1. 6. Let 1 = 1.5 m and 2 = 5.0 − 1.5 = 3.5 m. We denote the tension in the cable closer to the window as
F1 and that in the other cable as F2 . The force of gravity on the scaffold itself (of magnitude ms g) is
at its midpoint, 3 = 2.5 m from either end.
(a) Taking torques about the end of the plank farthest from the window washer, we find
mw g 2+ ms g 3 (80 kg) 9.8 m/s2 (3.5 m) + (60 kg) 9.8 m/s2 (2.5 m)
F1 = = = 8.4 × 102 N .
1+ 2 5.0 m
(b) Equilibrium of forces leads to
F1 + F2 = ms g + mw g = (60 kg + 80 kg) 9.8 m/s2 = 1.4 × 103 N
which (using our result from part (a)) yields F2 = 5.3 × 102 N.