Fluid Statics

The study of fluids at rest - pressure, buoyancy, and hydrostatic equilibrium

Introduction

Fluid statics (or hydrostatics) deals with fluids at rest. In static equilibrium, there are no shear stresses acting on fluid elements - only normal stresses (pressure). This simplification makes the analysis tractable while providing essential insights applicable to dams, submarines, atmospheric pressure, and more.

The fundamental principle is that pressure at a point in a static fluid acts equally in all directions (Pascal's principle) and varies only with depth in a gravitational field.

Pressure Fundamentals

Definition of Pressure

Pressure is the normal force per unit area exerted by a fluid on a surface:

$$p = \frac{F}{A} = \lim_{\Delta A \to 0} \frac{\Delta F_n}{\Delta A}$$

Units: Pa (Pascal) = N/m² = kg/(m·s²). Also: 1 atm = 101,325 Pa = 760 mmHg = 14.7 psi

Pascal's Law

Pressure at a point in a static fluid is the same in all directions (isotropic):

$$p_x = p_y = p_z = p$$

This is proven by considering force equilibrium on a small fluid wedge. Shear stresses are zero in a static fluid, leaving only normal pressure forces.

Pressure Transmission

A pressure change applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the container. This principle is the basis for hydraulic systems:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies F_2 = F_1 \cdot \frac{A_2}{A_1}$$

Hydraulic lifts, brakes, and presses exploit this mechanical advantage.

Hydrostatic Pressure Variation

Basic Hydrostatic Equation

For a fluid at rest in a gravitational field, the pressure increases with depth:

$$\frac{dp}{dz} = -\rho g$$

Where z is measured upward. For incompressible fluids (constant ρ):

$$p_2 - p_1 = -\rho g (z_2 - z_1) = \rho g h$$

where h = z₁ - z₂ is the depth below point 1.

Pressure in the Atmosphere

For gases, density varies with pressure and temperature. For an isothermal atmosphere:

$$p = p_0 \exp\left(-\frac{gz}{RT}\right) = p_0 \exp\left(-\frac{z}{H}\right)$$

where H = RT/g ≈ 8.5 km is the scale height for Earth's atmosphere.

Gauge Pressure

Pressure relative to atmospheric: p_gauge = p_abs - p_atm

Absolute Pressure

Measured from absolute zero pressure (perfect vacuum)

Manometry

Manometers use the hydrostatic equation to measure pressure differences by observing liquid column heights.

Simple U-Tube Manometer

$$p_A - p_B = \rho_{man} g h$$

The height difference h of manometer fluid (mercury, water, oil) indicates pressure difference.

Inclined Manometer

For small pressure differences, an inclined tube amplifies the reading:

$$\Delta p = \rho g L \sin\theta$$

where L is the length along the incline and θ is the angle from horizontal.

Multi-Fluid Manometer

For complex systems with multiple fluids, apply hydrostatic equation through each section:

$$p_1 + \sum_i \rho_i g h_i = p_2$$

Barometer

Measures absolute atmospheric pressure. Mercury barometer:

$$p_{atm} = \rho_{Hg} g h$$

At sea level: h ≈ 760 mm Hg = 29.92 in Hg

Forces on Submerged Surfaces

Plane Surfaces

The hydrostatic force on a submerged plane surface is:

$$F_R = \int_A p \, dA = \int_A (\rho g h) \, dA = \rho g \bar{h} A = p_c A$$

where p_c is the pressure at the centroid and A is the surface area.

The center of pressure (where resultant acts) is below the centroid:

$$y_{cp} = \bar{y} + \frac{I_{xc}}{\bar{y} A}$$

where I_xc is the second moment of area about the centroidal axis.

