Paradox of balance scales

[swansont]

I haven't read all of the posts in depth, but it seems to me that it simply to do with the location of the centre of gravity of the balance beam when unloaded.

If it's below the pivot, then the beam will naturally return to the horizontal, because any tilt on the beam causes the centre of gravity to rise.

So the beam is levelled by gravity acting through the c of g.

 

If the c of g was above the pivot, the beam would be unstable, and want to flip to whatever side it's tilted to on release.

The CoM has to be displaced to the side from the center line, which is accomplished by having the arms be angled or curved, as described by Tim88. You can have a balance whose CoM is below the pivot in equilibrium. Even the straight arm balance can have a CoM below the pivot, because the pans have mass.

[DimaMazin]

I haven't read all of the posts in depth, but it seems to me that it simply to do with the location of the centre of gravity of the balance beam when unloaded.

If it's below the pivot, then the beam will naturally return to the horizontal, because any tilt on the beam causes the centre of gravity to rise.

So the beam is levelled by gravity acting through the c of g.

 

If the c of g was above the pivot, the beam would be unstable, and want to flip to whatever side it's tilted to on release.

Levers and pans have many different g. We should take only g of support point.

[mistermack]

The CoM has to be displaced to the side from the center line, which is accomplished by having the arms be angled or curved, as described by Tim88. You can have a balance whose CoM is below the pivot in equilibrium. Even the straight arm balance can have a CoM below the pivot, because the pans have mass.

That's what I was saying. You can design the beam so that the centre of mass is below the pivot, and this will naturally return to the horizontal.

COM at the pivot, and the beam will be neutral, remaining where you put it.

COM above the pivot, and the beam is unstable.

 

Obviously, any manufacturer will design a balance with its COM below the pivot for convenience, so that it self-levels.

When the balance is tilted, the COM will rotate, rising from it's lowest position, so gravity will pull it back down when released.

[swansont]

That's what I was saying. You can design the beam so that the centre of mass is below the pivot, and this will naturally return to the horizontal.

COM at the pivot, and the beam will be neutral, remaining where you put it.

COM above the pivot, and the beam is unstable.

 

 

But it's not because the CoM rises; that explanation does not suffice. It's because it deviates to the side, so there is a torque on it. If you could engineer a device whose CoM remained centered but was below the pivot, it would not return to horizontal.

[mistermack]

 

 

But it's not because the CoM rises; that explanation does not suffice. It's because it deviates to the side, so there is a torque on it. If you could engineer a device whose CoM remained centered but was below the pivot, it would not return to horizontal.

I agree about the torque. But equally, if you could engineer a device whose CoM didn't rise, then it wouldn't return to the horizontal either. The torque, and rising CoM are inevitably going to be there in any simple beam, with CoM below the pivot.

The energy to return the beam to horizontal position gets put in when the CoM is raised.

[swansont]

I agree about the torque. But equally, if you could engineer a device whose CoM didn't rise, then it wouldn't return to the horizontal either. The torque, and rising CoM are inevitably going to be there in any simple beam, with CoM below the pivot.

The energy to return the beam to horizontal position gets put in when the CoM is raised.

It may be that it's not possible to have my scenario, but if the CoM is not off of the center line, there is no torque to get the balance to move.

 

It's not obvious to me that the CoM of a straight beam balance with pans below it is displaced from center, even though it's below the pivot, if the pivot is in the CoM of the bar.

[DimaMazin]

 

 

Here is one type that doesn't automatically return.

 

beam2.jpg

I have tried some experiments, but I can't get non-return like that. Are the beam scales in vacuum on surface of white dwarf ?

Is that due to thick rod?Part of mass of top lever is inclined to other side.

Edited July 18, 2017 by DimaMazin

[mistermack]

It's not obvious to me that the CoM of a straight beam balance with pans below it is displaced from center, even though it's below the pivot, if the pivot is in the CoM of the bar.

The CoM has to be displaced on a simple beam arrangement, if it's above or below the pivot.

If the pans are fixed then the beam and pans are a single object who's shape doesn't alter. The CoM will remain in the same place and will have to rotate with the beam.

If the pans are loose, then they just cancel each other out, so the centre of mass location will be unaffected, and the CoM of the beam will still rotate around the pivot, unless it's exactly AT the pivot point.

[TakenItSeriously]

Posting before reading some earlier replies but I wanted to post before getting some sleep.

 

I think I dierrived the equation.

 

So I started where I left off with the three variables on either side of the equation. where L canceled out so Δm ∝ Δy somehow with both being 0 at the horizontal.

 

Funny thing was I did have a flash back to this problem in a physics lecture at one point, probably physics 101 and the lechturer said something about this kind of problem. I asked him if there shouldnt be a y offset in the problem and not just a force and lever arm. He told me I was right then left it at that with out showing how it fit in the formula. It seemed strange but I guess I assumed it was in the text and I forgot all about it until now. That's why I remembered those three terms in a formula but couldnt remember the formula. lol.

 

I tried to remember the formula kind of derriving it in my head thinking it would come to me and it couldnt be that hard so I incorrectly assumed the sine and cosine functions were part of the formula. But I correctly assumed that the lever force was perpendicular to the lever so it had to be a component of gravity.

 

After I saw that it was never derrived, I changed my assumption that it couldnt be that easy after all, but still with three terms I figured I could find them in a diagram which I started here.

