Answer
$$T_1=220 N$$
$$T_2=255.6 N$$
Work Step by Step
$\Sigma\tau=T_1 sin\theta(3 m) - (w_m)(0.5) +(w_r)(1.5)=0$
$T_1=\frac{(90)(0.5)+(190)(1.5)}{(3)sin(30)}=220 N$
$\Sigma F_y=T_1sin\theta+T2sin\theta-(W_m+W_r)$
$\quad T_2sin=90+190-(220)sin(30)=170 N $
$\Sigma F_x=T_1cos\theta-T_2cos\theta=0$
$T_2cos\theta=(220)cos(30^{\circ})=190.5 N$
$\frac{T_2sin\theta}{T_2cos\theta}=\frac{170}{190.5}=0.892=tan\theta$
$\theta=tan^{-1}(0.892)=41.7^{\circ}$
$T_2sin\theta=170 N$
$T_2=\frac{170 N}{sin(41.7^{\circ})}=255.6 N$