Curved Surfaces

For curved surfaces, decompose into horizontal and vertical components:

Horizontal Component

F_H = Force on vertical projection of curved surface

Vertical Component

F_V = Weight of fluid above (or below) the curved surface

Buoyancy and Archimedes' Principle

Archimedes' Principle

A body immersed in a fluid experiences an upward buoyant force equal to the weight of displaced fluid:

$$F_B = \rho_{fluid} g V_{displaced}$$

The buoyant force acts through the centroid of the displaced volume (center of buoyancy).

Floating Bodies

For a floating body in equilibrium:

$$F_B = W \implies \rho_{fluid} g V_{submerged} = \rho_{body} g V_{body}$$

$$\frac{V_{submerged}}{V_{body}} = \frac{\rho_{body}}{\rho_{fluid}}$$

Stability of Floating Bodies

For rotational stability, the metacenter M must be above the center of gravity G:

$$GM = \frac{I_{waterline}}{V_{submerged}} - BG$$

where I_waterline is the second moment of the waterline area about the tilt axis. GM > 0 indicates stable equilibrium.

Applications

Dam Design

Calculate hydrostatic forces on dam walls to ensure structural stability against overturning and sliding.

Submarines

Buoyancy control through ballast tanks, hull design for pressure at depth.

Hydraulic Systems

Pascal's law applied in lifts, brakes, presses, and actuators.

Atmospheric Science

Pressure variation with altitude, barometric formula, weather systems.

Interactive Simulations

Hydrostatic Pressure Distribution in a Water Tank

Python

script.py31 lines

import numpy as np

# Hydrostatic Pressure Distribution
# P(z) = P_atm + rho * g * h, where h is depth below surface

rho = 1000.0      # water density (kg/m^3)
g = 9.81           # gravitational acceleration (m/s^2)
P_atm = 101325.0   # atmospheric pressure (Pa)
H = 10.0           # tank depth (m)

depths = np.linspace(0, H, 11)
pressures = P_atm + rho * g * depths
gauge_pressures = rho * g * depths

print("Hydrostatic Pressure in a Water Tank")
print("=" * 55)
print(f"Fluid density: {rho} kg/m^3")
print(f"Surface pressure: {P_atm/1000:.1f} kPa (atmospheric)")
print()
print(f"{'Depth (m)':>10} {'Gauge P (kPa)':>14} {'Abs P (kPa)':>14}")
print("-" * 42)
for d, gp, ap in zip(depths, gauge_pressures, pressures):
    print(f"{d:10.1f} {gp/1000:14.2f} {ap/1000:14.2f}")

# Force on a submerged vertical plate (width = 1 m)
width = 1.0
F_total = 0.5 * rho * g * H**2 * width
print(f"\nTotal hydrostatic force on a {width}m-wide vertical plate:")
print(f"F = {F_total/1000:.2f} kN")
print(f"Center of pressure: {2*H/3:.2f} m below surface")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Buoyancy Force on Submerged Objects

Fortran

program.f9045 lines

program buoyancy
  implicit none
  real(8) :: rho_fluid, g, pi
  real(8) :: radius, V_sphere, F_buoy, W_obj, F_net
  real(8) :: rho_obj
  integer :: i
  real(8), dimension(5) :: densities
  character(len=12), dimension(5) :: materials

rho_fluid = 1000.0d0
  g = 9.81d0
  pi = 4.0d0 * atan(1.0d0)
  radius = 0.1d0
  V_sphere = (4.0d0/3.0d0) * pi * radius**3

materials = (/ 'Wood        ', 'Ice         ', 'Aluminum    ', &
                 'Steel       ', 'Lead        ' /)
  densities = (/ 600.0d0, 917.0d0, 2700.0d0, 7800.0d0, 11340.0d0 /)

