Study Companion: S23 — Fluid Statics

Water behind a dam, oil in a tank, air in a tyre: a fluid at rest presses on every surface it touches, with an intensity that grows with depth. That pressure is the last distributed force of this part, and it is handled exactly like the others: find the intensity, integrate it into a resultant, and place the resultant by the principle of moments. The same reasoning, applied to a body surrounded by fluid, gives Archimedes' principle and the stability of floating ships.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 5/9 ("Fluid Statics"), with Sample Problems 5/19–5/23. Harper, Solving Statics Problems in MATLAB, "Fluid Statics" (Problem 5/170). Meriam's Sample Problem 5/20 is in US units, converted here: 1 lb/in² = 6895\,\mathrm{Pa}, 1 in = 25.4\,\mathrm{mm}. Fresh water \rho = 1000\,\mathrm{kg/m^3}, g = 9.81\,\mathrm{m/s^2}. Depth h is measured downward from the free surface.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

A home aquarium holds water 0.6\,\mathrm m deep behind a front glass pane 1.2\,\mathrm m long.

How hard does the water push on the glass, and how far above the bottom does that push act, which is where the frame must be strongest? The plausible guesses are "about the weight of a few buckets" and "at mid-depth, 0.3\,\mathrm m up". The answers are in Nugget 03.

Nugget 01: A Fluid at Rest Pushes Only Normally, and Equally in Every Direction

Claim. A fluid at rest cannot carry shear, so it exerts only normal forces on any surface it touches. At any point in it, the pressure is the same in every direction (Pascal's law).

The picture. Meriam's Figs. 5/30 and 5/31: a small triangular prism of fluid, and a vertical column of it.

On the prism, each face carries pressure times its area, normal to it. On the column, the pressures on top and bottom differ by the column's weight.

Check with numbers. The prism's faces: dy vertical, dx horizontal, ds sloping, all of depth dz into the page. Horizontally, p_1\,dy\,dz = p_3\,ds\,dz\sin\theta, and ds\sin\theta = dy, so p_1 = p_3. Vertically, p_2\,dx\,dz = p_3\,ds\,dz\cos\theta with ds\cos\theta = dx, so p_2 = p_3. For a prism with dx = 3\,\mathrm{mm}, dy = 4\,\mathrm{mm}, ds = 5\,\mathrm{mm} in water at 50\,\mathrm{kPa}, each face's force is 50\,\mathrm{kPa} times its own area, and the sloping face's 50(5) resolves into exactly 50(4) horizontally and 50(3) vertically.

The idea. Meriam (5/9): a fluid is any continuous substance which, at rest, cannot support shear force. Shear, a force tangent to a surface, arises only when adjacent layers move at different speeds; so a fluid at rest exerts only normal forces on a bounding surface. Fluids may be liquids or gases; their statics is hydrostatics or aerostatics. Pascal's law, from the prism: p_1 = p_2 = p_3 = p, and rotating the element shows p_4 equal too. The prism's weight, \rho g times its volume, is a third-order quantity that vanishes in the limit beside the second-order pressure forces, so it does not upset the argument.

The equation (physical law: equilibrium of a fluid element, with the material model of a fluid as unable to carry shear at rest):

Worked number. A force of 50\,\mathrm{kPa} on a 1\,\mathrm{cm^2} patch of a submarine's window is 50\times10^{3}(10^{-4}) = 5\,\mathrm N, whether the patch faces up, down or sideways.

Real example (one you can see). Water squirts from a leaking hose at right angles to the hose wall, wherever the hole is: the pressure pushes normal to the surface.

Plausible wrong answer. "Water presses down, so a vertical wall feels nothing." The pressure acts equally in all directions; on a wall it pushes horizontally.

Check. Can a fluid at rest exert a force along a surface? (No: that would be shear, which it cannot carry at rest.)

Narration spine. A tiny wedge of water floats in a tank; arrows push on its three faces. The sloping face's arrow splits into horizontal and vertical parts that match the other two exactly. The wedge spins round, and every face's arrow keeps the same length per unit area.

Bridge. The pressure is the same in every direction at a point; it is not the same at every depth.

