What is upthrust (buoyant force)?
When you lift a friend in a swimming pool, they feel light. The water is helping you. Every liquid and gas pushes up on anything inside it. This upward force is called upthrust or buoyant force. Its unit is the newton (N).
Why does it happen? Pressure in a liquid grows with depth. The bottom of an object is deeper than its top, so the liquid pushes harder on the bottom than on the top. The difference is an upward push.
Apparent weight = real weight − upthrust. This is what a spring balance shows when the object is under water.
Archimedes' principle
Archimedes' principle: when an object is fully or partly in a fluid, it feels an upthrust equal to the weight of the fluid it displaces (pushes aside).
Weight of displaced fluid = mass × g = (ρfluid × V) × g, so
Fb = ρfluid g Vsub
- ρfluid = density of the liquid (kg/m³), not of the object.
- Vsub = volume of the object that is under the liquid (m³).
- g ≈ 9.8 m/s² (often 10 m/s² in sums).
Upthrust does not depend on how heavy the object is, or on how deep it is once it is fully under. It depends only on the liquid's density and the volume under.
Checking it by experiment
- Hang a stone from a spring balance. Read its weight in air, W₁.
- Fill an overflow (Eureka) can up to the spout. Put an empty beaker of known weight under the spout.
- Lower the stone fully into the water. Read the balance again, W₂. Collect the water that spills out.
- Weigh the spilled water, Ww.
Result: loss in weight W₁ − W₂ = Ww. The upthrust equals the weight of displaced water. Repeat with salt water: the loss is bigger.
This also gives the volume of an odd-shaped stone (the spilled volume) and its relative density = W₁ ÷ (W₁ − W₂).
Floating and sinking
Two forces act: weight W (down) and upthrust Fb (up).
- ρobject > ρliquid: weight is more than the biggest upthrust, so the object sinks.
- ρobject = ρliquid: weight = upthrust when fully under, so it hangs at any depth.
- ρobject < ρliquid: the object rises and floats with only part of it under.
Law of flotation: a floating object displaces its own weight of liquid. So fraction under = ρobject ÷ ρliquid. Ice (920 kg/m³) in water: about 92% is under, which is why most of an iceberg is hidden.
Ships, submarines, balloons, hydrometers
A ship is hollow, so its average density (steel + air inside) is less than water. A submarine fills tanks with water to sink and blows them out with air to rise. A balloon filled with helium or hot air floats in air. A hydrometer floats lower in less dense liquids, so its scale reads the density of milk or battery acid.
Extra: forces between molecules (surface tension)
This is not buoyancy, but it is often taught next to it. Molecules of a liquid pull on each other. At the surface they are pulled only inwards and sideways, so the surface acts like a stretched skin. This is surface tension. It lets a steel needle or a water strider sit on water even though steel is denser, makes drops round, and lets a drop hang from a tap until its weight beats the pull.
Key formulas and definitions
- Upthrust F_b = ρ_liquid × g × V_sub
- Apparent weight = real weight − upthrust
- Floating: upthrust = weight of object
- Fraction under = ρ_object ÷ ρ_liquid
- Relative density = weight in air ÷ loss of weight in water
Worked examples
1. A 0.002 m³ stone is fully under water (ρ = 1000 kg/m³, g = 10 m/s²). Find the upthrust.
Step 1: F = ρgV. Step 2: F = 1000 × 10 × 0.002. Answer: 20 N.
2. A metal piece weighs 30 N in air and 22 N in water. Find the upthrust and its volume. (g = 10 m/s²)
Step 1: upthrust = 30 − 22 = 8 N. Step 2: 8 = 1000 × 10 × V, so V = 8 ÷ 10000 = 0.0008 m³ = 800 cm³.
3. For the piece in example 2, find its relative density and its density.
Relative density = 30 ÷ 8 = 3.75. Density = 3.75 × 1000 = 3750 kg/m³. It sinks in water.
4. A wooden block of density 600 kg/m³ floats in water. What fraction is under water?
Fraction = 600 ÷ 1000 = 0.6, so 60% is under and 40% is above.
5. A 1500 kg boat floats in a lake. What volume of water does it displace? (ρ = 1000 kg/m³)
Floating: weight of displaced water = weight of boat, so mass displaced = 1500 kg. V = 1500 ÷ 1000 = 1.5 m³.
6. The same boat sails into the sea (ρ = 1025 kg/m³). Does it rise or sink a little? Find the new displaced volume.
It still displaces 1500 kg. V = 1500 ÷ 1025 ≈ 1.46 m³. Less volume under, so the boat rises a little in sea water.
Common mistakes
- Using the object's density in F = ρgV. Use the density of the liquid.
- Thinking heavier objects get more upthrust. Upthrust depends on the volume under the liquid, not on the object's weight.
- Thinking upthrust grows as a fully sunk object goes deeper. Once fully under, V does not change, so upthrust stays the same.
- Saying 'heavy things sink'. A heavy steel ship floats; density (or a shape that makes the average density low) is what matters.