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
F
1
and that in the other cable as F
2
. The force of gravity on the scaffold itself (of magnitude m
s
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
F
1
=
m
w
g
2
+ m
s
g
3
1
+
2
=
(80 kg)
9.8 m/s
2
(3.5 m) + (60 kg)
9.8 m/s
2
(2.5 m)
5.0 m
= 8.4
× 10
2
N .
(b) Equilibrium of forces leads to
F
1
+ F
2
= m
s
g + m
w
g = (60 kg + 80 kg)
9.8 m/s
2
= 1.4
× 10
3
N
which (using our result from part (a)) yields F
2
= 5.3
× 10
2
N.