Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: MikeS on 28/11/2011 19:06:00
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A weighing balance (jewellers balance etc.) operates through gravity comparing the torque of the weighed object with a known mass.
The balance arm comes to rest where the acting forces presumably balance. But what are the forces and why do they balance? As one end of the balance drops it gets nearer to the Earth's centre of gravity where gravity is stronger. So why does it not keep dropping?
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In addition to the two pans there is normally a pointer arm at right angles to the beam that points down to the centre of the Earth when the pans are equally loaded.
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In addition to the two pans there is normally a pointer arm at right angles to the beam that points down to the centre of the Earth when the pans are equally loaded.
True, but that's just a technicality of how the balance is made and does not address the question.
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The weight of the pointer produces a restoring force that tends to maintain the bar of the balance horizontal, this restoring force is much greater than any differential gravity effects.
The balance suspension point may also be vertically above the centre of the beam which has the same effect.
if you design your balance to avoid these restoring forces it will be truly unstable and either of the equally weighted pans can stay down.
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The weight of the pointer produces a restoring force that tends to maintain the bar of the balance horizontal, this restoring force is much greater than any differential gravity effects.
The balance suspension point may also be vertically above the centre of the beam which has the same effect.
if you design your balance to avoid these restoring forces it will be truly unstable and either of the equally weighted pans can stay down.
I don't think this is true. The balance will work equally well without the pointer (but maybe more difficult to read).
I can't see this being true either. If the pointer (above the beam) is off centre its weight would tend to make it go further off centre, not restore it.
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The weight of the pointer produces a restoring force that tends to maintain the bar of the balance horizontal, this restoring force is much greater than any differential gravity effects.
The balance suspension point may also be vertically above the centre of the beam which has the same effect.
if you design your balance to avoid these restoring forces it will be truly unstable and either of the equally weighted pans can stay down.
I don't think this is true. The balance will work equally well without the pointer (but maybe more difficult to read).
I can't see this being true either. If the pointer (above the beam) is off centre its weight would tend to make it go further off centre, not restore it.
No it won't.
Have a look at pages 21 et seq here
http://books.google.co.uk/books?id=hYZC-k4mo-YC&pg=PA23&lpg=PA23&dq=%22sensitivity+of+a+balance%22+knife&source=bl&ots=BLB1Qa5EqB&sig=jKzHvhGBb_VWFf4WbhmncMI3zrM&hl=en&ei=JvrTTpXZNMmd-waYu8z6Dg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBwQ6AEwAA#v=onepage&q=%22sensitivity%20of%20a%20balance%22%20knife&f=false
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The basic principle of the balance is that the centre of gravity of the beam and pans is below the point of suspension so that as one pan goes down there is a slight restoring force that allows equally loaded pans to Sit in the centre. the change in gravitational attraction as one pan goes down a centimetre or two is far smaller than this and does not have any effect
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Are the fulcrums of the pans slightly lower than the centre fulcrum? (I think they are, but I can't remember!) If so, the the pans will contribute a small restoring torque.
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I still can't see the thing about restoring forces.
I came across this http://en.wikipedia.org/wiki/Roberval_Balance. A Roberval balance is frequently used in the retail trade for weighing fruit, veg etc. It's essentially a balance beam scale that is not suspended. Quote "Since the vertical beams are always exactly vertical, and the weighing platforms always horizontal, the potential energy lost by a weight as its platform goes down a certain distance will always be the same, so it makes no difference where you put the weight."
"the potential energy lost by a weight" is presumably referring to gravitational potential energy. So a balance beam weighing machine is actually comparing the GPE of an object in comparison to a known weight (mass).
But this leads back to my original question. " As one end of the balance drops it gets nearer to the Earth's centre of gravity where gravity is stronger. So why does it not keep dropping?" In other words if the weights are slightly unequal, the heaviest end of the balance will tip down slightly until equilibrium is established. But why is equilibrium established? Why does the beam not continue to tip? Can anyone give me a non-mathematical answer to this?
Let me just add this in order to try to clarify what I am getting at.
A simple beam balance can be made from a strip of wood suspended by thread at the centre. If coffee cans are suspended from either end by adjusting the suspension (fulcrum) points, the scale can be made to balance. Any very slight discrepancy in weight distribution will cause the beam to rotate a certain amount around the centre fulcrum until it reaches a point of balance. Why does it reach a point of balance?
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I have just realised I think I have answered my own question.
Say the left hand side of the beam is slightly heavier, the beam will tilt down on the left until equilibrium is established. The decreased GPE of the left hand (heavier) mass is equaled by the increased GPE of the right hand (lighter) mass.
So the balance beam scale is actually comparing minute differences in GPE over potentially very small distances.
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Mike,
I think it's all about "stable" and "unstable" equilibrium. If the centre of mass of the system happens to be below the fulcrum, the system will be stable, meaning it will self-correct.
If the centre of mass happens to be above the fulcrum, the system will be unstable, and it will simply fall over.
The trick with the balance is to arrange for the system to be just stable enough that a very small difference between the masses on the pans is amplified by the pointer.
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Mike,
I think it's all about "stable" and "unstable" equilibrium. If the centre of mass of the system happens to be below the fulcrum, the system will be stable, meaning it will self-correct.
