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the problem as in the book
"Two guy wires run from the top of a pole 2.6 m tall that supports a volleyball net. The two wires are anchored to the ground 2.0 m apart and each is 2.0m from the pole. The tension in each wire is 95 N. What is the tension in the net, assumed horizontal and attached at the top of the pole?"  I've attached an image.
What I tried to do was figure out the distance from one wire to the top of the pole.
[tex]X = (2.6m^2 + 2.0m^2)^{\frac{1}{2}} = 3.28m[/tex]
Then using the sum of torques (which is zero in equilibrium) at the first rope as the pivot point
[tex]\sum\tau = \tau_{2} + F_{tensioninnet}X = 0 \Rightarrow F_{tensionnet} = \frac{\tau_{2}}{X} = \frac{95N \cdot 2.0m}{3.28m}[/tex]
I get 60N, but the answer in the back of the book is 100 newtons. I also noticed if I multiplied the answer by [tex]tan(60)[/tex] (equilateral triangle at the bottom), and I get 100 N
"Two guy wires run from the top of a pole 2.6 m tall that supports a volleyball net. The two wires are anchored to the ground 2.0 m apart and each is 2.0m from the pole. The tension in each wire is 95 N. What is the tension in the net, assumed horizontal and attached at the top of the pole?"  I've attached an image.
What I tried to do was figure out the distance from one wire to the top of the pole.
[tex]X = (2.6m^2 + 2.0m^2)^{\frac{1}{2}} = 3.28m[/tex]
Then using the sum of torques (which is zero in equilibrium) at the first rope as the pivot point
[tex]\sum\tau = \tau_{2} + F_{tensioninnet}X = 0 \Rightarrow F_{tensionnet} = \frac{\tau_{2}}{X} = \frac{95N \cdot 2.0m}{3.28m}[/tex]
I get 60N, but the answer in the back of the book is 100 newtons. I also noticed if I multiplied the answer by [tex]tan(60)[/tex] (equilateral triangle at the bottom), and I get 100 N
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