 

 

Before I finished diagraming I realized I was doing the light clock derrivation or at least some close variant i.e. similar right triangles Pythagorean theorem. I honestly don't know the derrivation by heart but I have it somewhere. I know it was in my intuitive model of SR post.

 

The gravity force has two components: the lever force which must be perpendicular to the lever arm and the y offset was shown as the second component through a similar triangles across the cross bar.

 

Any way if you look at the boundary conditions at 0° and 90° the lever force transitions to Y offset. it seems to be analogous to space transition to time.

 

I'm nearly certain that the balance equation is the lorentz transformation but I need to get some sleep before I take on that transformation which is kind of tricky math.

 

I'll finish it up tomorrow.

Edited July 18, 2017 by TakenItSeriously

[swansont]

The CoM has to be displaced on a simple beam arrangement, if it's above or below the pivot.

If the pans are fixed then the beam and pans are a single object who's shape doesn't alter. The CoM will remain in the same place and will have to rotate with the beam.

If the pans are loose, then they just cancel each other out, so the centre of mass location will be unaffected, and the CoM of the beam will still rotate around the pivot, unless it's exactly AT the pivot point.

 

 

How can the loose pans cancel? They are both below the pivot point.

[Manticore]

I would treat the loose pans simply as forces that act vertically downwards from their suspension points.

[TakenItSeriously]

Manticore has it right I think. Pans will end up hanging down from cross beam by something like a chain like the scales of justice as the stereotype I was going for, Also needed some clean lines to show vectors.

 

OP did meantion old style and I'm all about old school

 

Sorry, you were referencing another post, never mind.

Edited July 18, 2017 by TakenItSeriously

[J.C.MacSwell]

 

 

How can the loose pans cancel? They are both below the pivot point.

Because the torque from each just cancels the other. The horizontal distance from the pivot always matches, unlike pans fixed to the cross beam. This is assuming the pivot point is centred on the beam, and the pans hang from points equidistant from and aligned with the pivot point (at same height when beam is level).

Edited July 18, 2017 by J.C.MacSwell

[swansont]

Because the torque from each just cancels the other. The horizontal distance from the pivot always matches, unlike pans fixed to the cross beam.

 

 

We were talking about contributions to center-of-mass with regard to the vertical.

[mistermack]

 

 

We were talking about contributions to center-of-mass with regard to the vertical.

I think the same applies as the horizontal. Assuming that the sides are symmetrical, the movement of the left side up will be exactly matched by the right side down, resulting in no change in the position of the CoM.

The exact same amount of mass rises the same distance, as what falls on the other side.

 

So in the case of loose pans, they will have no overall effect in returning the balance to the horizontal.

It will just be the centre of mass of the beam, rotating sideways and upwards, that provides the return force.

 

If the pans are loose, their own collective centre of mass will remain central, unlike that of the beam.

Edited July 18, 2017 by mistermack

[swansont]

I think the same applies as the horizontal. Assuming that the sides are symmetrical, the movement of the left side up will be exactly matched by the right side down, resulting in no change in the position of the CoM.

The exact same amount of mass rises the same distance, as what falls on the other side.

 

 

 

 

OK. No change, even though the CoM is below, just from movement. But now put equal masses on the pans. The CoM has shifted lower. Where is the restoring force/torque?

[DimaMazin]

Manticore has it right I think. Pans will end up hanging down from cross beam by something like a chain like the scales of justice as the stereotype I was going for, Also needed some clean lines to show vectors.

 

OP did meantion old style and I'm all about old school

 

Sorry, you were referencing another post, never mind.

IMG_0206.PNG

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?

[mistermack]

 

 

 

OK. No change, even though the CoM is below, just from movement. But now put equal masses on the pans. The CoM has shifted lower. Where is the restoring force/torque?

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.

[J.C.MacSwell]

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.

Right. It is the position of the pivot/hang points of the pans that is the key. If halfway between them is the pivot point of the cross beam, any equal weights will have no effect on the balance. If it lies below that there will be a restoring force, and if above it will be the opposite and be unstable.

[Janus]

Because the torque from each just cancels the other. The horizontal distance from the pivot always matches, unlike pans fixed to the cross beam. This is assuming the pivot point is centred on the beam, and the pans hang from points equidistant from and aligned with the pivot point (at same height when beam is level).

Whether the pans swivel or not will not determine if the beam will balance out. It will depend on where the anchor point of the pans to the beam are in relation to the pivot point of the beam.

 

Here we have a scale set up so that the swiveling pans are anchored on a line with the pivot

 

In this case, equal weights on the pans will allow the beam to hold any position.

 

Next we have it so that the anchors are not on a line with the pivot but below them

 

 

Now the scale will always balance with equal weights in the pan.

 

Last we have the anchors points above the pivot.

 

Now the scale is unstable and any disturbance will cause it to tip to one side or the other even with equal weights.

[J.C.MacSwell]

...and that's why we always have Janus batting fourth!

[mistermack]

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.

[Manticore]

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.

 

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.

[Tim88]

 

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.

 

That is not useless, as one is supposed to add weights anyway. However, common scales are much more practical for a quick determination of weight.

[Manticore]

 

That is not useless, as one is supposed to add weights anyway. However, common scales are much more practical for a quick determination of weight.

 

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).