Derivation

Fluid Statics: Pressure, Depth and Archimedes' Principle

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Statement

For a fluid at rest in a gravitational field, the pressure \(p\) is isotropic at any point and varies with depth \(h\) below a free surface according to \(p = p_0 + \rho g h\), where \(\rho\) is the fluid density, \(g\) is the gravitational acceleration, and \(p_0\) is the pressure at the surface. A body fully or partially submerged in the fluid experiences an upward buoyant force equal to the weight of the fluid it displaces, \(F_b = \rho_{\text{fluid}} g V_{\text{disp}}\), directed through the centre of buoyancy.

Why it matters

Fluid statics provides the foundation for understanding how pressure transmits through liquids and gases, which underpins hydraulic machinery, barometers, manometers, and the behaviour of floating structures. The result \(p = p_0 + \rho g h\) is one of the most frequently used relations in engineering and physics, appearing in dam design, submarine ballast calculations, and atmospheric science.

Archimedes' principle, derived from the same pressure field, explains why ships float, why icebergs sit mostly submerged, and how hot-air balloons and helium balloons rise. The principle is remarkable in its generality: it depends only on the displaced volume and the fluid density, not on the shape, material, or depth of the submerged object.

Assumptions

The fluid is at rest (static equilibrium).If the fluid is accelerating or flowing, pressure gradients must balance inertial terms as well as gravity, and the simple hydrostatic relation \(dp/dz = -\rho g\) no longer holds; the full momentum equation \(\rho \frac{D\vec v}{Dt} = -\nabla p + \rho \vec g\) must be used.

The fluid is incompressible (constant density \(\rho\)).For compressible fluids such as gases over large height ranges, \(\rho\) varies with pressure and temperature, and the exponential barometric formula \(p = p_0 e^{-Mgz/RT}\) replaces the linear relation; the linear form is only valid when \(\Delta h \ll RT/(Mg)\).

The gravitational field \(\vec g\) is uniform over the region of interest.On astronomical scales or near compact masses, \(g\) varies with position, and the pressure profile must be integrated with \(g(z)\) as a variable; the constant-\(g\) assumption is excellent for laboratory and engineering scales (\(\Delta h \ll R_{\text{Earth}}\)).

Surface tension effects are negligible.At small length scales (capillaries, droplets, bubbles with radii of order \(10^{-4}\) m or less), the Laplace pressure jump \(\Delta p = 2\gamma/r\) across a curved interface becomes comparable to or larger than \(\rho g h\), and the pressure is no longer solely determined by depth.

The fluid is a continuum.The derivation assumes pressure is a well-defined local field, valid when the molecular mean free path is much smaller than the characteristic length scale; in rarefied gases (Knudsen number \(\gtrsim 1\)) the continuum picture breaks down and kinetic theory is required.

Derivation

\[ p \equiv \frac{F_{\perp}}{A} \]

Definition of pressure as the normal force per unit area exerted by the fluid on a surface; tangential (shear) forces vanish in a fluid at rest. A

\[ p = p(x,y,z) \quad \text{is isotropic: } p_{xx} = p_{yy} = p_{zz} \equiv p \]

Consider an infinitesimal fluid wedge with faces of area \(A\), \(A\cos\alpha\), \(A\sin\alpha\) and angles \(\alpha\), \(\pi/2-\alpha\). Force balance in the \(x\)-direction: \(p_x A\cos\alpha = p A\cos\alpha\), giving \(p_x = p\). By symmetry \(p_y = p_z = p\). A fluid at rest cannot sustain shear, so the stress tensor is purely isotropic. C

\[ \sum F_z = 0: \quad p(z)\,A - \left[p(z) + \frac{dp}{dz}\,dz\right]A - \rho g\, A\,dz = 0 \]

