Tip of the Iceberg Is it possible to approximate the volume of an entire iceberg if one happens to know the volume of the visible tip? (Assume the water is perfectly still.) If so, how may that be done?
 A: An iceberg is freshwater ice. The density of pure ice is about 920 kilograms per cubic meter, or 57 pounds per cubic foot, and the density of sea water is about 1025 kilograms per cubic meter, or 64 pounds per cubic foot. So approximately 10% of the iceberg is above water.
A: The upward buoyancy force on the iceberg is equal to
$$F_{b}=g\rho_{{\rm water}}V_{{\rm under}},$$
in terms of the acceleration of gravity, the density of the liquid phase of water and the volume of water that is displaced by the iceberg (equal to the volume $V_{{\rm under}}$ of the iceberg that is underwater). If the iceberg is floating in equilibrium, this must be equal to the downward gravitational force on the iceberg,
$$F_{g}=g\rho_{{\rm ice}}(V_{{\rm under}}+V_{{\rm above}}).$$
Equating these gives an equation for the underwater volume in terms of the other quantities (all of which are known, including $V_{{\rm above}}$),
$$\rho_{{\rm water}}V_{{\rm under}}=\rho_{{\rm ice}}(V_{{\rm under}}+V_{{\rm above}}),$$
The solution for the unseen volume is straightforward,
$$V_{{\rm under}}=\frac{\rho_{{\rm ice}}}{\rho_{{\rm water}}-\rho_{{\rm ice}}}V_{{\rm above}}.$$
Note that, as expected, this has no physical solution unless the ice is less dense than the water.
