1. If we had a jar of marbles or something else of different densities and shook it, the most dense ones would go to the bottom and the less dense ones to the top. enter image description here(Image Source)
  2. If I put a cube of lead in water it would sink all the way to the bottom.

But for ice : what I am trying to understand is why doesn't the water (being denser than the ice) seek to reach the bottom, and the ice sit flat on top of it (as in the left image)? Instead, some part of the ice is submerged in the water (as in the right image), and some sits on top it.

Ice and Water

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    $\begingroup$ Archimedes principle $\endgroup$ Oct 29, 2016 at 2:24
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    $\begingroup$ Hi Alfred. So is the only difference between the top picture and the bottom right picture, that in the bottom right picture the ice is a solid and hence can't expand to the sides of the container? But in the topic picture the red liquid can do that and hence float completely on top? $\endgroup$
    – K-Feldspar
    Oct 29, 2016 at 2:29
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    $\begingroup$ Yes, liquids settle at equipotential levels and disperse up to the solid boundary of the container. They stratify depending on the density. $\endgroup$
    – anna v
    Oct 29, 2016 at 4:33
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    $\begingroup$ A "bubble" of a lighter liquid does act like the ice cube, pressing down in the middle more than the surface of the water. But then the liquid spreads and reaches the sides of the container. At this point, it has pushed down all the water below it and is the same height at all points. $\endgroup$
    – ErikE
    Oct 29, 2016 at 6:33
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    $\begingroup$ If you put the various liquids in small plastic bags then they would behave like ice. $\endgroup$
    – slebetman
    Oct 29, 2016 at 17:55

8 Answers 8


When put in water, an objects sinks to the point where the volume of water it displaces has the same weight as the object. Archimedes was the one who discovered this.

When you put lead in water, the weight of the lead is much greater than that of the same volume of water. Hence it sinks to the bottom. As ice only weighs about 90% of its volume of water, 90% of the ice will be under water, the rest above. The actual figure is 91.7%, given by the specific gravities of water (0.9998) and ice (0.9168) at 0C.

Actually, in the case of lead, if the water were deep enough, the lead would sink to the point where its weight equals that of the water under pressure at depths. As lead will compress as well as the water, that may never happen, but for other objects and/or fluids it might.

This is also the reasons why helium-filled balloons float up: their weight is less than that of the same volume of air. As they float up, the balloon expands, while the air gets rarer and hence lighter. At a certain altitude the two will be equal and the balloon will stop rising.

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    $\begingroup$ Thanks for the detailed answer hdhondt. Would you please be able to also explain why in the first image, the liquids of different densities have distinct boundaries based on their densities (whereas for ice, it is not a distinct boundary but half in and half out)? I.e. why doesn't what is true for the ice hold for my first image on top of the coloured liquids? If it did hold wouldn't the liquids be partially mixed? $\endgroup$
    – K-Feldspar
    Oct 29, 2016 at 3:19
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    $\begingroup$ The density difference results in a force (buoyant force) that pushes the less dense object up. In the case of the ice cube, it's a solid, so it sinks until the buoyant force cancels out the force exerted by gravity. In the case of a liquid, the forces will cause it to deform and spread out. If there is a container, the liquid will spread to fill the container. If not, it will spread out indefinitely. Oil on water will spread out until it is extremely thin, on the order of a few wavelengths of light, resulting in 'negative rainbow' colors due to interference effects. $\endgroup$ Oct 29, 2016 at 6:02
  • $\begingroup$ And actually, the ice does have a distinct boundary, it's just not a nice flat plane. When you remove the ice cube, you don't expect a hole to remain in the water, right? Gravity pulls down any high spots and the fluid redistributes to form a flat surface. This happens on any fluid boundary - fluid/gas and fluid/fluid of different densities. $\endgroup$ Oct 29, 2016 at 6:05
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    $\begingroup$ The density of water doesn't change much at depth though, lead would always sink all the way down. $\endgroup$
    – Firebug
    Oct 29, 2016 at 20:01
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    $\begingroup$ Water does not appreciably change in density as a function of pressure. Under extreme pressures it will simply solidify into Ice VI, VII, or X. $\endgroup$
    – J...
    Oct 31, 2016 at 19:10

I like to answer by reinterpreting your question: if you expect the ice to be completely atop the water because ice is less dense than water (as indicated in your left image), then you would also expect the ice to be completely below air because ice is more dense than air (in order for this to be true, think pushing your ice cube down into the water so that it's top surface is level with the water surface). Both is not possible at the same time, obviously. So you see that your reasoning does not give you a well-defined solution, and hence your reasoning (should it not sit completely atop water?) must be false.

