Pressure & Buoyancy

Fill a plastic bottle with water, screw the cap on, and hold it under the surface of a bath. It fights you. Let go and it sits there, going nowhere. Empty the bottle, cap it again, and now you cannot keep it down: it squirts out of your hand and jumps clear of the water. The bottle is the same size and the same shape both times. Only what is inside changed.

The water outside cannot see inside the bottle. Whatever pushes up on it depends only on the shape and depth of the region the bottle occupies, so that upward push is the same both times, and the difference in behaviour is entirely a difference in weight. That one observation is the whole of this topic. In the last lesson you found equilibria by asking where a potential energy curve flattens out. Here they come from a force balance in a fluid, and once you can write that balance down, almost every fluid statics problem is arithmetic.

Pressure, and why it grows with depth

Pressure is force per unit area: $p = F/A$, in pascals, where 1 Pa = 1 N/m². In a fluid at rest it has a property forces do not: no direction. Rotate a gauge at fixed depth and it reads the same. Pressure at a point is one number, and the force it makes on any small surface is $pA$ perpendicular to that surface, whichever way the surface faces.

Now find how it varies with depth. Imagine an invisible column of the fluid itself, top at the surface, bottom at depth $h$, cross-sectional area $A$. It is not moving, so the forces on it cancel. The atmosphere pushes down on the top with $p_0 A$, where $p_0$ is atmospheric pressure. Gravity pulls down on its mass. The fluid underneath pushes up on the bottom face with $pA$, where $p$ is the pressure at depth $h$. The sides push horizontally and cancel in pairs.

The mass of the column is its density times its volume, $\rho \cdot hA$, so its weight is $\rho g h A$. Setting up equals down:

$pA = p_0 A + \rho g h A$

Divide out the area, which never mattered:

$p = p_0 + \rho g h$

Read this carefully. $p$ is the absolute pressure, the real total push. The piece $\rho g h$ is the gauge pressure, the amount by which you beat atmospheric; tyre gauges, depth gauges and manometers all read it, because they compare against the air around them. If a problem says "gauge", drop the $p_0$. If it says "absolute" or "total", keep it.

Notice what cancelled: the area, so nothing about the width or shape of the container appears. Only the vertical drop $h$, the density, and $g$. Two points at the same height in the same connected fluid are therefore at the same pressure, whatever path the fluid takes between them.

Two consequences. A mercury barometer has vacuum above its sealed top, so $p_0 = \rho g h$ with no offset and the column stands at $h = p_0/(\rho g)$. And in fresh water one atmosphere of gauge pressure arrives at $h = 1.01\times10^5 / (1000 \times 9.8) = 10.3$ m: roughly another atmosphere every 10 metres.

Same level, same pressure

That last fact does more work than the formula itself. The trick is always the same: pick a horizontal line that passes through connected fluid of one kind, write the pressure at that line going down the left branch, write it again going down the right branch, and set them equal.

Take a U-tube holding water and pour a column of oil of height $h_1$ and density $\rho_1$ into the left arm, where it floats as a slug on top. Draw the line through the oil-water interface. Down the left arm you pass air then oil: $p_0 + \rho_1 g h_1$. Down the right arm you pass air then water, a height $h_2$ from the free surface to the same line: $p_0 + \rho_2 g h_2$. Equal, so $p_0$ cancels and

$\rho_1 g h_1 = \rho_2 g h_2 \quad\Longrightarrow\quad \rho_1 h_1 = \rho_2 h_2$

The heavier liquid needs less height to produce the same pressure. The water level in the right arm stands above the interface by $h_2 = (\rho_1/\rho_2) h_1$.

The same line-of-equal-pressure argument gives you the hydraulic lift. Two pistons of areas $A_1$ and $A_2$ sit at the same height on a sealed body of oil. Same height, same pressure, so $F_1/A_1 = F_2/A_2$, and

$F_2 = F_1 \frac{A_2}{A_1}$

Push the small piston with 150 N over $2.00\times10^{-3}\ \mathrm{m}^2$ and the pressure is $150 / 2.00\times10^{-3} = 7.50\times10^4$ Pa. The big piston, area $5.00\times10^{-2}\ \mathrm{m}^2$, feels $7.50\times10^4 \times 5.00\times10^{-2} = 3750$ N. Energy is not free here: the big piston moves 25 times less far, since the oil is incompressible and the volumes pushed must match.

