water buoyancy

If I weigh a cube of metal using a spring balance, then submerse the metal in water, the weight is less. Given that the metal is denser than water, and gravity stays the same, why is this?

This is called the “Archimedes Principle” http://en.wikipedia.org/wiki/Archimedes_Principle . The cube of metal displaces is volume of water. The displaced weight is subtracted from the weight of the metal (buoyancy).

One way to analyze the problem is to look at the pressure distribution on various parts of the metal cube. If it’s upright, the pressure on each of the 4 vertical faces is equal. The pressure on the bottom is greater than on the top (because the bottom is deeper in the water) - there is thus a net upward force.

Thanks, I actually conducted the experiment and found out the decrease in weight is approx. 15%. Then I started to wonder----is the loss of weight, equal to the we9gvolume of water displaced.

Thanks. As I said, I actually did the experiment and found that the loss of weight is approx. 15%. Is this loss equal to the weight of the water displaced.

It is.

So does this mean that if you weighed the cube in a vacuum (but under the same gravity), it would be fractionally heavier, by an amount equal to the weight of its own volume of air?

Yes. That’s why hot air balloons float: They take this feature to its logical conclusion and displace a weight of air greater than their weight.


Beat me to it. :wink:

Dagnabbit! I thought I had this displacement stuff figured out, until bmleff99 conked me with that metal cube. Now, I have some questions.

The OP didn’t say what metal the cube was made of. If the experiment were done with identical size cubes of, say, aluminum and lead, wouldn’t the percentage loss of weight be different? I thought that was Archimedes’s original puzzle.

Start with a vessel that isn’t full, but has enough water to cover the sunken cube without overflowing. Weigh the vessel & water. Add the cube, and let it sink to the bottom. Is the added weight, a) the same as the cube weighs in air, or b) the “air” weight, minus the weight of the water displaced?

Archimedes: Eureka!

Mrs. Archimedes: You don’t smell so good, either. Get back in the bath.

The percentage loss will be different, since the total weights are different. In fact, for some materials, the percentage loss can be 100% or greater (in which case, the material floats). But the amount of weight loss will depend only on the volume.

The additional weight will be the same as the cube weighs in the air.

Both the cube and the water still have the same mass, right? In addition, the Earth is still pulling on both the cube and the water with the same acceleration. Mass and gravitational acceration produce weight, and that weight doesn’t change whether the cube is in, out, over, or under the water.

In fact (and here’s where the confusion might be), if bmleff99 did his experiment by hanging the cube from a spring scale, and placed his bucket of water on a second scale, he’d see the “water scale” weight increase as the “hanging scale” weight increased.

I’m having a very hard time imagining this, but I’m curious. Could you elaborate a bit on how this would be set up?

Perhap’s I’m misunderstanding this (along with jackalope?), but shouldn’t the final word be “decreased”? As the cube is lowered into the water, it’s apparent weigh decreases by an amount equal to the weight of water displaced; the “water scale” reading would simultaneously increase by the same amount.

The percentage would indeed, be different, but the weight displaced would be the same. Example:

One cc of aluminum would be 2.70 grams.
One cc of lead would be 11.34 grams.

Immersed in water and weighed, the aluminum would be 1.70 grams (about 62% different) and the lead would be 10.34 grams (about 9% different).

Archy’s problem was determining the volume of the crown he was asked to verify, not the weight. The crown weighed the same as the amount of gold the jeweler had been given, but if, as suspected, some of the gold had been pocketed and the missing weight made up with the copper, the crown would weigh the same, but the volume would be higher than a pure gold crown. Mashing the crown down to a cube would make the problem trivial, but that was not an option

Say the gold crown weighed exactly one kilo. At a density of 19.3 grams per cc, its volume should be 51.8 cc He filled a basin with water to the brim, immersed the crown in it, carefully capturing the water displaced, and weighed the displaced water. If it was more than 51.8 grams, the jeweler was in trouble*. Likewise, he could have measured the volume of the water directly, if he had any graduate cylinders he could trust.

*IIRC the volume was too big and the jeweler was terminated with extreme prejudice.

Ah crap. Yes, the final word should be “decreased,” and you’re understanding it perfectly.

Ok, I think I’ve followed this thread so far, but now lets make it a little more complex. What if the cube was resting on a conveyor such that…

I was explaining this to a friend, and said that, because of the increase in density of water, there were more molecules of water hitting the bottom of the cube than the top. Thus creating the decrease in weight. He responded that he understood that water cannot be compressed, therefore there could not be more molecules hitting the bottom of the metal cube. How do I respond to this?

Water is nearly incompressible, so there’s no meaningful increase in density. But there is an increase in pressure with depth, which can be explained as the deeper water supporting the weight of the water above it.