Hydrostatic Forces & Buoyancy

Hydrostatic force and buoyancy/stability are reliable GATE ME marks (~1–2). The plane-surface force, centre of pressure, and metacentric-height stability criterion are the core.

Force on submerged surfaces

For a plane surface of area A with centroid at depth h̄:

• Total force F = ρg·h̄·A (acts normal to the surface).
• Centre of pressure (depth) h_cp = h̄ + I_G/(A·h̄), where I_G = second moment of area about the centroidal axis. The centre of pressure always lies below the centroid (deeper) because pressure increases with depth.

For a curved surface, resolve into horizontal component (= force on the vertical projection) and vertical component (= weight of fluid above the surface).

Buoyancy & stability

Archimedes' principle: buoyant force = ρg·V_displaced, acting up through the centre of buoyancy B (centroid of displaced volume). A body floats when weight = buoyancy.

Stability of a floating body: the metacentre M is where the buoyancy line crosses the axis. Metacentric height GM = BM − BG, with BM = I/V (I = second moment of the waterplane area, V = displaced volume).

• GM > 0 (M above G): stable.
• GM = 0: neutral; GM < 0 (M below G): unstable (capsizes).

Exam Tricks & Tips

• 🎯 F = ρg·h̄·A uses the CENTROID depth; the force acts at the deeper centre of pressure, not the centroid.
• 🎯 h_cp = h̄ + I_G/(A·h̄): the centre of pressure is always below the centroid; the offset shrinks as depth increases.
• 🎯 Buoyant force = weight of displaced fluid (ρgV) — independent of the body's own weight or depth (for full submersion).
• 🎯 GM = BM − BG, BM = I/V; GM > 0 means stable. A wider waterplane (larger I) raises BM and improves stability.
• 🎯 Curved surface: horizontal force = force on the vertical projection; vertical force = weight of fluid above — split, don't integrate directly.
• ❌ Common mistake: placing the hydrostatic force at the centroid — it acts at the centre of pressure (below the centroid), which matters for moment/tipping calculations.

Expected exam pattern

A 1-mark buoyancy or hydrostatic-force NAT, and a 2-mark centre-of-pressure or metacentric-height/stability problem. The "GM > 0 = stable" rule and the centroid-vs-centre-of-pressure distinction are common conceptual points.

Quick recap

Plane force F = ρg h̄ A at centre of pressure h_cp = h̄ + I_G/(A h̄) (below centroid). Curved surface: split into horizontal (projected area) and vertical (fluid weight above). Buoyancy = ρgV_displaced through B. Stability: GM = BM − BG, BM = I/V; GM > 0 stable.

Cover the answer, recall, then check. 11 cards on hydrostatics for GATE ME.

Q1. Total hydrostatic force on a submerged plane surface?
A1. F = ρg·h̄·A, where h̄ = depth of the centroid and A = area. It acts normal to the surface.

Q2. Depth of the centre of pressure?
A2. h_cp = h̄ + I_G/(A·h̄). It always lies below the centroid.

Q3. Why is the centre of pressure below the centroid?
A3. Pressure increases with depth, so the resultant shifts toward the deeper (higher-pressure) part of the surface.

Q4. State Archimedes' principle.
A4. Buoyant force = weight of displaced fluid = ρg·V_displaced, acting upward through the centre of buoyancy.

Q5. Where does the buoyant force act?
A5. Through the centre of buoyancy B — the centroid of the displaced fluid volume.

Q6. Formula for metacentric height.
A6. GM = BM − BG, where BM = I/V (I = second moment of the waterplane area, V = displaced volume).

Q7. Stability condition for a floating body?
A7. GM > 0 (metacentre above centre of gravity) → stable; GM < 0 → unstable (capsizes); GM = 0 → neutral.

Q8. How do you handle hydrostatic force on a curved surface?
A8. Horizontal component = force on the vertical projection; vertical component = weight of the fluid column above the surface.

Q9. How does waterplane width affect stability?
A9. A wider waterplane increases I, raising BM (= I/V) and GM — so wide, flat hulls are more stable.

Q10. Condition for a body to float?
A10. Its weight equals the buoyant force (weight of displaced fluid); it sinks to the depth where these balance.

Q11. A cube floats with half its volume submerged. Its specific gravity?
A11. 0.5 — the submerged fraction equals the ratio of body density to fluid density.

Dam walls, submerged gates and floating ships all come down to hydrostatic force and buoyancy. GATE tests the force on a submerged surface, where it acts (centre of pressure), and the stability of floating bodies through metacentric height.

