Answer
$\tau = -3.85~N\cdot m$
Work Step by Step
We can write a general expression for the torque about the pivot:
$\tau = r\times F = r~F~sin~\theta$
$r$ is the displacement vector from the pivot to the point where the force is applied
$F$ is the force vector
$\theta$ is the angle between these two vectors
We can find the net torque about the pivot:
$\tau = r_1~F_1~sin~\theta_1-r_2~F_2~sin~\theta_2$
$\tau = (1.30~m)(4.20~N)~sin~75.0^{\circ}-(2.15~m)~(4.90~N)~sin~60.0^{\circ}$
$\tau = -3.85~N\cdot m$
Note that the negative sign shows that the net torque tends to make the object rotate in the clockwise direction.
