Iron Ball Floating in Mercury

In Mathematical Discovery, George Polya presents the following problem:

An iron sphere is floating in mercury. Water is poured over the mercury and covers the sphere. Will the sphere sink, rise, or remain at the same depth?




Setting up

If we take the specific gravity of a substance and multiply by the volume, we will get the weight. Set

a = specific gravity in the upper "fluid" (air in the first stage and water in the second stage)

b = specific gravity in the lower "fluid" (mercury in both stages)

c = specific gravity of the floating solid (iron in both stages)

v = total volume of the floating solid

x = volume of the floating solid that is in the upper fluid

y = volume of the floading solid that is in the lower fluid

We can write two equations in terms of x and y:

x + y = v (The total volume of the floating body is the sum of the two parts.)

ax + by = cv (The weight of sphere -- cv -- is balance by the combined weights of the two displaced fluids

Solve for x and y in terms of a, b, c, and v

Approximate specific gravities for our question:

air = 0
Water = 1
Iron = 8
Mercury = 14

Therefore CASE 1 -- air over mercury -- x = (6/14)v and y = (8/14)v

Therefore CASE 2 -- water over mercury -- x = (6/13)v and y = (7/13)v

Interpretation: The iron ball will rise as the water is added.