Pressure and Buoyancy, Explained
After reading this you can compute the pressure at any depth in a fluid, find the buoyant force on a submerged object, and decide whether that object floats or sinks, all from two short formulas.
Dive to the bottom of a 3 metre swimming pool and your ears hurt. Drop a sealed empty bottle in the same pool and it shoots back to the surface. Both effects come from the same fact: a fluid at rest pushes on everything in it, and that push grows with depth. This article derives the two equations behind those experiences and shows how to read the numbers the calculator returns.
What the calculator does
The tool works in two modes. The first, pressure at depth, gives the hydrostatic pressure a fluid exerts a distance h below its surface. The second, buoyant force, gives the upward force on a submerged object and tells you whether it floats.
Both rest on the idea that pressure is force spread over area, measured in pascals (Pa), where 1\ \text{Pa} = 1\ \text{N/m}^2. Atmospheric pressure at sea level is about 101325 Pa, usually written as 1 atm. Keep that number handy; it is the yardstick for every pressure result you will read here.
When to use it, and when not
Use the hydrostatic formula when the fluid is at rest, has roughly constant density, and you want the pressure at a single still depth. Water in a tank, a lake, or the sea all qualify. Use the buoyancy formula for a solid object fully or partly submerged in such a fluid.
The model breaks down in three common situations. First, fast-moving fluids: once water flows, pressure also depends on speed (that is Bernoulli's domain, not this one). Second, gases over large height changes: air density falls with altitude, so ρgh with a fixed density is only good for a few tens of metres. Third, extreme depth in water: water compresses very slightly, so density at 10 km down is about 5 percent higher than at the surface, and the linear formula underestimates the pressure there.
The pressure from ρgh is the pressure added by the fluid. The total, or absolute, pressure includes the atmosphere pressing on the surface. A diver at 10 m feels about 1 atm from the water plus 1 atm from the air above, so roughly 2 atm total.
The hydrostatic pressure formula
Picture a column of fluid of cross-sectional area A reaching from the surface down to depth h. The fluid below must hold up the weight of everything above it. That column has volume A h, so its mass is \rho A h and its weight is \rho A h g. Pressure is that weight divided by the area it sits on:
The area cancels, which is why a thin straw and a wide reservoir show the same pressure at the same depth. Here \rho is the fluid density in kg/m³, g is gravity in m/s², and h is depth in metres. The result P comes out in pascals.
A useful check: for fresh water, \rho g = 1000 \times 9.81 = 9810 Pa per metre of depth. Ten metres gives 98100 Pa, which is close to 1 atm. That is the origin of the footnote rule: pressure rises about 1 atm for every 10 m of water.
The buoyancy formula
Now submerge an object. The fluid pressure on its bottom face is larger than on its top face, because the bottom is deeper. That pressure difference produces a net upward force. Archimedes found its value exactly: the buoyant force equals the weight of the fluid the object pushes aside.
Here V is the volume of fluid displaced (the submerged volume of the object), \rho_{\text{fluid}} is the fluid density, and g is gravity. Notice the object's own density does not appear. A litre of lead and a litre of cork, both fully submerged, feel the same buoyant force.
Whether the object floats depends on comparing that upward force to the object's weight, W = \rho_{\text{obj}} V g. Divide one by the other and the V g cancels:
So the rule is purely about densities. If \rho_{\text{obj}} \lt \rho_{\text{fluid}} the object floats. If it is denser, it sinks. If they are equal, it hovers at neutral buoyancy.
Worked example with the demo data
The demo button loads the field defaults. Run them through both modes.
Pressure at 10 m of fresh water
- Inputs: \rho = 1000 kg/m³, g = 9.81 m/s², h = 10 m.
- Apply P = \rho g h = 1000 \times 9.81 \times 10.
- Result: P = 98100 Pa.
- Convert to atmospheres: 98100 / 101325 \approx 0.968 atm. So the water alone adds just under 1 atm, matching the rule of thumb.
Buoyant force on a 10 L object at 500 kg/m³
- Convert volume to cubic metres: 10\ \text{L} = 0.01\ \text{m}^3.
- Buoyant force if fully submerged: F_b = 1000 \times 0.01 \times 9.81 = 98.1 N.
- Object weight: W = \rho_{\text{obj}} V g = 500 \times 0.01 \times 9.81 = 49.05 N.
