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Concept module

Pressure and Hydrostatic Pressure

Use one piston-and-tank bench to connect force per area, pressure acting in all directions, and the way density, gravity, and depth build hydrostatic pressure.

Interactive lab

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Starter track

Step 1 of 50 / 5 complete

Fluid and Pressure

Next after this: Continuity Equation.

1. Pressure and Hydrostatic Pressure2. Continuity Equation3. Bernoulli's Principle4. Buoyancy and Archimedes' Principle+1 more steps

This concept is the track start.

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Why it behaves this way

Explanation

Pressure starts as force per unit area. If the same push is spread over a larger area, each square meter gets a smaller share of that push. On this bench, the piston at the fluid surface makes that idea visible before depth is added.

A fluid at rest transmits pressure throughout the connected fluid, and at one point that pressure acts equally in all directions. The probe arrows do not point only downward because hydrostatic pressure is not a downward force rule. It is a scalar pressure field that can later create net forces only when pressure differs from one place to another.

Hydrostatic pressure grows with depth because deeper points support more fluid above them. In this bounded model, the probe reading is the surface pressure from $F / A$ plus the depth-dependent gain $ρ g h$ . That keeps force, area, density, gravity, depth, graphs, compare mode, prediction prompts, worked examples, and challenge checks attached to one honest static-fluid story without expanding into a full fluid-mechanics platform.

Key ideas

01Pressure from a surface load is force divided by area, so the same force gives a smaller pressure when it is spread across a wider piston.

02In a resting connected fluid, pressure at one point acts equally in all directions and points at the same depth share the same hydrostatic contribution.

03Hydrostatic pressure increases linearly with depth because each extra meter adds another $\rho g$ of pressure.

04Density and gravity set the slope of the pressure-depth graph, which is why denser fluids or stronger gravity produce larger pressure differences over the same height.

Frozen walkthrough

Step through the frozen example

Frozen walkthrough

Use the current tank directly. The same piston load, fluid properties, and probe depth now on screen drive both worked examples, so the algebra stays tied to the live visualization.

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Frozen valuesUsing frozen parameters

For the current tank with $F = 720$ , $A = 0.15$ , $ρ = 1 e 3$ , $g = 9.8$ , and $h = 1$ , what pressure should the probe read?

F

Surface force

720 N

A

Piston area

0.15 m^2

ρ

Fluid density

1e3 kg/m^3

g

Gravity

9.8 m/s²

h

Probe depth

1 m

1. Turn the piston load into surface pressure

The piston sets

P_{s} = F / A = 4.8 kPa

, so the same force is currently being spread across a moderate piston area.

2. Add the hydrostatic contribution from depth

The fluid column contributes

P_{h} = ρ g h = 9.8 kPa

at depth

1 m

3. Combine the two contributions

So the probe reads

P_{probe} = P_{s} + P_{h} = 14.6 kPa

, with the hydrostatic contribution currently larger.

Current probe pressure

P_{s} = 4.8 kPa, P_{h} = 9.8 kPa, P_{probe} = 14.6 kPa

The fluid column is doing most of the work here, so moving deeper or changing density would matter more than changing the same already-moderate surface load.

Fluid statics checkpoint

Two pressure probes are held at the same depth in the same resting fluid, but one is near the wall and the other is near the center. Which reads higher?

Make a prediction before you reveal the next step.

Decide whether sideways position matters once fluid, depth, gravity, and surface loading are all the same.

Check your reasoning against the live bench.

They read the same pressure.

In one connected resting fluid, the hydrostatic part depends on depth, density, and gravity. Sideways position does not create a second pressure law. Later buoyancy comes from different depths on the same object, not from a special sideways pressure rule.

Common misconception

Fluid pressure only pushes downward because the fluid is above the point you are looking at.

The fluid weight above a point explains why pressure grows with depth, but the pressure at that point acts equally in all directions.

That is why the same-depth line matters, and why later buoyancy comes from pressure being larger on deeper parts of an object than on shallower parts.

Quick test

Variable effect

Question 1 of 5

Answer from the tank, not from a slogan. The goal is to keep pressure causal and visual.

At the same force, what happens if the piston area is increased?

Use the live bench to test the result before moving on.

Accessibility

The simulation shows one piston pressing on a fluid tank and one probe at an adjustable depth. The piston width represents area, the downward arrow represents surface force, the probe depth marker shows how far below the surface the point is, and equal-length arrows around the probe show pressure acting equally in all directions at that point.

The readout card reports force, area, surface pressure, density, gravity, depth, hydrostatic pressure, and total probe pressure. Compare mode ghosts one second setup so the same tank can show two different ways to reach similar pressures.

All graphs on this page are response graphs for a static fluid. One graph varies depth, one varies density, one varies force, and one varies piston area while the other variables stay fixed.

Graph summary

The pressure-depth graph is the main hydrostatic graph. The surface-pressure line stays flat while the hydrostatic and total curves rise linearly with depth.

The density, force, and area graphs isolate the other levers. Density changes only the hydrostatic part, force changes only the surface-pressure part, and area changes the same surface load by spreading it out more or less.

Carry this into buoyancy and fluids

Keep this idea moving

Open the next concept, route, or track only when you want the current model to widen into a larger branch.

Move from static pressure to steady flowFluids

Pressure and Hydrostatic Pressure

Fluid and Pressure

Explanation

Step through the frozen example

For the current tank with F=720, A=0.15, ρ=1e3, g=9.8, and h=1, what pressure should the probe read?

At the same force, what happens if the piston area is increased?

Keep this idea moving

Continuity Equation

Bernoulli's Principle

Buoyancy and Archimedes' Principle

For the current tank with $F = 720$ , $A = 0.15$ , $ρ = 1 e 3$ , $g = 9.8$ , and $h = 1$ , what pressure should the probe read?