# Understanding Buoyancy Using Archimedes’s Principle

In physics, *Archimedes’s principle* says that any fluid exerts a buoyant force on an object wholly or partially submerged in it, and the magnitude of the buoyant force equals the weight of the fluid displaced by the object. An object that’s less dense than water floats because the water it displaces weighs more than the object does.

If you’ve ever tried to push a beach ball underwater, you’ve felt this principle in action. As you push the ball down, it pushes back up. In fact, a big beach ball can be tough to hold underwater. As a physicist in a bathing suit, you may wonder, “What’s happening here?”

What is the buoyant force, *F*_{b}*, *the water exerts on the beach ball? To make this problem easier, you decide to consider the beach ball as a cube of height *h* and horizontal face with area *A. *So the buoyant force on the cubic beach ball is equal to the force at the bottom of the beach ball minus the force at the top:

*F*_{buoyancy}* = F*_{bottom}* – F*_{top}

And because *F* = *PA, *you can work pressure into the equation with a simple substitution:

*F*_{buoyancy}* = (P*_{bottom}* – P*_{top}*)A*

You can also write the change in pressure, *P** _{bottom}* –

*P*

_{top}*,*as delta

*P:*

Note that *hA* is the volume of the cube. Multiplying volume, *V,* by density,

gives you the mass of the water displaced by the cube, *m,* so you can replace

with *m:*

*F*_{buoyancy}* = mg*

You should recognize *mg *(mass times acceleration due to gravity) as the expression for weight, so the force of buoyancy is equal to the weight of the water displaced by the object you’re submerging:

*F*_{buoyancy}* = W*_{water displaced}* *

That equation turns out to be Archimedes’s principle.

Here’s an example of how to use Archimedes’s principle. Suppose the designers at Acme Raft Company have hired you to tell them how much of their new raft will be underwater when it’s launched. You can see the new Acme raft in the figure.

The density of the wood used in their rafts is 550 kilograms/meter^{3}, and the raft is 20 centimeters high.

You get out your clipboard and reason that to make the raft float, the weight of the raft must equal the buoyant force the water exerts on the raft.

Say the raft is of height *h* and horizontal surface area *A; *that would make its weight equal to the following:

Now what’s the buoyant force that the water exerts on the raft? The buoyant force is equal to the weight of the water that the submerged part of the raft displaces. Say that when the raft floats, the bottom of the raft is a distance *y* underwater. Then the submerged volume of the raft is *Ay*. That makes the mass of the water displaced by the raft equal to the following:

The weight of the displaced water is just its mass multiplied by *g,* the acceleration due to gravity, so multiplying both sides of the equation by *g* gives you the weight, *W*_{water displaced}*,* on the left side of the equation. The displaced weight of water equals the following:

For the raft to float, the weight of the displaced water must equal the weight of the raft, so set the values for raft weight and water weight equal to each other:

*A* and *g* appear on both sides of the equation, so they cancel out. The equation simplifies to

Solving for *y* gives you the equation for how much of the raft’s height is underwater:

Plugging in the densities tells you how far the raft is submerged in terms of the raft’s height:

That means that 55 percent of the raft will be underwater. So if the raft is 20 centimeters (or 0.20 meters) high, how much is underwater when it’s floating? You can plug in the value for the raft’s height to find the answer:

*y* = 0.550(0.20 m) = 0.11 m

So 11 centimeters of the raft’s height will be underwater.