Vertical force balance on a small cylindrical fluid element of cross-section \(A\) and height \(dz\): upward pressure force minus downward pressure force minus weight equals zero in static equilibrium. B

\[ -\frac{dp}{dz}\,A\,dz - \rho g\,A\,dz = 0 \quad\Longrightarrow\quad \frac{dp}{dz} = -\rho g \]

Divide by \(A\,dz\) (nonzero) and rearrange; this is the hydrostatic equation, the static limit of the momentum equation with \(\vec v = 0\). B

\[ h \equiv z_{\text{surf}} - z \;\Rightarrow\; \frac{dp}{dh} = +\rho g, \qquad \int_{p_0}^{p} dp = \rho g \int_{0}^{h} dh' \quad\Longrightarrow\quad p - p_0 = \rho g h \]

Measure depth \(h\) downward from the free surface, so \(dh = -dz\) and the pressure rises with depth. Integrate from the surface (\(h=0\), pressure \(p_0\)) to depth \(h\); \(\rho\) and \(g\) are constant under the incompressibility and uniform-\(g\) assumptions. A

\[ p = p_0 + \rho g h \]

The pressure at depth \(h\) is the surface pressure plus the weight per unit area of the fluid column above; this is the hydrostatic pressure law. A

\[ \vec{F}_{\text{buoy}} = -\oint_{\partial V} p\,\hat{n}\,dA \]

The net force on a submerged body of volume \(V\) bounded by surface \(\partial V\) is the integral of pressure times the outward normal over the entire surface; the minus sign accounts for pressure acting inward. B

\[ \vec{F}_{\text{buoy}} = -\oint_{\partial V} p\,\hat{n}\,dA = -\int_V \nabla p\,dV \]

Gradient form of the divergence theorem, \(\oint_{\partial V} p\,\hat{n}\,dA = \int_V \nabla p\,dV\) (apply Gauss to \(p\,\vec c\) for an arbitrary constant vector \(\vec c\)). The surface pressures do not care what is inside, so we may evaluate the integral as if the body were replaced by fluid in equilibrium: that is Archimedes' own argument. C

\[ \nabla p = -\rho g\,\hat{z} \quad\Longrightarrow\quad \vec{F}_{\text{buoy}} = -\int_V (-\rho g\,\hat{z})\,dV = \rho g V\,\hat{z} \]

Substitute the hydrostatic gradient \(\nabla p = -\rho g\,\hat{z}\) (only the \(z\)-component is nonzero); the volume integral of a constant vector over \(V\) is the vector times the volume. B

\[ F_b = \rho_{\text{fluid}}\, g\, V_{\text{disp}} \]

The magnitude of the buoyant force equals the weight of the displaced fluid; \(V_{\text{disp}}\) is the volume of fluid displaced by the submerged portion of the body. A

\[ \rho_{\text{body}}\,g\,V_{\text{body}} = \rho_{\text{fluid}}\,g\,V_{\text{disp}} \quad\Longrightarrow\quad \frac{V_{\text{disp}}}{V_{\text{body}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}} \]

For a floating body in equilibrium, weight equals buoyant force; cancel \(g\) and solve for the fraction submerged, which depends only on the density ratio. B

\[ p(z) = p_0\,\exp\!\left(-\frac{M g z}{R T}\right) \]

For an ideal gas in hydrostatic equilibrium with constant temperature \(T\): combine \(dp/dz = -\rho g\) with the ideal gas law \(\rho = pM/(RT)\) to get \(dp/p = -(Mg/RT)\,dz\); integrate to obtain the barometric formula with scale height \(H = RT/(Mg)\). C

Result

\[ p = p_0 + \rho g h \qquad\qquad F_b = \rho_{\text{fluid}}\, g\, V_{\text{disp}} \]

Reading. Pressure in a static fluid increases linearly with depth, and the buoyant force on any submerged object equals the weight of the fluid it displaces. The first result governs how pressure builds up in liquids; the second explains floating and sinking without reference to the object's material or shape.

