Pressure, buoyancy, and fluid statics

Start with a submerged window

A tank has a vertical rectangular window 0.60 m wide and 0.80 m high. Its top is 0.40 m below the liquid surface, so its bottom is 1.20 m below. The liquid is at rest. Does the pressure push equally at every part of the window? Where would one equivalent force have to act to reproduce the distributed push and its turning effect?

Supply a uniform liquid density ρ = 1000 kg/m³ and gravitational acceleration g = 9.81 m/s². Density means mass per volume; gravity gives a mass m the downward weight mg. Set the gas pressure above the liquid and on the window’s dry side to the same assigned absolute value, 100000 Pa. These are illustrative inputs. We neglect gas-density changes, surface curvature and interfacial stresses.

Begin with the depth table, window figure and pressure-force balances. Units, standards, and physical constants and Newton’s laws and free-body diagrams provide useful preparation. Geometry, trigonometry, and coordinate systems helps with the window dimensions. The needed pressure and moment ideas are introduced here. Integrals and multivariable calculus is an optional return path for the final derivations; Center of mass and many-particle systems and Torque, angular momentum, and gyroscopes provide background for the stability comparison.

Define pressure and its reference

Pressure p is normal force per area locally: a small patch of area dA experiences pressure-force magnitude p dA. Its SI unit is the pascal, Pa = N/m². Positive pressure is compressive. For an ordinary pressure-only fluid at rest, there is no shear traction: the fluid pushes perpendicular to a contacting surface. At one point the pressure is the same for every orientation. This isotropy does not mean that pressure is the same at different positions.

A force also needs a direction. If n is a unit normal pointing outward from a solid into the surrounding fluid, the pressure force on the solid patch is dF = −p n dA. The minus sign makes the force point inward. A vertical tank wall receives a horizontal push; a horizontal bottom receives a downward push. Both are normal forces, even though gravity is vertical.

Absolute pressure uses the vacuum reference. Gauge pressure subtracts a stated reference pressure: p_g = p − p_ref. With p_ref = 100000 Pa, a gauge reading of −12000 Pa means an absolute pressure of 88000 Pa. A negative gauge reading says that the pressure is below that reference; it does not give a negative absolute value in this example.

For a surface with fluid on both sides, use both pressures with their opposite force directions. Gauge pressure can directly represent the net load only when its reference equals the actual pressure on the other side. Our static model excludes extra shear or elastic stresses in yield-stress materials and additional electromagnetic or interfacial stresses. Those require a fuller stress balance.

Balance a small fluid column

Take a narrow vertical column of resting liquid with horizontal face area A and height Δz. Its bottom pressure pushes upward, its top pressure pushes downward, and its weight points downward. Force balance gives (p_bottom − p_top)A = ρgAΔz for uniform density. Cancel A: the increase in pressure over a downward distance Δz is ρgΔz. A wider column has more weight and proportionally more pressure-force area, so width does not change this pressure difference.

Choose height z positive upward and put the free surface at z = 0. Define depth h = −z, positive downward. The local version of the balance is dp/dz = −ρg, or dp/dh = ρg. The minus sign with upward height is essential: pressure decreases as z increases.

If the surface pressure is p_s and ρ is constant, integration gives p(h) = p_s + ρgh. Our gauge reference equals p_s, so p_g(h) = ρgh. Here ρg = 9810 Pa/m. This law applies to gases too, but their density may change appreciably with pressure and temperature. For a supplied density profile that varies only with depth, ρ(h), the static pressure is p_s + g∫₀ʰρ(s) ds; an equation of state or a specified density profile is then needed.

The mechanism is a local balance of surface force and body force. A body force acts throughout the fluid volume; gravity contributes ρg per unit volume. In vector notation, a possible pressure-only static equilibrium satisfies ∇p = ρb, where b is the body acceleration and ∇p describes the spatial pressure gradient. Under our uniform gravity, b points downward and horizontal pressure gradients vanish.

