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This question was previously asked 11 years ago here, but I believe it does not answer my question for the following reasons:

(A) The first answer takes a different system, mine looks like this.

(B) The second answer says that tangential torques are different but doesn't explain why, which is my question. The tangential torques are both $mg\sinθr$, where $r$ is the distance of the weights from the pivot and $θ$ is the angle of deflection with the horizontal. The centre of mass is also vertically below the pivot as no force is applied in the horizontal direction.

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  • $\begingroup$ @ACB No, in that question the masses and the distances from the pivot are different, in mine they are the same. $\endgroup$
    – CallousCalculus
    Commented Feb 21, 2023 at 14:23
  • $\begingroup$ But it is balanced and the answers there have explained the same thing as here. $\endgroup$
    – ACB
    Commented Feb 21, 2023 at 14:31
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    $\begingroup$ As @ACB correctly points out, my answer on that other question applies even if the distances and the masses are the same. Just go through the same logic with $x = y$ and $M = m$. $\endgroup$
    – Michael Seifert
    Commented Feb 21, 2023 at 20:54

2 Answers 2

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balance beam diagram

Torque about a given axis equals force times distance of line of action of the force from the given axis.

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  • $\begingroup$ You use a system in which the pivot is above the beam, I have linked and am talking about a system in which the pivot is in the same line, but the masses are still hanging. $\endgroup$
    – CallousCalculus
    Commented Feb 21, 2023 at 11:44
  • $\begingroup$ @CallousCalculus If the pivot is on the line between the points from which the loads hang, then the setup does not act as a measuring balance. Such a setup would indeed be in equilibrium at any angle. $\endgroup$
    – Andrew Steane
    Commented Feb 21, 2023 at 12:09
  • $\begingroup$ ...But, In order to create a high-precision balance, you want the system to be as close as possible to neutral, while still being just a tiny-bit stable. The closer it is to neutral, the greater will be the deflection of the pointer for a given difference between the weight of the two pans. So for any actual measuring instrument, if it looks as if everything is "in the same line" it's probably because everything is very close to being "in the same line." $\endgroup$
    – Solomon Slow
    Commented Feb 21, 2023 at 17:32
  • $\begingroup$ @SolomonSlow Yes that's quite right. The other issues to think about are friction and the fact that you don't want a situation where the balance settles out-of-line even when the two weights are equal. $\endgroup$
    – Andrew Steane
    Commented Feb 21, 2023 at 21:01
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The centre of mass of the system, beam and scale pans, is below the pivot point and when the beam is displaced from the horizontal the centre of mass moves to one side of the pivot.
The system then moves towards the position of stable equilibrium under the influence of the restoring torque produced by the centre of mass.

For the type of balance shown in the video remove the scale pans.

enter image description here

You might find it difficult to get the beam to balance horizontally as the pivot is roughly coincident with the position of the centre of mass of the beam.

Now make a new hole in the beam "above" the central hole .
Try to get the beam to balance when the beam is pivoted at the new hole when the new hole is below the old hole, and then when the new hole is above the old hole.
In one position it will be a position of unstable equilibrium, new hole below old hole, and move towards the other position which is one of stable equilibrium, new hole above centre of mass.

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