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

We can find the net torque about the axle. Note that the clockwise direction is negative. $\tau_{net} = \sum \tau$ $\tau_{net} = (r_1\times F_1) + (r_2\times F_2)+(r_3\times F_3)$ $\tau_{net} = (0.075~m)(15~kg)(9.80~m/s^2)-(0.075~m)(5.0~kg)(9.80~m/s^2)-(0.225~m)(10~kg)(9.80~m/s^2)$ $\tau_{net} = -14.7~N~m$ The net torque about the axle is -14.7 N m.