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Buoyancy Force Calculator

Physics

The body is entirely under the surface, so it displaces its own volume and F_b = ρ_fluid · V · g.

Inputs

Fills the fluid density in kg/m³.
Fills the object density in kg/m³.
Leave blank to use the mass instead.
Used only when the volume is blank.
Enables the hydrostatic pressure panel.

Result

SINKS
Buoyant force
19.6133 N
Weight
153.9644 N
Apparent weight
134.3511 N
Net force (down)
134.3511 N

F_b = ρ_f · V_disp · g

The body drawn fully below the surface, with an upward buoyancy arrow of 19.61 N and a downward weight arrow of 153.96 N, drawn to scale against each other.surfaceF_b 19.61 NW 153.96 N

Every quantity

Displaced volume0.0020 m³
Displaced mass2.0000 kg
Object volume0.0020 m³
Object mass15.7000 kg
Object density7850.0000 kg/m³
Fluid density1000.0000 kg/m³
Density ratio (specific gravity)7.8500
Gravity used9.8066 m/s²
Pressure at the top face4903.3250 Pa
Pressure at the bottom face5883.9900 Pa
F_b from (P_bottom − P_top)·A19.6133 N

The same force in every unit

UnitValue
N19.6133
kN0.0196
lbf4.4092
kgf2.0000
dyn1961330.0000

Independent cross-check

Every mode recomputes its answer by a second route. A residual of zero means the two agree to the last bit a double can hold.

W − (m − m_displaced)·g, i.e. the weight actually lost on the scale = 19.6133 N · residual 1.066e-14

Push it under and watch

Drag the waterline to submerge more of the body. Buoyancy grows in step with the volume you push under, and the equilibrium is wherever the net force reaches zero.

Held under
100.00 %
Buoyant force
19.6133 N
Net force (down)
134.3511 N

Where these densities sit

A logarithmic ladder from air to gold. Anything lighter than the fluid marker floats in it; anything heavier sinks.

Air1.23 kg/m³
Gasoline720.00 kg/m³
Ethanol789.00 kg/m³
Diesel850.00 kg/m³
Ice917.00 kg/m³
Oil920.00 kg/m³
Water1000.00 kg/m³
This fluid1000.00 kg/m³
Seawater1025.00 kg/m³
Glycerin1260.00 kg/m³
Honey1420.00 kg/m³
Concrete2400.00 kg/m³
Aluminum2700.00 kg/m³
Steel7850.00 kg/m³
This object7850.00 kg/m³
Copper8960.00 kg/m³
Silver10490.00 kg/m³
Lead11340.00 kg/m³
Mercury13534.00 kg/m³
Gold19320.00 kg/m³

1. Fully submerged, so the displaced volume equals the object's own volume: V_disp = V = 0.002 m³.

2. Displaced mass = ρ_f · V_disp = 1000 × 0.002 = 2 kg

3. F_b = ρ_f · V_disp · g = 1000 × 0.002 × 9.80665 = 19.6133 N

4. W = ρ_o · V · g = 7850 × 0.002 × 9.80665 = 153.964405 N

5. W_apparent = W − F_b = 153.964405 − 19.6133 = 134.351105 N

6. Because F_b = (displaced mass) · g, the buoyant force in kgf is numerically the displaced mass: 2 kgf ↔ 2 kg.

7. Pressure view — cross-sectional area A = V / h = 0.002 / 0.1 = 0.02 m²

8. P_top = ρ_f · g · d_top = 1000 × 9.80665 × 0.5 = 4903.325 Pa

9. P_bottom = ρ_f · g · (d_top + h) = 1000 × 9.80665 × 0.6 = 5883.99 Pa

10. F_b = (P_bottom − P_top) · A = 980.665 × 0.02 = 19.6133 N — the same number ρ_f·V·g gives, because (d_bot − d_top)·A is exactly V.

Flotation does not care about gravity
Switch the celestial body and watch every force change while the submerged fraction stays exactly where it was. Standard gravity here is 9.80665 m/s²; the ratio ρ_object/ρ_fluid has no g in it at all.

About This Tool

Buoyancy Force Calculator – Archimedes’ Principle, Floating and Apparent Weight

Lower a rock into a bucket and it suddenly feels lighter. Push a beach ball under and it fights back. Both are the same effect: a fluid pushes up on anything placed in it, with a force equal to the weight of the fluid that thing shoves out of the way. That is Archimedes’ principle, and this buoyancy force calculatorworks in every direction around it — buoyant force, sink-or-float verdicts, how deep a floating body sits, what a scale reads underwater, and how much a balloon can lift.

