Paradox of balance scales

[mistermack]

 

Neutral scales would be useless - they would just keep tipping towards the heavier side until they were vertical. With no weight, they would remain in whatever position they had been left in.

It's hardly likely that scales would be designed without a platform just below the pans. So that's not really a practical problem.

In the practical world, I would think a small amount of self-levelling would be tolerable without losing too much sensitivity.

[swansont]

Can you really refer to the centre of mass of an apparatus with freely moving links? It has an instantaneous centre of mass, but doesn't necessarily have the same CoM as things move.

In this case, with added equal weights on either side, the two pans and weights will cancel each other out in the same way as empty pans.

The restoring force will be purely provided by the moving CoM of the rigid beam, as before. The swinging pans, with any weights you care to choose, will always cancel each other out.

 

 

Because of the CoM being offset in the horizontal direction, not the vertical. Adding masses to the pans doesn't shift the CoM in the horizontal direction, only the vertical, and does not supply a restoring torque unless one is already present. Which is not there if the CoM is centered in the horizontal direction.

[Tim88]

 

It is useless. Which ever side had the most weight would rotate downwards until it reached it's limit (vertically down if the mechanism would allow it). equalising the weights would result in no movement whatsoever. (whatever position the beam was in).

 

It simply works; the disadvantage is that one has to add weights up to the precision that one desires, and without an indication of how much one has to add each time. But I did experience on the market once that a sales person used an ordinary balance much the same as such a primitive one: she just added fruit until the scale tipped.

[Janus]

Yes, that's a variable I hadn't considered.

 

In practice though, would you design a self-levelling setup anyway? Surely you lose some sensitivity or accuracy, if the scales are self-levelling.

And they would be frustrating to use if they were inherently unstable.

 

I would want scales that are neutral, so that the slightest inequality results in clear unrestricted movement towards the heavier side.

Maybe a small amount of self-levelling would be tolerable, if there was a visual aid like a pointer to clearly show when the scales are balanced.

The trick is to make sure that the scale is balanced without weights. Once it is properly calibrated, then it will only perfectly balance if the weight on the pans are equal. Something like this:

 

The hanging arrow is affixed to the pivot point. There is a weight on a threaded rod which can be adjusted from one side to the other. First you adjust the weight so that the arrow points straight down with empty pans, you have now "zeroed" scale and it is ready for use.

[mistermack]

 

 

Because of the CoM being offset in the horizontal direction, not the vertical. Adding masses to the pans doesn't shift the CoM in the horizontal direction, only the vertical, and does not supply a restoring torque unless one is already present. Which is not there if the CoM is centered in the horizontal direction.

Agreed. That's basically what I said.

[Janus]

Agreed. That's basically what I said.

Of course this only applies if the hanging points of the Pans are in line with the pivot point of the arm.

 

In the case where the hanging points are lower than the pivot, you get the following

 

 

The red lines mark the center lines of the weights. If you try to tilt the scale, the right weight weight goes down and to the left and the left weight goes up and to the right. However, the Left weight/pan does not move as far right as the Right weight/pan moves to the left.

There is a net movement of the CoG's of the weights to the left. If the weights were equal and the scale balanced before tilting, this will supply the torque that acts to return the scale to the balance point.

[swansont]

Agreed. That's basically what I said.

 

 

No, it's not.

[mistermack]

 

 

No, it's not.

Tis

In the case where the hanging points are lower than the pivot, you get the following

Yes, I could see that from your previous post.

If you mentally picture an extreme case, of the arms sloping at 45 deg, it's obvious that the arm that dips is moving closer towards the centre line, and the arm that rises is moving away, so one loses torque and the other gains, giving the resultant return force.

[TakenItSeriously]

The Balance Paradox.

Axiomatic Logical Model explained.

The restoring force is demonstrated by the two lever forces that must always act perpendicularly to the lever and equal to each other.

 

On the left side the gravitational force is the larger force that is composed of the Lever Force component and ΔF component.

 

On the right side the Lever Force is the combination of the gravity force and ΔF

 

Pushing on one side causes the lever force to rotate with the cross bar while keeping the lever forces equal at all times. After releasing the balance then the lever forces must be must still be equal but without the delta comopnents meaning the offsets must drop to 0.

