Solve Mass and Dist. of Bar Weights: x-m

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A beam of mass 10.0 kg is in static equilibrium, suspended by a rope with a tension of 637 N, while a 40.0 kg mass is attached at one end and an unknown mass m1 is at the other. The beam's length is 3 m, and the torque must be calculated about the left end. The correct distance x from the left end to the rope attachment point is critical for solving the problem, with the center of mass of the beam assumed to be at 1.5 m. Calculations should account for the uniform distribution of mass and the sum of torques equaling zero. Accurate calculations and clear presentation of work are essential for resolving the issue.
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A beam of mass mb = 10.0 kg, is suspended from the ceiling by a single rope. It has a mass of m2 = 40.0 kg attached at one end and an unknown mass m1 attached at the other. The beam has a length of L = 3 m, it is in static equilibrium, and it is horizontal, as shown in the figure above. The tension in the rope is T = 637 N.

a) Determine the distance, x, from the left end of the beam to the point where the rope is attached. Note: take the torque about the left end of the beam.

I know that M1 is 15 kg. I keep getting 4 m from the point of rotation but that's wrong. here is the help they told me, but i just don't think it makes sense at all:

HELP: We know the sum of the torques will be zero. Since we are taking the torque about the left end of the beam, there are only three torques we will need to deal with.

HELP: We can assume the beam is uniform in distribution of mass and therefore take the torque of the beam at the center, 1.5 meters. Remember that the torque is the force time the perpendicular distance, i.e. be careful with the distances you choose. The torque is zero about every point, and therefore you can take the torque about another point as a check.
 
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It would be helpful if you showed your calculations. We have no way of knowing what you did otherwise.
 
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