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

In summary: T is the origin, and the scale is going down the left side 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.
  • #1
mahela007
106
0
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.
 
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  • #2
The balance is constructed with the pivot point above the arms.
 
  • #3
Could you please explain how that works?
 
  • #5
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).
 
  • #6
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?
 
  • #7
2f03878.png

(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.
 
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  • #9
Malle said:
2f03878.png

(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).
 
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1. Why do simple balances always come to rest in a horizontal position?

Simple balances come to rest in a horizontal position because of the law of gravity. This law states that all objects with mass are attracted to each other, and the force of gravity causes them to move towards each other. In the case of a simple balance, the two sides of the balance are equal in weight, so they are equally attracted to the center of the Earth. This results in the balance settling in a horizontal position.

2. Is there a specific reason why simple balances are designed to come to rest in a horizontal position?

Yes, the design of simple balances is based on the principle of equilibrium. This means that when the two sides of the balance are equal, there is no net force acting on the balance, and it remains in a state of rest. A horizontal position ensures that the balance is in equilibrium, making it easier to accurately measure the weight of an object.

3. Do simple balances always come to rest in a horizontal position, regardless of the weight of the objects being weighed?

Yes, simple balances will always come to rest in a horizontal position, regardless of the weight of the objects being weighed. As mentioned earlier, the design of the balance ensures that it is in equilibrium, regardless of the weight of the objects. However, if the weight on one side of the balance is significantly heavier than the other, it may cause the balance to tilt slightly, but it will still come to rest in a horizontal position once the weight is removed.

4. Can simple balances come to rest in a position other than horizontal?

Technically, yes, simple balances can come to rest in a position other than horizontal. This can happen if there is an external force acting on the balance, such as wind or vibration. However, the balance is designed to come to rest in a horizontal position, and any external forces would need to be significant to disrupt this equilibrium.

5. Are there any factors that can affect the ability of a simple balance to come to rest in a horizontal position?

Yes, there are a few factors that can affect the ability of a simple balance to come to rest in a horizontal position. These include the levelness of the surface the balance is placed on, the accuracy of the balance's calibration, and any external forces acting on the balance, as mentioned earlier. It is important to ensure that the balance is placed on a level surface and regularly calibrated to ensure accurate measurements.

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