Question

A window washer is standing on a scaffold supported by vertical ropes at each end. The scaffold weighs 200. N and is 3.00 m long. What is the tension in each rope when the 700. N worker stands 1.00 m from one end?

Answer 1

Let the tension on the 2 left ropes be T₁ and 2 right ropes be T_2.

Applying force balancing equation

T₁ + T₂ = W + W^'

T₁ + T₂ = 200 + 700

T₁ + T₂ = 900 .........(1)

Since scaffold is in rotational equillibrium appying balances of torque about center of mass of scaffold

-T₁(1.5) + T₂(1.5) - W(0.5) = 0

1.5T₂ - 1.5T₁ = 0.5(700)

1.5T₂ - 1.5T₁ = 350 ........................(2)

Multiplying (1) by 1.5 and adding to (2) we have

3T₂ = 1700

T₂ = 566.7 N

T₁= 333.3 N

Now tension in each rope to the left is T₁/2 = 166.67 N

tension in each rope to the right is T₂/2 = 283.4 N