If the centre of mass happens to be above the fulcrum, the system will be unstable, and it will simply fall over.
The trick with the balance is to arrange for the system to be just stable enough that a very small difference between the masses on the pans is amplified by the pointer.
Geezer
Exactly, that accounts for how to make the balance stable and easy to read but doesn't really answer the question. I am pretty sure I gave the correct answer in my last post.
"Say the left hand side of the beam is slightly heavier, the beam will tilt down on the left until equilibrium is established. The decreased GPE of the left hand (heavier) mass is equaled by the increased GPE of the right hand (lighter) mass ."
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"Say the left hand side of the beam is slightly heavier, the beam will tilt down on the left until equilibrium is established. The decreased GPE of the left hand (heavier) mass is equaled by the increased GPE of the right hand (lighter) mass ."
This seems to be saying that the heavier mass weighs less deeper within the gravity well and the lighter mass weighs more higher within the gravity well.
If this is correct then it could be tested by weighing a mass at different heights in the gravitational field using a very accurate type of 'spring' (not balance beam) weighing machine. I understand observation confirms this.
So to re-phrase the answer.
A simple balance beam comes to rest at the angle where equilibrium is reached. That is, where both objects have the same weight within the gravity well.
Edit
The part of this post that has strike through is completely wrong, please ignore it.
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I cannot believe that you really believe this nonsense about gravity wells, at least four senior correspondents have explained in words of one symbol how and why balances work as they do.
I think you are just practicing as a devils advocate to see if you can convince anyone of your ridiculous idea.
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syphrum
I realised my last post was wrong almost immediately after posting it but I was not near a computer so couldn't delete it.
I cannot believe that you really believe this nonsense about gravity wells, at least four senior correspondents have explained in words of one symbol how and why balances work as they do.
I think you are just practicing as a devils advocate to see if you can convince anyone of your ridiculous idea.
I don't believe it has been explained how a balance works. Perhaps you would be good enough to explain in words of one syllable how they work.
If a balance does not work through the effects of differential gravity in a gravity well (field) then I would certainly like to know how they work. Please enlighten me.
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May I refer you to the posts by Geezer, Soul Surfer, and myself also the article cited by Bored Chemist.
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Mike,
As Syhprum pointed out in his second post, any differential gravitational effect is negligible. In fact, the beam balance operates on the basis that gravitational effects are uniform.
The restoring torque is simply a consequence of the fact that the system is stable when it is at, or nearly at, equilibrium. If it was an unstable system in equilibrium (meaning all the forces summed to zero) there would be no restoring force and any "noise" in the system would cause one of the pans to hit the deck.
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It is not difficult to do a calculation of the net rotational forces acting on the beam taking in the effect of various positions of the fulcrum and the differential gravitational effect.
scientific arguments always look more convincing if accompanied by a mass of calculations.
to work out the differential gravitational effect (which is infinitesimal) I will have to make the assumption that the Earth is flat and of infinite extent so if MikeS considers this invalidates the argument I will not bother or perhaps some more competent mathematician will take over the task
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It's a quite nice question Mike. Seems it has to do with where the 'centre of gravity' is placed with the scale you use, at least when using equal arm balances. If it is under the pivot you will be able to to have unequal weights and see it tilt without it losing its balance.
This one shows some different balance instruments. Types of Mechanical Scales. (http://www.gilai.com/article_21/Hanging-in-the-Balance---Antique-Scales)
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Thank you all for your replies but I still don't think the original question has been answered.
A lot of the answers concentrate on how to make a balance beam stable, not why "The balance arm comes to rest where the acting forces presumably balance."
The articles referred to by Bored chemist and Geezer are of no help.
"A weighing balance (jewellers balance etc.) operates through gravity comparing the torque of the weighed object with a known mass. The balance arm comes to rest where the acting forces presumably balance."
Please ignore for a moment that most balance arm scales are designed to be balanced with the beam in the horizontal position.
If the weight to be measured is slightly heavier than the known weight the beam will reach stable equilibrium with the heavier end slightly lower. The beam is at an angle not horizontal. "But what are the forces and why do they balance?"
Agreed that a balance beam scale works by comparing an unknown weight to a known weight when the beam is horizontal. But what is it telling us when it reaches stable equilibrium at an angle?
I phrased it like this in a previous post.
"A simple beam balance can be made from a strip of wood suspended by thread at the centre. If coffee cans are suspended from either end by adjusting the suspension (fulcrum) points, the scale can be made to balance. Any very slight discrepancy in weight distribution will cause the beam to rotate a certain amount around the centre fulcrum until it reaches a point of balance. Why does it reach a point of balance?
This was the explanation that I posted.
"Say the left hand side of the beam is slightly heavier, the beam will tilt down on the left until equilibrium is established. The decreased GPE of the left hand (heavier) mass is equaled by the increased GPE of the right hand (lighter) mass ."
A few of you have said that the differential gravitational effect is negligible. But is it? How else can a balance reach stable equilibrium with the beam at an angle. If it's not a differential gravity effect what is it?
The difference in the strength of gravity at that scale, if difficult to measure, is certainly not insignificant.
The latest atomic clocks can measure gravitational time dilation down to about one centimetre in height. Therefore how can you say the differential gravitational effect is negligible? It must be significant.
