# Archimedes' principle - Buoyant force on a partially submerged object

Problem statement:

A gold nugget of mass m (and density ρAu) hangs from a can of radius r and negligible mass. When it floats, the fraction of the can immersed in water has a height h (see figure). If the nugget is put in the can, find the quotient between h’ (the height that is now underwater) and h.

Givens: ρ = 1.03 103 kg/m3; ρAu = 1.93 104 kg/m3 Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

Solution:

We are going to apply Archimedes’ principle to solve this problem. The difference between the situations (1) and (2) in the figure is the displaced volume of water. In (1) water is displaced by the can and the gold nugget, whereas in (2) water is displaced only by the can. The submerged part of the can is larger in (2) because the total weight is the same in both cases but in the second one the nugget doesn’t displace water, so the volume displaced by the can has to compensate for it.

In both situations the magnitude of the weight must be equal to the magnitude of the buoyant force. Below are the equations for both situations.

(1) We calculate the magnitude of both the weight and the buoyant force and equalize:  Since the volume of the gold nugget is unknown, we can write it in terms of its density: Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

On the other hand, the volume of the submerged fraction of the can is given by: Substituting and isolating m we obtain: (2) Now, we calculate the magnitude of the buoyant force in this case and follow the same procedure:   Equalizing (1) and (2) and substituting the givens: As expected, the can sinks more in the second case. The expression above allow us to calculate the immersed fraction of the can as a function of the density of the object attached to its bottom.

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