Edited July 20, 2017 by TakenItSeriously

[J.C.MacSwell]

 

 

No, it's not.

All Janus's post were pretty clear. Much clearer than anyone else, myself included, but I don't see anything mistermack wrote as contradictory. Maybe I'm missing something, but he pretty much implied the same IMO, if you follow his thinking and context of what he meant.

Edited July 21, 2017 by J.C.MacSwell

[swansont]

All Janus's post were pretty clear. Much clearer than anyone else, myself included, but I don't see anything mistermack wrote as contradictory. Maybe I'm missing something, but he pretty much implied the same IMO, if you follow his thinking and context of what he meant.

 

 

The CoM moving vertically does not provide a restoring torque. Period. Saying that it does because it also displaces the CoM laterally misrepresents the physics, and I gave an example where that doesn't happen.

[J.C.MacSwell]

 

 

The CoM moving vertically does not provide a restoring torque. Period. Saying that it does because it also displaces the CoM laterally misrepresents the physics, and I gave an example where that doesn't happen.

Sorry. I must have missed where that was claimed.

 

The vertical displacement of mass provides the energy. Restoration does not happen without it.

 

Was adding mass to change the CoM your example? Or was it the "if you could engineer a device" example...

Edited July 21, 2017 by J.C.MacSwell

[swansont]

Sorry. I must have missed where that was claimed.

 

The vertical displacement of mass provides the energy. Restoration does not happen without it.

 

Was adding mass to change the CoM your example? Or was it the "if you could engineer a device" example...

 

 

Yes. There's no restoring torque before or after. So moving the CoM is not the impetus.

[TakenItSeriously]

We should consider beam scales with very long levers and with massive pans on asteroid. Can there a torque be less than distinction of gravitational attractions as a condition of non-return?

Interesting question, although I don't know what you mean by distinction.

 

If you look at the simple form of the solution that I posted then it ahould be apparrent that forces scale in the near field of gravity.

 

if you are talking about scales that go beyond the scale of our solar system for example then I'd be afraid of sounding too stupid to venture a guess.

 

Or if Iwas too stupid to not venture a crazy guess, for example, I would try to think in terms of a circuit board to try and consider how the near field and the far field forces intersect. in that environment.

 

... Edit to add

 

So apparantly there are still real world problems in science that havent been solved yet or was that the last one?

 

Is anyone aware of any other real world paradoxes that haven't been solved yet? I'm only speaking of physical problems in common human experience.

[Mike-from-the-Bronx]

 

 

The CoM moving vertically does not provide a restoring torque. Period. Saying that it does because it also displaces the CoM laterally misrepresents the physics, and I gave an example where that doesn't happen.

I accidentally voted this post up when I wanted to vote J.C.MacSwell up.

 

swansont, you have displayed a lot of understanding in a lot of posts, but in this one you are dead wrong. J.C.MacSwell is dead right. Try re-posting this discussion on a classical physics or mechanical engineering forum and you will quickly be told that.

[swansont]

I accidentally voted this post up when I wanted to vote J.C.MacSwell up.

 

swansont, you have displayed a lot of understanding in a lot of posts, but in this one you are dead wrong. J.C.MacSwell is dead right. Try re-posting this discussion on a classical physics or mechanical engineering forum and you will quickly be told that.

 

 

Show me the torque

[TakenItSeriously]

Show me the torque

I already posted it with my solution to the paradox but here it is again.

 

You need to show the gravitational forces in terms of their forces perpendicular to each lever arm.

so if you say

F₁ > F₂

 

then there must be a right triangle where we see a lever component that is perpendicular to the lever.

We also see a ΔF component which is perpendicular to the lever force. which creates the three sides of the triangle.

 

offset due to ΔF > 0

 

On the right side the right angle is between the smaller gravitational force and ΔF

 

so the ΔF is representational of the offset.

 

Note that ΔF is not the same on both sides of the level but its very close.

 

As the rotation of the system rotation gets closer to 90° then it is more significant and you can calculate it using the lorentz transformation.

