Orientation · Module 7 of 10
Fluids and Pressure
Fluids push on everything they touch. This module introduces pressure, how it grows with depth, the force it puts on a surface, and why things float.
Readiness check
This module opens fluid mechanics. Tick what you can do comfortably.
- Multiply three numbers together.
- Recall that pressure is force per unit area.
- Recall the density of water is about 1000 kg/m3.
- Multiply a pressure by an area to get a force.
- Recall that heavy things can still float.
The core idea
Pressure is force spread over area, p = F / A. In a still fluid it grows with depth as p = ρgh, so deeper means higher pressure. That pressure pushes on any surface with a force F = pA, and a submerged body is buoyed up by the weight of the fluid it displaces.
pressure p = F / Ahydrostatic p = ρghforce on a surface F = pAA fluid, liquid or gas, cannot resist being sheared, so it flows, and it pushes on every surface it touches with a pressure, force per unit area. In a fluid at rest the pressure is not uniform: it increases with depth, because deeper fluid must hold up the weight of everything above it. That gives the hydrostatic law p = ρgh, where ρ is the fluid density, g is gravity, and h is the depth below the surface. Five metres down in water the gauge pressure is already about 49 kilopascals. This pressure is what a design must withstand, and the force on a submerged surface is simply the pressure times the area, F = pA, which is why a deep dam is built thick at the bottom. The same depth-pressure idea explains buoyancy: the pressure on the bottom of a submerged body exceeds that on its top, giving a net upward force equal to the weight of the fluid displaced, Archimedes' principle, which decides whether something floats. When the fluid moves, pressure, speed, and height trade off, and drag appears, the subjects the Fluid Mechanics course develops in full.
The skills, taught in order
Five skills introduce how fluids load a design.
7.1 Pressure
Pressure is force per unit area, in pascals. A fluid exerts it on every surface it touches, perpendicular to that surface. Small pressures over large areas make large forces, which is the theme of the module.
7.2 Hydrostatic pressure
In a still fluid, pressure increases with depth as p = ρgh, because deeper fluid supports more weight above it. This is why your ears hurt at the bottom of a pool and why deep tanks need thick walls.
7.3 Force on a surface
The force a pressure exerts on a flat surface is F = pA. A modest pressure on a large hatch or window becomes a large force, which is why submarine and dam surfaces are engineered carefully.
| Quantity | Relation | Grows with |
|---|---|---|
| Depth pressure | p = ρgh | depth |
| Surface force | F = pA | pressure and area |
| Buoyant force | F = ρgV | displaced volume |
Three relations built from pressure. Each is a product, so the pieces are easy to track.
7.4 Buoyancy
Because pressure is higher on the bottom of a submerged body than the top, there is a net upward force equal to the weight of the displaced fluid, F = ρgV. If it exceeds the body's weight, the body floats.
7.5 Fluids in motion
When a fluid flows, pressure, speed, and elevation trade off, and moving past a body it creates drag. These effects, central to pumps, pipes, and vehicles, are where Fluid Mechanics goes next.
Engineering connection: sizing a tank wall or a submersible hatch means finding ρgh at the depth and multiplying by area, exactly the workflow of Fluid Mechanics.
Worked example 1: pressure with depth
Find the gauge pressure 5 m below the surface of water, with ρ = 1000 kg/m3 and g = 9.81 m/s2.
- ProblemFind the gauge pressure at the point in Figure 1.
- Given / findρ = 1000 kg/m3, g = 9.81 m/s2, h = 5 m. Find p.
- AssumptionsStill water, constant density, gauge pressure (above atmospheric).
- Modelp = ρgh.
- Equationsp = 1000 × 9.81 × 5
- Solvep = 49050 Pa ≈ 49.05 kPa.
- CheckWater adds about 9.81 kPa per metre, and 5 × 9.81 ≈ 49 kPa, matching.
- ConclusionFive metres down, the water already presses at about 49 kPa, roughly half an atmosphere.
Worked example 2: force on a hatch
A flat hatch of area 0.5 m2 sits at that depth, where the gauge pressure is 49.05 kPa. Find the force the water exerts on it.
- ProblemFind the force on the hatch in Figure 2.
- Given / findPressure 49.05 kPa, area 0.5 m2. Find the force.
- AssumptionsUniform pressure over the hatch, gauge pressure acting outward.
- ModelF = pA.
- EquationsF = 49050 × 0.5
- SolveF = 24525 N ≈ 24.5 kN.
- Check24.5 kN is about 2.5 tonnes of force from only half an atmosphere, showing how area multiplies pressure.
- ConclusionThe hatch must resist about 24.5 kN, which is why underwater doors are heavily built and latched.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Pressure treated as force | Units of pascals used as newtons | "Per unit area, or total?" | Multiply pressure by area to get force. |
| Forgetting depth | Same pressure assumed everywhere | "How deep is this point?" | Pressure grows with depth as ρgh. |
| Wrong density | Pressure far off | "Which fluid, and its density?" | Use the fluid's density; water is 1000 kg/m3. |
| Gauge versus absolute | Atmospheric pressure double counted | "Above vacuum or above atmosphere?" | Gauge pressure is measured above atmospheric. |
Practice ladder
Find the gauge pressure 3 m below the surface of water.
Show answer
p = 1000 × 9.81 × 3 = 29430 Pa ≈ 29.4 kPa.
A window of area 0.2 m2 is at a depth where the gauge pressure is 49 kPa. Find the force on it.
Show answer
F = 49050 × 0.2 = 9810 N ≈ 9.81 kN.
A body submerged in water displaces 0.01 m3. Find the buoyant force on it.
Show answer
F = ρgV = 1000 × 9.81 × 0.01 = 98.1 N.
Estimate the force on a circular submarine porthole 0.3 m in diameter at a depth of 20 m in seawater (ρ ≈ 1025 kg/m3).
What good work looks like
Pressure p = 1025 × 9.81 × 20 ≈ 201 kPa. Area A = π(0.15)2 ≈ 0.0707 m2. Force F = pA ≈ 14.2 kN. A good answer computes p from ρgh, the circular area, and multiplies, then comments on how the force climbs with depth.
Working with AI, and proving it yourself
Use AI as a guide, not an oracle
Portfolio task
Pick a real submerged surface and estimate the pressure at its depth and the force on it.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write pressure.
p = F / A, force per unit area.
2. Write hydrostatic pressure.
p = ρgh, growing with depth.
3. Write the force on a surface.
F = pA.
4. State Archimedes' principle.
The buoyant force equals the weight of the displaced fluid.
5. Why do deep tanks need thick walls?
Pressure, and so the wall force, rises with depth.
Textbook mapping
This module follows Wickert and Lewis, An Introduction to Mechanical Engineering, 3rd edition. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| Properties of fluids | Wickert and Lewis, Section 6.2, Properties of Fluids |
| Pressure and buoyancy | Wickert and Lewis, Section 6.3, Pressure and Buoyancy |
| Fluid flow and drag | Wickert and Lewis, Sections 6.4 to 6.6 |
Section numbers refer to Wickert and Lewis, 3rd edition. Any edition with the same chapter titles is equivalent for study.