F_buoy = rho_fluid * g * V_sphere

write(*,'(A)') "Buoyancy Analysis: Sphere in Water"
  write(*,'(A)') "==================================="
  write(*,'(A,F8.4,A)') "Sphere radius: ", radius, " m"
  write(*,'(A,ES10.3,A)') "Volume: ", V_sphere, " m^3"
  write(*,'(A,F8.4,A)') "Buoyancy force: ", F_buoy, " N"
  write(*,*)
  write(*,'(A12,A10,A10,A10,A12)') &
    "Material", "rho", "Weight", "F_net", "Behavior"
  write(*,'(A)') repeat("-", 56)

do i = 1, 5
    rho_obj = densities(i)
    W_obj = rho_obj * V_sphere * g
    F_net = W_obj - F_buoy
    if (F_net < 0.0d0) then
      write(*,'(A12,F10.0,F10.4,F10.4,A12)') &
        materials(i), rho_obj, W_obj, F_net, "  Floats"
    else
      write(*,'(A12,F10.0,F10.4,F10.4,A12)') &
        materials(i), rho_obj, W_obj, F_net, "  Sinks"
    end if
  end do
end program buoyancy

Click Run to execute the Fortran code

Code will be compiled with gfortran and executed on the server

Detailed Derivations

Derivation: Hydrostatic Pressure from Force Balance

Consider a small cylindrical fluid element at rest with cross-sectional area A and height dz, oriented vertically (z positive upward).

Three forces act on the element in the vertical direction:

- Pressure on bottom face (upward): P(z) · A
- Pressure on top face (downward): P(z + dz) · A
- Weight (downward): ρg A dz

For static equilibrium the net force is zero:

$$P(z)\,A - P(z+dz)\,A - \rho g\,A\,dz = 0$$

Divide through by A dz and note P(z+dz) − P(z) = dP:

$$\frac{dP}{dz} = -\rho\,g$$

This is the fundamental hydrostatic equation. The negative sign indicates pressure decreases upward.

For an incompressible fluid (ρ = const), integrate from the free surface (z = 0, P = P₀) down to depth h:

$$\int_{P_0}^{P} dP' = -\rho g \int_{0}^{-h} dz' \implies P - P_0 = \rho g h$$

Final result — pressure at depth h below the surface:

$$\boxed{P(h) = P_0 + \rho\,g\,h}$$

Derivation: Archimedes' Principle from Pressure Integration

Consider a body of arbitrary shape fully submerged in a fluid of density ρ. The fluid exerts a pressure force on every infinitesimal surface element dA of the body.

The net pressure force on the body is:

$$\vec{F}_{\text{pressure}} = -\oint_S P\,\hat{n}\,dA$$

where n̂ is the outward unit normal to the body surface.

Replace the body with fluid. That replacement fluid is also in equilibrium, so:

$$-\oint_S P\,\hat{n}\,dA + \rho_f\,g\,V_{\text{body}}\,(-\hat{k}) = 0$$

The pressure forces on the replacement fluid exactly support its weight.

Since the pressure distribution on the surface depends only on position (not on what is inside), the pressure force on the original body is identical. Therefore the buoyant force is:

$$\boxed{F_B = \rho_f\,g\,V_{\text{displaced}}}$$

This upward force equals the weight of the displaced fluid, acting through the centroid of the displaced volume (center of buoyancy B).

Derivation: Manometer Equations

U-tube manometer

A U-tube connects two points A and B through a manometer fluid of density ρ_m. Start at A and walk through the tube to B, adding ρgh when going down and subtracting when going up:

$$P_A + \rho_1 g h_1 - \rho_m g h_m - \rho_2 g h_2 = P_B$$

For the simple case where both sides contain the same fluid of density ρ:

$$\boxed{P_A - P_B = (\rho_m - \rho)\,g\,h}$$

Differential manometer

For measuring small pressure differences, use two manometer fluids with densities ρ_m1 and ρ_m2:

$$P_A - P_B = (\rho_{m1} - \rho_{m2})\,g\,h$$

Inclined manometer

An inclined tube at angle θ from horizontal amplifies the reading. If the liquid travels distance L along the tube, the vertical rise is L sinθ:

$$\boxed{\Delta P = \rho_m\,g\,L\,\sin\theta}$$

For small θ, L is much larger than the vertical height, giving better resolution for small ΔP.

