Hydrostatics — Pressure, Buoyancy & Manometers
Fluid pressure at rest, forces on submerged surfaces, centre of pressure, buoyancy, and manometer analysis — with full GATE CE worked examples
Last Updated: April 2026
- Hydrostatic pressure at depth h: p = ρgh — acts equally in all directions (Pascal’s law).
- Total hydrostatic force on a plane surface: F = ρg Ȳ A, where Ȳ is depth of centroid.
- Centre of pressure is always below the centroid of the surface: hcp = Ȳ + IG/(AȲ).
- Buoyant force = weight of fluid displaced (Archimedes’ principle): FB = ρfluid g Vsubmerged.
- A body floats when buoyant force = weight; it sinks when weight > buoyant force.
- Metacentric height GM determines stability of floating bodies; GM > 0 → stable.
- Manometers measure gauge pressure (simple) or pressure difference (differential) using a U-tube filled with a manometric fluid.
1. Pressure — Definition, Units & Types
Pressure is defined as the normal force exerted by a fluid per unit area of the surface on which it acts.
p = F / A
SI unit: Pascal (Pa) = N/m²
1 kPa = 10³ Pa | 1 MPa = 10⁶ Pa
1 bar = 10⁵ Pa ≈ 1 atm (approximately)
1 atm = 101,325 Pa = 101.325 kPa
1.1 Types of Pressure
| Term | Definition | Relationship |
|---|---|---|
| Absolute pressure (pabs) | Pressure measured above perfect vacuum (zero pressure) | pabs = patm + pgauge |
| Gauge pressure (pg) | Pressure above atmospheric pressure | pg = pabs – patm |
| Vacuum pressure | Pressure below atmospheric (negative gauge pressure) | pvac = patm – pabs |
| Atmospheric pressure | Pressure exerted by the Earth’s atmosphere at sea level | patm ≈ 101.325 kPa = 10.33 m of water = 760 mm Hg |
1.2 Pascal’s Law
Pascal’s Law states: Pressure applied at any point in a confined, incompressible fluid is transmitted equally and undiminished in all directions throughout the fluid.
This is the operating principle behind hydraulic jacks, hydraulic brakes, and hydraulic presses. A small force on a small piston creates the same pressure, which produces a large force on a large piston: p = F₁/A₁ = F₂/A₂.
2. Pressure Variation with Depth
In a static fluid, pressure increases linearly with depth. This is derived from the hydrostatic equation — vertical equilibrium of a fluid element:
dp/dz = –ρg (z measured upward)
Integrating for an incompressible fluid (ρ = constant):
p₂ – p₁ = ρg(z₁ – z₂) = ρgh
where h = z₁ – z₂ = vertical depth from point 1 to point 2 (positive downward)
Gauge pressure at depth h below free surface:
p = ρgh = γh
γ = specific weight = ρg = 9810 N/m³ for water
2.1 Pressure Head
h = p / (ρg) = p / γ
Unit: metres of fluid
1 atm ≡ 10.33 m of water ≡ 0.76 m of mercury
2.2 Key Properties of Hydrostatic Pressure
- Pressure at a point is the same in all directions (isotropic) — proved by Euler’s equation for a fluid at rest.
- Pressure is the same at all points on the same horizontal level in a connected static fluid.
- Pressure does not depend on the shape or cross-sectional area of the container (hydrostatic paradox).
- Free surface of a static liquid is always horizontal (perpendicular to the gravity vector).
3. Total Force & Centre of Pressure on Submerged Plane Surfaces
3.1 Total Hydrostatic Force
For a plane surface inclined at angle θ to the free surface (θ = 90° for vertical plate):
F = ρg Ȳ A
where:
- F = total hydrostatic force (N)
- ρ = fluid density (kg/m³)
- g = 9.81 m/s²
- Ȳ = vertical depth of the centroid of the surface below the free surface (m)
- A = total area of the surface (m²)
Equivalently: F = p̄ × A where p̄ = ρgȲ is the pressure at the centroid.
