Physics questions are the most fun when people don’t immediately agree on the answer. What feels intuitive or obvious—sometimes isn’t. We can argue over the solution for hours of entertainment, and we might even learn something in the end.

Here’s one of these seemingly obvious questions that’s been around a long time: Suppose a large rock is on a boat that is floating in a very small pond. If the rock is dumped overboard, will the water level of the pond rise, fall, or remain unchanged?

Go ahead and debate it with your friends and family. While you convince them that your answer is correct, here is a picture of my boat with a rock in it:

OK, it’s not actually a boat, it’s part of a plastic bottle. Also, the “rock” is a lead weight and the “pond” is a beaker. But this way we can see what happens to the water level when we drop an object into it.

When a boat is floating on water, two forces are acting on it. First, there is the downward-pulling gravitational force, which is equal to the mass of the boat and everything on it (m) times the gravitational field (g=9.8 newtons per kilogram). We often call this product the “weight.”

The other force is the upward-pushing buoyancy interaction with the water. Two things are true about this buoyancy force. First, if the boat is floating, then the upward buoyancy must be equal to the weight of the boat. Second, the buoyancy force is equal to the weight of the water displaced by the boat.

We can calculate this buoyancy force by taking the volume of the water displaced (V_{d}) and using the density of water (ρ_{w}) along with the gravitational field (g).

With that, we can look at our tiny boat in a tiny pond. Just to make things as simple as possible, let’s assume the boat has massless walls—which is not a crazy approximation, since my boat is really a plastic bottle. That means the only weight is the rock. (Don’t worry, I will do a more realistic example later.) Here’s a force diagram:

Because the buoyancy force (F_{B}=ρ_{w} × V_{d} × g) is equal to the weight of the rock (m_{r} × g), we can find an expression for the volume of displaced water (V_{d}):

We need this volume because this is the amount the water level in the “pond” rises when the boat is added. It’s literally the definition of displaced water.

Now, let’s drop the rock into the water. The boat no longer matters and doesn’t displace any water since it doesn’t have a mass. Here’s the force diagram for just the rock:

Since this rock is much smaller than the boat, it displaces less water than when it was on the floating boat. This means that the upwards-pushing buoyancy force is also smaller—and now, it’s not a large enough force to balance the downward gravitational force. So instead, the rock rests on the floor of the pond (or beaker), which provides an extra upwards force (F_{f}).

With the smaller buoyancy force, there is less displaced water. That means it’s time to reveal the answer to our question: The water level will go down!

Are you surprised? People often think the answer is that the water level will rise, because the rock will displace water and force the water level upwards—but they are wrong, and that’s OK. We often base answers on our previous life experiences, and you’ve probably done something like adding marbles to a glass of water to make the level rise. It seems reasonable to imagine that the same thing will happen in this case.

But adding marbles to a glass is different than our boat scenario. Because the marbles are not floating in a boat, they’re not in the water in the first place. They’re probably in your pocket, or something like that—and taking a marble out of your pocket has no impact on the water level. When you plop that marble in the glass, the water has nowhere to go but up, and the water level rises. This is the same thing that would happen if you threw a rock into a pond while standing on the shore.

On the other hand, a rock on a boat is *already* displacing the water before it is dropped into it. That’s why the two cases are different, and that’s what makes this a fun physics question.

Here’s an actual picture using my beaker pond and plastic bottle boat:

We can even get an expression for the amount the water level dropped. Remember that we already calculated the volume displaced by the combination of the rock and the boat. Let’s call that V_{1} for the initial volume of water displaced. Now, with the rock on the bottom, it’s just going to displace an amount equal to the volume of the actual rock. Suppose this rock has a density of ρ_{r}. Then it will displace a volume V_{2}:

This gives a difference in volume, or the volume of water drop in the pond:

Just for fun, let’s put in the values from my tiny experimental version of this boat and rock. In this case, the rock is actually a lead weight with a mass of 130 grams. The density of water is 1 gram per cubic centimeter (g/cm^{3}), and the density of lead is 11.3 g/cm^{3}. Putting these values in gives a water level drop of 118 cm^{3}.

Looking at the readings on the beaker, with the mass in the boat the water is at 670 milliliters (which is 670 cm^{3}). When the mass is put in the water, it drops to 560 mL for a change of 110 mL. That’s fairly close to my calculation. *Nice*.

What About a Real Boat in a Real Pond?

OK, fine, let’s do the calculation for the real thing. Let’s imagine that I have a small boat with a mass of 100 kilograms (m_{b}). The boat is carrying a person with a mass of 70 kg (m_{p}) plus a 50-kg rock (m_{r}). The pond is a perfectly cylindrical pool with a radius of 3 meters and a depth of 2 meters.

First, I need to calculate the volume of water displaced when the rock is in the boat. The downward gravitational force (which equals the buoyancy force) would be equal to the gravitational field multiplied by the sum of the masses (boat plus person plus rock). We can use that to find V_{1}:

When the person throws the rock overboard, we have two volumes to consider. There’s the volume displaced by the boat plus the person, and then there’s the volume displaced by the rock on the bottom, which depends on the density of the rock.

When we take the difference in these two volumes, the part with the boat mass cancels.

See, I told you it was OK to use a massless boat! As long as the rock is denser than the water (which means it’ll sink), that expression on the right will be negative and the water level will drop when we toss the rock out of the boat.

If I put in our actual values for the masses and use a rock density of 4 g/cm^{3}, I get a volume drop of 0.03 cubic meters. If our cylindrical pond has a radius of 3 meters, the water level will drop a distance of 1 millimeter. Yes, that’s a super tiny drop in volume for our super tiny pond. But it’s still a drop in the water level—and it will always be a fun physics problem.