Nugget 02: Pressure Grows Linearly With Depth

Claim. In a fluid at rest, pressure increases with depth at the rate \rho g. In a liquid of constant density, p = p_0 + \rho gh; on open systems the atmospheric p_0 cancels and only the gauge pressure \rho gh matters.

The picture. The pressure on a vertical wall, and gauge pressure against depth.

Check with numbers. Fresh water at 10\,\mathrm m: p = \rho gh = (1.0\,\mathrm{Mg/m^3})(9.81\,\mathrm{m/s^2})(10\,\mathrm m) = 98.1\,\mathrm{kN/m^2} = 98.1\,\mathrm{kPa}, almost one atmosphere (101.3\,\mathrm{kPa}), which a column of water 101.3/9.81 = 10.33\,\mathrm m high would match.

The idea. Meriam (5/9): for a vertical column of fluid of cross-section dA, with h positive downward, the pressure on the upper face is p and on the lower face p + dp; the column's weight is \rho g\,dA\,dh; the side pressures are horizontal and do not enter. Vertical equilibrium, p\,dA + \rho g\,dA\,dh - (p + dp)\,dA = 0, gives

valid for liquids and gases alike: pressure increases with depth and decreases with height. Liquids are essentially incompressible, so \rho is constant and

with p_0 the pressure at the free surface. If p_0 is atmospheric and the instrument reads only the increment above it, it reads the gauge pressure \rho gh. Using Mg/m³ for \rho, m/s² for g and m for h, \rho gh comes out directly in kPa. Atmospheric pressure at sea level may be taken as 101.3\,\mathrm{kPa}. (Meriam's US example, 10 ft of fresh water at 62.4 lb/ft³, is 4.33\,\mathrm{lb/in^2}, that is 29.9\,\mathrm{kPa}, the same as \rho g(3.048\,\mathrm m).)

The equation (physical law: hydrostatic equilibrium; Eq. 5/24 adds the approximation of constant density):

Worked number. At the bottom of the puzzle's aquarium, h = 0.6\,\mathrm m: p = 1000(9.81)(0.6) = 5886\,\mathrm{Pa} = 5.89\,\mathrm{kPa} gauge; at the surface, zero.

Real example (one you can see). Your ears hurt at the bottom of a 3\,\mathrm m pool: the pressure there is 29.4\,\mathrm{kPa} above the air pressure inside your middle ear.

Plausible wrong answer. "A tall thin tank has a higher bottom pressure than a short wide one holding the same water." Pressure depends only on depth, not on the amount of water or the tank's shape.

Check. On an open tank's wall, why can atmospheric pressure be ignored? (It acts on both faces of the wall, inside through the water and outside directly, and cancels.)

Narration spine. Arrows on a dam wall grow linearly from nothing at the surface; a depth gauge descends beside them while a pressure readout climbs to 98.1\,\mathrm{kPa} at 10\,\mathrm m.

Bridge. A linearly growing pressure on a flat plate is a distributed load like S20's. Its resultant is an area and a centroid.

Nugget 03: On a Rectangular Plate, the Force Is the Pressure Diagram Times the Width

Claim. On a rectangular plate with a horizontal top edge, every vertical section sees the same trapezoidal pressure diagram. The resultant is that diagram's area times the plate's width, equivalently the pressure at the plate's centroid times its area, and it acts through the centroid of the pressure diagram, not of the plate.

The picture. Meriam's Sample Problem 5/19: a gate hinged at its top edge A.

Check with numbers (SP 5/19). Plate 4\,\mathrm m high, 6\,\mathrm m wide, water 3\,\mathrm m deep. The pressure diagram is a triangle from zero to \rho g(3); average pressure p_{av} = 1.000(9.81)(3/2) = 14.72\,\mathrm{kPa}; wetted area 3(6) = 18\,\mathrm{m^2}:

Moments about the hinge, R acting 3\,\mathrm m below A and B at 4\,\mathrm m: 3(265) - 4B = 0, B = 198.7\,\mathrm{kN}. (The plate's weight is vertical through A's line and drops out.)

The puzzle. The aquarium's glass: R = \frac12\rho gh^2b = \frac12(1000)(9.81)(0.6)^2(1.2) = 2119\,\mathrm N, the weight of 216\,\mathrm{kg}. It acts 0.6/3 = 0.2\,\mathrm m above the bottom, not at mid-depth: the frame's bottom rail carries most of it.