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For the purpose of calculation we will take an idealised scale of the following dimensions and calculate the rotational forces operating on the beam
The beam will have a length of two meters and height of four centimetre's and will have a suspension point in the dead centre for the fulcrum and similar suspension points precisely in line at each end for the pans .
The beam mass will be zero and the mass of the pans will be 1/9.82 Kg.
The beam will be suspended at the fulcrum point so that the pans are 10cm above the working surface.
Let us next push the left hand pan down to the surface, the right hand pan will rise and exert a torque of
(1-(.1^2))^.5 = 0.99498 newton meters tending to rotate he beam clockwise while the left hand pan will exert a similar torque tending to rotate the beam anti clockwise hence the system will be stable and the left hand pan will remain down.
Now let us calculate the effect of gravity, taking the radius of the Earth as 6,366,197.8 meters the left hand pan will be this distance from the centre of the Earth while the right pan will be 6,366,197.6 meters away.
Applying Newton's inverse square law the gravitational attraction on the right pan will be reduced by one part in (6,366,197.8/6,366,197.6)^2 =1.0000000631 hence there will be a net force of 0.99498*0.0000000631=0.0000000628 Newton's tending to hold the left hand pan down.
Now we come to the effect of raising the fulcrum point as it would normally be on any practical set of scales, If the fulcrum point is raised by one cm relative to the line of the pan suspension points when the left hand pan is pushed down the effective length of the left hand side the beam is reduced by one part in a ten thousand while that of the right hand beam is increased by the same amount hence a restoring force tending to move the beam to a horizontal position of 0.0002 Newton meters is generated vastly more than any gravitational effects.
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Here's a thought experiment for you.
I'm going to build a balance.
In a rather unorthodox way, I'm going to start with just the pointer.
Obviously, with nothing else there, it hangs straight downwards.
If I push it to one side and let go, it swings back and to, but it settled down to being vertical.
Now I put the beam on the balance and fix it to the pointer.
The pointer still wants to point downwards so, if I set the beam swinging it still ends up settling down with the beam horizontal, and the pointer vertical.
I can hang pans from the beam- the pointer still does its job and the beam is only stable when it's horizontal.
This still works in the case of a perfectly uniform gravitational field.
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MikeS
"If it's not a differential gravity effect what is it?"
Let me answer this specific point as I think it is something you have not considered.
When the beam is pivoted at a point above the line of the pan suspension points if for instance the left hand pan tends to drop the beam moves to the right reducing the effective length of the beam on the left hand side and increasing it on the right hand side.
This produces a negative feedback effect causing the beam to stabilise in the horizontal position if the masses in the two pans are equal.
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Bored Chemist
If we have a pointer the beam can be pivoted exactly in line with the pan suspension points and it will serve to stabilise the beam in a horizontal position but if the beam is pivoted above the line of the pan suspension points the system is still stable without a pointer.
PS although the matter is trivial and of little scientific interest I find this a good writing exercise
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If we have a pointer the beam can be pivoted exactly in line with the pan suspension points and it will serve to stabilise the beam in a horizontal position but if the beam is pivoted above the line of the pan suspension points the system is still stable without a pointer.
I agree. The pointer is, well, a pointer to tell you when the pans are balanced. Even if the pointer had zero mass, the balance would still work.
A diagram is necessary - stay tuned!
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I resisted the temptation add diagrams thinking it was a challenge to clarify the matter with text alone.
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In the next development of the thought experiment I explain that my pointer is of a non-traditional shape.
Specifically it is exactly the same shape as the beam. It just has a little arrow drawn on it labelled "Down".
There's nothing magical about what you call the mass that hangs downwards below the suspension point. A pointer will clearly do the job of ensuring that the beam comes to rest horizontal.
But the beam itself will also do that job.
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Look mum! No pointer.
[ Invalid Attachment ]
The forces (F) on the pan pivots are equal but the beam is not level. L1 is greater than L2, so the counter-clockwise torque is greater than the clockwise torque. The net torque is zero when L1 and L2 are equal, at which point the beam will be horizontal.
EDIT: BTW, this works when the pointer has zero mass, but it works just as well when the beam has no mass either.
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For the purpose of calculation we will take an idealised scale of the following dimensions and calculate the rotational forces operating on the beam
The beam will have a length of two meters and height of four centimetre's and will have a suspension point in the dead centre for the fulcrum and similar suspension points precisely in line at each end for the pans .
The beam mass will be zero and the mass of the pans will be 1/9.82 Kg.
The beam will be suspended at the fulcrum point so that the pans are 10cm above the working surface.
Let us next push the left hand pan down to the surface, the right hand pan will rise and exert a torque of
(1-(.1^2))^.5 = 0.99498 newton meters tending to rotate he beam clockwise while the left hand pan will exert a similar torque tending to rotate the beam anti clockwise hence the system will be stable and the left hand pan will remain down.
Now let us calculate the effect of gravity, taking the radius of the Earth as 6,366,197.8 meters the left hand pan will be this distance from the centre of the Earth while the right pan will be 6,366,197.6 meters away.
Applying Newton's inverse square law the gravitational attraction on the right pan will be reduced by one part in (6,366,197.8/6,366,197.6)^2 =1.0000000631 hence there will be a net force of 0.99498*0.0000000631=0.0000000628 Newton's tending to hold the left hand pan down.
