Question

A traffic light hangs from a pole as shown in (Figure 1) .The uniform aluminum pole AB is 7.30m long and has a mass of 10.0kg . The mass of the traffic light is22.0kg .

A. Determine the tension in the horizontal massless cable cd CD

CD

B. Determine the vertical component of the force exerted by the pivot

A on the aluminum pole.

Answer 1

The uniform aluminum pole AB is 7.30 m long and has a mass of 10 kg. The mass of the traffic light is 22.0 kg.


The weights of the traffic light and the aluminum pole AB produce clockwise torques on the vertical pole at point A.
The angle at A = 53?

Weight of traffic light = 22 * 9.8
The horizontal distance from the vertical pole to the traffic light = 8 * sin 53?
Torque = 22 * 9.8 * 7.30 * sin 53?

The weight of pole AB = 10 * 9.8. The center of mass is 4 (not given,let us consider) meters from point A. The angle at A = 53?.
The horizontal distance from the vertical pole to the center of mass = 4 * sin 53?
Torque = 10 * 9.8 * 4 * sin 53?

The tension in cable CD produces a counter clockwise torque on the vertical pole at point A. The vertical distance between cable CD and point A = 3.80 m.
Torque = T * 3.8

The counter clockwise torque must equal the clockwise torque.

T * 3.8 = 22 * 9.8 * 7.3 * sin 53? + 10 * 9.8 * 4 * sin 53?
T = 413 N


The vertical component must support the weight of the traffic light and pole AB.
Vertical component = sum of the 2 weights = (22+ 10) * 9.8 = 323.4 N
The tension in cable CD is the only horizontal force. So, the horizontal component must equal the tension.