You have a cube of ice floating in a glass of water. The question is what fraction of the ice will be above the water line? Assume that the ice is not bobbing.

Most of you have probably heard the answer to this before. But please provide a proof or solution, along with your assumptions.

I don't remember what the density of ice was, but it's slightly less than water, which is 1 g / mL.  Let's just call the density of ice d.

Now, for every unit of volume of the icecube beneath the water, one unit of water is displaced, and the weight of this water is pushing the ice cube up.  The total weight of the ice cube (above and below the water) is pushing down.  The fraction of ice above water will settle into an equilibrium where these two forces are equal.

Let the volume of the ice cube be V (it cancels out in the end).  Let the fraction of ice beneath water be x.

The force pushing down is equal to V*d.
The force pushing up is equal to x*V*1g/mL

At equilibrium, these two forces are equal:
x*V*1g/mL = V*d
x = d * mL/g

So the fraction of ice submersed in water is equal to the density in units of g/mL.

Interestingly enough, when an ice cube is added to water, the water level increases by the same amount as if the ice cube were melted before added.  This is easily verified:

Ice cube occupies V*x volume
Melted ice cube occupies V*d / (1 g/mL) = V*x volume

Posted by Tristan on 2005-04-22 01:28:27