## Physics: Principles with Applications (7th Edition)

(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 $\tau = -13~m\cdot N$
(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)