A scaffold of mass 52 kg and length 53 m is supported in a h

A scaffold of mass 52 kg and length 5.3 m is supported in a horizontal position by a vertical cable at each end. A window washer of mass 63 kg stands at a point 1.7 m from one end. What is the tension in (a) the nearer (relative to the person) cable and (b) the farther (relative to the person) cable?

Solution

As the scaffold and the window washer, both are in equilibrium, we can balance the forces and the torques to find the tensions in the wires. Let us assume that the tension in the string nearer to the person is T1 and in that farther from the person is T2

We will form equations involving the tensions and then resolve to find the values.

Part a.) Let us balance the torques about the end farther from the person.

Here, torque due to T1 = T1 x L = 5.3 T1

Torque due to weight of the person = mg x (L - 1.7) = 63 x 9.81 x (5.3 - 1.7) = 2224.908 N-m

Torque due to weight of the scaffold = Mg x L/2 = 52 x 9.81 x 5.3 / 2 = 1351.818 N-m

Now, balancing the net torque acting on the scaffold about the farther end we get:

5.3 T1 = 2224.908 + 1351.818 = 3576.726

or, T1 = 674.854 N is the required tension

Part b.) For string at the farther end, we balance the torques about the end closer to the person.

Torque due to the tension on the string at the other end = 5.3T2

Torque due to the weight of scaffold = Mg x L/2 = 52 x 9.81 x 5.3 / 2 = 1351.818 N-m

Torque due to the weight of the person = mg x (1.7) = 63 x 9.81 x (1.7) = 1050.651 N-m

Balancing the torques, we get:

5.3T2 = 1351.818 + 1050.651 = 2402.469

or, T2 = 453.296 N is the required tension