Calculating Mass of a Balanced Beam with Two People Using Torque Equations

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To determine the mass of a 4.5-meter beam balanced on a fulcrum, the torques created by a man weighing 886 N on one end and a girl weighing 393 N on the other end must be calculated. The torque from the man is 1807.44 N*m, while the girl's torque is 966.78 N*m. The beam's torque due to gravity is calculated using its center of mass, which is located at 2.25 meters from the left end. The correct equation for balance requires summing all torques about the fulcrum and setting them equal to zero, rather than assuming cancellation of equal distances on both sides. The calculated mass of the beam is found to be 204.242 kg, but further verification is needed to confirm its accuracy.
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Homework Statement


Determine the mass of a 4.5 meter beam, if it is balanced on a fulcrum 2.04 meters from the left end. A 886 N man stands on the very left end, and a 393 N girl stands on the right end. [/B]

Homework Equations


T = Fr

The Attempt at a Solution


The torque applied by the man would be 886 N * 2.04 m = 1807.44 N*m.
Since the length of the beam on the right side is 4.5-2.04 = 2.46 m, the torque applied by the girl is 393 N * 2.46 m = 966.78 N*m.
Since the fulcrum is slightly closer to the left, the torque applied by the beam is (2.46-2.04 m)*9.8*mass of the beam.
Since the torques must cancel, 1807.44 = 966.78 + 4.116*m
m = 204.242 kg Not the right answer though.
 
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EchoTheCat said:
(2.46-2.04 m)*9.8*mass of the beam.
Why is this the torque exerted by gravity on the beam? Where is the CM of the beam?
 
BvU this is what my sketch looks like.
 

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kuruman said:
Why is this the torque exerted by gravity on the beam? Where is the CM of the beam?
The torque from 2.04 meters on the left side and 2.04 meters on the right side would cancel out. Then, we are left with 0.42 meters on the right side. 0.42 meters*9.8m/s/s*mass of the beam, since T = Fr. The center of mass of the beam would be at 2.25 meters.
 
EchoTheCat said:
The center of mass of the beam would be at 2.25 meters.
Right. And the torque exerted by gravity is the weight of the plank times the distance from the CM of the plank to the fulcrum. This business of 2.04 m on the left cancelling 2.04 m to the right is nonsense. Just add the three torques about the fulcrum and set them equal to zero.
 
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