }=200$

$T_1\times \frac{1}{2}+\sqrt{\frac{3}{2}}{T}_1\times \frac{1}{\sqrt{2}}=200$

$T_1=146.41N$

$T_2=\sqrt{\frac{3}{2}}\times 146.41$

$T_2=179.32N$

Thus, the tension $T_1$ is $146.41N$ and the tension $T_2$ is $179.32N$.