Classical Mechanics

Archimedes’ Principle and Buoyancy

Definition

Archimedes’ principle states that any object placed in a fluid is pushed upward by a buoyant force equal to the weight of the fluid the object displaces. The force is found with F_b = ρVg, where ρ is the fluid’s density, V the displaced volume, and g gravity. An object floats when this upward force balances its weight.

Drop a steel spanner into a sink and it plunges straight to the bottom. Yet a steel ship the length of three football pitches sits calmly on the ocean, carrying thousands of cars. Same metal, opposite fate. The rule that explains both — and the floating ice in your drink, and the lift under a hot-air balloon — is one of the oldest in physics.

It came from a puzzle about a king’s crown more than two thousand years ago, and it still decides how submarines dive, how life jackets save lives, and how much of every iceberg hides beneath the waves. Once you can picture the fluid being pushed out of the way, the whole thing clicks.

What Is Archimedes’ Principle?

Picture lowering a sealed box into a bathtub. The water has to go somewhere, so it rises — the box has displaced a certain volume of water. Archimedes’ principle says the fluid pushes back: it shoves the box upward with a force exactly equal to the weight of the water that was moved aside.

That upward push is the buoyant force (also called upthrust). It does not care what the object is made of, only how much fluid the object displaces and how heavy that fluid is. This idea is often written simply as the Archimedes principle, and it applies to every fluid — water, oil, even the air around you.

So whether an object floats or sinks is a contest between two forces: gravity pulling it down by its weight, and buoyancy pushing it up by the weight of displaced fluid. Win the contest and it rises; lose and it sinks; tie and it hovers in place.

The man behind it, Archimedes of Syracuse (c. 287–212 BC), reportedly hit on the idea in his bath and shouted “Eureka!” The famous story has him testing a king’s gold crown for cheating. Whether he literally used a bathtub is debated by historians — a careful hydrostatic balance is the more likely method — but the physics he uncovered is rock-solid.

The Buoyant Force Formula

The size of the upthrust comes straight from the weight of displaced fluid. Since weight is mass times gravity, and the displaced mass is the fluid’s density times the displaced volume, the buoyant force is:

F_b = ρVg

Every symbol has a job, and each one carries an SI unit. Get the units right and the answer lands in newtons every time.

Symbol Quantity SI unit
F_bBuoyant force (upthrust)newton (N)
ρ (rho)Density of the fluidkg/m³
VVolume of fluid displaced
gGravitational field strength≈ 9.81 m/s²

One trap worth flagging now: ρ is the density of the fluid, never the object. Swap in the object’s density by mistake and the physics falls apart.

Apparent weight and the float fraction

When an object hangs submerged on a scale, the scale reads less than its true weight, because buoyancy lifts part of the load. That reduced reading is the apparent weight:

W_app = W − F_b

For something that floats, there is a neat shortcut for how deep it rides. The fraction of its volume sitting below the surface equals the ratio of the two densities:

V_sub / V = ρ_obj / ρ_fluid

This single ratio is why a denser object floats lower, and why ice — slightly less dense than water — barely pokes above the surface. You can run the numbers for any object and fluid with our Buoyancy Calculator.

How Buoyancy Works

Where does the upward push actually come from? Pressure. In any fluid, pressure grows with depth, because deeper layers carry the weight of everything above them. The relationship is P = ρgh — go deeper, and h grows, so the pressure climbs.

Now think about a submerged block. The fluid presses on every face, but the bottom face sits deeper than the top face. So the upward push on the bottom beats the downward push on the top. That difference is the buoyant force.

Put numbers on it. If the top face is at depth d and the block has height h and face area A, the bottom sits at depth d + h. The net upward force is the bottom force minus the top force:

F_b = ρg(d + h)A − ρg(d)A = ρg(hA) = ρgV. The depth d cancels, leaving exactly ρVg — the displaced volume V = hA times ρg. NASA’s Glenn Research Center walks through this same pressure-difference derivation in its Archimedes’ principle activity.

fluid surface object smaller pressure (top) larger pressure (bottom) F_b (buoyant force) W (weight)

Fluid pressure is greater on the deeper bottom face than on the top, and that imbalance produces the net upward buoyant force.

