Why Throwing a Rock Overboard Lowers the Pool

A boat is floating in a pool with a heavy rock aboard. Someone tips the rock over the side and it sinks to the bottom. Does the water level in the pool rise, fall, or stay where it was?

The rock never leaves the pool. Nothing is poured in, nothing is poured out, and the total mass of everything inside the walls is exactly what it was a second earlier. From that, most people conclude that the level cannot move. The level falls, and the reason it falls is worth more than the answer.

The trap starts from a true statement

Take the reflex seriously for a moment, because its premise is not wrong. The mass inside the pool really is conserved. What the reflex then does is treat the water level as if it were determined by that mass, and it is not. The level is set by the volume of water pushed aside by everything immersed in it, and a rock aboard a boat pushes aside a different amount of water than the same rock lying on the floor. So the correct reply does not deny the premise. It changes the quantity under discussion.

Those are the two halves of the same statement, and the whole problem lives in the gap between them. Set the density of water to $1$, so the rock has volume $V > 0$ and relative density $d > 1$ (it sinks, which is exactly what $d > 1$ means). While the rock rides in the boat, the boat has to carry it, so the boat sinks far enough to push aside water weighing as much as the rock. In our units that weight is $dV$, and a weight of water is a volume of water:

On the bottom, the rock is a submerged solid and nothing more. It pushes aside its own volume:

The boat itself drops out of the comparison. It got lighter when the rock left, so it floats higher, and it still displaces exactly its own weight of water, which has not changed by a gram. Its contribution is the same in both frames and cancels from the difference. Subtract (2) from (1):

Water that is no longer being held aside comes back into the pool and the surface settles. In a pool whose horizontal cross-section is $A$ over the range of levels involved, the level falls by

Both columns are volumes of water

There is a tempting way to draw this that is quietly incoherent, and it is worth naming because the first version of our own figure fell into it. Label the tall column "weight" and the short column "volume" and you have put a weight beside a volume, then asked the reader to subtract one from the other. Both columns are volumes of water. The tall one is measured by the rock's weight, the short one by the rock's size, and the argument is that the first measurement is the bigger number whenever the rock is denser than water.

The three-to-one ratio in that drawing is a choice, not a fact about rocks. It pins $d = 3$, which is roughly a heavy stone (granite sits near $2.7$), and it was picked so the difference between the two columns is visible at a glance. Nothing in equations (1) through (4) depends on the value. Any $d$ above $1$ gives the same sign.

The same answer from a force balance

Equation (3) is short but slightly abstract, because the phrase "displaced volume" is doing the work. Here is a second route with different machinery, and it has the advantage of showing exactly where the argument closes.

Look at the sunk rock alone and add up the forces on it. Its weight is $\rho_w\, d\, V g$ downward, where $\rho_w$ is the density of water and $g$ the acceleration of gravity. The buoyant force the water applies is only $\rho_w V g$, since the rock is fully submerged. Because $d > 1$, those two do not balance:

The shortfall $(d-1)\rho_w V g$ has to be supplied by something, and the only candidate is the floor. That is the whole point. While the rock was aboard, the water was holding up every last gram of it, through the boat. On the bottom, the floor takes over part of the load. Water that is supporting less weight is water that is standing aside for less, so the surface drops. The floor is where the argument closes, and no version of the reasoning that ignores the floor can be complete.

Why the reflex is a boundary case, read too widely

Set $d = 1$ in equation (3) and the change is exactly zero. A rock with the same density as water changes nothing when it goes over the side, because the weight of water it displaced while aboard is precisely the volume it displaces while submerged. The reflex answer is the correct answer to that case.

The same is true, for a different reason, of anything that floats on its own. Toss a sealed empty bottle out of the boat and it bobs beside it, still displacing its own weight of water, and the level does not budge. So "nothing changes" is right for neutrally buoyant objects and right for floating ones. It fails only for the case actually asked about, which is the one where the object sinks. The excess density $d - 1$is the entire reason the level moves at all, which is why "denser than water" has to be said out loud rather than smuggled in through the word "heavy".

What the answer quietly assumes

Four hypotheses sit under equations (1) to (4), and all four hold for the problem as stated.

• The boat still floats after the rock leaves. It does, because it got lighter. The only way to break this is a boat that was already sinking.
• The rock ends fully submerged. In a pool that is the intended reading, and it matters: equation (2) is about a submerged solid.
• The boat never touches the bottom, so its own displacement stays a floating displacement.
• The pool is prismatic over the range of levels involved, which is what lets us divide by a single cross-section $A$ in equation (4).

Only the last one is worth a second look, and it is harmless. If the pool has sloping walls, the cross-section is a function $A(h)$ instead of a constant, and the drop has to be found by solving $\int A(h)\,dh = (d-1)V$ over the interval swept by the surface. That changes the size of the drop. It cannot change the sign, because $A(h) > 0$ keeps the level monotone in the displaced volume. The pool gets shallower whatever shape it is.

Sources and further reading

• The principle behind equations (1) and (2) — Archimedes' principle
• Displacement as a measurement rather than a metaphor — Displacement (fluid)
• The quantity $d$ used throughout — Relative density
• The break-even case $d = 1$ — Buoyancy

The sign in equation (3) was checked symbolically, and then again by a numerical model that uses none of the reasoning above: it solves the boat's draft by bisection on its own buoyancy balance, solves the surface height by a second bisection on conservation of the water volume, and counts a rock on the floor only for the part of it below the surface. The level fell in all 192 grid configurations and in 4000 random draws over density, rock volume, pool area and boat mass, and the measured drop matched $(d-1)V/A$ to within $10^{-7}$ every time. Both boundary cases came out exactly flat: zero change at density 1, and zero change for a body at density 0.3, 0.6 or 0.95, which floats beside the boat and keeps displacing its own weight.