Units check. \([\rho g h] = \mathrm{kg/m^3 \cdot m/s^2 \cdot m = kg/(m\cdot s^2) = N/m^2 = Pa}\). For buoyancy: \([\rho g V] = \mathrm{kg/m^3 \cdot m/s^2 \cdot m^3 = kg\cdot m/s^2 = N}\). Both sides consistent.

Limiting cases

Surface of the fluid (\(h=0\)): \(p = p_0\), the ambient pressure (e.g. atmospheric pressure at an open tank surface).
Very small depth (\(\rho g h \ll p_0\)): the gauge pressure \(\rho g h\) is a small correction to \(p_0\); for water, 1 m of depth adds only \(\approx 10\,\mathrm{kPa}\), about 10% of atmospheric pressure.
Object density equals fluid density (\(\rho_{\text{body}} = \rho_{\text{fluid}}\)): the body is neutrally buoyant and floats fully submerged with \(V_{\text{disp}} = V_{\text{body}}\); it remains at any depth without rising or sinking.
Object density much less than fluid (\(\rho_{\text{body}} \ll \rho_{\text{fluid}}\)): the body floats with only a small fraction submerged; polystyrene foam (\(\rho \approx 30\ \mathrm{kg/m^3}\)) floats in water with about 3% of its volume below the waterline. A body that cannot float on a surface (a balloon in air) instead rises until the surrounding density falls to its own average density.
Compressible fluid, small height range (\(\Delta h \ll H\)): the exponential barometric formula reduces to the linear form \(p \approx p_0(1 - \Delta h/H)\), recovering the incompressible result to first order.

Breaks when

Fluid is accelerating or flowing. In a rotating fluid (e.g. a centrifuge) or a flowing river, the pressure field must satisfy the full Navier-Stokes or Euler equation; the simple \(dp/dz = -\rho g\) is replaced by \(\nabla p = \rho(\vec g - \vec a)\), and pressure is no longer a function of depth alone.
Compressible fluid over large height ranges. For the atmosphere, density varies by a factor of \(\sim e^{-1}\) every 8.4 km; using the incompressible formula over 10 km gives a pressure error of order 100% rather than the correct exponential decay.
Surface tension dominates (small scales). In a capillary of radius \(r = 0.1\,\mathrm{mm}\), the Laplace pressure \(\Delta p = 2\gamma/r \approx 2(0.072)/10^{-4} = 1440\,\mathrm{Pa}\), which exceeds \(\rho g h\) for depths less than \(\sim 15\,\mathrm{cm}\); the meniscus shape and pressure jump cannot be captured by hydrostatics alone.
Non-continuum regime. In a rarefied gas (Knudsen number \(\gtrsim 0.1\)), the concept of a local pressure field breaks down; molecular collisions are too infrequent to maintain local thermodynamic equilibrium, and the continuum pressure \(p = \rho k_B T / m\) is no longer meaningful.

Failure modes

Confusing gauge and absolute pressure. Writing \(p = \rho g h\) for the total pressure at depth, forgetting the \(p_0\) term; for a sealed tank or a gas column, the surface pressure is not zero and must be included.
Applying Archimedes' principle to the object's total volume when it is only partially submerged. Using \(V_{\text{disp}} = V_{\text{body}}\) for a floating object; only the submerged volume displaces fluid, and the correct relation is \(V_{\text{disp}}/V_{\text{body}} = \rho_{\text{body}}/\rho_{\text{fluid}}\).
Assuming pressure depends on the shape or area of the container. The hydrostatic paradox: pressure at a given depth depends only on \(\rho\), \(g\), and \(h\), not on the total volume of fluid or the container geometry; a tall narrow tube and a wide shallow basin at the same depth have the same pressure.
Using the incompressible formula for the atmosphere over large altitudes. Computing \(p = p_0 - \rho g h\) with constant \(\rho = 1.2\,\mathrm{kg/m^3}\) gives negative pressure above \(h \approx 8.6\,\mathrm{km}\); the exponential barometric formula is required.