Try it yourself. In the lab below, change the object’s density, its volume, and the fluid. Watch the weight, buoyant force, apparent weight and submerged percentage update live as the block settles to float or sink.

Buoyancy Lab

Real-World Examples of Buoyancy

Archimedes’ principle is not a textbook curiosity — it is doing quiet work all around you. Here are five places it shows up.

1. Why steel ships float

A solid lump of steel sinks because steel is far denser than water. A ship dodges this by being mostly hollow. Its hull encloses a huge volume of air, so the average density of the whole vessel — steel plus air — drops below that of water, and it floats. Overload it, and the average density climbs until it sinks.

2. How submarines dive and surface

Submarines play with their own weight on purpose. To dive, they flood ballast tanks with seawater, raising their average density above the surrounding water. To surface, compressed air blows that water back out, the average density falls, and buoyancy lifts them up.

3. Hot-air and helium balloons

Here the fluid is air, not water. A balloon rises when it displaces enough air to be lifted — that is, when its average density falls below the air around it. Helium does this because it is far lighter than air; a hot-air balloon does it by heating the air inside until it thins out.

4. Hydrometers

A hydrometer is a weighted float used to measure a liquid’s density. Drop it into a fluid and it sinks until it displaces its own weight. In a denser liquid it sits higher; in a thinner one it sinks lower — and the depth is read straight off a scale.

5. Icebergs

Ice is only slightly less dense than seawater, so an iceberg floats with just a sliver above the surface. Plug the densities into the float-fraction rule and the famous “tip of the iceberg” turns into a hard number — about 89.5% stays hidden underwater.

air seawater ≈ 10.5% above ≈ 89.5% below

An iceberg floats with roughly nine-tenths of its volume below the surface — the ratio of ice density to seawater density.

Iceberg showing the small portion above water and the large mass below, illustrating Archimedes' principle
Most of an iceberg’s volume sits below the surface, exactly as the density ratio predicts.

Common Misconceptions About Buoyancy

Buoyancy is one of those topics where intuition leads people astray. Four wrong beliefs come up again and again.

“The buoyant force depends on the object’s weight”

It does not. Buoyancy depends only on the fluid’s density, the displaced volume, and gravity — the object’s own weight and material never enter F_b = ρVg. As Georgia State University’s HyperPhysics notes, equal volumes of cork, aluminium and lead, fully submerged, all feel the same buoyant force. What differs is their weight, which decides whether they float.

“Heavy things always sink”

Weight alone settles nothing — average density does. A 100,000-tonne ship floats while a 5-gram nail sinks, because the ship’s average density (steel plus enclosed air) is below water while the nail’s is not. Compare the object’s average density with the fluid’s, never its raw mass.

“The buoyant force grows as the object sinks deeper”

For a rigid object in an incompressible fluid like water, it does not. Pressure rises with depth on both the top and bottom faces, but their difference stays the same, so F_b = ρVg is unchanged at any depth. (Gas-filled objects are the exception — they compress, shrinking V, and so lose buoyancy as they descend.)

“A sunken object displaces its own weight of fluid”

Only a floating object does that. A fully submerged object displaces its own volume of fluid, and the buoyant force equals the weight of just that displaced fluid — which, for anything that sinks, is less than the object’s own weight. Mixing up “displaces its weight” and “displaces its volume” is the single most common slip.

How Buoyancy Relates to Density, Pressure and Weight

Buoyancy never acts alone. It is really a story about density set against the fluid, pressure that builds with depth, and the weight it works against.

Density is the decider. If an object’s average density is below the fluid’s, it floats; above, it sinks; equal, it hovers. The table below shows where common materials land in fresh water.

