# Why does a balance balance?

1. Jan 15, 2006

### Chaos Pariah

I've been trying to wrap my mind around how a balance works, and I come up with a question I need help with...
If you have a traditional balance scale, and you put perfectly equal weights on both sides, of course the scale balances the two out, and the two objects being weighed end-up at equal distances from the ground, like a teeter-totter.
Why does the balance equal out, instead of merely remaining wherever one puts it?
It seems to me that if the weights are equal and the distance from the fulcrum is equal, that if I were to exert force to raise one side of the balance (lowering the other), the weights would remain wherever I adjusted them, instead of moving back to the point where the lever is once more parallel. What force causes the movement back to "center"?

2. Jan 15, 2006

### fargoth

lets assume the balance is not at equilibtium because you pushed one weight down, the rod connecting the two weights has the degree $$\Theta$$ from the horizon.
no, mg acts downward, meaning that between the rod and the force for the upper weight you got $$90-\Theta$$ and for the lower weight its $$90+\Theta$$.
now the torque on the rod is the product of r cross F ($$r x F$$) so its size is $$N=r*mg*sin(90+\Theta)-r*mg*sin(90-\Theta)$$ when r is the distance of weights from the rotation axis.

EDIT: sorry i just saw this one give N=0... im going to sleep right now, i'll give you the right answer tomorrow incase no one else comes first.

Last edited: Jan 15, 2006
3. Jan 15, 2006

### Dmstifik8ion

For the same reason it balances empty (when you calibrate it). The stability is due to the center of gravity being below the pivot point. This is why a pendulum seeks a straight downward position.

4. Jan 16, 2006

### fargoth

yeah, ok, i found the problem, if you got a rod with two equal masses on it, and the pivot point is on the center of mass, as in the exercise i tried to solve, the rod will be balanced for every angle.
if the center of mass is below the pivot point, it will indid return to $$\Theta=0$$ for equilibrium.
because the torque is $$2mgaSin(\Theta)$$ when "a" is the distance of the center of mass from the pivot point.
if you want instructions on how i got this solution id be happy to write it for you.

5. Jan 17, 2006

### Chaos Pariah

Thanks!

Thanks, everyone, for your help in explaining this to me. Actually, when I first found this forum, I searched for a thread that dealt with this question and came up empty - I guess I need to refine my skills in composing such searches. It turns out that immediately after posting my question, I was curious as to what others were asking about. I scrolled down the list of threads and found that someone had asked the identical question and had already received several very good answers. In any case, I now understand why the balance works. Thank you all very much for your efforts toward my continued education!