Why Does a Balance Regain its Equilibrium Position?

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A balance regains its equilibrium position due to the torque created by the position of the pivot relative to the center of mass. If the pivot is at the center of the beam, the balance remains in any position it is placed. However, if the pivot is above or below the center, a restoring torque is generated, returning the balance to equilibrium. The discussion clarifies that in an ordinary balance, the attached mass at the pivot contributes to this restoring force. Understanding the mechanics of how the pivot and center of mass interact is essential for grasping why balances behave this way.
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If an ordinary balance with two equal weights is displaced slightly from its equilibrium position then it regains its original position. Why?

There is no unbalanced torque at any position. So why should it regain its position? I am totally confused on this one. Can anyone help...
 
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hi ashishsinghal! :smile:

if you're talking about a see-saw balance with the pivot in the centre of the beam, then you're correct …

the beam will stay in whatever position you put it

but if the pivot is above (or below) the centre of the beam, there will be a torque, and it will return to the level position :wink:
 
I thought of the same thing but if the weights are tied to the see-saw balance with a string, then ? because that is an ordinary balance
 

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To restate tiny-tim:

If the pivot is in the centre of the beam, no matter where you position it, it will remain where you leave it.

If the pivot is above or below the centre of the beam, it will return to somewhere around the equilibrium position.

Do you understand this point?

What do you mean tied with a string? As in you tie them to each end of the beam with string? Because that isn't an ordinary balance. You usually have two pans that masses can be put into.
 
I understood that point. Maybe the term "ordinary balance" confused me.
Thanks
 
ashishsinghal said:
I thought of the same thing but if the weights are tied to the see-saw balance with a string, then ? because that is an ordinary balance

it makes no difference …

each weight will still be vertically under the point that its string is attached to …

the torque is exactly the same as if the weight was at that point :wink:
 
yeah, that is what my point was, here the center of mass of beam is below the pivot point but it will still not regain its position.
 
There is usually a small mass attached to the pivot in the case of an ordinary balance. This provides the restoring torque. Refer the figure:

[PLAIN]http://img11.imageshack.us/img11/9684/baler.jpg

Notice that the centre of mass of the attached mass is below the pivot.
There might be other constructions too but this is what I am familiar with.
Hope that helps :smile:
 
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That attached mass is generally a needle pointer and would not have adequate mass for restoring.
 
  • #10
ashishsinghal, in a balance the pans hang over pivots, the central pivot is raised slightly above the centerline between the two hang points. The metrology laboritory where I work has one on display. A beautiful work of art, circa 1950 I suppose, made of brass, glass enclosed, and full of knobs to remotely move around weights from outside the enclosure. The enclosure prevents air currents from disturbing the reading. I can't find an image online that is remotely equivalent.

Abdul Quadeer is correct. A center of mass of the beam, below the center pivot would have a restoring force, but one that is constant rather than proportional to the load. I would guess you'd want the restoring force proportional the the load, as the pivot friction should be proportional to the load, as well.
 
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  • #11
Thanks, now I know the reason for both the kinds of balances - one by Phrak and Abdul Quadeer and the other by tiny-tim and jarednjames. Thanks to all of you.
 
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