Now we come to the effect of raising the fulcrum point as it would normally be on any practical set of scales, If the fulcrum point is raised by one cm relative to the line of the pan suspension points when the left hand pan is pushed down the effective length of the left hand side the beam is reduced by one part in a ten thousand while that of the right hand beam is increased by the same amount hence a restoring force tending to move the beam to a horizontal position of 0.0002 Newton meters is generated vastly more than any gravitational effects.
But the left hand pan wont stay down. The beam will go back to horizontal.
If all three fulcrum points are in-line the balance will still work despite the fact that both arms remain the same length regardless of the beams inclination.
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MikeS
"If it's not a differential gravity effect what is it?"
Let me answer this specific point as I think it is something you have not considered.
When the beam is pivoted at a point above the line of the pan suspension points if for instance the left hand pan tends to drop the beam moves to the right reducing the effective length of the beam on the left hand side and increasing it on the right hand side.
This produces a negative feedback effect causing the beam to stabilise in the horizontal position if the masses in the two pans are equal.
It's true that I hadn't originally thought of the effective lengths of the arms changing due to the geometry of the fulcrum points.
But
If all three fulcrum points are in-line the balance will still work despite the fact that both arms remain the same length regardless of the beams inclination.
I think I am right in believing in this design torque is always equally balanced and therefore plays no part.
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This is the question that still hasn't been answered.
Let's consider a simple balance beam with all three pivot points in line. One pan is slightly heavier than the other. The beam will come to stable equilibrium with one pan lower than the other. Why?
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But it will not with all three pivot points in line and one pan heavier than the other the heavier pan will descend until it meets a stop.
This of course assumes their is no friction in the system and an equal mass of beam above or below the fulcrum points
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Mmmmm. Do you have any evidence for that?
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No I thought this was a theoretical discussion and I have not had access to any laboratory type balance since I left school seventy years ago.
Perhaps our Bored Chemist could confirm it.
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syphrum
"Practical balances are constructed with the central knife-edge lying a little
below the plane of the terminal knife-edge."
http://books.google.co.uk/books?id=hYZC-k4mo-YC&pg=PA23&lpg=PA23&dq=%22sensitivity+of+a+balance%22+knife&source=bl&ots=BLB1Qa5EqB&sig=jKzHvhGBb_VWFf4WbhmncMI3zrM&hl=en&ei=JvrTTpXZNMmd-waYu8z6Dg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBwQ6AEwAA#v=onepage&q=%22sensitivity%20of%20a%20balance%22%20knife&f=false
This is the opposite of the way that most of you have described the pivot points of a balance and the opposite of
Geezers diagram. L2 is greater than L1.
If a balance can work with the central pivot point either above or below the terminal pivot points then I see no reason why it should not work with all three pivot points in line.
In which case this is wrong
But it will not with all three pivot points in line and one pan heavier than the other the heavier pan will descend until it meets a stop.
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The balance will certainly work with all pivot points in line but with equal weights in each pan it will not automatically restore its self to a horizontal beam state if either pan is pushed down it will stay down.
As far as I recall our school balances had a lever to lift the central post after the pans were loaded, if they were equally loaded the beam would come up horizontal but of course they had some self restoring force built in.
Even in the absence of this feature they would still would still have come up in this manner but this state would have been unstable and if either were pushed down the would have stayed down.
Reading the article you quoted I notice that the central pivot is placed below the line of the pan pivots which in effect gives a positive feedback making the balance point unstable but stability is restored by the use of a pointer which gives a negative feedback effect to counteract this.
This combination is done to increase the sensitivity at small deflections.
We have of course been discussing the most basic form of balance but commercial designs are a little different
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syphrum
Ok so let's consider a simple balance beam with all three pivot points in line and the balance beam incorporates some kind of restoring force. One pan is slightly heavier than the other. The beam will come to stable equilibrium with one pan lower than the other.
Do you agree?
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With some sort of restoring force ? a pointer perhaps as your postioning of the pivots gives none.
In this case yes
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The question is why does the balance reach stable equilibrium with one pan lower than the other? Why does the heavier pan not continue until it reaches the stop?
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It depends how stability has been achieved, in the original case where stability had been achieved by the positioning of the pivots the non linear change in apparent length of the arms with the deflection angle would limit how far the heavier pan drops, the restoring force exerted by a pointer is also not in a linear relationship to the deflection angle .
The mathematics of the situation would involve trigonometric functions into which I don't really wish to go.
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syphrum
Ok so let's consider a simple balance beam with all three pivot points in line.
Would you consider the pans to contribute a restoring force without the requirement of the mass of a needle?
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No in this case there would be no restoring force so the heavier pan would go down to limit or equally weighted pans would just sit where they were placed, but let me continue
As stated by Bored Chemist there is no difference between a scale stabilised by the weight of a pointer to one stabilised by the positioning of the pivots so I will use the later to explain how trigmetric functions determine the deflection angle hence how far the weighted pan drops .
As the weighted pan drops the torque it develops varies as the cosine of the angle of the beam from the horizontal i.e starts high and falls away whereas the torque generated by the pointer being set at 90° to the beam varies as the sine of the angle of the beam from the horizontal i.e starts low and climbs.
when these two sources cancel out that is the angle at which the beam rests.