The idea. Meriam (5/9): on a plate 1–2–3–4 with its top edge horizontal, inclined at any angle, the gauge pressure varies linearly from edge 2–3 to edge 1–4, the same at every vertical section, so the problem is two-dimensional (Fig. 5/32b). With b the plate's width, a strip dA = b\,dy carries dR = p\,dA = bp\,dy, and p\,dy is a strip of the trapezoidal pressure area A':

where p_{av} = \frac12(p_1 + p_2) is the pressure at the depth of the plate's centroid O. Do not confuse the physical area A of the plate with the area A' of the pressure diagram. By moments about the top, RY = \int y(pb\,dy), so Y = \int y\,dA'/\int dA': R passes through the centroid C of the trapezoid (and of the prism of pressure in 3D). The easy way: split the trapezoid into a rectangle, R_2 = p_2A at the plate's centre, and a triangle, R_1 = \frac12(p_1 - p_2)A at its centroid (Fig. 5/32c). For open systems atmospheric pressure acts on all surfaces and gives zero resultant, so only gauge pressure is used.

Sample Problem 5/20, in SI. A closed tank's end is 0.635\,\mathrm m (25 in) wide and 0.965\,\mathrm m (38 in) deep: 0.203\,\mathrm m (8 in) of air at 5.52\,\mathrm{kPa} (0.80 lb/in²) gauge above 0.762\,\mathrm m (30 in) of water. The air pressure acts on the whole end, a rectangle; the water adds a triangle:

R = 5189\,\mathrm N (1166 lb). R_1 acts at half the full depth, 0.483\,\mathrm m below the top; R_2 at two thirds of the water depth, 0.203 + 0.508 = 0.711\,\mathrm m. Moments about the top: 5189h = 3381(0.483) + 1808(0.711), h = 0.562\,\mathrm m (22.1 in). Meriam's hint: splitting the pressure diagram is decidedly the simplest way.

The equation (mathematical identity: S20's resultant applied to the pressure diagram):

Worked number. For a 2\,\mathrm m deep pool wall per metre of length: R = \frac12(9810)(2)^2 = 19.6\,\mathrm{kN}, 0.67\,\mathrm m above the floor.

Real example (one you can see). Aquarium and pool-wall builders make the glass or panels thicker toward the bottom, and the strongest frame member is the bottom rail: the pressure, and so the load, is concentrated low.

Plausible wrong answer. "The water's force acts at mid-depth." The pressure grows with depth, so its resultant acts at two thirds of the depth, a third above the bottom.

Check. Double the aquarium's depth. By what factor does the force on the glass grow? (Four: R \propto h^2.)

Narration spine. The gate in its channel; pressure arrows grow down its face into a triangle, which shrinks to one arrow at a third of the depth from the bottom. The gate swings on its hinge until the ridge stops it, and the ridge's arrow grows to 198.7\,\mathrm{kN}.

Bridge. A curved surface's pressure arrows change direction from point to point. There is a cleaner way than adding them up.

Nugget 04: On a Curved Surface, Balance the Block of Liquid Above It

Claim. The resultant of pressure on a curved surface is found most easily from the equilibrium of the block of liquid between the surface and the free surface: the horizontal pressure on its vertical side, its weight, and the surface's reaction.

The picture. Meriam's Sample Problem 5/21: a quarter-cylinder dam face of radius 4\,\mathrm m and length 30\,\mathrm m.

The water block BDO above the curved face is held by the water beside it (horizontal P_x on OB), its own weight W through its centroid, and the dam's push. The water's force on the dam is that push reversed.

Check with numbers. P_x = \rho g\bar hA = 1.000(9.81)(4/2)(30)(4) = 2354\,\mathrm{kN}, acting r/3 above B; W = \rho gV = 1.000(9.81)(\pi 4^2/4)(30) = 3698\,\mathrm{kN} through the quarter circle's centroid, 4r/3\pi from OB. Equilibrium of the block: R_x = 2350\,\mathrm{kN}, R_y = 3700\,\mathrm{kN} (Meriam's rounding), and

Moments about B place it: 2350(4/3) + 3700(16/3\pi) - 3700x = 0, x = 2.55\,\mathrm m along the horizontal through B.