3.2 Centre of Pressure (hcp)
The centre of pressure is the point on the surface where the resultant hydrostatic force acts. It is always below the centroid (because pressure increases with depth, so the lower portion carries more force).
hcp = Ȳ + IG / (A Ȳ)
where:
- hcp = depth of centre of pressure below free surface (m)
- Ȳ = depth of centroid below free surface (m)
- IG = second moment of area of the surface about its centroidal axis (m⁴)
- A = area of surface (m²)
Note: The term IG/(AȲ) is always positive, confirming hcp > Ȳ.
3.3 Second Moment of Area — Common Shapes
| Shape | Area (A) | Centroid from top (ȳ) | IG (about centroidal axis) |
|---|---|---|---|
| Rectangle (b × d) | bd | d/2 | bd³/12 |
| Triangle (base b, height h) | bh/2 | h/3 from base; 2h/3 from apex | bh³/36 |
| Circle (diameter D) | πD²/4 | D/2 (radius) | πD⁴/64 |
| Semi-circle (diameter D) | πD²/8 | 2D/(3π) from diameter | 0.1098 R⁴ |
4. Force on Curved Submerged Surfaces
For curved surfaces, the direction of pressure varies along the surface, so we cannot integrate pressure directly as a single force. Instead, we resolve into horizontal and vertical components:
Horizontal component (FH):
FH = Force on the vertical projection of the curved surface
= ρg Ȳproj × Aproj
Acts at the centre of pressure of the projected area.
Vertical component (FV):
FV = Weight of fluid (real or imaginary) vertically above the curved surface up to the free surface
Acts through the centroid of the fluid volume above.
Resultant force:
F = √(FH² + FV²)
Angle with horizontal: θ = tan⁻¹(FV/FH)
Key insight: If the surface is concave upward (like the bottom of a tank), the vertical component acts upward (buoyancy-like). If convex upward (like an arch), FV acts downward.
5. Buoyancy & Archimedes’ Principle
5.1 Archimedes’ Principle
A body wholly or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the body.
FB = ρf × g × Vs
where:
- FB = buoyant force (N)
- ρf = density of the fluid (kg/m³)
- Vs = volume of the body submerged in the fluid (m³)
The buoyant force acts upward through the centre of buoyancy (B) — the centroid of the displaced fluid volume.
5.2 Conditions for Floating, Sinking & Neutral Buoyancy
| Condition | Relationship | Result |
|---|---|---|
| Floats | FB = W (weight of body); ρbody < ρfluid | Body partially submerged; Vs < Vtotal |
| Sinks | W > FB; ρbody > ρfluid | Body fully submerged and descends |
| Neutral buoyancy | W = FB; ρbody = ρfluid | Body fully submerged at rest anywhere in fluid |
5.3 Fraction Submerged for a Floating Body
For a floating body: Weight = Buoyant force
ρbody × Vtotal × g = ρfluid × Vsubmerged × g
Vsubmerged / Vtotal = ρbody / ρfluid = SGbody / SGfluid
Example: Ice (SG = 0.917) floats in water (SG = 1.0) with 91.7% submerged.
6. Stability of Floating Bodies — Metacentric Height
6.1 Key Points
| Point | Symbol | Definition |
|---|---|---|
| Centre of Gravity | G | Point through which the total weight of the body acts (downward) |
| Centre of Buoyancy | B | Centroid of the displaced fluid volume; buoyant force acts upward through B |
| Metacentre | M | Point where the line of action of the buoyant force (after small tilt) intersects the vertical through G |
| Metacentric Height | GM | Distance between G and M; GM = BM – BG |
BM = IWL / Vs
where:
- IWL = second moment of the waterplane area about the axis of tilt (m⁴)
- Vs = volume of fluid displaced (m³)
GM = BM – BG (if M is above G) or GM = BM + BG (if M is below G)
6.2 Stability Criteria
| Condition | GM | Stability |
|---|---|---|
| M above G | GM > 0 (positive) | Stable — restoring moment returns body to upright |
| M coincides with G | GM = 0 | Neutral — body remains in tilted position |
| M below G | GM < 0 (negative) | Unstable — overturning moment causes capsize |
For a fully submerged body (submarine, balloon), B is fixed (volume doesn’t change with tilt). Stability requires G to be below B — i.e., BG < 0 means G is below B.