Edited July 21, 2017 by TakenItSeriously

[Mike-from-the-Bronx]

 

 

Show me the torque

 

 

Fair enough. I can't seem to post images here so I will try to post a link to an MS Word document with images. Hope this works.

 

http://www.relativitysimulation.com/Documents/balancebeam3.docx

[TakenItSeriously]

Fair enough. I can't seem to post images here so I will try to post a link to an MS Word document with images. Hope this works.

 

http://www.relativitysimulation.com/Documents/balancebeam3.docx

I can't tell what your trying to represent, did you leave out some vectors that quantitatively cancel out? Edited July 21, 2017 by TakenItSeriously

[Mike-from-the-Bronx]

I can't tell what your trying to represent, did you leave out some vectors that quantitatively cancel out?

 

Draw a free body diagram of the balance beam.

Place a mark at the Center of Mass.

Place another mark at the Pivot Point.

Draw a vertical arrow through the center of mass. That’s the “f” in the formula, the force of gravity.

Draw another arrow from the pivot point to the center of mass. That’s the displacement (the “r” in the formula)

Torque is the cross product of those 2 arrows (vectors).

 

That’s all there is to it. (I have done so many of these kinds of analyses it is instinctive. But to someone who only saw this kind of thing in an introductory physics class I imagine it could get confusing.)

Edited July 21, 2017 by Mike-from-the-Bronx

[Janus]

I already posted it with my solution to the paradox but here it is again.

IMG_0213.PNG

 

You need to show the gravitational forces in terms of their forces perpendicular to each lever arm.

so if you say

F₁ > F₂

 

then there must be a right triangle where we see a lever component that is perpendicular to the lever.

We also see a ΔF component which is perpendicular to the lever force. which creates the three sides of the triangle.

 

offset due to ΔF > 0

 

On the right side the right angle is between the smaller gravitational force and ΔF

 

so the ΔF is representational of the offset.

 

Note that ΔF is not the same on both sides of the level but its very close.

 

As the rotation of the system rotation gets closer to 90° then it is more significant and you can calculate it using the lorentz transformation.

You have a problem with your image On the left you properly show F1 as being the vector sun of delta F and FL, but on the right, you show FL as being the vector sum of F2 and Delta F, which is incorrect, F2 is the vector sum of the respective forces of delta F and FL on that side.

 

The vectors, (here drawn as if the right weight was twice that of the left) would look like this:

 

Here Fg2>Fg1 and respectively F2>F1 and Fl2>Fl1 by the same factor.

 

F1 and F2 both act along the length of the beam and through the pivot and apply no torque to the beam. Fl1 and Fl2 apply torque to the beam with the net torque being in the direction of F2. In this situation the right weight will simply settle at the lowest point allowed by the scale. (If the cross beam had no restriction on its rotation limit, the final resting state would be with the beam perfectly vertical)

 

This is different than the case where the pivot point is above the points where the hanging pans exert their forces on the beam. Now F1 and F2 do not act through the pivot point, but through a point where, combined, they exert an torque on the beam. This torque will be opposed to the net torque exerted by F1 and F2, and as the beam rotates clockwise, the torque grows while the Net F1 and F2 torque diminishes. At some angle these torques equalize, and if it is within the limits of the scale's motion, this is where the scale will settle.(Even if the beam has full rotational freedom, this point will be somewhere before vertical.)

[Mike-from-the-Bronx]

Oh boy. I just realized something embarrassing. When analyzing the contributions of the pans it is important to define the hanging point of the pans to be at the same elevation as the pivot point. Otherwise there will be non-offsetting torques contributed by the pans. I just added that caveat in my document.

[TakenItSeriously]

Draw a free body diagram of the balance beam.

Place a mark at the Center of Mass.

Place another mark at the Pivot Point.

Draw a vertical arrow through the center of mass. Thats the f in the formula, the force of gravity.

Draw another arrow from the pivot point to the center of mass. Thats the displacement (the r in the formula)

Torque is the cross product of those 2 arrows (vectors).