The idea. Meriam (5/9): for a submerged cylindrical surface whose elements are parallel to the free surface, every vertical section is the same (Fig. 5/33), and a direct integration needs both components, R_x = b\int p\,dy and R_y = b\int p\,dx, plus a moment equation to locate R. The second method is usually much simpler: the block of liquid ABC above the surface is in equilibrium under the resultants P_y and P_x of the pressures on its flat faces, its weight W (area ABC times b times \rho g, through the area's centroid), and the reaction of the surface, which is -\mathbf R. SP 5/21's alternative solution integrates: with p = \rho gr\sin\theta and dA = b(r\,d\theta), R_x = \rho gr^2b\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta = \frac12\rho gr^2b and R_y = \rho gr^2b\int_0^{\pi/2}\sin^2\theta\,d\theta = \frac\pi4\rho gr^2b, so R = \frac12\rho gr^2b\sqrt{1 + \pi^2/4} = 4380\,\mathrm{kN}. Every dR is normal to the circle and so passes through O; hence R does too, and similar triangles give x = 2r/\pi = 2.55\,\mathrm m. Integration is feasible here mainly because the arc is simple.

The equation (physical law: equilibrium of the liquid block):

Worked number. The same dam with r = 3\,\mathrm m: P_x = \frac12(9.81)(9)(30) = 1324\,\mathrm{kN}, W = 9.81(\pi(9)/4)(30) = 2080\,\mathrm{kN}, R = 2466\,\mathrm{kN}, crossing 2r/\pi = 1.91\,\mathrm m from B.

Real example. Radial (Tainter) gates on river weirs are curved with their pivot at the circle's centre: since every pressure force passes through the centre, the water exerts no moment about the pivot, and a small hoist can raise a gate holding back thousands of tonnes.

Plausible wrong answer. "The vertical force on the curved face is zero; water only pushes sideways." The face carries the weight of all the water above it.

Check. Why does the water's force on a circular-arc gate pass through the arc's centre? (Each pressure element is normal to the arc, so along a radius.)

Narration spine. The curved dam; the water above it lifts out as a block outlined in dashes. Its weight arrow and the side push appear and combine into a slanting arrow; the arrow reverses and lands on the dam, its line passing through O.

Bridge. For a flat plate of any outline, the force is simple; where it acts is not.

Nugget 05: On a Plate of Any Shape, Average Pressure Times Area, Acting Below the Centroid

Claim. On a flat plate of any shape, the resultant is the pressure at the plate's centroid times its area, R = \rho g\bar hA. It acts at the centre of pressure, the centroid of the pressure volume, which lies deeper than the plate's centroid.

The picture. Meriam's Sample Problem 5/22: the semicircular end of a full tank.

Check with numbers. Solution I: a horizontal strip at depth y has width 2x = 2\sqrt{r^2 - y^2}: R = \int\rho gy(2x\,dy) = 2\rho g\int_0^r y\sqrt{r^2 - y^2}\,dy = \frac23\rho gr^3. Moments about the surface: \frac23\rho gr^3Y = 2\rho g\int_0^r y^2\sqrt{r^2 - y^2}\,dy = \rho gr^4\pi/8, so Y = 3\pi r/16 = 0.589r. Solution II by Eq. 5/25: \bar h = 4r/3\pi (S18), R = \rho g(4r/3\pi)(\pi r^2/2) = \frac23\rho gr^3. The centroid is at 0.424r; the resultant acts at 0.589r. For r = 0.5\,\mathrm m: R = \frac23(9810)(0.125) = 817.5\,\mathrm N at 0.295\,\mathrm m depth, against the centroid's 0.212\,\mathrm m.

The idea. Meriam (5/9): on a strip parallel to the surface the depth is constant, so dR = \rho gh\,dA, and R = \rho g\int h\,dA = \rho g\bar hA (5/25): \rho g\bar h is the pressure at the area's centroid O and the average pressure. Geometrically R is the volume of the truncated cylinder built on the plate with the pressure as height (Fig. 5/34b); where \bar h is not obvious, integrate R = \int\rho ghx\,dy with h and the strip length x in terms of y. The line of action follows from moments about the surface line:

the centroid of the pressure volume. Its point on the plate is the centre of pressure P, which is not the plate's centroid O. Meriam's hint to SP 5/22: be very careful not to assume R passes through the centroid of the area the pressure acts on.