Discussion

The hydrostatic equation \(dp/dz = -\rho g\) is the zero-velocity limit of the momentum equation for a fluid. In full generality, the Euler equation reads \(\rho\,D\vec v/Dt = -\nabla p + \rho\vec g\); setting \(\vec v = 0\) and \(D\vec v/Dt = 0\) leaves \(\nabla p = \rho\vec g\), which in one dimension with \(\vec g = -g\hat z\) gives the familiar \(dp/dz = -\rho g\). This connection makes clear that hydrostatics is not a separate theory but the static slice of fluid dynamics.

Archimedes' principle, derived here via the divergence theorem, is equivalent to the statement that the pressure gradient is uniform in a static fluid. The buoyant force is the integral of a constant vector field over the displaced volume, which is why it depends only on volume and fluid density, not on the object's shape or material. This is a powerful example of how a local differential law (\(\nabla p = \rho\vec g\)) produces a global integral result (buoyancy) through a topological identity (divergence theorem).

The distinction between gauge and absolute pressure is a source of persistent confusion. Gauge pressure \(p_{\text{gauge}} = p - p_{\text{atm}}\) is what a pressure gauge reads and what \(\rho g h\) gives directly; absolute pressure includes the atmospheric contribution. In most engineering applications (hydraulic presses, manometers), gauge pressure is the natural quantity because the atmosphere acts on both sides of the system. In thermodynamics and gas laws, absolute pressure is essential because the ideal gas law \(pV = nRT\) requires the true molecular pressure.

At a deeper level, the hydrostatic balance is an instance of a more general principle: in a static system, the stress tensor must be divergence-free (in the absence of body forces) or balance the body force density. For a fluid at rest, the stress tensor is \(\sigma_{ij} = -p\delta_{ij}\), and the equilibrium condition \(\partial_j \sigma_{ij} + \rho g_i = 0\) reduces to \(-\partial_i p + \rho g_i = 0\), i.e. \(\nabla p = \rho\vec g\). This is the same structure as the force balance in solid mechanics, where the Cauchy stress tensor balances body forces; the difference is that a fluid cannot sustain shear, so the stress is purely isotropic. The connection to elasticity theory and the theory of shells is direct and important in geophysics, where the pressure field in the Earth's mantle is computed by integrating \(\rho(r)g(r)\) over depth.

Common misconceptions. Pressure is not a force; it is force per unit area, and the total force on a surface depends on the area and orientation. A common error is to say “the pressure is greater at the bottom of the tank” as if pressure were a vector pointing downward; in fact, pressure acts equally in all directions at a point (isotropy), and the fluid pushes down on a horizontal bottom with force \(pA\) and outward on the side walls with the local pressure at each depth. Another misconception is that a heavier object sinks because it is “heavier”; in fact, it sinks because its density exceeds that of the fluid, and the relevant comparison is \(\rho_{\text{body}}\) vs. \(\rho_{\text{fluid}}\), not the total weight. A large steel ship floats because its average density (including the air inside) is less than that of water.

Worked examples

Example 1. A cylindrical tank of radius \(R = 0.50\,\mathrm{m}\) is filled with water (\(\rho = 1000\,\mathrm{kg/m^3}\)) to a depth \(h = 2.0\,\mathrm{m}\). The top is open to the atmosphere (\(p_0 = 1.013\times10^5\,\mathrm{Pa}\)). Find (a) the absolute pressure at the bottom, (b) the gauge pressure at the bottom, and (c) the total force on the circular bottom. Take \(g = 9.8\,\mathrm{m/s^2}\).

\[ p_{\text{bottom}} = p_0 + \rho g h \]