Core concept: the pressure on a submerged surface grows with depth, so the resultant force acts below the centroid (at the centre of pressure); a floating/submerged body is buoyed up by the weight of fluid it displaces (Archimedes).

Deep explanation

Beginner — force on a submerged plane

For a plane surface of area A with centroid at depth h̄ (measured to the centroid):
Resultant force F = ρ g h̄ A = (pressure at centroid) × area.
The force acts at the centre of pressure (CP), always below the centroid, at depth:
h_cp = h̄ + I_G sinθ ... more precisely h_cp = h̄ + I_G/(A h̄) for a vertical surface (θ = 90°), where I_G is the second moment of area about the centroidal axis. For a vertical rectangle of height H with top at the surface, CP is at 2H/3 from the top.

Intermediate — buoyancy (Archimedes)

Buoyant force F_B = ρ_fluid · g · V_displaced, acting upward through the centre of buoyancy (centroid of the displaced volume). A body floats when its weight equals the buoyant force; it sinks if denser than the fluid. Fraction submerged = ρ_body/ρ_fluid.

Advanced — stability of floating bodies

Stability depends on the metacentre M — the point where the line of buoyant force meets the body's centreline when tilted. The metacentric height GM = BM − BG, where:

• BM = I/V (I = second moment of the waterplane area about the tilt axis, V = displaced volume),
• BG = distance between centre of buoyancy B and centre of gravity G.
  Stable if GM > 0 (M above G) — a restoring couple returns the body upright. GM < 0 is unstable (capsizes). Larger GM means stiffer (faster, less comfortable) rolling.

Worked example

A vertical rectangular gate 2 m wide and 3 m high has its top edge at the water surface. Find the total hydrostatic force and the centre of pressure.
Centroid depth h̄ = 3/2 = 1.5 m. Area A = 2 × 3 = 6 m².
F = ρg h̄ A = 1000 × 9.81 × 1.5 × 6 = 88,290 N ≈ 88.3 kN.
h_cp = h̄ + I_G/(A h̄). I_G = bH³/12 = 2×3³/12 = 4.5 m⁴. h_cp = 1.5 + 4.5/(6×1.5) = 1.5 + 0.5 = 2.0 m below the surface (= 2H/3 ✓).

GATE relevance

Hydrostatic force and centre of pressure on gates/dams, buoyant force, and metacentric-height stability are standard Fluid Mechanics questions. The GM = BM − BG stability criterion and the 2H/3 CP result for surface-piercing rectangles recur frequently.

Exam tricks & shortcuts

• Force = pressure at centroid × area (ρg h̄ A) — quick and general.
• CP is always below the centroid; for a surface-piercing vertical rectangle it sits at 2/3 of the depth.
• Stability: compute GM = I/V − BG; positive means stable.
• Mnemonic: "Metacentre above G, the ship floats free."

Key formulas

• Pressure at depth: P = ρgh.
• Force on a submerged plane: F = ρg·h̄·A (h̄ = depth of centroid).
• Centre of pressure: h_cp = h̄ + I_G/(A·h̄).
• Buoyancy (Archimedes): F_B = ρg·V_displaced.
• Stability: metacentric height GM = BM − BG, BM = I/V; stable if GM > 0.

Hydrostatic force acts at the centre of pressure (below the centroid); floating stability needs a positive metacentric height.

Worked Example

Problem: A vertical rectangular gate 2 m wide and 3 m high has its top edge level with the free surface of water. Find the total hydrostatic force on the gate and the depth of the centre of pressure. (ρ = 1000 kg/m³, g = 9.81 m/s².)

Solution:
Gate area and centroid depth:
A = width × height = 2 × 3 = 6 m².
The centroid is at mid-height: h_c = 3/2 = 1.5 m below the surface.

Total hydrostatic force (acts at the centroid pressure):
F = ρ·g·h_c·A = 1000 × 9.81 × 1.5 × 6 = 88290 N ≈ 88.3 kN.

Depth of the centre of pressure:
h_cp = h_c + I_G/(A·h_c),
where I_G is the second moment of area about the centroidal horizontal axis:
I_G = b·h³/12 = 2 × 3³/12 = 54/12 = 4.5 m⁴.
h_cp = 1.5 + 4.5/(6 × 1.5) = 1.5 + 0.5 = 2.0 m.

Answer: Total force ≈ 88.3 kN, acting at a centre of pressure 2.0 m below the surface.