The equation (mathematical identity: integration of p\,dA over the plate, with the centroid's definition):

Worked number. A circular inspection hatch of diameter 0.6\,\mathrm m in a vertical tank wall, its centre 2\,\mathrm m below the surface: R = 9810(2)(\pi 0.3^2) = 5547\,\mathrm N, whatever its shape's details, because only \bar h and A enter.

Real example. Lock gates and the valves in dam outlets are designed for \rho g\bar hA, with their hinges or bolts placed for the centre of pressure, which is why the lower hinge of a lock gate is the heavier one.

Plausible wrong answer. "The resultant acts at the plate's centroid, where the average pressure is." The average pressure gives the size, not the place: more of the force comes from the deeper part, so it acts lower.

Check. A plate deep below the surface: does its centre of pressure approach its centroid? (Yes: when the depth variation across it is small against \bar h, the pressure is nearly uniform, and uniform pressure acts at the centroid.)

Narration spine. The tank's semicircular end; strips light up from the surface down, each with a pressure bar growing with depth. Two dots appear: blue at the centroid, orange at the centre of pressure, lower, and the resultant arrow pins to the orange one.

Bridge. Wrap the pressure all the way round a body and it no longer pushes it sideways: it lifts it.

Nugget 06: Buoyancy Equals the Weight of the Fluid Displaced

Claim. The resultant of the pressure on an immersed body equals the weight of the fluid it displaces and acts upward through the centroid of the displaced volume, the centre of buoyancy. A floating body is stable when the metacentre lies above its centre of mass.

The picture. Meriam's Fig. 5/36: a hull upright and listed.

Upright, the buoyancy F through B and the weight W through G are on one line. Listed, the displaced volume changes shape, B moves to B', and the vertical through B' meets the hull's centreline at the metacentre M.

Check with numbers (Meriam's Sample Problem 5/23). A pole buoy 8\,\mathrm m long, 0.2\,\mathrm m in diameter, 200\,\mathrm{kg}, anchored by a 5\,\mathrm m cable in 10\,\mathrm m of water. Let x be the distance from G (mid-length) to the waterline, so 4 + x metres are submerged; buoyancy B = 10^3(9.81)\pi(0.1)^2(4 + x)\,\mathrm N acts at mid-submerged length. Moments about the anchored end A (both arms carry \cos\theta, which cancels):

The submerged 7.14\,\mathrm m spans the 5\,\mathrm m of water above the anchor point: \theta = \sin^{-1}(5/7.14) = 44.5^\circ. The buoyancy is 2200\,\mathrm N, the weight 1962\,\mathrm N, and the cable pulls down with the difference, 238\,\mathrm N.

The idea. Meriam (5/9), after Archimedes: imagine a portion of fluid bounded by a closed surface. Removing it and replacing it by the forces it exerted on the boundary leaves the surrounding fluid undisturbed; and its own free-body diagram shows that the resultant of the pressure on its surface equals its weight mg, upward, through its centre of mass. A body occupying the same space receives the same surface forces:

through the centre of mass of the displaced fluid, the centroid of the displaced volume for a liquid of constant density. A body less dense than the fluid rises; in a liquid it comes to rest at the surface (the air above adds pressure that balances its effect on the part above water). Stability: in a ship's cross-section, B is the centre of buoyancy, F passes through it and balances W. If the ship lists through an angle, the displaced shape changes and B shifts to B'. The vertical through B' meets the centreline at the metacentre M; the distance \bar h from G to M is the metacentric height, practically constant for most hulls up to about 20^\circ of list. With M above G the couple of F and W rights the ship; with M below G it increases the list, an instability every design must avoid.

The equation (physical law: Archimedes' principle, from the equilibrium of the displaced fluid):

Worked number. Ice (917\,\mathrm{kg/m^3}) in sea water (1025\,\mathrm{kg/m^3}) floats with 917/1025 = 0.895 of its volume below the surface: nine tenths of an iceberg is hidden.

Real example (one you can see). A loaded cargo ship sits visibly deeper than an empty one: its draught marks (the Plimsoll line) measure the displaced volume, which must weigh as much as ship and cargo together.

