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The Physical Balance Problem

We all know the mechanism of physical balance. The balance is supported at the center and two masses are hung from the two ends. The masses being in the earth's gravitational field (which can be justifiably considered to be an uniform field acting vertically downwards at such a small scale), produce torques which are opposite to each other in direction. If the torques are equal in magnitude then the mass dipole (we can consider the balance beam simply a mass dipole) is at the equilibrium condition, i.e., perpendicular to the earth's gravitational field. If we additionally assume that the arm-lengths of the balance are equal then the above situation assures us that the counterpoised weights and hence masses are equal. Image 1 illustrates this situation. However the real problem is yet to come. Suppose a mischievous experimenter slightly displaces the balance beam from its equilibrium position making it no longer perpendicular to the field. Now as shown in Image 2, the downward acting forces are still equal(both are still mg, as the figure shows). And so are the arms of the forces about the axis of rotation(they no longer have their previous value though, but still, they are equal). That means the torques produced are still equal and opposite. So, my question is, why does the balance come down to its original equilibrium position as soon as it is released? From where does it get this torque to rotate back to its original configuration? Think and answer..;-)

Note by Kuldeep Guha Mazumder
1 year, 3 months ago

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Consider a balance beam that is “bent” (which just means that the center support and the end points of the arms are not strictly collinear). The graphic here shows a greatly exaggerated “bent” beam, the black showing the rest position and the blue showing the same but offset angularly by \(x\).

The downward force by equal weights on either end of the beam are shown in green. The torque force exerted on equal blue arms are shown in red. The net torque is proportional to

\(Cos\left( \theta -x \right) -Cos\left( \theta +x \right) \)

which simplifies to

\(2Sin\left( \theta \right) Sin\left( x \right) \)

which, to a good approximation, follows Hooke’s law


where \(k=2Sin\left( \theta \right) \)

Hence, when the balance beam is offset by small angle \(x\), it will oscillate as if against a spring, and friction losses will bring it to a rest where \(x=0\). Michael Mendrin · 1 year, 3 months ago

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@Michael Mendrin Suppose instead of a balance we take a perfectly straight mass-less rod with two equal masses at two ends and hold at its center. If then we make it offset by a small angle then also it comes back to its original position. Where does it get the torque from in this case? Kuldeep Guha Mazumder · 1 year, 3 months ago

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@Kuldeep Guha Mazumder I would be interested to know of an instance where something free to rotate about its center of gravity will rotate (in a divergence free field of gravity) to some other position and then stay there. A lot of motorcyclists that find themselves at a bad angle mid-air during a jump would like to know that. Michael Mendrin · 1 year, 3 months ago

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@Michael Mendrin Gravity is a curl-free field and not a divergence-free one..it is irrotational..it has high negative divergence..with field sinks at mass distributions..that is why we speak of a scalar and not a vector gravitational potential unlike magnetic potential which is vector (magnetic field is solenoidal).. Kuldeep Guha Mazumder · 1 year, 3 months ago

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@Kuldeep Guha Mazumder I'm referring to a local uniform gravitational force on all the parts of the system, i.e., not only irrotational, but in same direction with uniform magnitude. The simplest possible case "as a good approximation of the real thing"

Practical equal arm balance scales don't "right" themselves on account of the fact that gravity does typically have a [incredibly tiny local] divergence. It would take extremely sensitive instruments to be able to detect the divergence. Michael Mendrin · 1 year, 3 months ago

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Comment deleted Jul 16, 2015

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@Kuldeep Guha Mazumder Let's imagine that we're in orbit, like in the ISS. In fact, let's say that we're outside of it. Even the smallest of toy rockets, if placed on one end of the ISS, will be sufficient to rotate it. The work done is very tiny, which is converted into a very tiny angular momentum of the ISS. But it'll thereby keep rotating, it won't stop. It will take more work to make it stop.

In fact, it takes no net work to move the ISS from one angular position to another about its center of gravity. And regardless of what angular position the ISS is, it will stay in that angular position and not move (within a local window) That is the reason why many motorcylists are distressed to find out that they've ended up in a bad angle mid-air during a jump--it isn't easy to right yourself once you're in your own "free fall inertial frame".

The gravitational potential energy in an equal arm balance scale loaded with weights does not change with the angle of the beam, if the pivot point is right through the center of gravity of the beam. The pivot point, in the case of the bent beam, is slightly above it. Hence, in the case of the bent beam, the gravitational potential energy does change with the angle of the beam. (We can make the simplifying assumption that the weights "on the scale" can be point masses at either end of the balance beam, in this analysis). To see how the gravitational potential energy varies in both examples, with the straight balance beam, the weight one one end goes up by an equal distance as the the weight on the other end goes down. This is not the case with the bent balance beam.

