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

The net torque about the axle is $-0.40~N~m$
We can find the net torque about the axle. Note that clockwise is the negative direction. Let $F_1 = 30~N$ and let $F_2 = 20~N$. $\tau_{net} = \sum \tau$ $\tau_{net} = r_1~F_1 + r_2~F_2$ $\tau_{net} = -(0.040~m)(30~N)+(0.040~m)(20~N)$ $\tau_{net} = -0.40~N~m$ The net torque about the axle is $-0.40~N~m$.