Worked example 1: reading a manometer

A gas line is connected to one arm of an open water manometer. The water in the open arm stands 0.340 m higher than in the gas arm. What is the gauge pressure of the gas?

Draw the line of equal pressure at the lower water level, the one in the gas arm, because both branches are connected water at and below that line.

Above the line on the gas side there is only gas, whose own weight you ignore (air is about a thousand times lighter than water, so a third of a metre of it contributes a couple of pascals). The pressure at the line on that side is just $p_{\text{gas}}$. On the open side you start at the atmosphere and descend 0.340 m of water to the same level: $p_0 + \rho g (0.340)$. Set them equal:

$p_{\text{gas}} = p_0 + \rho g h$

Gauge pressure is what beats the atmosphere, so subtract $p_0$:

$p_{\text{gauge}} = 1000 \times 9.8 \times 0.340 = 3332\ \mathrm{Pa}$

Check the sign: the open side stands higher, so the gas is pushing the water down on its side, so the gas is above atmospheric. Had the open side been lower, the gauge pressure would have come out negative: a partial vacuum.

Archimedes: where the upward push comes from

Now put an object in the fluid. Every face feels pressure pushing inwards, perpendicular to that face. Side faces are at matched depths on opposite sides, so they cancel. Top and bottom do not: the bottom is deeper, the pressure there is larger, and the upward push wins.

Do it for a block of height $H$ and horizontal area $A$, top at depth $d$. Down on the top: $(p_0 + \rho g d)A$. Up on the bottom, at depth $d + H$: $(p_0 + \rho g (d+H))A$. Subtract, and both $p_0 A$ and $\rho g d A$ cancel, leaving a net upward force

$F_B = \rho g H A = \rho_{\text{fluid}}\, g\, V_{\text{disp}}$

since $HA$ is the volume of fluid pushed out of the way. This is Archimedes' principle: the buoyant force equals the weight of the fluid displaced. The density in it is the fluid's; the object's density is nowhere in it.

A slicker argument covers any shape. Scoop the object out and let the hole fill with fluid. That fluid sits in equilibrium, so the push from everything around it balances its own weight, $\rho_{\text{fluid}} g V$. Put the object back: the surrounding fluid touches the same surface at the same depths, so it pushes exactly as hard as before. The push does not know what is inside.

Floating: what fraction sits under

An object floating at rest has two vertical forces on it: its weight $\rho_o V g$, and buoyancy from the submerged part, $\rho_f V_{\text{sub}} g$. Balance them:

$\rho_f V_{\text{sub}} g = \rho_o V g$

The $g$ cancels and nothing else does, giving the most useful line here:

$\frac{V_{\text{sub}}}{V} = \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}$

Pine at 700 kg/m³ in water at 1000 floats with 0.7 of its volume under. Ice at 917 in seawater at 1025 floats with $917/1025 = 0.8946$ under, which is where the iceberg cliché comes from. A ratio bigger than 1 is a signal, not an answer: the object sinks.

Fully submerged and held by a string, the string takes the difference, and the scale reads the apparent weight

$T = mg - \rho_f V g = mg\left(1 - \frac{\rho_f}{\rho_o}\right)$

using $V = m/\rho_o$. For a 2.50 kg aluminium block at 2700 kg/m³ the real weight is $2.50 \times 9.8 = 24.5$ N and the factor is $1 - 1000/2700 = 0.6296$, so the scale reads $24.5 \times 0.6296 = 15.43$ N.

