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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.
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.
Quote from: syhprum on 28/11/2011 20:09:39The 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.
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.
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.
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.
For the purpose of calculation we will take an idealised scale of the following dimensions and calculate the rotational forces operating on the beamThe 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.
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.
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.
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?
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
No I think the buckets should be attached to the spokes so that you adjust the suspension points relative to the axle