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    $\begingroup$ I like this question but I am unsure how to answer it. The ice cube cannot be completely below the air as the air is everywhere (even if we push the ice cube down it could not be completely below the air). But for the water and ice, if we say, held the ice cube up, it could exist above the water). So I am not really sure how to relate both cases. $\endgroup$
    – K-Feldspar
    Oct 29, 2016 at 7:42
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    $\begingroup$ @K-Feldspar Why do you call my answer a question? And why do you want to answer it? :) My answer is an answer to the question "why doesn't it sit completely atop water" (= why does your left image make no sense), because I show that by the same argument, it should be completely below air. I will try to clarify that. $\endgroup$
    – bers
    Oct 29, 2016 at 9:05
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    $\begingroup$ Hi Ben. I didn't mean a question exactly, just a counter example to my original question to get me to understand the phenomenon. As to why I wanted to answer it, by answer I meant state the difference between the two cases so I could explain/understand the idea :) Sorry if I confused you. $\endgroup$
    – K-Feldspar
    Oct 29, 2016 at 9:10
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    $\begingroup$ This is a good argument, in that it shows that the OP's intuitive reasoning is flawed. However, it also implicitly suggests that it's the presence of air above the water that somehow pushes the ice cube down, which is not true -- even with no air, the ice cube would still be mostly submerged, just because that's the configuration that minimizes the combined potential energy of the ice and the water. (Of course, removing all the air would also make the water start boiling due to the pressure drop, but let's ignore that detail.) $\endgroup$ Oct 30, 2016 at 12:48
  • $\begingroup$ [...] Indeed, reducing the density of the air (say, by replacing it with helium) would actually make the ice sink slightly deeper, since the total buoyancy equals the combined weight of the water and the air displaced by the ice cube. With less air to displace, the cube must sink a little bit deeper into the water for the buoyancy to equal its own weight. $\endgroup$ Oct 30, 2016 at 12:52

I think the real question you're asking here is: why do less-dense fluids completely stay above more dense fluids, whereas less-dense solids partly sink?

If the ice covered the same area as the water below, then it would sit completely on top of the water. That's the case in a completely ice-covered pond.
But because the ice does have a non-negligible weight, the ice-water surface has higher than atmospheric pressure – this is what supports the ice layer from sinking. However, if there's any gap in the ice cover, the excess pressure will push water up until its own hydrostatic pressure compensates it. In particular, if most of the surface in not covered with ice, then there will be a lot of water rising above the bottom of the ice. This is more sensibly described the other way around: the ice sinks some way into the water.

Now, that gives you a state which is not quite optimal, energetically: there are portions of water sitting higher than portions of ice. If you could swap these portions, you'd get a more stable state.

Optimising energy of a floating ice block

The limit of this process is a state where all ice is sitting above water. However, since the ice is solid, you can't really do this without cutting / breaking off pieces of ice.

In the case of two fluids of different density, it is however easily possible to move portions of the lighter material around, and because it's energetically favourable it will actually happen automatically.


The difference between the picture of the liquids all nicely layered on top of each other and an ice cube in water is that the liquids can change their shape to fit the shape of the container while the ice cannot. If the ice could change its shape and flow to fill the container, you would see exactly what your intuition predicts: the ice sits fully on top of the water. In fact, if you visualize a lake that has frozen over, that's exactly what you get! However, your ice cube cannot flow like that. So, instead, it follows Archimedes's principle, as others have suggested, and "displaces" an amount of water equal to its mass. Since it is less dense than water, that amount of displacement will lead the ice cube to rest partially submerged.

As for the marbles shaking to the top, that's slightly different. It's the same principle, just with a twist. In the case of the container of marbles being shaken, it's a stochastic process. The statistical expectation is that the average shake will cause the less-dense objects to rise to the top and more dense objects to sink. However, we rarely shake them enough to fully separate the marbles into distinct crisp layers. Compare this to liquids. The equivalent case for a liquid would be lots and lots of little marbles, and you're shaking them continuously really really fast. Naturally, we see more stratification in liquids than we do in solids, simply because of this.