Follow pressure through connected fluid

A boundary pressure sets the starting value of the pressure field. Raising p_s by a uniform amount raises p everywhere by that amount if density, geometry and gravity remain the same. This is the static pressure-transmission idea behind Pascal’s law. It does not remove the existing depth variation. Raising the pressure equally on both sides of a window also leaves its net pressure difference unchanged.

At a free liquid surface, the liquid pressure matches the adjacent gas pressure in our model. With uniform gas pressure, uniform gravity and neglected surface tension, a connected resting free surface is horizontal. Curved menisci and capillary pressure need the separate surface-tension model. Two disconnected containers can have different surface pressures or levels; equal depth alone does not then establish equal pressure.

Within the connected resting liquid described here, points at the same height have equal pressure. To compare different heights, move down by Δh and add ρgΔh, or move up and subtract it. At a flat interface between two resting liquids, pressure is continuous when interfacial stresses are neglected. Use each liquid’s own density on its part of the route. This pressure chain is also the basis of a liquid-column manometer.

For example, prescribe an upper layer with density 800 kg/m³ and thickness 0.30 m, over a lower layer with density 1000 kg/m³ and thickness 0.50 m. The interface gauge pressure is (800 kg/m³)(9.81 m/s²)(0.30 m) = 2354.4 Pa. Add 4905 Pa across the lower layer to reach 7259.4 Pa at the bottom. Substituting one density for both layers would describe a different model.

Add the distributed window force

Return to the window. At depth h, the liquid-side absolute pressure is 100000 Pa + ρgh, while the dry-side gas pressure is 100000 Pa. Their net difference is Δp(h) = ρgh. It rises from 3924 Pa at the top to 11772 Pa at the bottom. The fluid therefore applies a distributed load: each patch contributes its own normal force Δp dA.

The area is A = (0.60 m)(0.80 m) = 0.48 m². Its geometric centre, or area centroid, is at depth h_c = 0.80 m. Because pressure changes linearly with depth and the window has constant width, the area-average net pressure equals the centroid value, 7848 Pa. The net force is F = ρgh_cA = 3767.04 N, directed from the liquid toward the dry side.

Using absolute pressures gives the same result when both sides are included: the liquid side contributes 51767.04 N, and the opposite gas contributes 48000 N in the other direction. Their difference is 3767.04 N. The large one-sided force is not the net window force. A different gas pressure on one side would add a real uniform pressure difference.

The first figure keeps depth positive downward and compares actual pressure values on one kPa axis. Its equivalent resultant reproduces the total force and moment of the distributed load. It does not turn the fluid contact into a concentrated force at one physical point. No material strength or allowable load has been supplied.

Locate the centre of pressure

The total force does not identify its turning effect. Measure depth from a horizontal line through the free surface, and take moments about that line. A force element at depth h contributes moment magnitude h Δp dA. The centre of pressure is the position of the equivalent resultant: Fh_cp must equal the sum of those moments. Thus h_cp = (∫_A h Δp dA)/F.

For this vertical rectangle, the pressure-weighted depth is h_cp = h_c + I_G/(h_cA). Here I_G is the second moment of the window area about its horizontal centroidal line: I_G = bH³/12 = 0.0256 m⁴, with width b = 0.60 m and height H = 0.80 m. It describes the spread of area with squared distance. It is an area quantity with units m⁴, not a mass moment of inertia.

The added depth is (0.0256 m⁴)/[(0.80 m)(0.48 m²)] = 0.0666667 m, so h_cp ≈ 0.866667 m. The deeper patches have larger forces and therefore pull the pressure-weighted average below the area centroid. The total moment is Fh_cp = 3264.768 N·m. The line of action also lies halfway across this symmetric window.

This centroid formula used Δp = ρgh. If a separate uniform difference Δp₀ is added, its force acts at the area centroid and shifts the combined centre of pressure toward that centroid. A horizontal plane has uniform pressure at one depth, so its centre of pressure is its area centroid. The final optional section derives the vertical formula and its inclined-plane version.

