Pressure in water and buoyancy

Stand in deep snow in ordinary shoes and you sink in. Strap on a pair of skis and you stay near the top. Your weight is the same both times; what changed is the area that carries it. A force spread over a larger area presses less hard on each square metre. Physics measures this with pressure: the force divided by the area it acts on.

Force is measured in newtons and area in square metres, so pressure is measured in newtons per square metre, a unit called the pascal: $1\ \mathrm{Pa} = 1\ \mathrm{N/m^2}$. A pascal is small, so we often use kilopascals, $1\ \mathrm{kPa} = 1000\ \mathrm{Pa}$.

Change the area under the same load of $600\ \mathrm{N}$ and watch how far it sinks into the soft snow.

In shoes with a total sole area of $0.05\ \mathrm{m^2}$, a weight of $600\ \mathrm{N}$ gives $p = 600/0.05 = 12\,000\ \mathrm{Pa} = 12\ \mathrm{kPa}$. On skis with $0.3\ \mathrm{m^2}$, the same weight gives $p = 600/0.3 = 2000\ \mathrm{Pa}$: six times less, because the area is six times larger.

Now think about water at rest in a tank. Pick a small flat patch somewhere under the surface, lying level, with area $A$ and at depth $h$. What presses down on it? The water directly above it. That water forms a column with the patch as its base and the surface as its top.

The column has the volume $Ah$: its base area times its height. Density, written $\rho$ (the Greek letter rho), is the mass of each cubic metre; one cubic metre of water has a mass of $1000\ \mathrm{kg}$, so water has $\rho = 1000\ \mathrm{kg/m^3}$. The column’s mass is therefore $\rho A h$, and its weight is $\rho A h g$, where $g = 9.81\ \mathrm{N/kg}$ is the weight of each kilogram. The column is at rest, so something holds it up: the water below pushes up on its base with a force equal to its weight. (The water beside the column pushes on its sides, sideways, so it cannot help to hold the column up.) Dividing that force by the area gives the pressure at depth $h$:

The area $A$ cancels, so the pressure does not depend on how large a patch we chose; it depends only on the depth, and it grows in proportion to it: twice as deep, twice the pressure. At $h = 1\ \mathrm{m}$, $p = 1000 \times 9.81 \times 1 = 9810\ \mathrm{Pa}$, about $9.8\ \mathrm{kPa}$.

Move the patch deeper and watch the column above it grow. The patch is $10\ \mathrm{cm} \times 10\ \mathrm{cm}$, so $A = 0.01\ \mathrm{m^2}$.

At $h = 2\ \mathrm{m}$ the column holds $0.01 \times 2 = 0.02\ \mathrm{m^3}$ of water, with a mass of $1000 \times 0.02 = 20\ \mathrm{kg}$ and a weight of $20 \times 9.81 = 196.2\ \mathrm{N}$. Spread over $0.01\ \mathrm{m^2}$, that is $196.2/0.01 = 19\,620\ \mathrm{Pa}$, the same as $\rho g h = 1000 \times 9.81 \times 2$.

We have left out the air. The atmosphere also presses on the water surface, with about $101\ \mathrm{kPa}$ at sea level, and that adds to the pressure at every depth: $p = p_0 + \rho g h$, where $p_0$ is the air pressure. Gauges usually show only the part caused by the water, $p - p_0 = \rho g h$, which is called the gauge pressure. Ten metres down the water adds $1000 \times 9.81 \times 10 = 98\,100\ \mathrm{Pa}$, almost as much again as the whole atmosphere, which is why divers feel it in their ears.

One more consequence: two points at the same depth in still water must have the same pressure. If they did not, the water between them would be pushed from the higher pressure towards the lower one and would start to flow, but it is at rest. So in still water the pressure depends only on the depth, whatever the shape of the container.

So far our patch lay level and was pushed from above. But water also presses on the walls of a tank and on the sides of a fish. Experiments show that at any point in still water the pressure pushes equally hard on a small surface facing any direction, up, down or sideways, and always straight at the surface, at right angles to it. The second part has a simple reason. Water cannot keep its shape: unlike a block of wood, it cannot resist a push that tries to slide one layer over another. So still water can only push straight at a surface; any sideways part would make it flow.

Now put something in the water: a box whose top and bottom faces are level, each with area $A$, and whose height is $H$. Its top face is at depth $h_1$ and its bottom face at depth $h_2 = h_1 + H$. Water presses on all six faces.

Look at the four sides first. Each side has another side directly opposite it, covering the same depths, and from chapter 2 the pressure depends only on depth. So the pushes on opposite sides are equal and point in opposite directions, and they cancel. The sides squeeze the box, but they do not move it.

The top face is pushed down by the pressure at depth $h_1$, and the bottom face is pushed up by the pressure at depth $h_2$. A force is a pressure times an area (chapter 1), so

The bottom is deeper, so the upward push wins. The difference $h_2 - h_1$ is just the height $H$ of the box, and $HA$ is its volume $V$. So the water pushes the box up with the net force $\rho g V$. Look at what this is: $\rho V$ is the mass of a box-sized amount of water, and $\rho V g$ is that water’s weight. The upward force from the water equals the weight of the water the box pushes aside. This net upward force is called buoyancy, and the rule is Archimedes’ principle. (We used gauge pressure. The air’s $p_0$ adds the same force $p_0 A$ to the top and to the bottom, so it cancels.)