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"Ok so let's consider a simple balance beam with all three pivot points in line. "
Why should we?
It's not as if anyone would actually make a balance like that (unless they were relying on a pointer to keep it in check).
If they did that, it wouldn't work.
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Ok so let's consider a simple balance beam with all three pivot points in line.
Would you consider the pans to contribute a restoring force without the requirement of the mass of a needle?
Absolutely not. If the weights are equal there is no torque in any position.
Your question is the same as "what would happen if I picked up a perfectly balanced bicycle wheel by it's axle?" Obviously, it won't rotate because there is no reason why it should rotate.
If the fulcrums on a scale are in line, and the center of mass of the entire beam assembly (including the pointer) is conincident with point of rotation of the beam, there will be no restoring force. If the weights are equal, the balance will remain in any position it is put in. If the weights are unequal, the heavier one will tip the balance as far as it can go.
A scale must have a restoring force because a balance is actually comparing the the restoring force with the difference in the weights on the pans. The displacement of the pointer from center actually calibrates that amount on a graduated scale.
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Bored Chemist
The balance in the article you directed us to actually had the fulcrum below the suspension points of the pans and relied on the weight of the pointer to provide stabilisation.
This combination of positive and negative feedback was done to increase the sensitivity to small deflections
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Geezer
The wheel of an inverted bicycle could well be used as a model balance, a small weight could be placed say where the valve comes out to emulate the pointer and pseudo pans attached either in line with the axle or symmetrically above or below it for experiment.
A simple and readally available model
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Geezer
The wheel of an inverted bicycle could well be used as a model balance, a small weight could be placed say where the valve comes out to emulate the pointer and pseudo pans attached either in line with the axle or symmetrically above or below it for experiment.
A simple and readally available model
Oh, you mean something like this?
[ Invalid Attachment ]
(The pseudo pans are calibrated galvanised buckets attached to a piece of string.)
(Astute correspondents will observe that the distance between the string and the axis of the wheel is constant.)
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No I think the buckets should be attached to the spokes so that you can adjust the suspension points relative to the axle
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No I think the buckets should be attached to the spokes so that you adjust the suspension points relative to the axle
Yes. That would make it more like a beam balance, but I like my version better!
The only thing that produces a restoring force is the weight of the valve which tends to prove that there are many ways to produce a restoring force, but you gotta have one.
We could even eliminate (or counterbalance) the unbalanced mass of the valve and replace it with a torsion spring and get the same result.
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Thank you all for taking part you have convinced me. A balance needs a restoring force.
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Geezer
The wheel of an inverted bicycle could well be used as a model balance, a small weight could be placed say where the valve comes out to emulate the pointer and pseudo pans attached either in line with the axle or symmetrically above or below it for experiment.
A simple and readally available model
Oh, you mean something like this?
[ Invalid Attachment ]
(The pseudo pans are calibrated galvanised buckets attached to a piece of string.)
(Astute correspondents will observe that the distance between the string and the axis of the wheel is constant.)
I like it!
Let's suppose the wheel is perfectly balanced, has no bearing friction and has no restoring force, the buckets perfectly balanced, the string has no mass and is very long.
If we pull one of the buckets almost as far down as possible the other bucket will be almost as high as possible. The lower bucket will now weigh marginally more than the higher bucket. If you bring the bucket to a stop and release hold of the bucket will it continue down or will it remain stationary?
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May I refer you to my calculation of the differential gravitational effect (post 2307), provided the lower bucket is still above the surface of he Earth there is a very small additional gravitational attraction to it so as you have specified that there is zero friction in the system presumably it would continue down but you would be hard put to realise such a system in practice.
PS are you a member of the legal profession ?
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PS are you a member of the legal profession ?
No, that would be me! And we only ask silly questions when we are being paid to do so :-)
I think Syhprum is quite correct - I estimated it to be in the order of 10^-7 to 10^-8N
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I think Syhprum is quite correct - I estimated it to be in the order of 10^-7 to 10^-8N
If we had a really long piece of string we might even be able to observe the effect.
[ Invalid Attachment ]
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the difficulty is finding string of zero mass
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You could use neutrally buoyant string and set the rig up underwater.
Of course, the variation of water density with depth...
The fact is that balances are actually built so that they come to rest with the beam horizontal.
The means by which this is achieved are simple enough to understand.
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the difficulty is finding string of zero mass
It might be even harder to find a skyhook to hang the wheel from.
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Eureka, use a loop of string then it cancels out.
A convenient place to carry out the experiment would be the Kreigmarine memorial in Laboe.
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Maybe they'd let us use that giant wheely thing they built beside the Thames?
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we would be shielded from the wind in Laboe
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We could use the Monument: one of its design parameters was for a astronomical and gravitational experiments (it has a secret(ish) underground lab at the bottom) and it is still used to this day.
http://en.wikipedia.org/wiki/London_Monument
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Looks like we might have to recalibrate the lower bucket.
[ Invalid Attachment ]
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To answer syphrums question, no I am not a member of the legal profession although it has been suggested, more than once that I think like one. (I still haven't worked out how to take that.) My primary background was as an engineer troubleshooting for BT. My brain seems to be wired that way, if you'll excuse the pun.
