Torque

[Quipzley]

http://www.geocities.com/tucan1_1/Physics.jpg (http://www.geocities.com/tucan1_1/Physics.jpg)
There is a picture of the problem.

Here are the formulas i can get from that:

The sum of the forces in the y direction with positive being upward
cos 36.9 * Tension 1 + cos 53.1 * Tension 2 - m * g = 0

The sum of the forces in the x direction with positive being to the right
sin 36.9 * Tension 1 + sin 53.1 * Tension 2 = 0

The sum of the torques with clockwise being positive
cos 53.1 * Tension 2 * (6.1 - x) + cos 36.9 * Tension 1 * x

This leads me with 4 variables and 3 equations.
Tension 1, Tension 2, m (mass), and x

Maybe this problem is more simple than the approach i'm taking?

[e(ho0n3]

Calculate the torque about the left end of the bar. That should help.

[Quipzley]

You have to take in consideration the torques of either side because they have to equal zero.

[Quipzley]

i got the equation x*m*g=Tension 2*cos 53.1*6.1

what is possible to solve for???

[e(ho0n3]

Well, now you have an expression for x in terms of m and Tension 2. Find out what these are and you're done (i.e. use the other equations you have).