(III) Two wires run from the top of a pole 2.6 m tall that supports volleyball net. The two wires are anchored to the ground 2.0 m apart, and each is 2.0 m from the pole (Fig. 9–66). The tension in each wire is 115 N. What is the tension in the net, assumed horizontal and attached at the top of the pole?

A body is supposed to be in translation equilibrium when it travels with a constant velocity. This means that the acceleration of the body should be zero.

Given data:

The tension in each wire is \({F_{\rm{T}}} = 115\;{\rm{N}}\).

Draw a triangle made by the pole and one of the wires.

The angle that the tension (along the wire) makes with vertical can be calculated as:

\(\begin{array}{l}{\theta _1} = {\tan ^{ - 1}}\left( {\frac{{2.0\;{\rm{m}}}}{{2.6\;{\rm{m}}}}} \right)\\{\theta _1} = 37.6^\circ \end{array}\)

The expression for the part of the tension that is parallel to the ground is as follows:

\({F_{{\rm{Th}}}} = {F_{\rm{T}}}\sin {\theta _1}\)

Draw a free body diagram that shows only the force parallel to the ground.

The horizontal parts of the tension lie as the sides of an equilateral triangle. So, each makes a 30° angle with the tension force of the net.

Now, apply the force equilibrium condition along the direction of the tension in the net.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{{\rm{net}}}} - 2{F_{{\rm{Th}}}}\cos \left( {30^\circ } \right) = 0\\{F_{{\rm{net}}}} - 2{F_{\rm{T}}}sin{\theta _1}\cos \left( {30^\circ } \right) = 0\\{F_{{\rm{net}}}} - 2\left( {115\;{\rm{N}}} \right)sin\left( {37.6^\circ } \right)\cos \left( {30^\circ } \right) = 0\\{F_{{\rm{net}}}} = 121.5\;{\rm{N}}\end{array}\)

Thus, the tension in the net is \(121.5\;{\rm{N}}\).