Find buoyancy from pressure

Hold a cube of side 0.10 m fully under the liquid, with its top at depth 0.40 m and bottom at 0.50 m. Each horizontal face has area 0.01 m². The top pressure pushes downward with force 1039.24 N; the bottom pressure pushes upward with force 1049.05 N. Equal-depth forces on opposite vertical faces cancel horizontally. The remaining hydrostatic force is 9.81 N upward.

The bottom-minus-top calculation gives [ρg(0.10 m)](0.01 m²) = ρgV, where V = 0.001 m³ is the displaced liquid volume. This is Archimedes’ principle for uniform density and gravity: buoyancy equals the weight of the displaced fluid. For an arbitrary fully wetted shape, adding the normal pressure forces over its closed surface gives the same result. In uniform density the buoyancy line passes through the centroid of the displaced volume, called the centre of buoyancy B.

Assign the cube mass 0.80 kg. Its weight is 7.848 N, so hydrostatic buoyancy minus weight is 1.962 N upward. A downward holding force of 1.962 N balances it at rest. This static comparison does not predict post-release fluid forces, acceleration or speed; motion can change the surrounding pressure field.

The uniform 100000 Pa offset cancels around a closed surface because opposite normal directions balance. Buoyancy comes from the pressure gradient. Moving this fully submerged cube deeper in the same uniform-density liquid leaves ρgV unchanged. If liquid cannot act beneath a body sealed against a floor, the required pressure field is missing and this closed-surface argument cannot simply be used. Contact forces, surface tension and varying fluid density also require the appropriate separate balance.

Find a floating draft

A freely floating body in static vertical equilibrium displaces enough liquid for buoyancy to equal its weight: ρgV_displaced = mg. The displaced volume is therefore m/ρ. For a partly submerged body, the common ambient pressure cancels when the wetted and dry parts of the closed body surface are accounted for; the remaining hydrostatic contribution comes from the immersed volume.

Supply a sealed rigid rectangular prism of length L = 0.80 m, width B_w = 0.60 m and height H_b = 0.40 m. Its mass is 115.2 kg. Its total volume is 0.192 m³ and mean density is 600 kg/m³. Keep it upright, with no other vertical support. The displaced volume is 0.1152 m³, giving draft T = V_displaced/(LB_w) = 0.24 m. Draft is the submerged depth from the bottom to the waterline; freeboard is the remaining 0.16 m above it.

The buoyancy and weight are both 1130.112 N. The submerged part is a rectangular prism, so its centre of buoyancy is halfway through the draft: 0.12 m above the bottom. A uniform mass distribution puts the centre of gravity G halfway through the whole body, 0.20 m above the bottom. Weight acts through G; buoyancy acts through B. Upright, the two lines are vertical and aligned.

For this upright prism, the submerged fraction T/H_b = 0.60 equals mean body density divided by liquid density. This fraction uses the body’s total mass and external displaced volume; cavities can lower the mean density without changing the density of the solid material. The force balance determines a draft, but it has not yet established how the body responds to a small tilt.

Separate floating balance from stability

A small sideways tilt changes which parts of a floating body are immersed. The centre of buoyancy shifts even when the total displaced volume stays the same. The vertical buoyancy line can then produce a moment opposing the tilt or increasing it. For a fully submerged freely equilibrated body in uniform liquid, its displaced shape is fixed: G below B gives a restoring couple, G above B gives an overturning couple, and coincidence removes that couple. A floating body requires the extra waterline geometry.

For a symmetric floating body at small heel about its long axis, define the metacentre M in a view relative to the body: it is the limiting intersection of the new buoyancy line with the body’s centreline as heel tends to zero. Buoyancy itself remains vertical in the laboratory frame. Its height above B is BM = I_w/V_displaced. The waterplane is the body’s horizontal cross-section at the waterline. Here I_w = ∫y² dA is the second moment of that waterplane area about the long centreline; y is sideways distance. For our rectangle, I_w = LB_w³/12 = 0.0144 m⁴, so BM = 0.125 m and M is 0.245 m above the bottom.