Run that backwards to measure density without measuring volume. If a sample weighs $W_{\text{air}}$ in air and $W_{\text{water}}$ submerged, the buoyant force is the difference, so $\rho_f V g = W_{\text{air}} - W_{\text{water}}$ while $\rho_o V g = W_{\text{air}}$. Divide, and $Vg$ disappears:

$\rho_o = \rho_f \frac{W_{\text{air}}}{W_{\text{air}} - W_{\text{water}}}$

A sample weighing 40.0 N in air and 32.0 N in water has $\rho_o = 1000 \times 40.0/8.0 = 5000$ kg/m³.

Worked example 2: how much can the raft carry

A raft is a slab of wood of area 2.00 m², thickness 0.200 m, density 600 kg/m³, floating in fresh water. What is the largest mass you can put on it before water reaches the top surface?

The phrase "before water reaches the top surface" is the whole problem. It names the limiting state: the raft is submerged right to its top, so $V_{\text{sub}}$ equals the full volume of the raft, $V = 2.00 \times 0.200 = 0.400\ \mathrm{m}^3$. At that limit the buoyant force is at its maximum:

$F_B = \rho_w V g = 1000 \times 0.400 \times 9.8 = 3920\ \mathrm{N}$

The weight it must support is the raft plus the load. The raft's mass is $\rho_{\text{wood}} V = 600 \times 0.400 = 240$ kg, so its weight is $240 \times 9.8 = 2352$ N. Balance:

$3920 = 2352 + M(9.8)$ $M = \frac{3920 - 2352}{9.8} = \frac{1568}{9.8} = 160\ \mathrm{kg}$

There is a shortcut. $g$ cancels throughout, so the load is the mass of water the raft could displace minus the raft's own mass: $(1000 - 600) \times 0.400 = 160$ kg. Lift per cubic metre is the density difference: worth remembering as its own idea.

Worked example 3: the beaker on the scale

A beaker of water sits on a scale; beaker plus water have mass 3.00 kg. A metal block of volume $2.50\times10^{-3}$ m³ hangs from a string, fully submerged, not touching the beaker. What does the scale read?

Two instincts fight here, and both are wrong: "the string holds the block, so the scale should not care", and "the block is in the water, so the scale reads its whole weight". Use Newton's third law. The water pushes up on the block with

$F_B = \rho_w V g = 1000 \times 2.50\times10^{-3} \times 9.8 = 24.5\ \mathrm{N}$

so the block pushes down on the water with 24.5 N. That is a real downward force on the beaker-plus-water system, and the scale has to supply it.

Take the beaker and the water and nothing else as your system. Forces on it: its own weight $3.00 \times 9.8 = 29.4$ N down, the block's reaction 24.5 N down, and the normal force $N$ from the scale up. Nothing accelerates, so

$N = 29.4 + 24.5 = 53.9\ \mathrm{N}$

The string carries the rest, and scale reading plus tension add to the total weight of everything, as they must. Whenever something dangles in the liquid, draw the system boundary first and ask which forces cross it.

Worked example 4: the bobbing cylinder

A wooden cylinder of density 600 kg/m³ and height 0.150 m floats upright in water with its axis vertical. Push it down slightly and release it. Find the period of the small vertical oscillation.

This one needs a real idea: look at the change in force rather than the force. At its floating position the buoyancy exactly cancels the weight: that is what floating means. Push it down an extra distance $x$ and the weight has not changed at all, while the buoyancy has grown by the weight of the extra sliver of water displaced. The net force is that increase, pointing back up.

The extra sliver has area $A$ (the cylinder's cross-section) and height $x$, so the extra buoyancy is $\rho_w g A x$, upward. Taking down as positive $x$:

$F_{\text{net}} = -\rho_w g A x$

That is Hooke's law with effective spring constant $k = \rho_w g A$, and any force proportional to displacement and pointing back to equilibrium gives simple harmonic motion with $T = 2\pi\sqrt{m/k}$. The mass is the cylinder's, $m = \rho_o A L$, with $L = 0.150$ m its full height. So

$T = 2\pi\sqrt{\frac{\rho_o A L}{\rho_w g A}} = 2\pi\sqrt{\frac{\rho_o L}{\rho_w g}}$

The area cancels: a wider cylinder has more mass to move but displaces proportionally more water for the same dip, and the two effects are equal. Now the numbers:

