Distance from left end of beam to rope attachment point

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Homework Statement


A beam of mass 10.0 kg, is suspended from the ceiling by a single rope. It has a mass 40.0 kg attached at one end and a mass of 14.9 attached at the other. The beam has a length of L = 3 m, it is in static equilibrium, and it is horizontal. The tension in the rope is T = 637 N.
Determine the distance, x, from the left end of the beam to the point where the rope is attached

Homework Equations


Torque = Force Distance
3 = x1 +x2

The Attempt at a Solution



40 (X1) = 14.9 (X2) (since it's in equilibrium)
40 (3-X2) = 14.9 (X2)
120 - 40(X2) = 14.9 (X2)
120 = 54.9 (X2)
2.186 = X2

I have the feeling that I'm going to feel like an idiot once someone points out the problem with this to me...
 
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Discussion

Since the beam itself has a mass of 10.0 kg, shouldn't it also provide a torque as the rope is NOT connected to the middle of it. (the torque due to the force acting through the Centre of mass of the beam)
 
Don't assume the rope is attached at the center of the beam. They did not say that in the problem.

You are going to have to assume that the rope is not attached at the center and thus the weight force on the beam itself will cause torque about the pivot.
 
The beam is horizontal, so the rope will be located at the center of gravity of the beam.

So try solving for the location of the center of gravity.
 
That makes sense. I have this question though? How do I solve for how much torque the beam provides if I don't know what position it is in? And how do I solve for position if I don't know the torque? It looks like chicken and egg syndrome to me, though I know there has to be a way to solve it...
 
I know (x1m1 + x2m2) / m1+m2 = Center of Gravity
And I know m1 = 14, m2 = 40, and x1+x2=3, but that leaves me with:

14x1 + 40 (3-x1) / 54 = Center of Gravity
There are two unknowns: x1 and Center of Gravity, and each is used to solve for the other. How do I solve?