Derivation: Hydrostatic Force on Submerged Plane Surfaces

Consider a flat plate of area A submerged at an angle θ to the free surface. Define y along the plate from the intersection with the surface. The depth of an element dA at position y is h = y sinθ.

The resultant hydrostatic force is obtained by integrating pressure over the surface:

$$F = \int_A P\,dA = \int_A \rho g\,h\,dA = \rho g \sin\theta \int_A y\,dA$$

Recognise that ∫ y dA = ȳ A (first moment of area), and h̄ = ȳ sinθ is the depth of the centroid:

$$\boxed{F = \rho\,g\,\bar{h}\,A}$$

The force equals the pressure at the centroid times the area.

The resultant acts at the center of pressure y_cp. Taking moments about the surface line:

$$y_{cp} = \frac{\int_A y^2\,dA}{\int_A y\,dA} = \frac{I_0}{\bar{y}\,A}$$

Apply the parallel axis theorem I₀ = I_G + ȳ² A:

$$\boxed{y_{cp} = \bar{y} + \frac{I_G}{\bar{y}\,A}}$$

Since I_G > 0, the center of pressure is always below the centroid.

Derivation: Stability of Floating Bodies — Metacentric Height

A floating body in equilibrium has its center of gravity G and center of buoyancy B on the same vertical line. The weight W acts at G downward; the buoyant force F_B = W acts at B upward.

When the body tilts by a small angle dθ, the submerged shape changes and B shifts to B'. The new line of action of F_B intersects the original vertical at the metacenter M.

The horizontal shift of the center of buoyancy is computed from the moment of the transferred volume wedges. For a waterplane area with second moment I about the tilt axis:

$$BB' = \frac{I}{V_{\text{sub}}}\,d\theta \implies BM = \frac{I}{V_{\text{sub}}}$$

The metacentric height GM is the distance from G to M. Measuring upward from B:

$$\boxed{GM = BM - BG = \frac{I}{V_{\text{sub}}} - BG}$$

Stability criterion: if GM > 0, the restoring moment opposes the tilt (stable). If GM < 0, the moment amplifies the tilt (unstable). The righting moment is M_R = W · GM · sinθ.

Pressure Distribution Diagram

Linear increase of hydrostatic pressure with depth in an incompressible fluid.

Advanced Simulations

Pressure Profiles: Incompressible vs Compressible Atmospheres

Python

script.py47 lines

import numpy as np

# Compare hydrostatic pressure profiles for
# (a) incompressible liquid  (b) isothermal atmosphere  (c) US standard atmosphere

g = 9.81
rho_water = 1000.0
P_atm = 101325.0
R_air = 287.0        # specific gas constant for air (J/(kg K))
T0 = 288.15          # sea-level temperature (K)
lapse = 0.0065       # temperature lapse rate (K/m)

# --- Water column (0 to 100 m depth) ---
depths = np.linspace(0, 100, 11)
P_water = P_atm + rho_water * g * depths

print("=== Hydrostatic Pressure in Water (incompressible) ===")
print("  Depth (m)   Gauge (kPa)    Abs (kPa)")
print("  " + "-" * 40)
for d, pw in zip(depths, P_water):
    print("  {:8.1f}   {:12.2f}   {:11.2f}".format(d, rho_water*g*d/1000, pw/1000))

# --- Isothermal atmosphere ---
altitudes = np.linspace(0, 30000, 11)
H = R_air * T0 / g   # scale height ~ 8435 m
P_iso = P_atm * np.exp(-altitudes / H)

print()
print("=== Isothermal Atmosphere (T = {:.1f} K, H = {:.0f} m) ===".format(T0, H))
print("  Alt (km)   P (kPa)   rho (kg/m3)")
print("  " + "-" * 40)
for alt, p in zip(altitudes, P_iso):
    rho = p / (R_air * T0)
    print("  {:8.1f}   {:9.2f}   {:10.4f}".format(alt/1000, p/1000, rho))