$\frac{\rho_o L}{\rho_w g} = \frac{600 \times 0.150}{1000 \times 9.8} = \frac{90.0}{9800} = 9.184\times10^{-3}\ \mathrm{s}^2$

Take the square root, $0.09583$ s, and multiply by $2\pi$:

$T = 2\pi \times 0.09583 = 0.602\ \mathrm{s}$

A hydrometer works off the same idea from the other end. Its weight is fixed, so $m = \rho_f V_{\text{sub}}$ fixes the submerged volume: in a lighter liquid $V_{\text{sub}}$ must grow, and all the growth happens along the thin stem. The extra depth is $\Delta x = \Delta V_{\text{sub}}/A_{\text{stem}}$. For $m = 0.0150$ kg going from 1000 to 900 kg/m³, $\Delta V = 0.0150(1/900 - 1/1000) = 1.667\times10^{-6}$ m³, so $\Delta x = 1.667\times10^{-6}/4.00\times10^{-5} = 0.0417$ m.

Traps

Gauge versus absolute. Adding $p_0$ when you should not, or dropping it when you must not, is the commonest loss here. The word in the question tells you which. Gas-law problems always want absolute pressure, because $pV$ = constant is about the real total push. A bubble rising from depth $h$ obeys $(p_0 + \rho g h)V_{\text{bottom}} = p_0 V_{\text{top}}$, and if the volume triples then $p_0 + \rho g h = 3p_0$, giving $\rho g h = 2p_0$ and $h = 2 \times 1.01\times10^5/(1000\times9.8) = 20.6$ m.

Which density. In $F_B = \rho V g$ the density is the fluid's. In $m = \rho V$ it is the object's. Swapping them gives answers off by exactly the density ratio, which looks plausible enough to slip through. And if the object is denser than the fluid there is no floating solution at all: the fraction formula is meaningless and the thing rests on the bottom.

Air counts. A balloon displacing 5.00 m³ of air at 1.225 kg/m³ feels $1.225 \times 5.00 \times 9.8 = 60.0$ N of lift. Archimedes does not care that the fluid is a gas. For solids in air the effect is tiny; for a balloon it is the entire point.

Released floats do not stop at the surface. A ball of density 300 kg/m³ released 1.20 m down has net upward force $(\rho_w - \rho_o)Vg$ against mass $\rho_o V$, so its acceleration is $g(\rho_w/\rho_o - 1) = 9.8 \times (1000/300 - 1) = 22.87$ m/s² up. Over 1.20 m it gains kinetic energy, then flies clear of the water and decelerates at $g$ for a further $1.20 \times 22.87/9.8 = 2.8$ m.

Recognising when this is the technique

Reach for hydrostatics when a problem hands you a density and a depth, or a density and a fluid. That pairing is the signature.

If it mentions a depth, a level difference, a tube, a barometer, a manometer, or two liquids that do not mix, you want $p = p_0 + \rho g h$ and the equal-level rule: draw a horizontal line through connected fluid of one kind and write the pressure twice.

If it mentions floats, submerged, apparent weight, what fraction, how much load, or a scale reading with something in the water, you want Archimedes. Write a force balance on the object, $\rho_{\text{fluid}} V_{\text{disp}} g$ up against $\rho_{\text{object}} V g$ down, and look for the factors that cancel: $g$ always does, and often the area or volume too.

If the object is displaced slightly from floating and released, or it is hinged, the balance alone is not enough. For a small displacement, compute the change in buoyancy and you have a spring constant. For a hinged rod, take torques about the hinge with the buoyant force acting at the midpoint of the submerged part and the weight at the midpoint of the whole rod; the submerged length depends on the angle, so you get one equation in $\theta$.

Check dimensionless answers against sense before writing them down. A fraction submerged over 1 means the thing sinks.

Next you leave still fluids and look at what happens when the frame you measure in is itself accelerating, in Non-Inertial Frames & Pseudo-Forces. Buoyancy comes back there in a strange guise: in an accelerating tank, the surface tilts and a helium balloon leans forward instead of back.