How do I do buoyancy problems? - MCAT

First, you must understand this principle: the buoyant force, which is directed upward, is equal by definition to the weight of the fluid displaced.

Problem type 1: the object is floating, in equilibrium, on the surface of a fluid.

If the density of the fluid above the surface can be neglected (i.e., it’s much lighter than the fluid below, usually a gas above a liquid), then you need to understand two things:

• The buoyant force, which is equal to the weight, mg, of the fluid displaced, also equals the weight of the object (because the object is in equilibrium); it follows that the mass of the object is exactly equal to the mass of the fluid displaced.
• The portion of the object that is below the fluid equals the ratio of the object’s density to that of the fluid.

The second principle follows from the first. Notably, it means that the way an object floats does not depend on how strong gravity is.

If the density of the fluid above the surface cannot be neglected (i.e., it’s a liquid or a very dense gas), then the buoyant force is the sum of the weights of the respective fluids displaced by the portions of the object above and below the surface. In practice, subtract the density of the lighter fluid from the density of the the heavier fluid, use the principle above for solving problems neglecting the upper fluid to get a density of the object, and then add the density of the lighter fluid back in to get the virtual density of the object. (If you’re finding how much is below the surface of the heavier fluid, subtract the density of the lighter one from both the heavier fluid and the object, and find the ratio as above.) An example shows it better:

Question: An object is floating at the surface of two fluids; two thirds of its volume is below the surface. If the densities of the two fluids are 1.0 g/cm^3 and 1.6g/cm^3, what is the density of the object?

Answer: Subtract the density of the lighter fluid from that of the heavier, and we get 0.6 g/cm^3. Two thirds of the object is below the surface, so that yields a density of 2/3 * 0.6 = 0.4 g/cm^3. Then we need to add the density of the first fluid back in: 0.4 + 1.0 = 1.4 g/cm^3.

Problem type 2: the object is below the surface of the fluid.

The easiest way to do these problems quickly is to do everything in terms of water: figure out what the buoyant force would be if the fluid were water, and then convert at the end. You need to have the density of water memorized in three different units: kg/m^3, g/cm^3, and kg/L. Note that a liter (L) is equal to 1000 cc’s; many MCAT objects are most easily considered in these units.

To solve the problem, you will want to find the buoyant force, which is the weight of the fluid dispaced. To do this, you find the mass of the fluid displaced, and at first you pretend it’s water. After finding the buoyant force, you subtract from the weight of the object to find its apparent weight; use that to get acceleration if needed. An example:

Question: A 20 cm x 40 cm x 5 cm object of density 0.4g/cm^3 is submerged with specific gravity of 1.2g/cm^3. Find the apparent weight of the object, and its acceleration.

Answer: The object’s volume is 4000cm^3, or 4L. If the fluid were water, the mass displaced would therefore be 4kg (because the density of water is 1kg/L), so the buoyant force would be 4kg x g = 40N. This fluid is 1.2 times as dense as water, so the buoyant force is 1.2 x 40N = 48N.

The object is 0.4 times as dense as water, so its weight is 40N x 0.4 = 16N (and its mass is therefore 1.6 kg; we use that below). (Note that again we do everything in terms of water, because it’s the easiest way. Note also that the volume of the object is, by definition, the same as that of the fluid displaced.) The apparent weight is therefore 16N - 48N = -32N. The negative weight means the net force on the object is upward; this makes sense because it’s lighter than the fluid, i.e., it wants to float.

To find the object’s acceleration, use F = ma –> a = F/m = 32N/1.6kg = 20m/s^2.

The reliance on water makes many calculations easier to do quickly, and also provides a running sanity check — you know it immediately when you are getting a nonsensical result — but it actually doesn’t change the method in a technical sense.