Plausible wrong answer. "Buoyancy acts through the body's centre of mass." It acts through the centroid of the displaced volume, which moves as the body tilts; that offset is what rights, or capsizes, a ship.

Check. A floating body's G is raised by stacking cargo high on deck. What happens to the metacentric height, and why does it matter? (It shrinks; if M falls below G the ship becomes unstable.)

Narration spine. A blob of water inside the sea outlined in dashes; arrows on its surface sum to an upward arrow equal to its weight. A stone of the same shape replaces it with the same arrows. Then a hull rocks: B slides sideways, a vertical from it meets the centreline at M above G, and a curved arrow rights the hull.

Bridge. Harper keeps the densities as parameters for a pole like SP 5/23's.

Nugget 07: A Floating Pole's Angle, and When the Formula Stops Applying

Claim. For a uniform pole of density \rho', length L, anchored at the bottom of liquid of density \rho and depth h, moments about the anchor give \sin\theta = (h/L)\sqrt{\rho/\rho'}. When that exceeds 1, the formula has no real solution: physically, the pole stands vertical.

The picture. Harper's Problem 5/170, plotted, beside Meriam's buoy.

Check with numbers. Submerged length h/\sin\theta, buoyancy B = \rho gA(h/\sin\theta) at half that length; weight W = \rho'gLA at L/2. Moments about C, both arms times \cos\theta:

with \psi = \rho'/\rho and \eta = h/L. For \psi = 0.5, \eta = 0.5: \sin\theta = 0.707, \theta = 45^\circ. For \psi = 0.2, \eta = 0.5: \sin\theta = 1.118 > 1, impossible: the pole floats vertical.

The idea. Harper (5.5) plots \theta against \psi for \eta = 0.25, 0.5, 0.75. MATLAB warns that it is ignoring imaginary parts: for small \psi, \sin^{-1} of a number above 1 is complex. Setting \sin\theta = 1 gives the transition \psi = \eta^2 (0.0625, 0.25, 0.5625): below it the moment equation cannot be satisfied by any tilt, the buoyancy always wins, and the pole stands vertical (held by the anchor, which then also supplies a vertical pull). The physically correct curves are horizontal at 90^\circ back to \psi = 0, which is what plotting only the real part happened to show, a coincidence Harper warns not to rely on.

The equation (physical law: moment equilibrium about the anchor, valid while the pole is tilted):

Worked number. Meriam's buoy fits the formula when its length below the anchor point is counted: its density is 200/(\pi(0.1)^2(8)) = 796\,\mathrm{kg/m^3}, \psi = 0.796, and with 5\,\mathrm m of water above its end, \eta = 5/8: \sin\theta = 0.625/\sqrt{0.796} = 0.700, \theta = 44.5^\circ, as found directly.

Real example. Spar buoys used to mark channels and moor instruments are long anchored poles; designers choose their density so they stay nearly upright in the water depth expected.

Plausible wrong answer. "MATLAB's real part is the answer." Here it happens to coincide with the physics; in general a complex result signals that the model's assumption, a tilted pole, has failed.

Check. A pole with \psi = 0.8 in water of depth h = 0.75L: what angle? (\sin\theta = 0.75/0.894 = 0.839, \theta = 57.0^\circ.)

Narration spine. A pole tethered at the tank bottom tilts as a density slider moves; lighter poles stand steeper. At \psi = \eta^2 the pole reaches vertical and stays there as the slider keeps going, while the plotted curve runs flat along 90^\circ.

Bridge. Part 5's distributed forces all came down to areas and their first moments. The second moments of area come next.

Synthesis

The aquarium's glass carries 2.12\,\mathrm{kN}, the weight of 216\,\mathrm{kg}, acting just 0.2\,\mathrm m above the bottom, not at mid-depth: a third of the way up the pressure triangle. Every result of this lesson was found the same way as in S20. The intensity is the pressure, set by depth alone, p = \rho gh; the resultant is the volume of the pressure figure over the surface, \rho g\bar hA; and its line passes through that figure's centroid, below the plate's own centroid. For a curved surface, the block of fluid above it carries the bookkeeping; for a body wholly surrounded, the same block argument gives Archimedes' buoyancy, and the shift of its line under a list decides whether a ship rights itself. A fluid at rest presses normally with p = \rho gh; its resultant on a surface is \rho g\bar hA through the centroid of the pressure figure, and on an immersed body the weight of the fluid displaced.