The one equation everything comes from

The upward force is the density of the fluid times the volume displaced times gravity:

F_b = ρ_fluid · V_displaced · g

Notice what is notin it. The object’s material does not appear. Its shape does not appear. Only how much fluid it pushes aside. A cubic metre of lead and a cubic metre of styrofoam held under water feel exactly the same upward push of about 9807 N— they differ only in how hard gravity pulls them down. Two litres of steel in fresh water displaces 2 kg of water, so F_b = 1000 × 0.002 × 9.80665 = 19.6133 N. Its dry weight is 153.96 N, so on an underwater scale it reads 134.35 N— it has lost exactly the weight of the water it replaced.

Why the submerged fraction is a pure density ratio

For a body floating freely, the upward force balances the weight: ρ_object · V · g = ρ_fluid · V_submerged · g. The same g sits on both sides, so it cancels exactly, and dividing through by the total volume leaves the whole answer:

V_submerged / V = ρ_object / ρ_fluid

This is the most useful line in the whole subject. Ice at 917 kg/m³ in seawater at 1025 kg/m³ floats with 917/1025 = 89.4634 %of itself below the surface — the famous tip of the iceberg. Nothing about the berg’s size, mass or the local gravity enters the result. The same iceberg on the Moon would ride at precisely the same waterline; only the forces involved would shrink.

Buoyancy is really a pressure difference
Pressure in a fluid grows with depth, so the bottom of a submerged block is pushed up harder than the top is pushed down. For a block of area A between two depths, that imbalance is (P_bottom − P_top)·A = ρ_f·g·(d_bot − d_top)·A, and (d_bot − d_top)·A is simply the volume. The pressure view and ρVg are the same statement.

Sink, float or hover

Compare the two densities and the verdict follows immediately. Denser than the fluid and it sinks, with a leftover net force of (ρ_object − ρ_fluid)·V·g. Lighter and it floats, rising until it displaces exactly its own weight. Equal, and it is neutrally buoyant— it hovers wherever you leave it, which is the trim condition every scuba diver spends the first day of training chasing.

This is also the answer to the question of why a steel ship floats when a steel bar sinks. What matters is average density: a hull is mostly air, so the mass of the whole ship divided by the volume of the hull comes out well below 1000 kg/m³. Load cargo and the average density rises, so the ship settles deeper until it again displaces its own weight.

Weighing something to find out what it is

Weigh an object dry, then weigh it hanging in water. The weight it seems to lose is precisely the buoyant force, which gives both its volume and its density:

ρ_object = ρ_fluid · W_air / (W_air − W_sub)

Gravity cancels out of that formula completely, so you never need to know the local g. This is the Archimedes crown problem in its original form: a crown weighing 25 N in air and 23.55 N in water works out at 17241 kg/m³— denser than lead, well short of gold’s 19320. The king was being cheated.

Two close readings lose precision fast
Because the method rests on the difference of two weighings, a light object in a light fluid loses only a sliver of its weight and that subtraction destroys most of your significant figures. Below about one percent weight loss the derived density is barely meaningful; the calculator flags that regime instead of printing decimals it cannot justify.

Draft, freeboard and lifting gases

Turn the fraction into a real depth and you get a hull’s draft. For a prismatic body, draft = h · ρ_object/ρ_fluid, and the freeboard is whatever height is left above the waterline. A 10 cm tall wooden block at 700 kg/m³ floats with 7 cm wet and 3 cm dry. The same displacement route, draft = V_displaced / A_waterplane, gives the identical answer and is how naval architects actually work.

Buoyancy in air behaves no differently, which is what makes balloons possible. A 1000 m³ helium envelope displaces 1225 kg of air while the helium inside weighs only 178.6 kg, leaving about 1046 kg of lifting capacity. A hot-air balloon achieves the same thing with plain air made thinner by heat: since the envelope is open at the bottom, the pressure inside matches the pressure outside and density falls in proportion to absolute temperature, ρ_hot = ρ_ambient · T_ambient / T_hot.

