# Why do simple balances always come to rest in a horizontal position ?

by mahela007
Tags: balances, horizontal, position, rest, simple
 P: 106 Why is it that simple balances always come to rest in a horizontal position when equal weights are placed on both ends? If the weights are equal, then (assuming that the distance from each weight to the pivot is the same) the torques produced by the weights are equal and opposite and the net torque about the pivot is zero. Since torque is the product of the force and the perpendicular distance from the pivot to the force, shouldn't the balance be able to come to rest in ANY position? Thanks.
 Mentor P: 21,662 The balance is constructed with the pivot point above the arms.
 P: 106 Could you please explain how that works?
P: 1,395

## Why do simple balances always come to rest in a horizontal position ?

 P: 969 I've read all the answers here and in the link, and only russ watters' solution (that the pivot is above the arms, assuming a finite thickness of the arms) and abdul quadeer's (a counterweight below the pivot) makes sense to me. One of the solutions in the link says the pivot can be below the arms. I'm not quite sure I understand that. If that is the case, then wouldn't the stable point be as far away from the horizontal position as possible, until the balance goes upside down and becomes stable (russ's solution).
 P: 969 I found a website that talks about this: http://metrology.burtini.ca/mechanical.html in the last 3 paragraphs of the page. To be honest it's looks quite complicated. Why did they draw the scale in such a funny shape, like a pair of glasses? It looks like the fulcrum point is upside down in all examples, as russ says. But why draw the weights at the two ends as rightside up pivots? They aren't really pivots are they? Anyways, what I found fascinating is the 3rd example, that you can have an upside down pivot but still have most of the beam's weight above the upside down pivot, making the beam again unstable. But somehow this is useful because if your weights are too heavy, the beam deflects downwards until most of the beam is below the upside down pivot, and then the beam becomes stable again. But isn't this silly: why not just have all the beam's weight below the upside down pivot to begin with?
 P: 8 (click for a larger version) This is supposed to be a balance, rotating around point A with loads at point B and point C. For a balance such as this to be at rest, we must have equilibrium for forces in the x-direction, the y-direction (both implicitly defined in the image by the forces Pax and PPay) as well as the torque around any point. We are interested in the case Pb = Pc = P x: Pax = 0 y: Pay = 2P rot around A (assume alpha is small): P(L2cos α + L1sin α) = P(L2cos α - L1sin α) => PL1sin α = 0 => P = 0 v L1 = 0 v α = 0 So, either we have no load, we have no extension L1 or the balance is in equilibrium only in the horizontal position. EDIT: For the link RedX gave in his last point, I'm assuming they drew the balance like that because it is one type of implementation. See for instance this image. Also, the triangles indicate the respective points and are above or below the line depending on in which direction a load should pull to make the balance work correctly. So in the 1st class you would have both the power and the load pulling downwards, while in the 2nd class your power would have to pull upwards.
 P: 460 I had posted the same question. I am posting its link. http://www.physicsforums.com/showthread.php?t=473638 Hope it helps.
P: 969
 Quote by Malle (click for a larger version) This is supposed to be a balance, rotating around point A with loads at point B and point C. For a balance such as this to be at rest, we must have equilibrium for forces in the x-direction, the y-direction (both implicitly defined in the image by the forces Pax and PPay) as well as the torque around any point. We are interested in the case Pb = Pc = P x: Pax = 0 y: Pay = 2P rot around A (assume alpha is small): P(L2cos α + L1sin α) = P(L2cos α - L1sin α) => PL1sin α = 0 => P = 0 v L1 = 0 v α = 0 So, either we have no load, we have no extension L1 or the balance is in equilibrium only in the horizontal position.
That'll certainly work (great picture btw). If the scale were shaped like a T, and you pin the bottom of the stem of the T, then as you tilt the T to the right, the right arm of the T gets lower, but is exactly compensated by the left arm of the T going higher. However, the whole T is going lower because the stem is getting lower.

Another way to look at it is imagine the point (0,1) on a unit circle and it's tangent line. As you tilt the (0,1) vector towards the right to the point (1/sqrt(2),1/sqrt(2)), then it's true that there is just as much of the tangent line above the horizontal as there is below. But the horizontal is now at y=1/sqrt(2) instead of y=1. So the potential energy decreases when you tilt the T to the right or left, so this T is unstable.

So what you need to do is have an upside down T-shape.

Although in the picture of the link you gave, the scale seems to use a counterweight that's used to measure the difference in weight of the objects on the scale, instead of having an upside down pivot (but it doesn't matter since the weights hang below the pivot).

 Related Discussions Introductory Physics Homework 1 General Physics 7 Introductory Physics Homework 1 Introductory Physics Homework 2 Classical Physics 1