• Central idea: hydrostatic pressure is a distributed load whose intensity grows linearly with depth.
• Key relationship: dp = \rho g\,dh; R = \rho g\bar hA through the centre of pressure; F = \rho gV.
• Most important assumption: fluid at rest (no shear), constant density for liquids, atmospheric pressure cancelling on open systems.
• Open question: the centre of pressure needed \int y^2\,dA-type integrals. Such second moments of area also govern beam bending. (S24.)

Computed views

The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Harper's programs; the Python is a translation, and both print the same numbers.

Python (numpy, scipy). S23_fluid_statics.py integrates p = \rho gh over the aquarium glass, the pool wall and the hatch, and over the quarter-circle dam in its two components. It solves Meriam's pole buoy with brentq and tabulates Harper's angle against the density ratio. The tilt is capped at 90^\circ where his MATLAB returned complex numbers.

Part 1: walls and hatches
  aquarium glass 0.6 m x 1.2 m: R = 2119 N (216 kg), 0.200 m up
  bottom pressure 5886 Pa
  pool wall 2 m deep: 19.6 kN per metre, 0.67 m above the floor
  0.6 m hatch, centre 2 m deep: R = 5547 N (rho g h A = 5547)
Part 2: quarter-circle dam, r = 3 m, 30 m long (kN)
  P_x = 1324, P_y = 2080 (water above: 2080), R = 2466 kN, line 1.91 m from B
Part 3: buoyancy
  ice in sea water: 0.895 of it below the surface
  SP 5/23 pole: 7.14 m submerged, tilt 44.5 deg, buoyancy 2199 N, cable 237 N
  by Harper's formula: psi = 0.796, theta = 44.5 deg
  Harper's theta (deg) against psi:
    psi     eta 0.25  eta 0.50  eta 0.75
    0.05      90.0      90.0      90.0
    0.25      30.0      90.0      90.0
    0.50      20.7      45.0      90.0
    0.80      16.2      34.0      57.0
    1.00      14.5      30.0      48.6

The buoy's unrounded buoyancy is 2199\,\mathrm N and its cable 237\,\mathrm N; the lesson rounds them to 2200 and 238. The vertical component of the dam's force equals the weight of the water above the gate, as Archimedes' argument says it must.

MATLAB (Octave-compatible). S23_fluid_statics.m does the same with integral and fzero. It caps \sin\theta at 1 with min, which is the physics Harper's real-part plot happened to show.

aquarium: R = 2119 N, 0.200 m up
dam: P_x = 1324, P_y = 2080, R = 2466 kN
SP 5/23: 7.14 m submerged, tilt 44.5 deg
   psi  eta 0.25  eta 0.50  eta 0.75
  0.05      90.0      90.0      90.0
  0.25      30.0      90.0      90.0
  0.50      20.7      45.0      90.0
  0.80      16.2      34.0      57.0
  1.00      14.5      30.0      48.6

Traps

• Letting a fluid push along a surface. At rest it pushes only normally. (From Nugget 01.)
• Making pressure depend on the amount of liquid. It depends only on depth. (From Nugget 02.)
• Including atmospheric pressure on an open tank wall. It acts on both sides and cancels. (From Nugget 02.)
• Confusing the plate's area with the pressure diagram's area. R = bA' uses the diagram's. (From Nugget 03.)
• Placing the resultant at mid-depth. It acts at the pressure diagram's centroid. (From Nugget 03.)
• Forgetting the weight of liquid above a curved surface. It is the vertical component. (From Nugget 04.)
• Putting the resultant at the plate's centroid. The centre of pressure is deeper. (From Nugget 05.)
• Placing buoyancy at the body's centre of mass. It acts at the displaced volume's centroid. (From Nugget 06.)
• Trusting a formula past its validity. \sin\theta > 1 means the pole is vertical. (From Nugget 07.)