Hydrostatic pressure at depth \(h\) below the free surface. A

\[ p_{\text{bottom}} = 1.013\times10^5 + (1000)(9.8)(2.0) = 1.013\times10^5 + 1.96\times10^4 = 1.209\times10^5\ \mathrm{Pa} \]

Arithmetic: \(\rho g h = 1000 \times 9.8 \times 2.0 = 19600\,\mathrm{Pa}\). A

\[ p_{\text{gauge}} = \rho g h = 1.96\times10^4\ \mathrm{Pa} \]

Gauge pressure is the excess over atmospheric pressure. A

\[ A = \pi R^2 = \pi(0.50)^2 = 0.785\ \mathrm{m^2} \]

Area of the circular bottom. A

\[ F = p_{\text{bottom}}\,A = (1.209\times10^5)(0.785) = 9.49\times10^4\ \mathrm{N} \]

Force the contents (water plus the atmosphere above it) exert on the bottom. If the tank stands in air, atmospheric pressure underneath cancels \(p_0A\), leaving a net \(\rho g h A = 1.54\times10^4\ \mathrm{N}\), the weight of the water. B

\[ p_{\text{bottom}} = 1.21\times10^5\ \mathrm{Pa},\quad p_{\text{gauge}} = 1.96\times10^4\ \mathrm{Pa},\quad F = 9.5\times10^4\ \mathrm{N} \]

Reading. The absolute pressure at the bottom is about 20% higher than atmospheric due to the 2 m water column. The total force on the inside of the bottom is about 95 kN, the weight of roughly 9.7 tonnes; most of it is atmospheric and is balanced from below.

Units check. \([\rho g h] = \mathrm{kg/m^3 \cdot m/s^2 \cdot m = Pa}\). \([pA] = \mathrm{Pa \cdot m^2 = N}\). Consistent.

Example 2. A solid block of wood with density \(\rho_{\text{wood}} = 600\,\mathrm{kg/m^3}\) and volume \(V = 0.020\,\mathrm{m^3}\) floats in seawater (\(\rho_{\text{sw}} = 1025\,\mathrm{kg/m^3}\)). Find (a) the fraction of the block's volume that is submerged, (b) the buoyant force, and (c) the weight of the block. Take \(g = 9.8\,\mathrm{m/s^2}\).

\[ \frac{V_{\text{disp}}}{V_{\text{body}}} = \frac{\rho_{\text{wood}}}{\rho_{\text{sw}}} = \frac{600}{1025} = 0.585 \]

Floating equilibrium: weight equals buoyant force, so \(\rho_{\text{wood}} g V = \rho_{\text{sw}} g V_{\text{disp}}\); cancel \(g\) and solve for the submerged fraction. B

\[ V_{\text{disp}} = 0.585 \times 0.020 = 0.0117\ \mathrm{m^3} \]

Submerged volume. A

\[ F_b = \rho_{\text{sw}}\, g\, V_{\text{disp}} = (1025)(9.8)(0.0117) = 117.6\ \mathrm{N} \]

Buoyant force equals the weight of displaced seawater. A

\[ W = \rho_{\text{wood}}\, g\, V = (600)(9.8)(0.020) = 117.6\ \mathrm{N} \]

Weight of the block; equals the buoyant force by equilibrium. A

\[ \text{Fraction submerged} = 0.585,\quad F_b = 118\ \mathrm{N},\quad W = 118\ \mathrm{N} \]

Reading. About 58.5% of the wood block is below the waterline. The buoyant force and the weight are equal (118 N), confirming equilibrium. The block floats higher in seawater than in freshwater because seawater is denser.

Units check. \([\rho g V] = \mathrm{kg/m^3 \cdot m/s^2 \cdot m^3 = N}\). Consistent.

