Why is pressure greater at the bottom end of a box submerged in water?

It is my understanding that the pressure acting on the bottom end of the box would be greater than on the top. But if we place a box under water, then the height of water above the top end of the box and the bottom end of the box is still the same, assuming there is no water inside the box, so why would the pressure be greater at the bottom?

1 Answer

I assume you talk about a water reservoir subject to the gravitational field of the earth, and that you consider a thermodynamic equilibrium situation (which implies also mechanical equilibrium within the water). To find an answer to your question, first look at a column of water with lateral cross section $$A$$ in the absence of the box. We cut a thin horizontal slice of height $$dz$$ out of the column at vertical position $$z$$. The pressure above the slice is $$p(z)$$, the pressure below the slice is $$p(z+dz)$$. The force pushing downward on the slice is $$F_\downarrow = p(z)A,$$ the force pushing upwards on the slice is $$F_\uparrow = p(z+dz)A.$$ In addition, a gravitational force $$F_\mathrm{G} = g\rho\cdot A\cdot dx$$ acts downwards on the slice. Here, $$\rho$$ is the mass density of water (we assume it to be independent of pressure here, which is a good approximation for water in most practical cases).

For the slice to be in mechanical equilibrium, we must have $$F_\downarrow + F_G = F_\uparrow.$$ Inserting our expressions from above, we find $$p(z)A + \rho\cdot A\cdot dz = p(z+dz)A\quad\Rightarrow\quad \frac{p(z+dz)-p(z)}{dz}=\frac{dp}{dz} = g\rho.$$ This means, with increasing depth below the water surface, there is a linear increase of the pressure, with the rate of increase given by $$\rho g$$. We find the pressure as a function of depth $$z$$ below the water surface $$p(z) = p_0 + g\rho z,$$ where $$p_0$$ is the pressure at the water surface.

Now that we know $$p(z)$$ of the water in the absence of the box, we can come back to your box problem: think of two points at depth $$z$$ below the bottom of the submerged box, one having a water column above it, the other having a bit of water column but also the box above it. The key insight is, that there cannot be a pressure difference between these two points in equilibrium. The absence of the pressure difference is clear from the fact that the water is in mechanical equilibrium; a lateral pressure gradient would indicate a non-equilibrium situation and start water to flow laterally at the same height.

Suppose, the box is a cuboid with lateral area $$A$$, height $$h$$ and negligible weight, filled with air. The top surface is at a depth $$z$$ below the water surface. The pressure acting onto the top surface of the box is $$p(z)$$, that at the bottom surface is $$p(z+h)$$. This leads to an imbalance of pressure forces on the box, which is $$p(z)A-p(z+h)A = -g\rho A h.$$ This force, called the buoyancy force, has a negative sign, because it pushes the box upwards. In order to keep the box submerged at constant depth, you have to push it with a force of exactly the same magnitude downwards to balance the buoyancy force.

If the box were filled with water, the pressure forces would be the same, but an additional gravitational force $$g\rho A h$$ would act downwards. You see that this force exactly balances the force due to the pressure difference. This is Archimedes' principle. The upward buoyant force is exactly equal in magnitude as the weight of the fluid that the submerged box displaces.