91影视

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.

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.

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.

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.