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 sme depth?
What is the argument for your conclusion?
Hint: Suppose the fluid we poured into the upper part of the vat was more mercury. What would happen to the ball?
Compute fhe fraction of the volume the iron ball that is above the top level of the mercury in each situation. Answering this quantitative question will help answer and justify the answer to the original question.
Hint: Review the Law of Archimedes
The Law of Archimedes states that a floating body is bouyed up by a vertical forec equal in magnitude to the weight of the displaced fluid.
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.