Material / fluid Density (kg/m³) In fresh water (1000 kg/m³)?
Helium (gas)0.18Rises (in air)
Air1.225
Cork≈ 240Floats
Ice917Floats
Vegetable oil≈ 920Floats
Fresh water1000Reference
Seawater1025Sinks slightly
Aluminium2700Sinks
Iron7870Sinks
Lead11,340Sinks
Mercury (liquid)13,534Sinks
Gold19,300Sinks

Pressure is the engine. The buoyant force exists only because fluid pressure climbs with depth, pushing harder on the bottom of an object than the top. This is the same depth-pressure idea that governs how fast things fall through a fluid, where buoyancy and drag both resist gravity — see our guide to terminal velocity.

Weight is the opponent. Floating is simply the balance point where the upthrust equals the object’s weight — a state of equilibrium, exactly the kind of force balance covered in Newton’s laws of motion. When the forces do not balance, the object accelerates, and the net force follows directly from Newton’s second law.

Worked Problems

Problem 1
A solid cube of volume 0.0025 m³ is fully submerged in fresh water (density 1000 kg/m³). Find the buoyant force on it. Take g = 9.81 m/s².
Show Solution
Solution: Step 1: Use Archimedes’ principle, F_b = ρVg. Step 2: Substitute with units: F_b = (1000 kg/m³)(0.0025 m³)(9.81 m/s²). Step 3: Solve: F_b = 24.525 N. Answer: F_b ≈ 24.5 N
Problem 2
A 12 kg metal block of volume 0.0015 m³ hangs from a spring scale and is lowered fully into water. What apparent weight does the scale read? Take g = 9.81 m/s².
Show Solution
Solution: Step 1: True weight W = mg = (12)(9.81) = 117.72 N. Buoyant force F_b = ρVg. Step 2: F_b = (1000)(0.0015)(9.81) = 14.715 N. Step 3: Apparent weight W_app = W − F_b = 117.72 − 14.715 = 103.005 N. Answer: W_app ≈ 103 N
Problem 3
A block of wood with density 650 kg/m³ floats in fresh water (1000 kg/m³). What fraction of its volume sits below the surface?
Show Solution
Solution: Step 1: For a floating object, V_sub / V = ρ_obj / ρ_fluid. Step 2: Substitute: V_sub / V = 650 / 1000. Step 3: Solve: V_sub / V = 0.65. Answer: 65% of the wood is submerged
Problem 4
Ice has a density of 917 kg/m³ and seawater 1025 kg/m³. What percentage of a floating iceberg lies below the surface?
Show Solution
Solution: Step 1: Use the float-fraction rule, V_sub / V = ρ_ice / ρ_seawater. Step 2: Substitute: V_sub / V = 917 / 1025. Step 3: Solve: V_sub / V = 0.8946. Answer: ≈ 89.5% is underwater (only about 10.5% shows above)
Problem 5
A crown weighs 27.5 N in air and 25.4 N when fully submerged in water. Find its density and decide whether it is pure gold (gold density 19,300 kg/m³). Take g = 9.81 m/s².
Show Solution
Solution: Step 1: Buoyant force F_b = weight in air − apparent weight = 27.5 − 25.4 = 2.1 N. From F_b = ρ_water·V·g, the volume V = F_b / (ρ_water·g). Step 2: V = 2.1 / (1000 × 9.81) = 2.14 × 10⁻⁴ m³. Mass m = W/g = 27.5 / 9.81 = 2.803 kg. Step 3: Density ρ = m/V = 2.803 / (2.14 × 10⁻⁴) ≈ 13,100 kg/m³. Answer: ρ ≈ 13,100 kg/m³ — far below gold’s 19,300 kg/m³, so the crown is not pure gold
Problem 6
A raft of volume 0.40 m³ and mass 120 kg floats in fresh water. What is the maximum extra load it can carry before it sinks? Take g = 9.81 m/s².
Show Solution
Solution: Step 1: Maximum upthrust occurs when the raft is on the point of full submersion: F_b(max) = ρVg = (1000)(0.40)(9.81) = 3924 N. Step 2: This supports a maximum total weight of 3924 N, i.e. a total mass of 3924 / 9.81 = 400 kg. Step 3: Subtract the raft’s own mass: 400 − 120 = 280 kg. Answer: Maximum load ≈ 280 kg
Problem 7
A helium balloon has a volume of 5.0 m³ and a fabric mass of 2.0 kg. Using air density 1.225 kg/m³ and helium density 0.1786 kg/m³, find the net upward force. Take g = 9.81 m/s².
Show Solution
Solution: Step 1: Buoyant force from displaced air: F_b = ρ_air·V·g = (1.225)(5.0)(9.81) = 60.09 N. Step 2: Total weight = weight of helium + fabric = (0.1786)(5.0)(9.81) + (2.0)(9.81) = 8.76 + 19.62 = 28.38 N. Step 3: Net upward force = 60.09 − 28.38 = 31.71 N. Answer: Net lift ≈ 31.7 N (about 3.23 kg of extra payload)
Problem 8
An object weighs 8.00 N in air, 6.00 N in water, and 6.40 N in oil. Find the object's density and the oil's density. Take water density 1000 kg/m³ and g = 9.81 m/s².
Show Solution
Solution: Step 1: In water, F_b = 8.00 − 6.00 = 2.00 N, so V = 2.00 / (1000 × 9.81) = 2.04 × 10⁻⁴ m³. Step 2: Object density ρ_obj = ρ_water × W_air / (W_air − W_water) = 1000 × 8.00 / 2.00 = 4000 kg/m³. Step 3: In oil, F_b = 8.00 − 6.40 = 1.60 N, so ρ_oil = ρ_water × (1.60 / 2.00) = 800 kg/m³. Answer: Object density = 4000 kg/m³; oil density = 800 kg/m³

