## College Physics (4th Edition)

(a) $F = 33.9~N$ (b) $T = 39.2~N$
(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. $T~sin~30^{\circ} = mg$ $T = \frac{mg}{sin~30^{\circ}}$ $T = \frac{(2.0~kg)(9.80~m/s^2)}{sin~30^{\circ}}$ $T = 39.2~N$