Answer
$\tau_{net} = 5.0~N~m$
Work Step by Step
We can find the magnitude of the net torque about the axle as:
$\tau_{net} = \sum \tau$
$\tau_{net} = r_1~F_1 + r_2~F_2$
$\tau_{net} = (0)(50~N)+(0.10~m)(50~N)$
$\tau_{net} = 5.0~N~m$
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