Answer
(a) The force exerted by the left-hand cable is $539~N$
(b) The force exerted by the right-hand cable is $392~N$
Work Step by Step
(a) Let's consider the torques about an axis at the point where the right-hand cable is attached to the board. The painter's weight and the board's weight exert a counterclockwise torque about this axis. The left cable exerts a clockwise torque about this axis.
Since the system is in equilibrium, the magnitudes of the clockwise and counterclockwise torques must be equal in magnitude. We can find the force exerted by the left-hand cable:
$\tau_{CW} = \tau_{CCW}$
$(5.0~m)~F_L = (3.0~m)(75~kg)(9.80~m/s^2)+(2.5~m)(20.0~kg)(9.80~m/s^2)$
$F_L = \frac{(3.0~m)(75~kg)(9.80~m/s^2)+(2.5~m)(20.0~kg)(9.80~m/s^2)}{5.0~m}$
$F_L = 539~N$
The force exerted by the left-hand cable is $539~N$
(b) Since the system is in equilibrium, the sum of the vertical forces must equal zero. That is, the forces exerted downward must be equal in magnitude to the forces exerted upward.
We can find the force exerted by the right-hand cable:
$F_L+F_R = (75~kg)(9.80~m/s^2)+(20.0~kg)(9.80~m/s^2)$
$F_R = (75~kg)(9.80~m/s^2)+(20.0~kg)(9.80~m/s^2)- F_L$
$F_R = (75~kg)(9.80~m/s^2)+(20.0~kg)(9.80~m/s^2)- (539~N)$
$F_R = 392~N$
The force exerted by the right-hand cable is $392~N$
