Answer
(a) We can see the free body diagram below.
(b) $T_A = 3.64~N$
$T_B = 1.68~N$
Work Step by Step
(a) We can see the free body diagram below.
(b) We can make an equation with the vertical components of the forces:
$\sum F_y = 0$
$T_A~sin~30.0^{\circ}-T_B~sin~30.0^{\circ}-mg = 0$
$T_A~sin~30.0^{\circ} = T_B~sin~30.0^{\circ} + mg$
$T_A = T_B+\frac{mg}{sin~30.0^{\circ}}$
The sum of the horizontal components of the two tension forces provide the centripetal force to keep the ball moving in a circle:
$T_A~cos~30.0^{\circ}+T_B~cos~30.0^{\circ} = m~\omega^2~r$
$T_A~cos~30.0^{\circ}+T_B~cos~30.0^{\circ} = m~\omega^2~(0.150~m)~cos~30.0^{\circ}$
$T_A+T_B = m~\omega^2~(0.150~m)$
$(T_B+\frac{mg}{sin~30.0^{\circ}})+T_B = m~\omega^2~(0.150~m)$
$2T_B = m~\omega^2~(0.150~m) - \frac{mg}{sin~30.0^{\circ}}$
$T_B = \frac{m~\omega^2~(0.150~m)}{2} - \frac{mg}{2sin~30.0^{\circ}}$
$T_B = \frac{(0.100~kg)(6.00\pi~rad/s)^2~(0.150~m)}{2} - \frac{(0.100~kg)(9.80~m/s^2)}{2sin~30.0^{\circ}}$
$T_B = 1.68~N$
We can find $T_A$:
$T_A = T_B+\frac{mg}{sin~30.0^{\circ}}$
$T_A = (1.68~N)+\frac{(0.100~kg)(9.80~m/s^2)}{sin~30.0^{\circ}}$
$T_A = 3.64~N$