# --- US Standard Atmosphere (troposphere with lapse rate) ---
P_std = P_atm * (1 - lapse * altitudes[:8] / T0) ** (g / (lapse * R_air))

print()
print("=== US Standard Atmosphere (lapse rate model, troposphere) ===")
print("  Alt (km)    T (K)     P (kPa)")
print("  " + "-" * 35)
for i in range(8):
    alt = altitudes[i]
    T = T0 - lapse * alt
    print("  {:8.1f}   {:7.1f}   {:9.2f}".format(alt/1000, T, P_std[i]/1000))

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Buoyancy & Metacentric Height vs Heel Angle

Python

script.py73 lines

import numpy as np

# --- Buoyancy calculations for various objects ---
g = 9.81
rho_w = 1025.0   # seawater density (kg/m3)

print("=== Buoyancy of Various Objects in Seawater ===")
print("  rho_sw = {:.0f} kg/m3".format(rho_w))
print()

objects = [
    ("Pine wood",   550.0, 0.001),
    ("Ice",         917.0, 0.001),
    ("Concrete",   2300.0, 0.001),
    ("Steel",      7850.0, 0.001),
    ("Aluminum",   2700.0, 0.001),
]

print("  {:12s} {:>8s} {:>10s} {:>10s} {:>10s} {:>10s}".format(
    "Material", "rho", "V (L)", "W (N)", "F_B (N)", "Status"))
print("  " + "-" * 65)
for name, rho_obj, vol in objects:
    W = rho_obj * vol * g
    Fb = rho_w * vol * g
    frac = rho_obj / rho_w
    status = "Floats ({:.0f}%)".format(frac*100) if frac < 1 else "Sinks"
    print("  {:12s} {:8.0f} {:10.3f} {:10.3f} {:10.3f} {:>10s}".format(
        name, rho_obj, vol*1000, W, Fb, status))

# --- Metacentric height for a rectangular barge ---
print()
print("=" * 60)
print("=== Metacentric Height: Rectangular Barge ===")
print()

B_barge = 10.0    # beam / width (m)
L_barge = 30.0    # length (m)
D_barge = 4.0     # depth / height (m)
rho_barge = 600.0 # average barge density (kg/m3)

# Draft (submerged depth)
draft = D_barge * rho_barge / rho_w
V_sub = B_barge * L_barge * draft

# Center of buoyancy B is at draft/2 from keel
KB = draft / 2.0
# Center of gravity G at D_barge/2 from keel (uniform)
KG = D_barge / 2.0
BG = KG - KB

# Waterplane second moment of area about longitudinal axis
I_wp = (L_barge * B_barge**3) / 12.0
BM = I_wp / V_sub
GM = BM - BG

print("  Beam = {:.1f} m, Length = {:.1f} m, Depth = {:.1f} m".format(B_barge, L_barge, D_barge))
print("  Draft = {:.3f} m".format(draft))
print("  KB = {:.3f} m, KG = {:.3f} m, BG = {:.3f} m".format(KB, KG, BG))
print("  I_wp = {:.1f} m4".format(I_wp))
print("  BM = I/V_sub = {:.3f} m".format(BM))
print("  GM = BM - BG = {:.3f} m".format(GM))
print("  Stability: {}".format("STABLE (GM > 0)" if GM > 0 else "UNSTABLE (GM < 0)"))

# Righting moment vs heel angle
print()
print("  Heel angle (deg)   Righting moment (kN m)")
print("  " + "-" * 42)
W_barge = rho_barge * B_barge * L_barge * D_barge * g
for theta_deg in range(0, 31, 5):
    theta = np.radians(theta_deg)
    M_right = W_barge * GM * np.sin(theta)
    print("  {:16.0f}   {:22.1f}".format(theta_deg, M_right/1000))

Click Run to execute the Python code

Code will be executed with Python 3 on the server

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