Answer
The net torque about the axle is $-0.40~N~m$
Work Step by Step
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 $-0.40~N~m$.