- Since W = 49.05\ \text{N} \lt F_b = 98.1\ \text{N}, the object floats.
- Submerged fraction at rest: 49.05 / 98.1 = 0.5, so it floats with exactly half its volume underwater. That matches the density ratio 500/1000 = 0.5.
Reading and interpreting the results
For pressure, the raw output is in pascals, which are small units. Divide by 1000 for kilopascals, or by 101325 for atmospheres. A car tyre runs near 220 kPa, so the pressure at 22 m of water roughly matches your tyre. That comparison keeps the numbers grounded.
For buoyancy, look at three numbers together: the buoyant force F_b, the object weight W, and their difference. If F_b \gt W, the difference is the net upward force that lifts the object until part of it breaks the surface. If F_b \lt W, the difference is the apparent weight you would feel holding it underwater. A 49.05 N object with 98.1 N of buoyancy is easy to hold down; the water pushes with almost half again its weight.
| Object density (kg/m³) | Weight (N) | Buoyant force full (N) | Submerged fraction | Verdict |
|---|---|---|---|---|
| 250 | 24.53 | 98.1 | 0.25 | Floats |
| 500 | 49.05 | 98.1 | 0.5 | Floats |
| 900 | 88.29 | 98.1 | 0.9 | Floats |
| 1000 | 98.1 | 98.1 | 1.0 | Neutral |
| 1200 | 117.7 | 98.1 | 1.0 | Sinks |
Common mistakes
The most frequent error is mixing units on volume. The field asks for litres, but the physics needs cubic metres. Always use 1\ \text{L} = 0.001\ \text{m}^3. If you enter 10 expecting cubic metres, your buoyant force comes out 1000 times too large.
Do not confuse the fluid density with the object density. The buoyant force uses the fluid density; the weight uses the object density. Swap them and a floating object appears to sink. In the demo, using 500 for the fluid would halve the buoyant force to 49.05 N and wrongly report neutral buoyancy.
A second mistake is reporting ρgh as the absolute pressure. It is only the part the fluid adds. Add atmospheric pressure (101325 Pa) if you need the total force on a submerged surface, such as a submarine hatch.
A third is assuming a floating object displaces its full volume. It does not. A floating object displaces only enough fluid to match its own weight, which is why the submerged fraction equals the density ratio. The full-volume buoyant force applies only when something holds the object completely under.
Related tools
Buoyancy and weight are both forces, so once you have them you often need Newton's laws. The Force Calculator turns a net force into an acceleration, useful for finding how fast that empty bottle rises. The Density Calculator gives you the object density from its mass and volume, which is the single input that decides floating. If the object then moves, the Kinematics Calculator handles its rise or fall, and the Kinetic and Potential Energy Calculator tracks the energy as it does so.
Frequently asked questions
Why does pressure not depend on the shape of the container?
Because the area cancels in P = \rho g h. Only the vertical depth of fluid above a point matters. A narrow tube and a wide tank filled to the same height show the same pressure at the bottom. This is called the hydrostatic paradox.
Does a heavier object sink faster because gravity is stronger on it?
Sinking or floating is decided by density, not weight. A 1 kg block of cork floats and a 1 g steel pellet sinks. Both feel the same buoyant-to-weight ratio as any other piece of the same material, because the shared volume cancels.
How deep before water pressure crushes a person?
The human body is mostly water and transmits pressure evenly, so uniform pressure alone is not the danger. The problem is air-filled spaces (lungs, sinuses) that get squeezed. At 30 m of seawater the added pressure is about 1025 \times 9.81 \times 30 \approx 302\,000 Pa, roughly 3 atm on top of the atmosphere, which is why divers manage air volumes carefully.
Why does the same object float higher in seawater than in fresh water?
Seawater is denser (about 1025 kg/m³ versus 1000). The submerged fraction is the density ratio, so an 800 kg/m³ object sits 800/1025 \approx 0.78 submerged in seawater but 800/1000 = 0.80 in fresh water. It rides about 2 percent higher in the sea.
Can I use this for air or other gases?
Yes, over short height changes. Set the fluid density to about 1.225 kg/m³ for air at sea level. Buoyancy on a helium balloon works the same way. But because air density drops with altitude, do not trust a single ρgh value across kilometres of height.
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