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by spacemule 1002 days ago
Thanks for answering. I'm not even an amateur in physics, so forgive me for this elementary follow-up question. Why does it not balance in water? It would seem to me that the same weight it on each side of the scale, so the same force is pushing down on each side of the scale.
2 comments

You're right there is the same force pushing down on each side. But there's also an additional force in water pushing up. It's called upthrust. It's equal to the weight of displaced water. If the densities are different the samples with the same weight will have different volumes and different upthrusts.

This can be quite a useful phenomenon practically. Anyone who has learned to scuba dive will know that if you take a deep breath in from your air tank you increase your volume without changing your weight and will be able to slightly change your bouyancy and float upwards or downwards without moving your limbs. Submarines also use the same phenomenon.

> Submarines also use the same phenomenon

For clarity, the submarines fix their voice but adjust their weight by pumping or flooding their ballast tanks

> Why does it not balance in water?

Technically, it doesn't balance an air either, but it's just harder to notice the effect.

For an exaggerated example, imagine a 1-kg cube of styrofoam and another 1-kg cube of steel, each placed on opposite trays of a balance-scale.

The scale will be truly balanced in vacuum, and in air will appear balanced to the unaided human eye.

However when you submerge the entire system underwater, the differences will be shockingly visible... especially when the styrofoam floats up and off the scale entirely! It isn't because of negative mass or negative weight, just buoyancy that can no longer be ignored.

By conducting the "weighing" experiment again in a denser fluid (water) any subtle differences in density/ displacement become easier to measure.

That is not how a balance scale works. The two sides balance, because the weigher makes them balance. That can be done in a vacuum, air, or water.

The trick being described has the weigher balance the scale in air, with unknown mass A and some mass of gold B. The two sides are equal in weight in air, different from mass. Then the entire balance system is submerged. The difference in density of A and B, and therefor volume, leads the balance to become unbalanced.

It would be wise to add mass to the lighter side to try and measure how imbalanced. I would bet most adulterants are about half the density of gold or less. In g/cm3: Gold is 19.3, silver 10.5, lead 11.3, copper 9.0, nickel 8.9.

> That is not how a balance scale works. The two sides balance, because the weigher makes them balance.

I think you're confusing the tool-calibration step with the actual measuring. As long as a scale is built symmetrically, an empty scale will read as balanced in any environment. Both sides are made from the same materials with the same densities and displacements etc.

Once you apply the dissimilar samples (such as a known 1kg mass of styrofoam versus a known 1kg mass of steel) it will cease to read as perfectly even, because the air-buoyancy of the samples will be different. Perhaps not enough to see easily, but it's there.

____

It may help to consider that for this experiment we do not actually need to see an "equal" weight-measurement from any kind of scale. What we're actually trying to check is that the readout doesn't change when swapping the surroundings from air to water, ex:

    diff_air = weight(real_gold, air) - weight(suspicious_crown, air)
    diff_water = weight(real_gold, water) - weight(suspicious_crown, water)
    assert(diff_air == diff_water, "Error, density mismatch detected")
A good counterfeiter will ensure diff_air==0, but that's just them trying to cheat a much-simpler "very similar mass" test, and it isn't a prerequisite for this "same density" test.
> ... consider that for this experiment we do not actually need to see an "equal" weight-measurement from any kind of scale.

Ah, you seem to be assuming that the mass of the crown is known and an equivalent mass of gold can be produced.

> What we're actually trying to check is that the readout doesn't change ...

I'm not sure what you mean by readout. If one side is A and the other B the balance only has three readings, A>B or A=B or A<B.

Typically I would expect the crown to be put on one side of a balance. Then the weigher would search for the amount of gold that makes the balance, well, balance (A=B). In effect the weigher chooses diff_air==0.

> A good counterfeiter will ensure diff_air==0

I don't see how the counterfeiter has any influence on diff_air, since the mass is not known beforehand.

Your equations seem correct.

> Ah, you seem to be assuming that the mass of the crown is known and an equivalent mass of gold can be produced.

The first is effectively true because you can weigh the crown in air, which for goldish-density crownlike objects is so relatively thin that you'll get similar results to doing the comparisons in vacuum.

The second assumption of a reference-sample was explicitly stated earlier in the thread, inside the Wikipedia quote.

> I don't see how the counterfeiter has any influence on diff_air, since the mass is not known beforehand.

The crooked crown-maker already knows (A) how much true-gold mass their customer is expecting them to deliver and (B) they have many opportunities to mass-measure and adjust the profitably-adulterated not-quite-pure-gold object they are creating.

You should use "/" instead of "-" in your equations. If you have two objects of the same density but with different sizes, the ratio of their (apparent) weights will be the same in air and in water, but the difference will be different!