 

Thats all there is to it. (I have done so many of these kinds of analyses it is instinctive. But to someone who only saw this kind of thing in an introductory physics class I imagine it could get confusing.)

 

I can see how that works for a static system but for a dynamic system, saying where the center of mass will be before reaching equilibrium seems circular.

 

Using my solution, you can explicitely show why the torque forces exist.

 

I guess I'm saying that because the torque forces do exist your solution should always be right but it doesn't necessarily explain the mechanism of why the torque forces exist. Its just confirming results that we already knew. That the balances returns to equilibrium.

You have a problem with your image On the left you properly show F1 as being the vector sun of delta F and FL, but on the right, you show FL as being the vector sum of F2 and Delta F, which is incorrect, F2 is the vector sum of the respective forces of delta F and FL on that side.The vectors, (here drawn as if the right weight was twice that of the left) would look like this:scales.pngHere Fg2>Fg1 and respectively F2>F1 and Fl2>Fl1 by the same factor.F1 and F2 both act along the length of the beam and through the pivot and apply no torque to the beam. Fl1 and Fl2 apply torque to the beam with the net torque being in the direction of F2. In this situation the right weight will simply settle at the lowest point allowed by the scale. (If the cross beam had no restriction on its rotation limit, the final resting state would be with the beam perfectly vertical)This is different than the case where the pivot point is above the points where the hanging pans exert their forces on the beam. Now F1 and F2 do not act through the pivot point, but through a point where, combined, they exert an torque on the beam. This torque will be opposed to the net torque exerted by F1 and F2, and as the beam rotates clockwise, the torque grows while the Net F1 and F2 torque diminishes. At some angle these torques equalize, and if it is within the limits of the scale's motion, this is where the scale will settle.(Even if the beam has full rotational freedom, this point will be somewhere before vertical.)

Consistently defining the lever and gravitational forces as far as which were components was my initial assumption as well but it never allowed for consistent results.

 

You can think of it this way.

F1 > F2

while

F1L = F2L = FL

this implies that FL is inbetween the two real forces

FL < F1

but

FL > F2

 

so the larger of the forces is what must be the hypoteneuse

 

It's asymetrical, but so is an unbalanced result.

 

edited out joke after I realized where I was posting this, LOL, no offense if you saw it.

 

edit to add:

after seeing your model, I noticed that you did not make the lever forces equal. i.e. no equilibrium.

 

That would imply the balance arrm is spinning and never stops spinning.

Edited July 21, 2017 by TakenItSeriously

[swansont]

I already posted it with my solution to the paradox but here it is again.

IMG_0213.PNG

You need to show the gravitational forces in terms of their forces perpendicular to each lever arm.

so if you say

F₁ > F₂

then there must be a right triangle where we see a lever component that is perpendicular to the lever.

We also see a ΔF component which is perpendicular to the lever force. which creates the three sides of the triangle.

offset due to ΔF > 0

On the right side the right angle is between the smaller gravitational force and ΔF

so the ΔF is representational of the offset.

Note that ΔF is not the same on both sides of the level but its very close.

As the rotation of the system rotation gets closer to 90° then it is more significant and you can calculate it using the lorentz transformation.

The masses are equal, and why is there a force not pointed straight down?

Fair enough. I can't seem to post images here so I will try to post a link to an MS Word document with images. Hope this works.

 

http://www.relativitysimulation.com/Documents/balancebeam3.docx

The pivot point is at the center of mass of the beam in my example.

[TakenItSeriously]

The masses are equal, and why is there a force not pointed straight down?

 

The real force is always pointed straight down. However the force that acts on the balance is the lever force which is always perpendicular to the cross bar. if the cross bar is crooked or its not perpendicular to the gravity force then the net force and the gravity force are seperated by a delta force

 

When the masses are equal then the cross bar is already horizontal and there is no need for a returning force.

 

I was showing what happens the instant after the unequal masses become equal when the cross bar is not horizontal. then the delta force is the returning force so as the cross bar approaches the horrizontal the delta force approaches zero. when the delta force is zero, then the gravity force is equal to the lever force and they both point straight down but that only happens when the bar is horrizontal.

Edited July 21, 2017 by TakenItSeriously