Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm talking about a Weighing Balance shown in the figure:

enter image description here

Press & Hold on onside of the horizontal beam and then release it. It makes some oscillations and comes back to equilibrium like shown in the figure.

Both the pans are of equal equal masses. When the horizontal beam is tilted by an angle using external force, the torque due to these pan weights are equal in magnitude & opposite in direction. Then why does it come back to it position? What's making it to come back?

share|cite|improve this question
I'm not gonna write up an answer, since there seem to already be a couple of correct ones, but I'll do my best to put it more succinctly. If the center of mass is above the rotation point, horizontal is an unstable equilibrium point. This is how you build a teeter-totter (or seesaw). If the center of mass is below the rotation point, horizontal is a stable equilibrium point. That's how you build a mass balance. For small mass imbalances, the deviation from horizontal is proportional to the imbalance. – Anonymous Coward Aug 16 '11 at 1:21
up vote 7 down vote accepted

The horizantal beam on such scales is intentionally placed below the rotational axis. As long as the weights are in equilibrium the torque is equal on both sides.

But as soon as the position changes e.g. tipping the left scale down, the torques differ because only the tangential part of the gravitational force vector in relation to the rotational axis contributes to the torque around it. When tipping down the left scale, torque on the left side gets smaller and torque on the right side gets bigger, therefore the right side moves down again until equilibrium is reached (besides some swings to accommodate for the temporary impulse energy).

This effect gets the more pronounced as the distance of the horizontal bar approaches the half length of the bar.

This effect would not be if the horizontal bar went exactly through the axis.

share|cite|improve this answer
this is the best answer by far people... it took me a while to get it but i believe this is correct. – Timtam Aug 14 '11 at 2:54
This answer is extremely confusing. I think it's also wrong, but it's so confusing I can't say for sure. – Mark Eichenlaub Aug 14 '11 at 3:49
I can confirm that you are confused... however the above answer is correct. Maybe you need to look at the geometry a little more? I can help you if you need it, but I'm surprised you don't get it as it's pretty simple. – Timtam Aug 14 '11 at 5:21
I wanted to add a drawing but it was 3 am local time and I was tired. Maybe I'm gonna add one today :) – oleschri Aug 14 '11 at 12:14
does it boil down to a pendulum with a complex (bar-shaped) weight? – IljaBek Aug 14 '11 at 18:40

If it would only be the weights exerting torque, the balance would be in equilibrium at all angles. What makes the balance go back to the horizontal position is the fact, that the center of mass is below the beam. consider this picture

enter image description here

The needle exerts a torque too, so you have more torque on the side, where the plate is higher. You can have more subtle configurations (like in your picture, where the beam is rounded below) but the mechanism is the same.

share|cite|improve this answer

It behaves this way because that's how it was built. By adjusting the mass distribution, we could make a scale that flops to one side, is roughly balanced at all angles, etc. However, those scales would not be useful, so the scale isn't built that way.

It might be assumed from the left/right symmetry of the picture that the system cannot decide which way to go, and so is at an equilibrium point. This equilibrium will be stable if a small perturbation (rotating the beam a small angle) raises the center of mass. It will be unstable if a small perturbation lowers the center of mass.

Beyond that, it is difficult to say how the center of mass moves simply by looking at your picture because we do not completely understand the mass distribution and the location of the pivot point.

When finding the center of mass, we can ignore any stationary pieces because we are only interested in the change of the height of the center of mass. Additionally, if the pans hang freely down, it appears as if one will rise by the same amount the other falls, and thus they will not change the height of their center of mass when considered jointly. They can also be ignored.

Let's assume the rest of the scale rotates rigidly. In that case, the center of mass of the rigid portion we're considering will be constrained to a circle with its center at the pivot point. If the center of mass is exactly at the bottom of the circle, we have a stable equilibrium. Otherwise, it is unstable.

share|cite|improve this answer
right, or you can just look at this problem generally from a torque perspective... which is the point which you seem to be missing – Timtam Aug 14 '11 at 5:13
@Tim I have actually heard of torque before, and understand arguments based on torque. As it turns out, it is possible to explain the same phenomenon different ways, and those different explanations may all be helpful. Also, in my other comment I was referencing the unclear writing in the other answer, not the fact that I am incapable of understanding freshman physics. Thank you for your extremely helpful and lucid contribution to this discussion. – Mark Eichenlaub Aug 19 '11 at 0:35

Nice question! if the following analogy applies : imagine a seesaw on a half-sphere fulcrum (top of the picture). if it inclines e.g. to left side (bottom) - the length from the right edge to fulcrum ($L_2$) increases, the lever rule kicks in ($F_2>F_1$) and the weight of the right side brings the seesaw back to equilibrium (top) (which is then broken again by inertia)

seesaw oscillation

share|cite|improve this answer
Sorry, I didn't get you. – claws Aug 13 '11 at 12:14
If i've got this right then: You mention in your question the pans of equal mass, but you also need to take into account the mass of the beam, the CoM of which in the above diagram shifts, leading to an imbalance and a restorig force... – Nic Aug 13 '11 at 14:06
this seems correct, why the downvote? – Timtam Aug 14 '11 at 2:50
This seems rather different from the system in the question, but I do understand what's being communicated and it's somewhat insightful. Reminds me of boat stability. Oh, and it's possible to build this in an unstable configuration as well if the fulcrum has a sharper curvature. – Alan Rominger Aug 15 '11 at 4:18

it seems to me that when I do the calculations for the torques considering the pivot as a single point in space they are equal. However I must assume that real scales are built such that any rotation through an angle $\theta$ will cause the pivot point to shift slightly from center causing unequal torques. The effect of this net torque is to restore the beam to the horizontal no matter which direction in theta you rotate (e.g. whether you push down on the left or right piece, both case result in a restoring torque). I would also add that the oscillation die down due to frictional damping, the whole system is approximated by simple harmonic motion with damping.

share|cite|improve this answer

Actually with the lever sticking up above the balance scale shown; balance axis point; it adds gravitational weight to whichever side is initially positioned downward, and also since this side is down, it should weigh more being closer to the ground... and should tend to keep this side down... However is doesn't...

If you put a weight, on some point of a wheel... to make it balance; you put an identical weight at an identical point equal and opposite... Spinning the wheel at high rpm will tell you quickly, if you did right. Then if idle, it should sit still, in any upright position...

If two identical boats were tied together with a rope and were sailing parallel, in a parallel steady wind, in friction-less water and etc... With one boat ahead, and off to the side; they... would maintain their positions and the friction-less rope is meaningless.

I don't really know the answer... but you might align the scale north & south; then again east and west; to see if there is a difference due to Earth's centrifugal effect? It seems to keep the oceans fairly balanced, (disregarding the moon.)

If it is true that matter has "matter waves"; they may be equaling out repelling against the ground... Whereas with the large heavy wheel above mentioned, this effect would probably be negated...?

If this experiment were laid upon a flat horizontal plane... and different forces were applied equally like the sail boats... I have an idea the results would be: the mechanism would hold still, wherever positioned; with similar circumstances.

share|cite|improve this answer
Please publish your answer as normal text, rather than as preformatted text, so it is more readable. – Joshua Sep 18 '14 at 17:33

protected by Qmechanic Sep 18 '14 at 17:49

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.