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Mmmm rabbit in a bucket, tasty.
Continuing with 'Geezers wheel and bucket balance' (less rabbit).
The lower bucket, as measured by a 'spring balance' weighs more than the higher bucket so we expect it to continue down to the full extent possible.
Both buckets when in the same horizontal plane weighed the same. The lower bucket now weighs more because it is deeper within the gravity well but has lost a certain amount of gravitational potential energy (GPE). The higher bucket has correspondingly lost weight but gained the same amount of GPE. Therefore both buckets still contain the same amount of mass/energy through the mass/energy equivalence principle.
The question is if both buckets have in effect the same mass would the lower bucket continue its journey down or would they both balance regardless of position?
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Even if you position the two buckets so that the gravitational attraction is the same on each the system is inherently unstable any infinitesimal disturbance could set them moving,rather like a pencil standing on its point.
Re-reading your post it would seem that you want to bring special relativity into the problem, I regret being only a retired technician lacking even "O" levels such mathematics are beyond me, perhaps one of the better educated correspondents will be able to take the matter further.
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The maths are beyond me as well unfortunately but it's still interesting to consider thought experiments. I tend to agree that the system is 'probably' inherently unstable. The instability presumably being due to the differential effects of gravity.
I hope someone will answer the question in my last post above.
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My maths isn't that good, but I have seen lots of adverts from balance makers.
A while ago (maybe 15 or 20 years) one of them produced an ad that was a picture of a gold bar next to a picture of a tall building with the caption something like
1.000000 grams on the ground floor 0.999997 grams on the 30th floor.
I forget what the numbers were but the point was that you could measure (just) the difference in the force of gravity between the top and bottom of a tall building.
Now imagine that I take that gold bar and drop it from the top of the building to the ground.
Most of the energy released will be turned into heat. The bar will warn up- but only very slightly.
So the gravitational energy is equivalent to a small rise in temeperature.
On the other hand, I know that it's still (15 or 20 years on) impossible to measure the change in mass due to changes in temperature that arise from relativity.
I conclude that the change in weight due to differential gravitation is much bigger than the change in mass due to potential energy.
I'm sure the calculation would confirm this.
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Thanks but.
Say you dropped it in a vacuum, there would be no warming on the way down. When it hits the ground it has gained kinetic energy (by converting GPE into kinetic) and that energy is converted into sound and heat upon contact. The energy gained is substantial. You wouldn't want it to drop on your head. Is it simply that a very large amount of energy is required to produce even a slight increase in mass hence it is difficult to measure the increase, or is there some other explanation?
I conclude that the change in weight due to differential gravitation is much bigger than the change in mass due to potential energy.
But isn't this the same thing?
At the top of the building the bar weighs less but has a higher GPE. At the bottom of the building the bar weighs more but has a lower GPE. The two being equal due to the energy/mass equivalence principle?
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Aaaah I've just realised the two probably arn't equivalent. I hadn't taken into account the energy lost in slowing the bar down to stationary. Does that make sense?
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OK lets do the maths.
Assume the gold bar is 1Kg because it makes the maths easy. Similarly, it falls 1 metre.
It lands at ground level and that is 6400Km (exactly) from the centre of the Earth where I choose to do the experiment and the local value of g is 9.8 m/s/s
It falls 1 metre so it converts potential energy =Mgh into kinetic energy just before it hits the ground and it converts that into thermal energy when it hits.
The energy released is 9.8J
From E=MC^2 we get a change in mass of
9.8=300,000,000* 300,000,000 M.
M= 1.1E-16 Kg
On the other hand, the change in weight (and an apparent change in mass) is given by the change in g
g= M(earth) X G/ (6,400,000^2)
whereas the value 1 metre further up is
M(earth) X G /(6,400,401^2)
So the ratio of them is (6,400,000^2 ) to (6,400,001^2)
So a 1 Kg mass would apparently weigh 3.125 E-7 Kg less
Since the two mass changes are different by a factor of about 3 billion, they are not the same.
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I could of course have made these simple calculations despite zero "O" levels but I think this question has been flogged to death.
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With gold at £25,000 per kilo the loss in value on the thirtieth floor would have been only a few pennies but if it had been Polonium or anti matter or some real expensive stuff it would be real serious
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I'm revising the diagram to include a polonium rabbit.
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Like a phoenix from the grave, some things just wont die.
From an earlier post.
"Continuing with 'Geezers wheel and bucket balance'" (less rabbit, polonium or otherwise).
"The lower bucket, as measured by a 'spring balance' weighs more than the higher bucket so we expect it to continue down to the full extent possible.
Both buckets when in the same horizontal plane weighed the same. The lower bucket now weighs more because it is deeper within the gravity well but has lost a certain amount of gravitational potential energy (GPE). The higher bucket has correspondingly lost weight but gained the same amount of GPE. Therefore both buckets still contain the same amount of mass/energy through the mass/energy equivalence principle.
The question is if both buckets have in effect the same mass would the lower bucket continue its journey down or would they both balance regardless of position?"
From an earlier post.
Quote from: Bored chemist on 04/12/2011 11:13:14
”I conclude that the change in weight due to differential gravitation is much bigger than the change in mass due to potential energy.”