7. Manometers — Simple, U-tube & Differential
A manometer is a device that uses the hydrostatic pressure–depth relationship to measure pressure or pressure difference. They contain a manometric fluid (often mercury, SG = 13.6, or coloured water) denser than the fluid whose pressure is being measured.
7.1 Simple (Piezometer) Tube
A vertical transparent tube open to atmosphere connected to the pipe. The liquid rises to a height h = p/(ρg). Only suitable for low positive gauge pressures; cannot measure negative (vacuum) pressures.
7.2 U-tube Manometer (Simple)
A U-shaped tube partly filled with a heavy manometric fluid (mercury). The pipe fluid is in one limb; the other is open to atmosphere.
Procedure — balance pressure at the datum (manometric fluid meniscus level):
Pressure at datum from left limb = Pressure at datum from right limb
pA + ρ₁g h₁ = patm + ρm g hm
Gauge pressure: pA(gauge) = ρm g hm – ρ₁ g h₁
where ρm = manometric fluid density; hm = height of manometric fluid above datum; h₁ = height of pipe fluid above datum
7.3 Differential U-tube Manometer
Used to measure the pressure difference between two points in a pipe (e.g., across a venturimeter). Both limbs connect to the two pipe points; the manometric fluid sits between them.
Pressure difference:
pA – pB = hm (ρm – ρ₁) g
where hm = difference in manometric fluid levels (m)
ρm = manometric fluid density (e.g., 13,600 kg/m³ for mercury)
ρ₁ = pipe fluid density (e.g., 1,000 kg/m³ for water)
7.4 Inverted U-tube Manometer
Used when the manometric fluid is lighter than the pipe fluid (e.g., oil or air used as manometric fluid in a water pipe). The inverted U sits above the pipe, with the light manometric fluid at the top.
pA – pB = hm (ρ₁ – ρm) g
Here ρm < ρ₁ (lighter manometric fluid)
7.5 Manometer Golden Rule
Start at the known pressure point (or one end), add pressure when going down through a fluid column, subtract when going up, and equate to the pressure at the other end. Keep consistent units throughout — work entirely in Pascals or entirely in metres of fluid head.
8. Worked Examples (GATE CE Level)
Example 1 — Total Force on a Vertical Rectangular Gate
Problem: A rectangular sluice gate 2 m wide and 3 m deep is hinged at the top and sits vertically with its top edge at 1 m below the water surface. Find (a) the total hydrostatic force on the gate and (b) the depth of the centre of pressure.
Given:
Width b = 2 m, Depth d = 3 m → Area A = 2 × 3 = 6 m²
Top edge at h₁ = 1 m below surface; bottom edge at h₂ = 4 m below surface
Depth of centroid: Ȳ = h₁ + d/2 = 1 + 1.5 = 2.5 m
ρ = 1000 kg/m³, g = 9.81 m/s²
(a) Total hydrostatic force:
F = ρg Ȳ A = 1000 × 9.81 × 2.5 × 6 = 147,150 N = 147.15 kN
(b) Centre of pressure:
IG = bd³/12 = (2 × 3³)/12 = 54/12 = 4.5 m⁴
hcp = Ȳ + IG/(AȲ) = 2.5 + 4.5/(6 × 2.5) = 2.5 + 4.5/15 = 2.5 + 0.3 = 2.8 m below the free surface
Answer: F = 147.15 kN; hcp = 2.8 m
Example 2 — U-tube Manometer (GATE CE 2018 type)
Problem: A pipe carries water and is connected to a U-tube manometer containing mercury (SG = 13.6). The centre of the pipe is 0.2 m above the datum. The mercury level in the limb connected to the pipe is 0.1 m above the datum, and the mercury in the open limb is 0.4 m above the datum. Find the gauge pressure at the pipe centre.
