Archimedes' Principle and Buoyancy

Pick up a bucket of water and a small stone. Feel the weight of the stone in your hand: it is whatever it is — a kilogram, maybe two. Now lower the stone slowly into the bucket until it is fully submerged but still hanging from your fingers. The stone feels lighter. Not by a lot, but measurably. Something in the water is pushing up on the stone, counteracting part of its weight.

This is buoyancy, and it is not an exotic effect. It is the same thing that lets a steel ship weighing twenty thousand tonnes sit placidly on the Arabian Sea. It is why the dabbawala's aluminium tiffin tower floats in the overflow bucket at Dadar station while it is being washed. It is why a spoon of ghee sits on top of a bowl of dal instead of sinking. It is why a submarine can rest neutrally at any depth of the Indian Ocean while its crew plays chess.

The remarkable thing is that buoyancy is not a new force. It is just the pressure-depth law of a static fluid, applied to the body's surface and added up. Once you see that sum, the formula that follows — the one Archimedes wrote down twenty-three centuries ago — is forced on you by arithmetic. The rest of this article derives the buoyant force from pressure, explains exactly what happens when a body floats, computes apparent weight problems cleanly, and ends with the geometric idea — the metacentre — that separates a ferry that rights itself from a ferry that capsizes.

What the fluid does to a submerged body

Take a solid block — mass m, shape a simple rectangular slab — and submerge it fully in a still fluid of density \rho_\text{fluid}. Call the top face area A and the block's height h. The top face is at depth h_1 below the fluid surface; the bottom face is at depth h_2 = h_1 + h. Hold the block at rest by hand for now.

The sideways forces cancel

The pressure at a given depth is the same everywhere in the fluid (for a horizontal slice — this is a consequence of the fluid being in equilibrium; see Pressure in Fluids). The left face of the block and the right face sit at the same range of depths. The pressure at each depth pushes horizontally — inward from each side, equal in magnitude. The net sideways force is zero.

The vertical forces do not cancel

The top face, at depth h_1, has pressure P_1 = P_0 + \rho_\text{fluid} g h_1, where P_0 is the pressure at the fluid's top surface. This pressure pushes the block downward with force

The bottom face, at depth h_2 = h_1 + h, has pressure P_2 = P_0 + \rho_\text{fluid} g h_2. This pressure pushes the block upward with force

The upward F_\text{bot} is larger than the downward F_\text{top} because h_2 > h_1. Their difference — the net upward force from the fluid — is the buoyant force:

But h A is exactly the volume of the block, which equals the volume of fluid the block displaces. Call that V_\text{displaced}. Then

Why: the atmospheric P_0 cancelled between the two faces because it is the same at both. What is left is just the weight of a column of fluid of cross-section A and height h — which is the fluid that would have been sitting where the block now sits if the block weren't there. That "missing column" is exactly what the block displaces.

This is Archimedes' principle for a rectangular block. The formula does not look like it depends on the shape of the block at all — the only geometric input is the volume of the submerged part, not how that volume is distributed. It turns out this is a general truth, and the derivation extends to any shape.

The shape does not matter — an invariance argument

Suppose the block weren't rectangular. Imagine replacing it with a fluid of the same density as the surrounding fluid, carved into the block's shape. That fluid parcel is part of the bulk fluid and is in equilibrium — it is not moving. Two forces act on it: its own weight W_\text{parcel} = \rho_\text{fluid} V_\text{displaced} g (downward) and the surface forces from the surrounding fluid (integrated over the parcel's surface). Newton's first law requires those two to balance, so the surface forces give an upward resultant of magnitude \rho_\text{fluid} V_\text{displaced} g.

Now remove the fluid parcel and replace it with the original solid body of the same shape. The surrounding fluid does not know or care — it still exerts the same pressure on every surface point at the same depth, because the pressure only depends on depth, not on what is in the middle. So the surface force on the solid body from the surrounding fluid is still exactly \rho_\text{fluid} V_\text{displaced} g directed upward.

Why: the buoyant force depends only on the pattern of pressure on the body's surface, and the surrounding fluid's pressure pattern at each depth is identical whether the "object" inside is a fluid parcel or a solid body of the same shape. Replace the body with its fluid-parcel shadow — you are now analysing a fluid parcel in static equilibrium — and the upward force from the surrounding fluid must equal the weight of the parcel. That weight is \rho_\text{fluid} V g.

This argument — replacing a solid with an imaginary fluid parcel of the same shape — is the classic trick that makes Archimedes' principle work for any shape, however ugly.

Three things to notice in the statement.

1. The fluid's density matters, not the body's. The buoyant force is set by the fluid the body is sitting in — water, mercury, air, oil — and by how much of that fluid gets pushed aside. The body's own density does not appear here. (It does appear when you compute whether the body floats or sinks — that is the balance of F_B against the body's weight, not the value of F_B itself.)