Problems

A U-tube manometer contains mercury (\(\rho_{\text{Hg}} = 13600\,\mathrm{kg/m^3}\)). One side is connected to a gas line, the other is open to the atmosphere. The mercury level on the open side is 150 mm higher than on the gas side. Find the gauge pressure of the gas. (\(g = 9.8\,\mathrm{m/s^2}\).)
Solution
The gauge pressure equals the pressure difference due to the mercury column: \(p_{\text{gauge}} = \rho_{\text{Hg}} g h = (13600)(9.8)(0.150) = 13600 \times 1.47 = 19992 \approx 2.00\times10^4\ \mathrm{Pa}\).
A submarine at a depth of 300 m in seawater (\(\rho = 1025\,\mathrm{kg/m^3}\)) has a circular porthole of radius 0.20 m. Find the total force on the porthole due to the water pressure (gauge). (\(g = 9.8\,\mathrm{m/s^2}\).)
Solution
Gauge pressure: \(p = \rho g h = (1025)(9.8)(300) = 1025 \times 2940 = 3.0135\times10^6\ \mathrm{Pa}\). Area: \(A = \pi r^2 = \pi(0.20)^2 = 0.1257\ \mathrm{m^2}\). Force: \(F = pA = (3.0135\times10^6)(0.1257) = 3.79\times10^5\ \mathrm{N}\).
A solid sphere of radius 0.10 m and density \(8000\,\mathrm{kg/m^3}\) is fully submerged in water (\(\rho = 1000\,\mathrm{kg/m^3}\)). Find the buoyant force and the apparent weight (weight minus buoyant force). (\(g = 9.8\,\mathrm{m/s^2}\).)
Solution
Volume: \(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(0.10)^3 = 4.189\times10^{-3}\ \mathrm{m^3}\). Buoyant force: \(F_b = \rho g V = (1000)(9.8)(4.189\times10^{-3}) = 41.05\ \mathrm{N}\). Weight: \(W = \rho_{\text{sphere}} g V = (8000)(9.8)(4.189\times10^{-3}) = 328.4\ \mathrm{N}\). Apparent weight: \(W - F_b = 328.4 - 41.05 = 287.3\ \mathrm{N}\).
An ideal gas at temperature \(T = 288\,\mathrm{K}\) and molar mass \(M = 0.029\,\mathrm{kg/mol}\) has a surface pressure \(p_0 = 1.013\times10^5\,\mathrm{Pa}\). Using the barometric formula, find the pressure at an altitude of 5.0 km. (\(R = 8.314\,\mathrm{J/(mol\cdot K)}\), \(g = 9.8\,\mathrm{m/s^2}\).)
Solution
Scale height: \(H = RT/(Mg) = (8.314)(288)/(0.029 \times 9.8) = 2394.4/0.2842 = 8425\ \mathrm{m} \approx 8.4\ \mathrm{km}\). Pressure: \(p = p_0 e^{-z/H} = (1.013\times10^5) e^{-5000/8425} = (1.013\times10^5) e^{-0.5935}\). Compute \(e^{-0.5935} = 0.5525\). Thus \(p = (1.013\times10^5)(0.5525) = 5.60\times10^4\ \mathrm{Pa}\).
A hydraulic press has a small piston of area \(A_1 = 0.010\,\mathrm{m^2}\) and a large piston of area \(A_2 = 0.50\,\mathrm{m^2}\). A force of 200 N is applied to the small piston. (a) Find the pressure in the fluid. (b) Find the force exerted by the large piston. (c) If the large piston lifts a load through 0.05 m, how far does the small piston move? (Assume incompressible fluid and no losses.)
Solution
(a) \(p = F_1/A_1 = 200/0.010 = 2.0\times10^4\ \mathrm{Pa}\). (b) \(F_2 = p A_2 = (2.0\times10^4)(0.50) = 1.0\times10^4\ \mathrm{N}\). (c) Volume conservation: \(A_1 d_1 = A_2 d_2\), so \(d_1 = (A_2/A_1) d_2 = (0.50/0.010)(0.05) = 50 \times 0.05 = 2.5\ \mathrm{m}\).