Two window washers, Bob and Joe, are on a 3.00-m-long, 315-N scaffold supported by two cables attached to its ends. Bob weighs 795 N and stands 1.00 m from the left end, as shown in the figure below. Two meters from the left end is the 500-N washing equipment. Joe is 0.500 m from the right end and weighs 880 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable?

Since the scaffold is in rotational and translational equilibrium, the net force and the net torque on the scaffold are both zero. Begin by drawing a free body diagram of the scaffold, which shows all the forces and their locations. There are six forces:

Tension force T₁ pulling up at the left cable
Bob's weight 795 N pulling down 1.00 m from the left
Scaffold's weight 315 N pulling down at the center, 1.50 m from the left
Equipment weight 500 N pulling down 2.00 m from the left
Joe's weight 880 N pulling down 0.500 m from the right, or 2.50 m from the left
Tension force T₂ pulling up at the right cable, 3.00 m from the left

The scaffold is in translational equilibrium, so the net force is zero.

∑F = 0

T₁ − 795 − 315 − 500 − 880 + T₂ = 0

T₁ + T₂ = 2490

The scaffold is in rotational equilibrium, so the net torque about the left end is zero. Remember that torque is force times perpendicular distance.

∑τ = 0

T₁ (0) − 795 (1.00) − 315 (1.50) − 500 (2.00) − 880 (2.50) + T₂ (3.00) = 0

T₂ (3.00) = 4467.5

T₂ = 1490 N

Substitute into the first equation.

T₁ + 1490 = 2490

T₁ = 1000 N

Therefore, the force on the left cable is 1000 N, and the force on the right cable is 1490 N.