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

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The discussion focuses on solving for the distance, x, from the left end of a beam to the point where a rope is attached, given a beam mass (mb) of 10.0 kg, an attached mass (m2) of 40.0 kg, and an unknown mass (m1) of 15 kg. The beam is in static equilibrium with a tension (T) of 637 N. Participants emphasize the importance of calculating torques around the left end of the beam, noting that the beam's mass distribution is uniform and that the torque from the beam should be considered at its center, 1.5 meters from the left end.

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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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