Chapter 8 - Rotational Motion and Equilibrium - Learning Path Questions and Exercises - Exercises - Page 306: 29

$T_{2}=40N$ $T_{1}=49N$

Since m=1.5 kg is in equilibrium, horizontally, $T_{1}sin45^{\circ}=T_{2}cos30^{\circ}$ $T_{1}=\sqrt \frac{3}{2}T_{2}$ ...... equation (1) Also, vertically, $T_{1}cos45^{\circ}=mg+T_{2}sin30^{\circ}$ $\frac{1}{\sqrt 2}T_{1}=14.7+\frac{1}{2}T_{2}$ ...... equation (2) From equation (1) & (2), we have $T_{2}=40N$ $T_{1}=49N$