On a finely made laboratory equal arm balance beam, the non-colinearity is very subtle, but it is there. In fact, adjustment knobs in some such balances are provided to adjust the "k" factor--how quickly or slowly the balance beam oscillates. Michael Mendrin · 1 year, 3 months ago

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@Michael Mendrin And yes, if we can have a mass dipole with equal arms and equal point masses at both end such that the center of mass perfectly coincides with the center of rotation, then I believe it will stay in equilibrium irrespective of its configuration in the uniform gravitational field, at least statics permits so. Kuldeep Guha Mazumder · 1 year, 3 months ago

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@Michael Mendrin The fact that the balance always goes back to its equilibrium position with its beam horizontal is simply due to its design (the center of mass is below the fulcrum) which gives it a static equilibrium. You will see, that double pan balances (not the ones with suspended pans, but those with erected pans standing above their beams) are never horizontal. Whenever the weights are removed, like a see-saw, one of their pans goes down and touches the base, while the other one remains high up there. In this case the center of mass being above the fulcrum, the equilibrium is always unstable (i.e., there is a tendency of toppling than getting restored to its original horizontal alignment). Kuldeep Guha Mazumder · 1 year, 3 months ago

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@Michael Mendrin I am extremely sorry to have written such a nuisance. I have the habit of writing wrong things so please just ignore them. The correct answer (which I had thought of when I framed the problem) is given below: It is a problem of elementary statics. When for a mass distribution system, the center of mass lies above the point of rotation the system has an unstable equilibrium. This is the case of a see-saw, where the two people always ride on the see-saw beam and hence the center of mass of the effective system is a bit above the center of rotation, i.e., the see-saw fulcrum. So, in case of a see-saw, we never see an equilibrium where the see-saw beam has a horizontal alignment. Whenever the two participating parties stop, the see-saw doesn't come to rest at a horizontal position (that's funny even to imagine). Instead, one end touches the ground, while the other end remains high up. This is an example of unstable equilibrium. However, in case of a beam balance, the scales are always hanging downwards from the two ends, and hence the center of mass is always below the fulcrum. In this case thus when we tip one end of the balance beam to disturb its natural horizontal configuration, it comes back to its natural position following the principle of stable equilibrium once the tipping force is withdrawn. Now that is the answer I had actually thought of. But I forgot it in course of time and posted something which is nothing but garbage and which I am going to delete in a couple of seconds..:-) Kuldeep Guha Mazumder · 1 year, 3 months ago

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@Kuldeep Guha Mazumder Actually, I thought this was an intriguing subject because I think most people are not aware that even in a precision equal arm balance beam, the pivot point and the points of load attachment aren't exactly colinear. It just LOOKS like it ought to be.

As a matter of fact, there is a type of equal arm balance that is based on a parallelogram, with trays on top, that does not "naturally" come to a level rest either.

Edit: Here's a graphic of a mechanical analysis of a Roberval balance, as compared to an equal arm balance beam. Notice that in both cases, the pivot point is not exactly colinear with the loading.

Michael Mendrin · 1 year, 3 months ago

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The answer to this problem lies at the center of the balance. Look that a free mass dipole never rotates in an uniform gravitational field (unlike electric or magnetic dipoles). This is simply because unlike electric charge (or magnetic poles), mass is not of dual nature. Mass is simply mass, there is nothing seen as positive mass or negative mass (at least within the purview of classical mechanics). Why do electric dipoles rotate in uniform electrostatic fields? Simply because the electrostatic forces on the two end charges of the dipole (which are of opposite nature) are anti-parallel, resulting in a couple which rotates the dipole. But this is not the case for a mass dipole. A free mass dipole always translates in an uniform gravitational field. Then why does our physical balance rotate? It is, after all a mass dipole. This is simply because it is not free, it is clamped at the center. Think and figure out the rest of the problem. Kuldeep Guha Mazumder · 1 year, 3 months ago

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@Kuldeep Guha Mazumder Kuldeep, it took me a while to find that picture of a cheap scale. It is so cheaply made it gives away the reason why balance scales behave the way you've described. I'm waiting to see if anybody else catches on.

Edit: Here's a website that gives more precise details on how to build an equal arm balance scale (apparently accurate enough for use in amateur rocketry)

Construction of a Simple Balance Beam Scale

If one looks at the details closely enough, one should be able to see why balance beam scales behave as they do. Michael Mendrin · 1 year, 3 months ago

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No no..no question of gravity and warped space-time..it is preposterous even to imagine the effects of space-time warping in a daily-needs store (well, most of the local shops here in Kolkata still use physical balances).. Think simpler.. Kuldeep Guha Mazumder · 1 year, 3 months ago

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I've got it! General Relativity! Gravity and warped spacetime!

Pix of cheap balance Michael Mendrin · 1 year, 3 months ago

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@Michael Mendrin What, relativity? Kishore S Shenoy · 10 months, 1 week ago

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@Kishore S Shenoy GENERAL Relativity Michael Mendrin · 10 months ago

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