A scaffold of mass 77 kg and length 5.0 m is supported in a horizontal position by a vertical cable at each end. A window washer of mass 78 kg stands at a point 1.8 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

A) T2 = 912.88 N

B) T1 = 607.12 N

Explanation:

First of all, we see that the sum of the tensions is equal to the total weight.

Now, for the scaffold, weight; W_s = 77 x 9.8 = 755.6 N

For the window washer, Weight; W_w = 78 x 9.8 = 764.4 N

Total weight;W_t = W_s + W_w

W_t = 755.6 N + 764.4 N = 1520 N

Thus,

T1 + T2 = 1520

Where T1 and T2 are the tensions in farther and nearer cables respectively.

Now, we need to do a torque problem.

The window washer is 1.8m from the right end of the scaffold and so the weight of the scaffold is at its center. This is 2.5 m from either end. Let the pivot point be at right end of the scaffold.

For the window washer, counter clockwise torque = 764.4 x 1.5 = 1146.6 N.m

For the scaffold, counter clockwise torque = 755.6 x 2.5 = 1889 N.m

Total Torque; T = 1146.6 + 1889 = 3035.6 N.m

For the cable at the left end of the cable, clockwise torque = T1 x 5

Set this equal to the total counter clockwise torque and solve for T1.

Thus,

T1 x 5 = 3035.6

T1 = 3035.6 ÷ 5 = 607.12 N

T2 = 1520 – 607.12 = 912.88 N

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.