# Why Do Balance Scales Stop Even If Unbalanced

## Homework Statement

Quote from the question source:
In times past or in other countries, people purchase fresh produce from piles. The vendor puts a mass on one side of the scale - same 1 kg - and some produce on the other side to weigh out 1 kg of produce for the customer. When there is a little less than 1 kg of produce, the scale tilts downward on the 1 kg but stops at some point rather than continuing to rotate. If the weights/torques are not balanced, why doesn't the scale continue to rotate? Why do you want the scale fixed directly parallel to the ground or perpendicular to the force of gravity to know you have just the right weight? Why not just have the scale stationary - no longer rotating? Do NOT consider friction because it is not relevant to the answer.

## Homework Equations

None given, but maybe:
Torque = lever arm x force

## The Attempt at a Solution

I drew a diagram which you can see here: http://i.minus.com/ibzw9bF4vAnaJ.png (The weight differences are probably very exaggerated.)

But I am not sure of the answer: I thought that the scale stopped because the lever arm decreases as the scale becomes more vertical, but that happens to both sides so both torques decrease at the same rate (right?).

So why do scales stop when slightly unbalanced? It seems like friction is not the answer, because sometimes the scales tip backwards and then forwards until it stops.

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Related Introductory Physics Homework Help News on Phys.org
Since the weights are unchangeable, what must be done to change the torques?

Since the weights are unchangeable, what must be done to change the torques?
I thought that somehow the force must decrease on the heavier side a bit, but I'm not sure how. Is some of the force of gravity of the heavier object being supported by the bar (which is in turn supported by the fulcrum)? If so, how would I diagram/calculate that?

I know the lever arm decreases as the bar becomes more vertical, but that decreases the torque on both sides equally (I think).

gneill
Mentor
On real scales the pivot point is generally above the points of attachment of the weight pans on the lever arms. Think of a coat hanger with the pivot at the hook and the pans suspended from the ends of either arm.

On real scales the pivot point is generally above the points of attachment of the weight pans on the lever arms. Think of a coat hanger with the pivot at the hook and the pans suspended from the ends of either arm.
If you had a scale like this one (click), how would it work? The points of attachment seem to be level with the pivot. My physic textbook says that even if the objects are hanging from the pivot bar, it doesn't change the torque (it illustrated that with a seesaw hanging someone over a cliff, but the idea should still apply for scales).

rcgldr
Homework Helper
If you had a scale like this one (click), how would it work?
It would only work if the weight on each scale was the same. As gneill mentioned, by having the pivot point at the above the center of the balance bar, then the center of mass of the system is raised when the scale is angled from horizontal, creating an torque to oppose the imbalance in weight.

There 2 types of equilibrium, stable and unstable equilibrium
In the case of scale it is a stable equilibrium .

In both of your questions the answer lies on the position of the centre of mass,CM.
The scale will return to a position where the CM is directly under the pivot.

If both of equal mass, then it is perpendicularly under the pivot.
If the left is heavier than the right mass, then the CM is bit left to the centre thus it will turn anticlockwise such that finally it is below the pivot(resulting in no more torque, since the force is now in line with the axis of rotation)

Anyway we need friction to dampen the oscillation.

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