Given:
ρwater = 1000 kg/m³; ρHg = 13,600 kg/m³
Pipe centre at 0.2 m; Hg meniscus (connected limb) at 0.1 m above datum
Hg meniscus (open limb) at 0.4 m above datum
Balance at lower Hg meniscus (0.1 m datum level):
Left side: pA + ρw g × (0.2 – 0.1) = pA + 1000 × 9.81 × 0.1
Right side (open to atm): 0 + ρHg g × (0.4 – 0.1) = 13600 × 9.81 × 0.3
pA + 981 = 13600 × 9.81 × 0.3 = 40,021.2
pA = 40,021.2 – 981 = 39,040.2 Pa ≈ 39.04 kPa (gauge)
Answer: pA ≈ 39.04 kPa (gauge)
Example 3 — Buoyancy & Fraction Submerged
Problem: A wooden block (SG = 0.75) floats at the interface of oil (SG = 0.85) on top and water (SG = 1.0) below. The block is 0.4 m × 0.4 m × 0.2 m. Find the depth of the block in water.
Given:
Block volume V = 0.4 × 0.4 × 0.2 = 0.032 m³
Weight of block W = ρwood × V × g = 750 × 0.032 × 9.81 = 235.44 N
Let x = depth submerged in water (m); then (0.2 – x) = depth in oil
Volume in water = 0.4 × 0.4 × x = 0.16x m³
Volume in oil = 0.4 × 0.4 × (0.2 – x) = 0.16(0.2 – x) m³
Equilibrium: W = FB,water + FB,oil
235.44 = 1000 × 9.81 × 0.16x + 850 × 9.81 × 0.16(0.2 – x)
235.44 = 1569.6x + 1334.16(0.2 – x)
235.44 = 1569.6x + 266.832 – 1334.16x
235.44 – 266.832 = (1569.6 – 1334.16)x
–31.392 = 235.44x
x = –31.392 / 235.44 = –0.1333 m
The negative value means the block does not reach the water layer — it floats entirely within the oil. Let us verify by checking if buoyancy in oil alone supports the block:
FB,oil at full submergence = 850 × 9.81 × 0.032 = 266.83 N > 235.44 N ✓
So the block floats fully within the oil layer.
Correct approach — float in oil only:
W = ρoil × g × Vsubmerged in oil
235.44 = 850 × 9.81 × (0.4 × 0.4 × d)
d = 235.44 / (850 × 9.81 × 0.16) = 235.44 / 1334.16 = 0.1764 m in oil
Depth in water = 0 m (block does not reach water layer)
Answer: Block floats entirely in oil; depth in oil = 0.176 m; depth in water = 0 m.
Example 4 — Metacentric Height of a Floating Vessel
Problem: A rectangular barge 6 m wide, 12 m long, and 3 m deep floats in water with a draught (depth of submergence) of 2 m. The centre of gravity G is 1.8 m above the keel. Find the metacentric height GM and comment on stability.
Given:
Width B = 6 m, Length L = 12 m, Draught d = 2 m
Centre of buoyancy B is at midpoint of submerged volume above keel = d/2 = 1.0 m
G is 1.8 m above keel → BG = 1.8 – 1.0 = 0.8 m (G is above B)
Second moment of waterplane area about longitudinal axis:
IWL = LB³/12 = 12 × 6³/12 = 12 × 216/12 = 216 m⁴
Displaced volume:
Vs = 6 × 12 × 2 = 144 m³
BM:
BM = IWL/Vs = 216/144 = 1.5 m
GM:
GM = BM – BG = 1.5 – 0.8 = +0.7 m
Answer: GM = +0.7 m > 0 → Stable floating condition.
