ends. Bob weighs 730 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 910 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable?
Let: Rr = reaction force at the right cable Rl = reaction force at the left cable Summing moments @ Rr, clockwise rotation is positive: Rl(3m) = (910 N)(0.5 m) + (500 N)(1 m) + (730 N)(2 m) Rl = 805 N Summing moments @ Rl, clockwise rotation is positive: Rr(3 m) = (730 N)(1 m) + (500 N)(2 m) + (910 N)(2.5 m) Rr = 1,335 N
Rainwater forgot about the weight of plank itself... So: (0.5 m)×(910 N) + (1 m)×(500 N) + (1.5 m)×(295 N) + (2 m)×(730 N) = (3 m)× T? T? = 952.5 N (1 m)×(730 N) + (1.5 m)×(295 N) + (2 m)×(500 N) + (2.5 m)×(910 N) = (3 m)× T? T? = 1482.5 N Check: (952.5 N) + (1482.5 N) ?=? (730 N) + (295 N) + (500 N) + (910 N) (2435 N) !=! (2435 N) Seems OK ;-)
