91Ó°ÊÓ

For equilibrium, the net force in the x and y directions should be zero.For this problem, find the component of the tensions of the wires and then compare the force components in the vertical and horizontal directions.

The mass of the traffic light is \(m = 33\;{\rm{kg}}\).

The angle of the right wire is \(\theta = {37^ \circ }\).

The angle of the left wire is \(\phi = {53^ \circ }\).

Let \({T_1}\) and \({T_2}\) be the tensions in the wires.

The free-body diagram of the problem is given below.

The condition of the equilibrium for the horizontal forces is

\(\begin{array}{c}{T_2}\cos \phi = {T_1}\cos \theta \\{T_2} = {T_1}\frac{{\cos \theta }}{{\cos \phi }}\end{array}\). … (i)

The condition of the equilibrium for the vertical forces is

\(\begin{array}{c}{T_1}\sin \theta + {T_2}\sin \phi = mg\\{T_1}\sin \theta + {T_1}\frac{{\cos \theta }}{{\cos \phi }}\sin \phi = mg\\{T_1}\left( {\sin \theta + \cos \theta \tan \phi } \right) = mg\\{T_1} = \frac{{mg}}{{\sin \theta + \cos \theta \tan \phi }}\end{array}\).

Now, substituting the value in the above equation,

\(\begin{array}{c}{T_1} = \frac{{mg}}{{\sin \theta + \cos \theta \tan \phi }}\\ = \frac{{\left( {33\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)}}{{\sin {{37}^ \circ } + \cos {{37}^ \circ }\tan {{53}^ \circ }}}\\ = 194.6\;{\rm{N}}\\ \approx 190\;{\rm{N}}\end{array}\).

Now, from equation (i),

\(\begin{array}{c}{T_2} = \left( {194.6\;{\rm{N}}} \right)\frac{{\cos {{37}^ \circ }}}{{\cos {{53}^ \circ }}}\\ = 258.3\;{\rm{N}}\\ = 260\;{\rm{N}}\end{array}\).

Hence, the tension in the left wire is 260 N, and the tension in the right wire is 190 N.