Fluid Statics
Pressure, Pascal’s Law, Hydrostatic Force, Buoyancy & Manometers — Complete Guide
Last Updated: March 2026
📌 Key Takeaways
- Pressure in a static fluid increases linearly with depth: P = P₀ + ρgh.
- Pascal’s law: Pressure applied to an enclosed fluid transmits equally in all directions — the basis of hydraulic systems.
- Hydrostatic force on a submerged surface: F = ρgh̄A, where h̄ is the depth of the centroid.
- Archimedes’ principle: Buoyant force = weight of fluid displaced = ρfluidgVdisplaced.
- Manometers measure pressure differences using the hydrostatic pressure equation applied to each fluid column.
- Fluid statics has no velocity — all analysis involves only pressure, gravity, and geometry.
1. Pressure Variation with Depth
In a fluid at rest, the only forces acting are gravity and pressure forces. Balancing these forces on a small fluid element gives the fundamental equation of fluid statics:
Hydrostatic Pressure Equation
P = P₀ + ρgh
Where: P = pressure at depth h (Pa), P₀ = pressure at the free surface (Pa), ρ = fluid density (kg/m³), g = 9.81 m/s², h = depth below the free surface (m)
Key observations:
- Pressure increases linearly with depth in an incompressible fluid (constant ρ).
- Pressure at any point depends only on vertical depth, not on the shape or size of the container (this is the hydrostatic paradox).
- All points at the same depth in a continuous, connected fluid have the same pressure.
- The pressure increase per metre of water depth is ρg = 1000 × 9.81 = 9,810 Pa ≈ 9.81 kPa/m.
- Every 10.33 metres of water adds approximately 1 atmosphere of pressure.
Pressure Head
h = P / (ρg)
Pressure expressed as an equivalent height of fluid column. Used extensively in hydraulics.
2. Pascal’s Law & Hydraulic Systems
Pascal’s Law
Pressure applied to an enclosed fluid is transmitted equally and undiminished to every point in the fluid and to the walls of the container.
This principle makes hydraulic systems possible. In a hydraulic press or lift:
Hydraulic Advantage
P₁ = P₂ → F₁/A₁ = F₂/A₂
F₂ = F₁ × (A₂/A₁)
A small force on a small piston creates a large force on a large piston.
Mechanical advantage = A₂/A₁
Important: While force is multiplied, work is conserved. The small piston must move a larger distance than the large piston: d₁ × A₁ = d₂ × A₂ (volume of fluid is constant). You gain force but sacrifice distance — energy conservation is never violated.
Applications: hydraulic brakes (cars), hydraulic lifts (garages), hydraulic presses (manufacturing), earth-moving equipment, aircraft control systems.
3. Hydrostatic Force on Submerged Surfaces
Force on a Plane Surface
Resultant Hydrostatic Force
F = ρgh̄A
Where: F = total force (N), h̄ = vertical depth of the centroid of the surface (m), A = area of the surface (m²)
The force acts at the centre of pressure, which is always below the centroid (deeper than the centroid). For a vertical rectangular surface of height d with its top edge at depth ht:
Centre of Pressure (Vertical Rectangle)
hcp = h̄ + IG/(h̄A)
Where IG = second moment of area about the centroidal axis. For a rectangle: IG = bd³/12.
Force on a Curved Surface
For curved surfaces, resolve the force into horizontal and vertical components:
- Horizontal component FH = hydrostatic force on the vertical projection of the curved surface.
- Vertical component FV = weight of the fluid directly above (or that would be above) the curved surface.
- Resultant: F = √(FH² + FV²), acting at angle θ = tan⁻¹(FV/FH).
4. Buoyancy & Archimedes’ Principle
Archimedes’ Principle
FB = ρfluid × g × Vdisplaced
The buoyant force equals the weight of the fluid displaced by the immersed body.
FB acts vertically upward through the centre of buoyancy (centroid of the displaced volume).
| Condition | What Happens |
|---|---|
| FB > Weight of body (ρbody < ρfluid) | Body floats — rises until displaced fluid weight = body weight |
| FB = Weight of body (ρbody = ρfluid) | Neutral buoyancy — body stays at any depth |
| FB < Weight of body (ρbody > ρfluid) | Body sinks to the bottom |
For a floating body, the weight of fluid displaced equals the weight of the body. This gives:
Floating Body — Draft
ρbody × Vbody = ρfluid × Vsubmerged
Fraction submerged = ρbody/ρfluid = SGbody
5. Stability of Floating Bodies
A floating body can be in stable, unstable, or neutral equilibrium depending on the relative positions of the centre of gravity (G) and the metacentre (M):
| Condition | Stability | What Happens When Tilted |
|---|---|---|
| M above G (GM > 0) | Stable | Restoring moment returns body to upright position |
| M below G (GM < 0) | Unstable | Overturning moment tips body further — capsizes |
| M coincides with G (GM = 0) | Neutral | No restoring or overturning moment — stays tilted |
Metacentric Height
GM = BM − BG
Where: BM = I/Vdisplaced
I = second moment of area of the waterplane about the longitudinal axis, Vdisplaced = volume of fluid displaced
Ship design requires a positive GM for roll stability. Typical values: large ships GM = 0.3–1.0 m; small boats GM = 0.5–1.5 m.
6. Manometers — Measuring Pressure
Manometers are devices that measure pressure by balancing it against a column of liquid. They use the hydrostatic equation P = ρgh at each interface.