From an earlier post.
"But isn't this the same thing?
At the top of the building the bar weighs less but has a higher GPE. At the bottom of the building the bar weighs more but has a lower GPE. The two being equal due to the energy/mass equivalence principle?"
Let’s do the maths for the above experiment.
From an earlier post by Bored chemist.
“Assume the gold bar is 1Kg because it makes the maths easy. Similarly, it falls 1 metre.
It lands at ground level and that is 6400Km (exactly) from the centre of the Earth where I choose to do the experiment and the local value of g is 9.8 m/s/s
It falls 1 metre so it converts potential energy =Mgh into kinetic energy just before it hits the ground and it converts that into thermal energy when it hits.
The energy released is 9.8J
From E=MC^2 we get a change in mass of
9.8/300,000,000* 300,000,000 M.
M= 1.1E-16 Kg”
I work it out to be M=1.088888888889e-10 not the above figure.
Continuing the earlier post by Bored chemist
”On the other hand, the change in weight (and an apparent change in mass) is given by the change in g
g= M(earth) X G/ (6,400,000^2)
whereas the value 1 metre further up is
M(earth) X G /(6,400,401^2)
So the ratio of them is (6,400,000^2 ) to (6,400,001^2)
So a 1 Kg mass would apparently weigh 3.125 E-7 Kg less
Since the two mass changes are different by a factor of about 3 billion, they are not the same.”
Continuing the maths.
The two buckets have the same mass (1kg) when horizontal, so remain horizontal. If we manually position one bucket (a) 1 meter lower then the other bucket(b) is one meter higher.
Bucket a now weighs 1.00000015625kg
Bucket b now weighs 0.9999998437500244kg
That’s a difference of 1.563600000587e-7
Both buckets are now stationary but bucket a is now heavier than bucket b so we would expect bucket a to continue to fall whilst bucket b continues to rise and thereby increasing the weight differential.
But this is only half of the story. As bucket a falls and increases in weight it looses GPE. Likewise as bucket b rises and looses weight it increases its GPE. The buckets were positioned manually. Any loss in GPE of bucket a is equaled by the equivalent gain in GPE of bucket b.
Continuing the earlier post by Bored chemist
"It falls 1 metre so it converts potential energy =Mgh into kinetic energy just before it hits the ground and it converts that into thermal energy when it hits.
The energy released is 9.8J
From E=MC^2 we get a change in mass of
9.8/300,000,000* 300,000,000 M.”"
=1.088888888889e-10 (my calculated figure)
Bucket a has lost 1.088888888889e-10kg so now weighs 0.9999999998911111kg less in GPE
Bucket b has gained 1.088888888889e-10kg so now weighs 1.000000000108889kg more in GPE
That’s a difference of 1.563500000579e-7
That’s a discrepancy of 1.000000080001e-11
Quote from earlier post.
"But isn't this the same thing?
At the top of the building the bar weighs less but has a higher GPE. At the bottom of the building the bar weighs more but has a lower GPE. The two being equal due to the energy/mass equivalence principle?"
You can’t get much closer than that. The minute difference in results is probably due to the slight differences in the strength of gravity not being taken into account at the different heights.
From an earlier post.
"The question is if both buckets have in effect the same mass would the lower bucket continue its journey down or would they both balance regardless of position?"
From the above it seems obvious to me that any difference in weight of the buckets caused by differential gravity is equally balanced by a corresponding change in GPE thereby negating the effect caused by the change in weight. In other words, the buckets will remain perfectly balanced, regardless of differences in height in any position.
Going back to previous posts a neutrally balanced, balance beam should remain stable (ignoring any instability caused by construction) in any position due to any differences in pan weigh due to differential gravity being balanced by a corresponding change in GPE. The two being equivalent through E =mc2
Unless you know differently of course. [;)]
Added
The figure given by Bored chemist is correct, I don't know how I managed to get it wrong using an on-line calculator. I must have entered it wrongly I guess. The strange thing is the wrong figure gave a result that when 'plugged' into everything else gave exactly the answer that I was expecting. Guess that was why I didn't re check it. Obviously I need to go over it all again.
Just gone over my figures and discovered I squared 300,000 not 300,000,000
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"9.8/300,000,000* 300,000,000 M.”"
=1.088888888889e-10 (my calculated figure)"
No arithmetic error !!!
9.8/300000000^2 = 1.088888*10^-14 Wrong again back to infants school
Deduct another brownie point it should of course be 1.0888*10^-16
I think some maintenance on sliderules needs to be done.
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It falls 1 metre so it converts potential energy =Mgh into kinetic energy just before it hits the ground and it converts that into thermal energy when it hits.
The energy released is 9.8J
From E=MC^2 we get a change in mass of
9.8/300,000,000* 300,000,000 M.
M= 1.1E-16 Kg”
I work it out to be M=1.088888888889e-10 not the above figure.
Well, try doing it again until you get a better answer.
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You're dividing 9.8 by 9*1016. The answer has to be something times 10-16.
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Do you think we have the answer to the Neutrino anomaly here .
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I'll add it to the poll.
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Looking back through my last post it looks like I made two completely separate mathematical mistakes.
The first being From E=MC^2 we get a change in mass of
9.8/300,000,000* 300,000,000 M.”"