Frequently Asked Questions

What is Archimedes' principle in simple terms?
Archimedes’ principle says that any object in a fluid is pushed up by a force equal to the weight of the fluid it pushes out of the way. If that upward push is at least as large as the object’s weight, it floats; if not, it sinks. The principle works in liquids and in gases.
What is the formula for buoyant force?
The buoyant force is F_b = ρVg, where ρ is the fluid’s density in kg/m³, V is the volume of fluid displaced in m³, and g is gravity (about 9.81 m/s²). The result is in newtons. Note that ρ is the density of the fluid, not the object.
Why do heavy ships float but a small nail sinks?
Floating depends on average density, not raw weight. A ship is mostly hollow, so its hull plus the air inside has an average density below water, and it floats. A solid nail has no trapped air, so its density stays well above water and it sinks.
Does the buoyant force depend on depth?
For a rigid object in an incompressible fluid such as water, no — the buoyant force is the same near the surface as it is deep down. Pressure rises with depth on both the top and bottom of the object, but the difference between them stays constant, so F_b = ρVg does not change. Compressible objects, like gas bubbles, are the exception.
How much of an iceberg is underwater?
About 89.5% of an iceberg sits below the surface, leaving only around 10.5% visible. The figure comes from the density ratio: ice (917 kg/m³) divided by seawater (1025 kg/m³) gives roughly 0.895, the submerged fraction.
Is buoyancy the same in air as in water?
The principle is identical, but the size of the force differs hugely. Air is about 800 times less dense than water, so it provides a far smaller buoyant force for the same displaced volume. That is why buoyancy is obvious in water yet noticeable in air only for very light objects like balloons.
How did Archimedes use his principle on the king's crown?
The story goes that Archimedes needed to check whether a crown was pure gold without melting it. By comparing the crown’s weight in air with its apparent weight in water, he could find its volume and therefore its density. A density below gold’s would reveal that cheaper metal had been mixed in.

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