Self-check

1. Find the gauge and absolute pressure 50\,\mathrm m down in sea water (1030\,\mathrm{kg/m^3}).
2. Show from the prism of Fig. 5/30 why the prism's weight can be ignored.
3. SP 5/19's gate with the water only 2.5\,\mathrm m deep: find R and B.
4. SP 5/20 in SI with the air space vented to the atmosphere: find the resultant on the tank end and its depth below the top.
5. SP 5/21's dam with r = 3\,\mathrm m: find P_x, W, R and where R crosses the horizontal through B.
6. SP 5/22 with r = 0.5\,\mathrm m: find R and the depth of the centre of pressure; compare with the centroid's depth.
7. The puzzle's aquarium filled to 0.8\,\mathrm m: find the force on the glass and the factor by which it grew.
8. SP 5/23's buoy anchored with a 6\,\mathrm m cable instead: find \theta.
9. Harper's pole: find \theta for \psi = 0.5, \eta = 0.5, and for \psi = 0.2, \eta = 0.5.
10. What fraction of an iceberg floats below sea level (\rho_{\text{ice}} = 917, \rho_{\text{sea}} = 1025\,\mathrm{kg/m^3})?
11. A ship's cargo is moved from the hold to the deck. Explain what happens to its stability.
12. Why is gauge pressure enough for the force on SP 5/19's gate?

Fully worked solutions

1. p = \rho gh = 1030(9.81)(50) = 505.2\,\mathrm{kPa} gauge; absolute 505.2 + 101.3 = 606.5\,\mathrm{kPa}.

2. The pressure forces scale with face areas, dy\,dz (second order); the weight with the volume, \frac12dx\,dy\,dz (third order). As the prism shrinks, the weight vanishes faster than the pressure forces, so it cannot affect p_1 = p_2 = p_3 in the limit.

3. R = \frac12\rho gh^2b = \frac12(9.81)(2.5)^2(6) = 183.9\,\mathrm{kN}, acting 2.5/3 = 0.833\,\mathrm m above the bottom, so 4 - 0.833 = 3.167\,\mathrm m below A. \Sigma M_A: 183.9(3.167) = 4B, B = 145.6\,\mathrm{kN}.

4. Only the water's triangle: R = R_2 = 1808\,\mathrm N, acting at two thirds of the water depth below the water surface, 0.203 + \frac23(0.762) = 0.711\,\mathrm m below the top.

5. P_x = \frac12\rho gr^2b = \frac12(9.81)(9)(30) = 1324\,\mathrm{kN}; W = \rho g(\pi r^2/4)b = 9.81(7.069)(30) = 2080\,\mathrm{kN}; R = \sqrt{1324^2 + 2080^2} = 2466\,\mathrm{kN}, through O, crossing the horizontal through B at 2r/\pi = 1.91\,\mathrm m.

6. R = \frac23\rho gr^3 = \frac23(9810)(0.125) = 817.5\,\mathrm N; Y = 3\pi r/16 = 0.295\,\mathrm m; centroid 4r/3\pi = 0.212\,\mathrm m, 82\,\mathrm{mm} higher.

7. R = \frac12(9810)(0.8)^2(1.2) = 3767\,\mathrm N, acting 0.267\,\mathrm m above the bottom; it grew by (0.8/0.6)^2 = 1.78.

8. The moment equation does not involve the cable, so 4 + x = 7.14\,\mathrm m still; now 4\,\mathrm m of water lies above the pole's end: \theta = \sin^{-1}(4/7.14) = 34.1^\circ.

9. \sin\theta = \eta/\sqrt\psi = 0.5/0.707 = 0.707, \theta = 45^\circ. Second: 0.5/0.447 = 1.118 > 1; \psi = 0.2 < \eta^2 = 0.25, so the pole stands vertical, \theta = 90^\circ.

10. Floating: \rho_{\text{sea}}gV_{\text{sub}} = \rho_{\text{ice}}gV, so V_{\text{sub}}/V = 917/1025 = 0.895: about 89.5\,\%.

11. Raising the cargo raises G; the metacentre, fixed by the hull's shape and draught, does not move up with it, so the metacentric height GM shrinks. If M falls below G, a list produces a moment that increases it: the ship is unstable.

12. The water's surface and the gate's dry side are both at atmospheric pressure, which acts on both faces of the gate and cancels; only the excess \rho gh produces a net force.