=1.088888888889e-10 (my calculated figure)
I was thinking kilometers not meters so squared 300,000 not 300,000,000.
Working out the bucket weight difference with one bucket dropping one meter whilst the other bucket rose one meter, the difference in weight due to the difference in GPE (using my wrong figure) was 1.563500000579e-7 kg
My second mistake
“M(earth) X G /(6,400,401^2)
So the ratio of them is (6,400,000^2 ) to (6,400,001^2)”
For some reason I read ˆ as multiply not square.
This resulted in a weight difference of the buckets due to height, of 1.563600000587e-7 kg
That’s one heck of a coincidence. One digit difference in the 11 decimal place.
Added 12-12-11 0850
I need to change my glasses, just noticed one digit difference in the fourth decimal place, oh well.
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I wasn't sure whether to start this in a new thread or not, as I seem to have worn this one out.
If you weigh an object within a gravitational gradient (at different heights) you get different weights. The object weighs more the deeper it is within the gravity well.
Likewise the object has different GPE depending upon its height within the gravity well.
As the object is moved lower within the gravity well it looses some of its GPE but gains weight (mass).
The question is why aren’t these the same? Why does the objects weight plus GPE at one height not equal its weight plus GPE at another height? Surely, because of the equivalence principle, E=mc2, they should be the same?
If GPE is much less over a given distance than weight difference (by many orders of magnitude) then why is dropping a 1kg weight on your head from 1m noticeable whereas its weigh difference over 1m is not?
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"The question is why aren’t these the same? "
Because one of them is much bigger than the other.
A simplistic answer would be "Ask Newton"
He would be able to calculate the change in weight with altitude but not the change in mass with energy.
They are simply different things.
Why should they be the same?
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To keep this bizarre thread going even longer perhaps we should consider the case where we have a very long string and the lower bucket is lowered down the gravity train tunnel.
I rather worry for the rabbit, firstly irradiated by the Polonium now subjected to the Earth centre temperatures and the effects of zero Gravity (at the centre).
We have not yet considered the effects of centrifugal Centripetal force ! No doubt the shades of Newton, Einstein, and Mach will have to be invoked.
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Quote Bored chemist.
"The question is why aren’t these the same? "
"Because one of them is much bigger than the other.
A simplistic answer would be "Ask Newton"
He would be able to calculate the change in weight with altitude but not the change in mass with energy.
They are simply different things.
Why should they be the same?"
You say “they are simply different things. Why should they be the same?”
As I see it, a change in weight with altitude should be equivalent to a change in GPE with altitude. Mass and energy being equivalent. What are the factors to consider here? The gravitational gradient, which in turn leads to a time dilation gradient and mass/energy. GPE and weight are both affected by the gravitational gradient (caused by a change in height). When you change the height you change the weight. Why does the GPE not change by the same amount? GPE is dependent upon weight and height. GPE and weight seem to look like ‘different faces of the same coin’. However, the difference in weight due to a change in height is not equivalent to the change in equivalent mass of the change in GPE.
For an object to ‘weigh’ more deeper within a gravity well it must have gained mass which is equivalent to a gain of energy. This seems to make sense, as it requires more energy to do anything the deeper you go within the gravity well. In other words, gravity ‘ties up’ energy rendering it unusable.
The same argument seems to apply to GPE. The deeper the object is within the gravity well the less GPE it has. Energy has been ‘tied up’ rendering it unusable.
"He (Newton) would be able to calculate the change in weight with altitude but not the change in mass with energy."
Surely,"the change in weight with altitude", should be equivalent to "the change in mass with energy" But the change in mass due to a change in height is many orders of magnitude greater than the change in mass due to the change in GPE. I still don’t understand why?
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Mike you're mixing your theories. You want to use general relativity, because that's the theory that tells you that energy effects gravity, but in general relativity, gravitational potential energy isn't a thing, so your arguments don't hold.
But there are two effects in GR, similar to to "position effects gravity" and "GPE effects gravity." If a small mass is near a big mass, the big mass curves space-time, which influences how the small mass will move under gravity. This is how position effects gravity. However, the small mass also has a very small effect on the space-time around it, which alters the total way space-time curves around the big mass by a very tiny bit and in turn effects how the small mass feels gravity. Since this effect is tiny, it will mostly be unimportant, but if you really want to do the calculations of how this tiny distortion in space-time effects its overall motion under gravity, you can do so.
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"As I see it, a change in weight with altitude should be equivalent to a change in GPE with altitude. "
The real world doesn't see it like that so nor should you.
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A balance requires a restoring torque. The restoring torque is provided by a centre of mass of the balance being below than the horizontal plane of the beam knife edge. When a pan is pushed down, there c.m. moves away, on the opposite side, out of the vertical plane of beam knife edge, but remaining below the horizontal plane of the beam knife edge. This creates restoring torque.
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To make it clearer, think of a balance beam with no pan - say a straight edge. Pass a pin through its centre of mass. This balance is stable at any angle. If you hang equal masses on either end, it becomes a balance with a restoring torque. The centre of mass shifts below the pivot. On the other hand, if you stick equal masses at ends of the straight edge, on the line of symmetry, you will not change the centre of mass, (it still passes through the pivot), and you will have no restoring torque.