2. The g in the formula is the local gravitational acceleration. On the Moon, where g is 1.6 \text{ m/s}^2, buoyant forces are weaker. This is why astronauts float more easily in a water-tank during neutral-buoyancy training on Earth, but would sink faster in the same tank on the Moon. In the International Space Station where g_\text{eff} \approx 0, there is no buoyant force at all — you cannot make a bubble "rise" in a spacecraft because there is no "up."

3. The principle applies to gases as well. The atmosphere is a fluid. A hot-air balloon rises because the heated air inside is less dense than the surrounding cooler air, making the buoyant force larger than the weight of balloon plus hot air plus payload. A hydrogen-filled weather balloon — like those ISRO releases from the Thumba Equatorial Rocket Launching Station — works the same way, with hydrogen replacing hot air.

Floating bodies — where the density ratio appears

Now drop the body into the fluid and let it find its own level. Three things can happen.

• \rho_\text{body} > \rho_\text{fluid}: the body sinks. The buoyant force (with the body fully submerged) is smaller than the weight; the net force is downward until the body hits the container's bottom.
• \rho_\text{body} = \rho_\text{fluid}: the body is neutrally buoyant — it stays at whatever depth you release it at. Fish and submarines regulate their density to achieve this.
• \rho_\text{body} < \rho_\text{fluid}: the body floats, with part of it above the fluid surface.

The third case is the one that produces a clean formula. Let the body have volume V_\text{body} and density \rho_\text{body}. Let a fraction f of its volume be submerged. In equilibrium, the buoyant force balances the weight:

The V_\text{body} and g cancel, giving

Why: the fraction submerged must adjust itself so that the weight of the displaced fluid exactly balances the body's weight. If the body is half as dense as the fluid, it needs to push aside only half its volume worth of fluid — so half of it is submerged.

Numbers to calibrate the intuition

• A block of sal wood has density \rho \approx 850 \text{ kg/m}^3. In fresh water (\rho = 1000 \text{ kg/m}^3), the submerged fraction is f = 850/1000 = 0.85 — 85% under, 15% above.
• A hollowed-out brass pot (full body of brass would sink — \rho = 8400 \text{ kg/m}^3 — but the hollow form has average density around 200 \text{ kg/m}^3 including the air inside): f = 200/1000 = 0.2. Twenty percent submerged.
• A cork in sea water: \rho_\text{cork} \approx 240 \text{ kg/m}^3, \rho_\text{seawater} \approx 1025 \text{ kg/m}^3. f = 240/1025 \approx 0.23 — almost a quarter submerged.
• A human body in fresh water: \rho_\text{body} \approx 1010 \text{ kg/m}^3 (slightly denser than water). The person sinks slowly. But in sea water (\rho = 1025 \text{ kg/m}^3), \rho_\text{body} < \rho_\text{fluid}, so the person floats — faintly. This is why swimming in the sea is easier than in a pool. In the Dead Sea (\rho \approx 1240 \text{ kg/m}^3), f = 1010/1240 \approx 0.81 — you sit half-out of the water like a beach ball.

Apparent weight — why you feel lighter underwater

Hold the block in your hand while it is fully submerged in water. The forces on the block are: weight W = \rho_\text{body} V g pulling down, buoyant force F_B = \rho_\text{fluid} V g pushing up, and the tension T in your arm's grip pulling up (or a scale reading, if you hang the block from a spring balance under water).

Newton's first law for the stationary block:

T is what your hand (or the spring balance) actually reads. This is the apparent weight:

For a 1 \text{ kg} iron block (\rho \approx 7800 \text{ kg/m}^3) submerged in water: volume V = 1/7800 \approx 1.28 \times 10^{-4} \text{ m}^3; buoyant force F_B = 1000 \times 1.28 \times 10^{-4} \times 9.8 \approx 1.26 \text{ N}; true weight W = 9.8 \text{ N}; apparent weight W_\text{app} = 9.8 - 1.26 = 8.54 \text{ N}. It feels about 13% lighter.

This is how you determine the density of an irregular object without a geometrical measurement: weigh it in air (gives you W_\text{true}), weigh it submerged in a fluid of known density (gives you W_\text{app}, and hence F_B). Then

This is the operation Archimedes legendarily performed in his bath to test the goldsmith's honesty. The logic is exact even if the anecdote is embellished.

Worked examples

Stability of floating bodies — why ships don't tip over

Newton's first law says that a floating body in equilibrium stays in equilibrium — but equilibrium can be stable (tip it a little and it returns) or unstable (tip it a little and it goes further). The difference matters: a ship, an aluminium tiffin tower, and a Godavari ferry are all floating bodies, and whether they capsize in a wave depends on whether their equilibrium is stable.