Example 5 — Differential Manometer Pressure Difference
Problem: A differential mercury manometer connects two points A and B in a water pipeline. Point A is 0.5 m higher than point B. The mercury level in limb A is 0.3 m below the centreline of A, and the mercury level in limb B is 0.15 m below the centreline of B. Find pA – pB. (ρHg = 13,600 kg/m³; ρw = 1000 kg/m³)
Set up datum at lower mercury meniscus (limb A side).
Let the lower Hg meniscus be at elevation 0 (datum).
Height of Hg column = difference in Hg levels = Let hm be found from geometry.
Pipe A centreline is 0.3 m above lower Hg meniscus.
Pipe B centreline: B is 0.5 m below A → B is at –0.5 + 0.3 = –0.2 m relative to datum.
Upper Hg meniscus (limb B): 0.15 m below B centreline = –0.2 – 0.15 = –0.35 m below datum.
Wait — upper Hg meniscus must be above lower. Let us set datum at lower Hg meniscus (in limb A).
hm = difference in Hg levels.
Using the standard differential manometer formula directly:
Balancing at the lower meniscus level:
pA + ρwg(0.3) = pB + ρwg(hAB + 0.15) + ρHgg × hm – ρwg × hm
Taking hm = 0.3 + 0.5 – 0.15 = 0.65 m (Hg column height from geometry):
pA – pB = ρwg(0.5) + (ρHg – ρw)g × hm – ρwg(0.3 – 0.15)
= 1000×9.81×0.5 + (12600)×9.81×0.65 – 1000×9.81×0.15
= 4905 + 80,374.5 – 1471.5
pA – pB = 83,808 Pa ≈ 83.81 kPa
Answer: pA – pB ≈ 83.81 kPa
9. Common Mistakes
Mistake 1 — Using the Wrong Depth for Force Calculation (Using Bottom Edge Instead of Centroid Depth)
Error: Computing hydrostatic force as F = ρg hbottom × A, using the depth to the bottom edge of the surface instead of the depth to its centroid.
Root Cause: Confusing centre of pressure (where force acts) with centroid (what determines magnitude). The total force depends on the average pressure — which acts at the centroid depth Ȳ, not at the bottom.
Fix: Always use F = ρgȲA where Ȳ is the depth of the centroid. Then separately calculate hcp = Ȳ + IG/(AȲ) for where that force acts.
Mistake 2 — Forgetting that Centre of Pressure is Below the Centroid
Error: Writing hcp = Ȳ – IG/(AȲ), placing the centre of pressure above the centroid.
Root Cause: Sign confusion in the derivation. Since pressure increases downward and the term IG/(AȲ) is always positive, the centre of pressure is always deeper than the centroid.
Fix: Memorise: hcp = Ȳ + IG/(AȲ). As the surface goes deeper (Ȳ increases), the term IG/(AȲ) → 0, meaning hcp → Ȳ. Very deep surfaces have the centre of pressure almost at the centroid — this is physically intuitive.
Mistake 3 — Confusing Gauge and Absolute Pressure in Manometer Problems
Error: Mixing absolute and gauge pressures in the same manometer balance equation — adding atmospheric pressure on one side but not the other.
Root Cause: Careless substitution. In a manometer open to atmosphere at one end, the atmospheric pressure appears on both sides of the balance equation and cancels out, giving gauge pressure directly.
Fix: Be explicit about the datum. If the open limb is at atmospheric pressure patm, write it in the equation. It will cancel, confirming you get gauge pressure. If you need absolute pressure, add patm at the end.
Mistake 4 — Wrong BG Sign in Metacentric Height Calculation
Error: Applying GM = BM – BG without checking whether G is above or below B, leading to a wrong sign for stability assessment.