Move the box up and down and change its height. The box is $0.5\ \mathrm{m} \times 0.5\ \mathrm{m}$ across, so $A = 0.25\ \mathrm{m^2}$. The arrows’ lengths show the pushes.

With $H = 0.4\ \mathrm{m}$ and the top at $h_1 = 0.6\ \mathrm{m}$, $F_{\text{down}} = 1000 \times 9.81 \times 0.6 \times 0.25 = 1471.5\ \mathrm{N}$ and $F_{\text{up}} = 1000 \times 9.81 \times 1.0 \times 0.25 = 2452.5\ \mathrm{N}$, a difference of $981\ \mathrm{N}$. The box’s volume is $0.4 \times 0.25 = 0.1\ \mathrm{m^3}$, and $0.1\ \mathrm{m^3}$ of water weighs $1000 \times 0.1 \times 9.81 = 981\ \mathrm{N}$. Lower the box by $0.5\ \mathrm{m}$: the pushes become $2697.75\ \mathrm{N}$ and $3678.75\ \mathrm{N}$, and the difference is still $981\ \mathrm{N}$.

Notice that $h_1$ has disappeared. Lower the box deeper and both pushes grow by the same amount, so their difference stays the same: once the box is completely under water, its buoyancy does not depend on how deep it is. If the top sticks out of the water, no water pushes on the top, and the upward push is $\rho g h_2 A$: the weight of the water that would fill just the part of the box below the surface.

A box was easy because its top and bottom are flat and level. A ball or a fish has curved surfaces, where the pushes point in many directions. Instead of adding them all up, we use an argument that works for any shape. Imagine the object taken out and its space filled with water instead: the same shape, in the same place. That water is part of the still water in the tank, so it does not move. The total push of the surrounding water on it must therefore exactly balance its weight: it points upwards and equals the weight of that water, $\rho V g$. Now put the object back. The surrounding water is at the same depths as before, so it pushes on the object’s surface exactly as it pushed on the water. The buoyancy on any object equals the weight of the water it pushes aside:

Here $V_{\text{sub}}$ is the volume of the part of the object that is under water, because only that part pushes water aside.

Will an object rise or sink? Compare its weight with the largest buoyancy it can get. An object with density $\rho_{\text{obj}}$ and volume $V$ has the mass $\rho_{\text{obj}} V$ and the weight $W = \rho_{\text{obj}} V g$. Completely under water, its buoyancy is $\rho_{\text{water}} V g$. Both contain $Vg$, so the comparison comes down to the densities. If $\rho_{\text{obj}} > \rho_{\text{water}}$, the weight wins and the object sinks; if $\rho_{\text{obj}} < \rho_{\text{water}}$, the buoyancy wins and it rises. If the densities are equal, a fully submerged object has equal buoyancy and weight. We call that neutral buoyancy; with no other vertical force and no initial motion, it stays at that depth.

As a light object emerges, it displaces less water and its buoyancy decreases. The depth where buoyancy equals weight is its floating equilibrium. It may still be moving when it passes this depth, so it can bob above and below it. Water resistance gradually removes that motion. Once it floats at rest, with no other support or applied force, $B = W$:

In the second line we divided both sides by $g$ and by $\rho_{\text{water}} V$. So a floating object has the fraction $\rho_{\text{obj}}/\rho_{\text{water}}$ of its volume under water. Wood with a density of $650\ \mathrm{kg/m^3}$ floats with $650/1000 = 0.65$ of its volume submerged. A steel ship floats because its hull encloses a large volume of air: its average density, its total mass divided by its total volume including the air inside, is less than that of water.

Now we test all of this in a tank. A sphere with a radius of $0.43\ \mathrm{m}$ sits above the water. The density slider changes what the sphere is made of without changing its size. The gold probe measures the gauge pressure at the depth you set. With the pointer you can grab the sphere, push it under the water and let go.

The sphere’s volume is $V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi \times 0.43^3 = 0.333\ \mathrm{m^3}$. At a density of $650\ \mathrm{kg/m^3}$ its mass is $650 \times 0.333 = 216.5\ \mathrm{kg}$ and its weight is $216.5 \times 9.81 = 2124\ \mathrm{N}$. Completely under water it pushes aside $0.333\ \mathrm{m^3}$ of water, so its buoyancy is $1000 \times 0.333 \times 9.81 = 3267\ \mathrm{N}$. That is more than its weight, so it rises when released and settles with $0.65$ of its volume under water.

Before you press Play: what will happen at a density of $1200\ \mathrm{kg/m^3}$?

Press Play, then grab the sphere, push it under and let go. Try densities above and below $1000\ \mathrm{kg/m^3}$, and move the probe deeper to check $\rho g h$. During motion, water resistance also acts. Equal buoyancy and weight at an instant do not by themselves mean that the sphere has stopped. A floating sphere at rest has equal buoyancy and weight; a heavy sphere resting on the bottom also needs an upward force from the tank floor.