The centre of gravity G and the centre of buoyancy B

The centre of gravity G is the point where the body's weight effectively acts — the geometric centre of the body's mass distribution. It does not move when the body tilts (relative to the body itself).

The centre of buoyancy B is the geometric centre of the submerged part of the body — the centroid of the volume that is below the waterline. When the body tilts, the shape of the submerged part changes, so B moves.

In equilibrium, G and B both lie on the vertical axis of symmetry of the body, and the weight (acting downward at G) and the buoyant force (acting upward at B) are equal and opposite. No net force, no net torque.

The metacentre M

Tilt the body by a small angle. The submerged part changes shape — on one side it gets deeper, on the other side shallower — so B shifts sideways, to a new position B'. Draw a vertical line through B'. That line intersects the body's original central axis at a point called the metacentre M.

For small tilts, M is approximately a fixed point of the body (independent of the tilt angle). And its position relative to G determines stability:

• M above G — stable. When the body tilts, the buoyant force at B' and the weight at G form a couple that produces a righting torque — rotating the body back to upright.
• M below G — unstable. The same couple now produces a capsizing torque — rotating the body further away from upright. A small disturbance grows into full overturn.
• M = G — neutral.

The distance GM (with sign: positive if M is above G) is called the metacentric height. Ship designers try to keep GM positive and large enough for safety, but not so large that the righting torque snaps the ship back violently (which would make it uncomfortable and could shift cargo). For cargo ships, GM \approx 0.3–1.0 m is typical. A Godavari passenger ferry likely has GM of around 0.5 m; a stainless-steel tiffin tower, being wide and squat with most of its mass low, also has M above G and rights itself if tilted.

The quick intuition: if a body's heavy parts are low and its buoyant part is spread wide at the waterline, M tends to sit above G, and the body is stable. A tall, narrow body with mass concentrated high up tends to have M below G — a recipe for capsizing. This is why rowing boats are wide and shallow, and why cruise ships have their engines and fuel tanks at the bottom.

Common confusions

• "A heavier object sinks faster in a fluid." Only if the fluid's viscosity is large enough to matter. In an ideal, non-viscous fluid, what determines sinking versus floating is density, not weight. A 1-gram piece of lead sinks; a 1000-gram piece of cork floats. The acceleration with which something sinks does depend on the difference \rho_\text{body} - \rho_\text{fluid} relative to \rho_\text{body}, but not on absolute weight.

• "Buoyant force only acts in water." It acts in any fluid, including gases. A helium balloon rises because of the buoyant force from air. You yourself are being pushed up by the atmospheric buoyancy, though the effect is tiny (since \rho_\text{air}/\rho_\text{body} \approx 0.001).

• "A fully submerged body experiences more buoyant force the deeper it goes." No. The buoyant force F_B = \rho_\text{fluid} V g depends only on displaced volume and fluid density — and for an incompressible fluid and a rigid body, neither changes with depth. A submarine feels the same buoyant force at 10 m as at 300 m (assuming the hull doesn't crush and the water's density is the same, which is nearly true for the ocean).

• "Floating bodies have a fraction \rho_\text{body}/\rho_\text{fluid} submerged — this must be less than 1." The formula only applies when the body does in fact float, which requires \rho_\text{body} < \rho_\text{fluid}. If \rho_\text{body} > \rho_\text{fluid}, the body sinks and the formula does not apply — the body is fully submerged (fraction = 1) and still has unbalanced weight.

• "Apparent weight is the 'real' weight of an object in a fluid." No — the real weight is always mg. The apparent weight is the reading on a scale that is pulling up on the object from inside the fluid; it is the net downward force minus the buoyant force, which is also what the scale reads.

• "Archimedes' principle doesn't apply to a body touching the container bottom." If the body rests on a waterproof seal against the bottom — so fluid cannot reach its underside — then the pressure at the underside is missing, and there is no upward P_2 A force. The body is then held down by atmospheric pressure plus water pressure on its top. This is an edge case that produces real engineering surprises (e.g., a suction-cup stone stuck to a river bed). For a body with fluid all around it, including underneath, the principle applies as stated.

Where this leads next

• Pressure in Fluids — the pressure-depth law that is the only input to Archimedes' principle.
• Pascal's Law and Hydraulic Machines — the same pressure law, applied to confined fluids with pistons.
• Atmospheric Pressure and Barometers — buoyancy in air, and the Torricelli column that measures the atmosphere.
• Surface Tension and Capillarity — the next level of fluid surface physics, where small objects float despite \rho > \rho_\text{fluid}.
• Equation of Continuity — what happens when the fluid starts flowing.