Root Cause: BG is always a positive distance, but its effect on GM depends on relative positions. If G is above B (common in ships): GM = BM – BG. If G is below B (heavy keel): GM = BM + BG (always stable).
Fix: Draw a clear diagram. Mark keel (K) at bottom, B above K at d/2, and G at its known height. Compute KB, KG, then BG = KG – KB (positive if G above B, negative if G below B). Then GM = KM – KG = (KB + BM) – KG.
Mistake 5 — Applying the Simple Pressure Formula (p = ρgh) to Inclined Surfaces as Depth Rather Than Vertical Height
Error: For an inclined submerged surface, using the slant length (distance along the inclined surface) instead of the vertical depth h when computing pressure p = ρgh.
Root Cause: Hydrostatic pressure depends only on vertical depth, not on the geometry of the path from the surface to the point. This is a consequence of the hydrostatic equation dp/dz = –ρg (z is vertical).
Fix: Always resolve to vertical depth: h = L sinθ, where L is the slant distance and θ is the angle of inclination to the horizontal.
10. Frequently Asked Questions
Q1. Why does the centre of pressure approach the centroid as the depth of immersion increases?
The distance between the centre of pressure and the centroid is given by the term IG/(AȲ). As Ȳ (depth of the centroid) increases, this term decreases inversely. Physically, at very large depths, the pressure across the entire surface is nearly uniform — there is relatively little variation between the top and bottom of the surface compared to the absolute pressure. When pressure is nearly uniform, the resultant force acts at the geometric centre (centroid), just as a uniform distributed load on a beam acts at the midpoint. This limiting behaviour has a practical implication: deeply submerged dams carry nearly uniform pressure near their base, so the centre of pressure shift matters most for shallower structures like lock gates and sluice gates.
Q2. What is the difference between centre of buoyancy and centre of gravity, and why does their relative position determine stability?
The centre of gravity (G) is where the weight of the body acts downward; it is fixed for a given load distribution. The centre of buoyancy (B) is the centroid of the displaced fluid volume — it acts upward and shifts position when the body tilts, because the shape of the submerged volume changes. When a floating body tilts by a small angle, B moves to the side of the tilt (more volume is now submerged on that side). If the shifted buoyant force passes through a point M (the metacentre) that is above G, it creates a restoring couple that pushes the body back upright — this is stable equilibrium. If M is below G, the couple is overturning — unstable. The metacentric height GM quantifies this margin of stability: larger positive GM means greater stability but also a stiffer, more uncomfortable roll in a ship.
Q3. Why is mercury commonly used as the manometric fluid and not water?
Mercury (SG = 13.6) is used because its very high density means that a small difference in mercury level corresponds to a large pressure difference. If water were used as the manometric fluid, the U-tube would need to be 13.6 times taller to measure the same pressure — impractical for most engineering installations. Mercury is also immiscible with water and most oils, has a well-defined density at standard conditions, and has a clearly visible meniscus. The main disadvantage is toxicity (environmental and health hazard), which has driven adoption of alternative high-density fluids and electronic pressure transducers in modern practice. For low-pressure differences, lighter manometric fluids (light oil, SG ≈ 0.8) in inverted U-tube manometers are used to improve sensitivity.
Q4. How does Archimedes’ principle apply to the design of dams and retaining walls?
While Archimedes’ principle is primarily associated with floating bodies, it appears in dam and retaining wall design through uplift pressure. Water seeping under a dam foundation or through cracks in a retaining wall exerts an upward pressure on the base — effectively a buoyant force trying to lift the structure. This uplift reduces the effective weight pressing down, which is critical for sliding stability (sliding factor of safety = resisting friction / net horizontal force) and overturning stability. IS 6512 (criteria for design of solid gravity dams) requires uplift pressure to be included in the stability analysis. Drainage galleries and relief drains are built into dam bases to reduce uplift by dissipating seepage pressure before it builds up under the full dam width.