two 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 2m apart, and each is 2m from the pole. the tension in each wire is 115N. what is the tension in the net, assumed horizontal and attached at the top of the pole?

What I tried to do was figure out the distance from one wire to the top of the pole. Which I got 3.28m then multiplied my answering by the equilateral triangle. I got 120N, you can check if my answer is correct.

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aristocles aristocles · Answer 2 · 2019-09-19T03:54:15+02:00

Answer:

[tex]T_x = 140.8 N[/tex]

Explanation:

As per the given figure we can find the angle of each wire with the given pole

so we have

[tex]tan\theta = \frac{2}{2.6}[/tex]

[tex]tan\theta = 0.77[/tex]

[tex]\theta = 37.56^o[/tex]

now let the top of the pole is our origin

then we have point on the ground at the base of the string connected is given as

[tex]x_1 = 2cos30 = \sqrt3[/tex]

[tex]y_1 = 2sin30 = 1[/tex]

[tex]z_1 = -2[/tex]

for other wire we have

[tex]x_2 = 2cos30 = \sqrt3[/tex]

[tex]y_2 = -2sin30 = -1[/tex]

[tex]z_2 = -2[/tex]

now unit vector along the length of the string is given as

[tex]\hat r_1 = \frac{\sqrt3 \hat i + \hat j - 2\hat k}{\sqrt{(3 + 1 + 4)}}[/tex], [tex]\hat r_2 = \frac{\sqrt3 \hat i - \hat j - 2\hat k}{\sqrt{(3 + 1 + 4)}}[/tex]

now tension in two threads is given as

[tex]T_1 = 115 \hat r_1[/tex]

[tex]T_2 = 115\hat r_2[/tex]

total tension in the thread

[tex]T = 115\frac{(2\sqrt3 \hat i + 0 \hat j - 4\hat k)}{\sqrt{(3 + 1 + 4)}}[/tex]

now horizontal component of tension is given as

[tex]T_x = 115 \frac{2\sqrt3}{\sqrt8}[/tex]

[tex]T_x = 140.8 N[/tex]