Answer
(a) $\tau = 14~m\cdot N$ (counterclockwise)
(b) $\tau = 13~m\cdot N$ (clockwise)
Since the net torque is clockwise, we can express this as $\tau = -13~m\cdot N$
Work Step by Step
(a) $\tau = \sum r\cdot F~sin(\theta)$
$\tau = (1.0~m)(52~N)~sin(58^{\circ}) - (1.0~m)(56~N)~sin(32^{\circ})$
$\tau = 14~m\cdot N$ (counterclockwise)
(a) $\tau = \sum r\cdot F~sin(\theta)$
$\tau = (2.0~m)(56~N)~sin(32^{\circ}) - (1.0~m)(65~N)~sin(45^{\circ})$
$\tau = 13~m\cdot N$ (clockwise)
