## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can find the net torque about the axle. Let $F_1$ be the 50-N force which is directed downward. $\tau_{net} = \sum \tau$ $\tau_{net} = r_1~F_1 + r_2~F_2$ $\tau_{net} = (0)(50~N)+(0.10~m)(50~N)$ $\tau_{net} = 5.0~N~m$ The net torque about the axle is 5.0 N m.