Calculating the Angle of Tension in Tightrope Walking: A Physics Problem Solved

[Meowzers]

Hi, I have tried to solve the following problem, and I have my work for it.

A tightrope walker is practicing. She is balanced on a rope at the exact midpoint of the rope. the tension in the rope is 4000N, and the mass of the tightrope walker is 68kg. Consider the left side of the rope. What angle with the horizntal does it make? State this as a positive number in degrees.

So first off, I calculated the force of the tightrope walker:
F=mg
F=(68kg)(9.8m/s2)
F=666.4N

Then, since the walker is in the middle of the rope, I assumed that the tension of each side of the rope would be 2000N. Plus, the triangle that she forms with the rope because she's in the center is isosceles. Therefore, the angle on both sides formed with the horizontal will be equal.

Next, I solved for the y-component of the diagram. The sum of the forces must equal 0 since there is no acceleration.

2000*sin(x) + 2000*sin(x) + 666.4N*sin(270) = 0
2000*sin(x) + 2000*sin(x) = 666.4
4000*sin(x) = 666.4
sin(x) = .1666
x = 9.59 degrees

Where am I making a mistake? Did I assume something incorrectly?

 

[Meowzers]

Nevermind, I figured out my problem.