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.