Note: there is also a second effect in the case of the marbles. The above example occurs only if every object is of the same size. In the case of mixed media (such as a container of mixed nuts), this effect is overshadowed by another effect. Small objects can pile in underneath large ones, so in general the large ones (such as the macadamia nuts) rise to the top.

Even in this case, with an effect which is more pronounced than density, we don't see a perfect stratification of the mixed nuts. There's always a small number of them that are not in the "right place," such as a peanut that manages to get thrown on top of a sea of walnuts. So when you compare three cases, the stratified liquids, the ice on top of water, and the container of marbles, you actually don't see the stratified liquids and marbles "acting the same" and ice acting strange. You really see the liquids being most stratified, then ice on top of water, and the marbles actually provide the least stratification of the three examples. If your intuition tells you otherwise, that may suggest that your intuition is accounting for some of these random factors for you, and biasing your results.


I'll try to explain this using some mathematics. Let us have an ice cube floating in water. Let the density of water be $ \rho_1 $ and that of ice be $\rho_2$. Let the volume of the ice cube be $v$. Let the submerged volume be $v'$. If you consider the forces on the ice block: $$ \rho_2 \cdot v \cdot g = \rho_1 \cdot v^\prime\cdot g $$ Cancelling g from both the sides, $$ \rho_2\cdot v = \rho_1\cdot v^\prime$$ Now, $$ v^\prime / v = \rho_2 / \rho_1 $$ Clearly, $\rho_2 < \rho_1$ . That means, $$\begin{align} v^\prime / v &< 1 \\ \implies~~~~~~ v^\prime & < v\end{align}$$

This means that the submerged volume (v') is less than the cube's total volume. That also means that the cube will not fully submerge in water, provided no external force is applied.


Lots of good answers already, but I think this is the kind of intuitive answer that you're looking for:

Understand that the "information" that the water has to "know" for this behavior to occur is not the density of the ice. Put another way, the water supporting the ice doesn't "know" the density of the ice above. It doesn't think, "Oooh, this is one un-dense solid! I'll push it aaall the way up!!" Rather, the water only feels the weight of the object.

It sounds silly until you realize it conveys the crucial point: if you take the part of the ice that is not touching the water (which is practically all of it except the molecules at the submerged surface) and replace it with some other material of equal weight and center of gravity but different density, then, intuitively, nothing should change -- even if the density of the object is higher now.

This intuition should tell you that low density is neither necessary nor sufficient for flotation. Rather, it's only necessary to prevent complete submersion, which of course the ice is not doing. How an object actually floats would depend on other factors, like the shape of the object, its center of gravity, and how it was initially submerged in the water.

Hope this addresses the crux of the issue for you!


People have been answering based on force balance, but there's an equivalent (and, in my opinion, more satisfying) answer based on energy conservation.

Suppose the ice cube is initially just on top of the surface of the water. If you lower it a tiny bit into the water, a tiny bit of water will be forced to move up, increasing the gravitational potential energy. But the entire ice cube will move down, significantly decreasing the energy. So the minimum energy configuration always has the object sinking into the water at least a little bit.

Note that this argument doesn't apply to two fluids, where the equilibrium configuration really does have the lighter fluid on top. Why? Consider moving a cube of the lighter fluid down. The same arguments above hold, except that when the heavier fluid moves up, all of the lighter fluid is forced to move up too. So the completely separated configuration really does have minimum energy.

In the ice cube example, the 'lighter fluid' being pushed up is air, not ice. Since air is much lighter than water, the ice cube sinks.

  • $\begingroup$ Actually the crucial thing is not that only a tiny bit of water will be forced up when lowering the ice cube infinitesimally into the water, but that the water that's initially forced up also only has to travel a tiny distance up. Thus $\Delta \!\!U_\text{water} \propto \Delta\!\!h^2$, but $\Delta\!\!U_\text{ice} \propto \Delta\!\!h$. $\endgroup$ Oct 30, 2016 at 11:27
  • $\begingroup$ @leftaroundabout I left that detail out, because everything moves approximately the same tiny distance. In $\Delta U = mgh$, the real difference is in $m$, not $h$. $\endgroup$
    – knzhou
    Oct 30, 2016 at 17:45

I would say what is shown in your second image out of three, where the piece of ice is on top of the water, is statically impossible because the pressure on the water surface under ice is greater than the pressure on the water surface between the ice and the walls of the container, so water is squeezed up between the walls and the ice. On the other hand, if the ice fit tightly in the container, it could "float" on top of water.


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