Define the signed metacentric height GM = z_M − z_G, using heights above the bottom. Let heel φ be positive in one chosen direction and let positive torque increase φ. The first-order hydrostatic torque is τ ≈ −W GM φ, where W = mg and φ is in radians. Positive GM gives torque opposing either small tilt; negative GM gives torque increasing it. GM = 0 eliminates only this first-order term. Higher-order geometry is needed to judge a finite tilt in that case.

The uniform reference prism has GM = 0.045 m. Compare two stipulated internal mass redistributions with the same total mass and exterior shape: put G at 0.245 m or at 0.260 m above the bottom. All three retain the same upright draft and equal buoyancy and weight, but their first-order moments differ. At +5°, the linear formulas give approximately −4.437939 N·m, 0 and +1.479313 N·m respectively. These are evaluations of the first-order model, not exact finite-angle vessel responses.

The third figure uses one geometric scale for the upright sections and one torque scale for the three linear curves. No internal free liquid, shifting contents, waves or external torque are included. This local stability calculation supplies neither a roll period nor a capsize range; those require additional dynamics and geometry.

Optional return: integrals and body force

On a first reading, the column balance, pressure tables and three comparisons provide the main route. Return here after Integrals and multivariable calculus for the compact derivations. For the window, take horizontal strips dA = b dh. Integrate F = ρgb∫_(0.40 m)^(1.20 m) h dh = ρgb(1.20² − 0.40²)m²/2. The moment is ρgb∫_(0.40 m)^(1.20 m) h² dh = ρgb(1.20³ − 0.40³)m³/3. Dividing the moment by force gives h_cp = 13/15 m.

More generally, define h_c = (∫_A h dA)/A and write h = h_c + ξ on a vertical plane. Since ∫_A ξ dA = 0, ∫_A h² dA = Ah_c² + ∫_A ξ² dA. The second integral is I_G, yielding h_cp = h_c + I_G/(Ah_c). For a plane at angle θ to horizontal, the vertical offset is ξ sinθ, so the depth correction is I_G sin²θ/(Ah_c). I_G is then about the centroidal line parallel to the plane’s intersection with the free-surface plane. At θ = 0, pressure is uniform and the correction vanishes.

For a closed solid surface in a uniform hydrostatic field, F = −∮p n dA. The divergence theorem changes this into F = −∫_V∇p dV = ρgV e_z, with e_z pointing upward. A spatially uniform pressure offset has zero gradient and contributes no net force. If the undisturbed hydrostatic density varies with height, the corresponding buoyancy is g∫_Vρ(z) dV upward; the simple ρgV expression needs uniform density over that displaced volume.

The small-angle waterplane result also follows from added and removed wedges. A tilt changes local immersion by an amount proportional to φy, so the first-order sideways volume moment contains φ∫y² dA. Dividing by unchanged displaced volume gives the displacement of B and BM = I_w/V_displaced. Subtract the height of G above B to obtain GM. This is why a waterplane area second moment, rather than the mass moment of inertia, enters the static torque.

Relative rest in an accelerating container changes the body-force balance. In a frame translating with constant acceleration +a e_x, effective body acceleration is −a e_x − g e_z. A constant-pressure free surface then satisfies a dx + g dz = 0, or dz/dx = −a/g. For a prescribed a = 2 m/s², its height decreases by about 0.101937 m across 0.50 m along +x. This assumes an established relative equilibrium, with no transient sloshing; the liquid is not at rest in an inertial frame. Non-inertial frames, fictitious forces, tides, and constraints provides the return path for that frame change.

Moving fluids require a momentum balance with acceleration and, when relevant, shear stress. Conservation of mass, momentum, and energy and Euler’s equation and Bernoulli’s principle develop those neighbouring topics. The present balances answer static pressure and support questions under their stated conditions.

Worked examples

Try it yourself

1. A pressure instrument uses reference 100000 Pa and reads −12000 Pa. What is the absolute pressure? Separately, in the assigned connected liquid, what is the lower-minus-upper pressure difference between depths 0.65 m and 0.20 m?