Simple U-Tube Manometer
A U-shaped tube with one end connected to the pressure source and the other end open to atmosphere. A manometric fluid (usually mercury, SG = 13.6) fills the bottom of the U-tube.
U-Tube Manometer Equation
PA + ρ1gh₁ − ρmghm = Patm
Start from one side, add ρgh going down and subtract ρgh going up, until you reach the other side.
Differential Manometer
Measures the pressure difference between two points. Both ends are connected to the fluid system.
Differential Manometer
PA − PB = (ρm − ρfluid) × g × h
Where h = manometric fluid height difference, ρm = manometric fluid density, ρfluid = working fluid density
Manometer calculation method: Start at one known pressure point. Move through the manometer step by step — add ρgh when going down into a fluid, subtract ρgh when going up. Equate to the pressure at the other end. This systematic approach works for any manometer configuration.
7. Worked Numerical Examples
Example 1: Pressure at Depth
Problem: Find the absolute pressure at a depth of 15 m in a lake. Atmospheric pressure = 101.3 kPa, ρwater = 1000 kg/m³.
Solution
P = P₀ + ρgh = 101,300 + 1000 × 9.81 × 15 = 101,300 + 147,150
P = 248,450 Pa = 248.5 kPa ≈ 2.45 atm
Example 2: Hydraulic Lift
Problem: A hydraulic lift has a small piston diameter of 5 cm and a large piston diameter of 30 cm. What force must be applied to the small piston to lift a 1,500 kg car?
Solution
Weight of car = 1500 × 9.81 = 14,715 N
A₁ = π(0.025)² = 1.963 × 10⁻³ m², A₂ = π(0.15)² = 7.069 × 10⁻² m²
F₁/A₁ = F₂/A₂ → F₁ = F₂ × A₁/A₂ = 14,715 × (1.963 × 10⁻³)/(7.069 × 10⁻²)
F₁ = 408.5 N ≈ 409 N
A force of only 409 N (about 42 kg-force) lifts a 1,500 kg car — a mechanical advantage of 36:1.
Example 3: Buoyancy — Will It Float?
Problem: A wooden block has density 600 kg/m³ and volume 0.1 m³. It is placed in water (ρ = 1000 kg/m³). Find the buoyant force and the fraction of the block submerged.
Solution
Weight = ρblock × V × g = 600 × 0.1 × 9.81 = 588.6 N
Since ρblock < ρwater, the block floats.
For floating equilibrium: FB = Weight → ρwater × Vsub × g = 588.6
Vsub = 588.6 / (1000 × 9.81) = 0.06 m³
Fraction submerged = Vsub/Vtotal = 0.06/0.1 = 60%
Alternatively: fraction = ρblock/ρwater = 600/1000 = 0.6 = 60% ✓
Example 4: U-Tube Manometer
Problem: A U-tube manometer connected to a pipe carrying water shows a mercury column height difference of 200 mm. The centre of the pipe is 300 mm above the mercury surface on the high-pressure side. Find the gauge pressure at the pipe centre. (ρHg = 13,600 kg/m³, ρwater = 1000 kg/m³)
Solution
Starting from pipe centre (point A), going down 0.3 m through water, then up 0.2 m through mercury to atmosphere:
PA + ρwg(0.3) − ρHgg(0.2) = 0 (gauge)
PA = ρHgg(0.2) − ρwg(0.3)
= 13,600 × 9.81 × 0.2 − 1000 × 9.81 × 0.3
= 26,683.2 − 2,943 = 23,740 Pa = 23.74 kPa (gauge)
8. Common Mistakes Students Make
- Measuring depth incorrectly for hydrostatic force: The depth h̄ in F = ρgh̄A is the vertical depth of the centroid of the surface below the free surface — not the distance along the inclined surface. For inclined surfaces, h̄ = y̅ sin θ where y̅ is the distance along the surface from the free surface to the centroid.
- Forgetting that centre of pressure is below the centroid: The resultant force acts at the centre of pressure, which is always deeper than the centroid for submerged surfaces. Setting them equal loses marks.
- Sign errors in manometer calculations: When traversing a manometer, always add ρgh when moving downward into a fluid and subtract ρgh when moving upward. Inconsistent signs are the most common manometer error.
- Using weight instead of mass in buoyancy problems: FB = ρgVdisplaced gives force in Newtons. Comparing this directly to mass in kg is dimensionally wrong — compare to weight (mg) instead.
- Confusing gauge and absolute pressure: Gauge pressure = absolute pressure − atmospheric pressure. Manometers open to atmosphere measure gauge pressure. Always state which you are calculating.
9. Frequently Asked Questions
What is Pascal’s law?
Pascal’s law states that any pressure change applied to an enclosed fluid is transmitted equally to every part of the fluid. This means a small force on a small area creates the same pressure as a large force on a large area — enabling hydraulic systems where small inputs produce large outputs. Hydraulic brakes, lifts, and presses all operate on this principle.
What is Archimedes’ principle?
A body partially or fully immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces: FB = ρfluidgVdisplaced. If this force exceeds the body’s weight, it floats; if less, it sinks. This explains how steel ships (overall density less than water due to hollow hull) float despite steel being denser than water.
How does pressure vary with depth in a fluid?
In a static, incompressible fluid, pressure increases linearly with depth: P = P₀ + ρgh. For water, this means roughly 9.81 kPa increase for every metre of depth, or about 1 atmosphere for every 10.33 metres. Pressure at any depth depends only on the vertical distance below the surface, not on the shape of the container (hydrostatic paradox).