Answer
The center of mass of the paint cans is $~~0.70~m~~$ from the right cable.
Work Step by Step
To find the center of mass of the paint cans, we can consider the torque about a rotation axis at the left cable.
There is a clockwise torque from the weight of the paint cans, a clockwise torque from the weight of the scaffold, and an opposing counterclockwise torque from the tension in the right cable.
Let $x$ be the horizontal distance from the left cable to the center of mass of the paint cans. We can find $x$:
$\tau_{net} = 0$
$(2.00~m)(722~N)-(1.00~m)(50.0~kg)(9.8~m/s^2)-(75.0~kg)(9.8~m/s^2)~x = 0$
$(75.0~kg)(9.8~m/s^2)~x = (2.00~m)(722~N)-(1.00~m)(50.0~kg)(9.8~m/s^2)$
$x = \frac{(2.00~m)(722~N)-(1.00~m)(50.0~kg)(9.8~m/s^2)}{(75.0~kg)(9.8~m/s^2)}$
$x = 1.30~m$
Since the center of mass of the paint cans is $1.30~m$ from the left cable, the center of mass of the paint cans is $~~0.70~m~~$ from the right cable.