   Show the answer and reasoning

   Absolute pressure is 88000 Pa. The liquid pressure difference is +4414.5 Pa.

   Add the reference to the gauge reading: 100000 Pa − 12000 Pa = 88000 Pa. For the liquid comparison, the lower point is 0.45 m deeper, so ρgΔh = (1000 kg/m³)(9.81 m/s²)(0.45 m) = 4414.5 Pa. A common reference cancels in this difference.

2. The upper liquid layer is 0.30 m thick with density 800 kg/m³; the lower layer is 0.50 m thick with density 1000 kg/m³. With a common surface reference and g = 9.81 m/s², find interface and bottom gauge pressures.

   Show the answer and reasoning

   Interface pressure is 2354.4 Pa; bottom pressure is 7259.4 Pa.

   The upper layer adds 800 × 9.81 × 0.30 Pa = 2354.4 Pa. Pressure is continuous at the flat interface under the stated no-interfacial-stress assumption. The lower layer adds 1000 × 9.81 × 0.50 Pa = 4905 Pa. Add the two contributions rather than treating both layers as the denser liquid.

3. For the same window, add a uniform pressure difference Δp₀ = 3000 Pa toward the dry side, while leaving the hydrostatic part unchanged. Find the combined force and centre-of-pressure depth. Would increasing ambient pressure equally on both sides produce this change?

   Show the answer and reasoning

   F = 5207.04 N and h_cp ≈ 0.848230 m. An equal ambient increase on both sides produces no such extra net load.

   The added force is Δp₀A = (3000 Pa)(0.48 m²) = 1440 N and acts at h_c = 0.80 m. Add its moment 1152 N·m to the hydrostatic moment 3264.768 N·m, obtaining 4416.768 N·m. Divide by 3767.04 N + 1440 N = 5207.04 N. The position moves toward the centroid because the new contribution is uniform. Equal pressure offsets on the two sides cancel.

4. Move the held 0.10 m cube downward by 1.00 m in the same uniform liquid, while keeping liquid access to every face. Does its buoyancy change? Can the same reasoning automatically be used if the cube seals against a floor?

   Show the answer and reasoning

   Its buoyancy remains 9.81 N. The sealed-floor case requires a new contact and pressure-access balance.

   Both horizontal-face pressures increase by 9810 Pa, so each absolute face-force magnitude increases by 98.10 N. Their difference remains 9.81 N because displaced volume and density are unchanged. A sealed floor prevents the assumed liquid pressure from acting on the bottom face; the closed-surface hydrostatic argument then lacks its required pressure field.

5. Redistribute the prism’s internal mass while preserving total mass and exterior geometry, placing G at 0.260 m above the bottom. What happens to upright draft, GM and the first-order torque at +5°? What does G at 0.245 m establish?

   Show the answer and reasoning

   Draft stays 0.24 m. GM becomes −0.015 m, giving approximately +1.479313 N·m at +5°, which increases that small tilt. G at 0.245 m gives GM = 0 and only a zero first-order torque term.

   The same mass still displaces 0.1152 m³, so B and the small-angle M remain at 0.120 m and 0.245 m above the bottom. Subtract z_G from z_M: GM = −0.015 m. Substitute into −W GMφ with W = 1130.112 N and φ = π/36 rad. At GM = 0, higher-order immersion geometry is needed; the first-order result alone is not a finite-angle neutral-stability conclusion.

What to carry forward

• In the stated pressure-only fluid at rest, local pressure is isotropic, while its spatial gradient balances the body force.
• With height positive upward, dp/dz = −ρg. Boundary pressure and density determine the hydrostatic pressure field.
• Net surface load uses the pressure difference across the surface. A gauge reference represents that difference only when it matches the opposite-side pressure.
• A distributed load has both a total force and a moment. The centre of pressure locates an equivalent resultant with both effects.
• Uniform-density buoyancy equals displaced-fluid weight when the required hydrostatic pressure field acts on the body. It is a pressure-gradient effect.
• Equal buoyancy and weight determine floating force balance. Small-angle rotational response also depends on the mass distribution and waterplane geometry.