Answer
The magnitude of the force from the left pedestal is $~~1160~N$
Work Step by Step
We can consider the torque about a rotation axis at the right pedestal.
We can find $F_L$, the magnitude of the force from the left pedestal:
$\tau_{net} = 0$
$d~F_L - (3.0~m)(580~N) = 0$
$F_L = \frac{(3.0~m)(580~N)}{d}$
$F_L = \frac{(3.0~m)(580~N)}{1.5~m}$
$F_L = 1160~N$
The magnitude of the force from the left pedestal is $~~1160~N$.
