Question

A traffic light hangs from a pole, as shown in Fig. 9–59. The uniform aluminum pole AB is 7.20 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg. Determine (a) the tension in the horizontal massless cable CD and (b) the vertical and horizontal components of the force exerted by pivot A on the aluminum pole.

Short answer

1. The tension in the cable is 410 N.
2. The horizontal and vertical forces exerted by the pivot point on the aluminum pole are 410 N and 328 N, respectively.

Step-by-step solution

Concepts

In equilibrium, the net force in the x and y direction should be zero also; the torque about any point is zero.For this problem, you should calculate the torque about the pivot point.

Given data

The length of the pole is \(AB = L = 7.20\;{\rm{m}}\).

The mass of the pole is \(m = 12.0\;{\rm{kg}}\).

The mass of the traffic light is \(M = 21.5\;{\rm{kg}}\).

The angle of the pole is \(\theta = {37.0^ \circ }\).

Let T be the tension in the cable.

The cable is joined at \(h = 3.80\;{\rm{m}}\) height from the pivot point.

Let \({F_{\rm{x}}}\) and \({F_{\rm{y}}}\) be the horizontal and vertical components of the force exerted by pivot A on the aluminum pole.

Calculation of part (a)

Part (a)

At equilibrium, the torque about the pivot point is zero. Then,

\(\begin{array}{c}Th - \left( {mg \times \frac{{L\cos \theta }}{2}} \right) - \left( {Mg \times L\cos \theta } \right) = 0\\Th = \left( {\frac{m}{2} + M} \right)gL\cos \theta \end{array}\).

Now, substituting the values in the above equation,

\(\begin{array}{c}T \times 3.80\;{\rm{m}} = \left( {\frac{{12.0\;{\rm{kg}}}}{2} + 21.5\;{\rm{kg}}} \right)\left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {7.20\;{\rm{m}}} \right)\cos {37.0^ \circ }\\T = 407.8\;{\rm{N}}\\T \approx 410\;{\rm{N}}\end{array}\).

Hence, the tension in the cable is 410 N.

Calculation of part (b)

At equilibrium, for the horizontal forces,

\(\begin{array}{c}{F_{\rm{x}}} = T\\ = 407.8\;{\rm{N}}\\ \approx 410\;{\rm{N}}\end{array}\)

At equilibrium, for the vertical forces,

\(\begin{array}{c}{F_{\rm{y}}} = mg + Mg\\ = \left( {m + M} \right)g\\ = \left( {12.0\;{\rm{kg}} + 21.5\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\ = 328\;{\rm{N}}\end{array}\).

Hence, the horizontal and vertical forces exerted by the pivot point on the aluminum pole are 410 N and 328 N, respectively.