Answer
(a) $F = 33.9~N$
(b) $T = 39.2~N$
Work Step by Step
(a) The applied force $F$ is equal in magnitude to the horizontal component of the tension.
$F = T~cos~30^{\circ}$
The vertical component of the tension is equal in magnitude to the ball's weight.
$mg = T~sin~30^{\circ}$
We can divide the first equation by the second equation:
$\frac{F}{mg} = \frac{T~cos~30^{\circ}}{T~sin~30^{\circ}}$
$F = mg~cot~30^{\circ}$
$F = (2.0~kg)(9.80~m/s^2)~cot~30^{\circ}$
$F = 33.9~N$
(b) The vertical component of the tension is equal in magnitude to the ball's weight.
$T~sin~30^{\circ} = mg$
$T = \frac{mg}{sin~30^{\circ}}$
$T = \frac{(2.0~kg)(9.80~m/s^2)}{sin~30^{\circ}}$
$T = 39.2~N$
