Answer
(a) The net torque is $F~d$
(b) The net torque is $F~d$
Work Step by Step
(a) The torque is $\tau = r\times F$, where $r$ is the smallest distance between the force vector and the rotation axis, and $F$ is the perpendicular component of the force vector.
The upward force exerts a counterclockwise torque. We can find the magnitude of this torque:
$\tau_1 = x_1~F$
The downward force exerts a clockwise torque. We can find the magnitude of this torque:
$\tau_2 = x_2~F$
We can find the magnitude of the net torque:
$\vert \tau \vert = \vert \tau_2-\tau_1 \vert$
$\vert \tau \vert = \vert x_2~F-x_1~F \vert$
$\vert \tau \vert = \vert F~(x_2-x_1) \vert$
$\vert \tau \vert = F~d$
The net torque is $F~d$
(b) The upward force exerts a counterclockwise torque. We can find the magnitude of this torque:
$\tau_1 = x_1~F$
The downward force exerts a clockwise torque. We can find the magnitude of this torque:
$\tau_2 = x_2~F$
We can find the magnitude of the net torque:
$\vert \tau \vert = \vert \tau_2-\tau_1 \vert$
$\vert \tau \vert = \vert x_2~F-x_1~F \vert$
$\vert \tau \vert = \vert F~(x_2-x_1) \vert$
$\vert \tau \vert = F~d$
The net torque is $F~d$.