The melting ice cube

A floating ice cube displaces its own weightof water, not its own volume. Melt it and it becomes exactly that weight of water — precisely the volume it was already displacing — so the level in a glass of fresh water does not move at all. In salt water the cube was displacing slightly less volume of denser fluid, so melting does raise the level a little. That is the same reasoning behind why melting sea ice contributes almost nothing to sea level while melting land ice, which was displacing nothing to begin with, contributes all of it.

Frequently Asked Questions

Is the Buoyancy Force Calculator free?

Yes, Buoyancy Force Calculator is totally free :)

Can I use the Buoyancy Force Calculator offline?

Yes, you can install the webapp as PWA.

Is it safe to use Buoyancy Force Calculator?

Yes, any data related to Buoyancy Force Calculator only stored in your browser (if storage required). You can simply clear browser cache to clear all the stored data. We do not store any data on server.

How does this buoyancy calculator work?

Everything comes from Archimedes' principle: the upward force a fluid exerts equals the weight of the fluid the body pushes aside, F_b = ρ_fluid · V_displaced · g. Pick the question you are actually asking — buoyant force on a submerged body, sink or float, how much of a floating body sits below the waterline, the draft of a hull, what a scale reads underwater, an unknown density from two weighings, one object across several fluids, or balloon lift — and the tool converts every input to SI, solves in SI, and rounds only for display. Each mode also recomputes its answer by a second, independent route and shows the residual between the two, so you can watch the calculation check itself instead of trusting a single number.

Why does a steel ship float when a steel block sinks?

Because flotation is decided by average density, not by the material. A solid steel block is 7850 kg/m³ against water's 1000, so it sinks. A ship is mostly air: its hull encloses a volume far larger than the steel it is built from, and the mass of ship divided by the volume of hull is well under 1000 kg/m³. The steel has not changed — the volume it displaces has. The same reasoning explains why the ship rides deeper as cargo is loaded, and why it floats slightly higher in salt water than in a fresh-water river.

Why is exactly 89.46 % of an iceberg underwater, whatever its size?

For a freely floating body the weight and the buoyant force are equal, so ρ_object · V · g = ρ_fluid · V_submerged · g. Gravity appears on both sides and cancels, and so does the total volume when you divide through, leaving submerged fraction = ρ_object / ρ_fluid. For ice at 917 kg/m³ in seawater at 1025 kg/m³ that is 917/1025 = 0.894634, or 89.4634 % below the surface. Nothing about the iceberg's size, its mass, or the local gravity enters the answer — a berg the size of a house and one the size of a county hide the same proportion, and both would hide the same proportion on the Moon.

How do I find an object's density by weighing it in water?

This is Archimedes' crown problem. Weigh the object dry, then weigh it hanging in a fluid of known density. The weight it appears to lose is exactly the buoyant force, so its volume is V = (W_air − W_sub)/(ρ_fluid · g) and its density is ρ_object = ρ_fluid · W_air/(W_air − W_sub). Gravity cancels out of the density formula entirely, so the answer is the same anywhere in the universe and you never need to know the local g. Enter 25 N in air and 23.55 N in water and you get 17241 kg/m³ — denser than lead, lighter than gold, so whatever that object is, it is not solid gold.

How accurate are the results, and when should I distrust them?

The arithmetic is exact to double precision and each mode is cross-checked by a second route, but the density mode has a genuine measurement hazard: it depends on the difference of two nearly equal weighings. When the object loses less than about 1 % of its weight on immersion — a light object in a light fluid — that subtraction throws away most of your significant figures and the derived density becomes noise, so the tool warns you rather than printing ten confident decimals. Results also assume a still, uniform, incompressible fluid, so they do not cover surface tension on very small objects, gas compression with depth in deep water, or a partly filled hollow body taking on water.

If a floating ice cube melts, does the glass overflow?

In fresh water, no — and the arithmetic shows exactly why. A 50 g cube of ice occupies 54.53 cm³ but floats displacing only 50.00 cm³ of water, because it displaces its own weight, not its own volume. When it melts it becomes 50 g of water, which is 50.00 cm³ — precisely the volume it was already displacing, so the level does not move by even a fraction of a millimetre. In seawater the answer changes: the cube displaced only 48.78 cm³ of the denser salt water but still melts into 50.00 cm³ of fresh water, so the level rises slightly. That difference is why melting sea ice barely affects sea level while melting land ice, which was displacing nothing at all, raises it fully.