Week 9 In-Depth — Why Pressure Becomes a Single Force

Read the cheatsheet first for the recipes. This note explains why each formula is the shape it is, with one substantial worked example per topic.

1 — Why the resultant force uses the centroid pressure

Pressure on a submerged plane varies linearly with depth: $p(y)=\rho gy$. The total force is the integral of pressure over the surface:

$F_R={\int }_{A}pdA={\int }_A\rho gydA=\rho g{\int }_{A}ydA.$

The integral ${\int }_{A}ydA$ is, by definition, the first moment of area of the surface about the free-surface axis, and that equals $y_C\cdot A$ (the centroid depth times the area). So

$F_R=\rho g\cdot (y_{C}A)=\rho g{y}_{C}A.$

That’s the entire derivation. The reason the centroid depth appears — and not, say, the depth of the top or bottom edge — is that the centroid is, by construction, the depth that averages $y$ over the surface (weighted by area). Pressure scales linearly with depth, so its average across the surface is exactly the pressure at the average depth.

Why the resultant is perpendicular to the surface

Fluid pressure is isotropic — it pushes equally in every direction. But at any solid boundary, only the component normal to the surface does net work. The tangential components from neighbouring fluid parcels cancel out along the wetted face. So the resultant is normal to the plate, regardless of plate orientation.

Worked example — the rectangular gate, all steps

A tank gate $2.5$ m wide and $1.5$ m tall, hinged at the bottom, is held closed by a horizontal force $F$ applied $1.5$ m above the hinge. Water surface is $3$ m above the top of the gate. Find $F_R$ and $F$.

Step 1 — geometric properties. The rectangle’s own centroid sits at $h/2=0.75$ m down from its top edge. The top edge is $3$ m below the free surface, so

$y_C=3+0.75=3.75\text{ m}.$

Area:

$A=b\times h=2.5\times 1.5=3.75{\text{ m}}^2.$

Step 2 — resultant force.

$F_R=\rho g{y}_{C}A=1000\times 9.81\times 3.75\times 3.75\approx 137950\text{ N}\approx 138\text{ kN}.$

Step 3 — centre of pressure. We need $I_C$ for the rectangular gate:

$I_C=\frac{b{h}^3}{12}=\frac{2.5\times 1.5^3}{12}=0.70{\text{ m}}^4.$

Now apply the centre-of-pressure formula (the why of which is the next section):

$y_R=y_C+\frac{I_C}{y_{C}A}=3.75+\frac{0.70}{3.75\times 3.75}\approx 3.80\text{ m}.$

So the line of action of $F_R$ is at depth $3.80$ m — that’s $0.05$ m below the centroid, which makes sense (small correction for a not-very-deep plate of moderate size).

Step 4 — moment arm of $F_R$. The bottom of the gate is at depth $3+1.5=4.5$ m. The line of action is at $3.80$ m. So $F_R$ acts at

$d=4.5-3.80=0.70\text{ m above the hinge.}$

Step 5 — sum moments about the hinge. $F$ acts horizontally at $1.5$ m above the hinge, opposing $F_R$. Taking counter-clockwise positive:

$\sum M_H=0⟹F\cdot 1.5=F_R\cdot 0.70⟹F=\frac{137950\times 0.70}{1.5}\approx 64.4\text{ kN}.$

The geometry to internalise: $F_R$ acts at $y_R$ below the free surface, which is above the centroid by less than half the gate height — so for a gate hinged at the bottom and supported near the top, the moment arm of $F_R$ is roughly gate height / 3. That’s why answers tend to come out in the tens of kN for residential-scale gates.

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2 — Why $y_R>y_C$ (the pressure prism argument)

Visualise the pressure on a submerged vertical plate as a 3-D prism sticking out from the plate, with height equal to the local pressure at each point. Because pressure grows linearly with depth, the prism is wedge-shaped: thin on top, fat on the bottom.

The total force $F_R$ is the volume of this prism. Its line of action passes through the prism’s centroid.

Now: the centroid of a wedge-shaped prism (more volume on the bottom) sits lower than the centroid of its footprint (the plate). That’s where $y_R>y_C$ comes from. It’s purely a geometric consequence of the linear pressure law.

Derivation of $y_R=y_C+I_C/(y_{C}A)$

Equating the moment of the resultant about the free-surface axis to the moment of the distributed pressure:

$F_R\cdot y_R={\int }_{A}p\cdot ydA={\int }_A\rho gy\cdot ydA=\rho g{\int }_A{y}^{2}dA.$

The integral ${\int }_A{y}^{2}dA$ is the second moment of area about the free-surface axis (call it $I_O$). We can shift this to the centroidal axis using the parallel-axis theorem:

$I_O=I_C+y_C^{2}A.$

Substitute and divide both sides by $F_R=\rho g{y}_{C}A$:

$y_R=\frac{\rho g(I_C+y_C^{2}A)}{\rho g{y}_{C}A}=\frac{I_C}{y_{C}A}+y_C=y_C+\frac{I_C}{y_{C}A}.$

So:

• The first term is where the centroid is.
• The correction $I_C/(y_{C}A)$ is always positive, hence $y_R>y_C$.
• At great depth $y_C\gg \sqrt{I_C/A}$, the correction vanishes — for a deeply submerged small plate, the prism is nearly a uniform slab and its centroid coincides with the plate’s centroid. (Think: at $1$ km depth, the difference in pressure across a $0.5$ m plate is negligible compared to the average pressure.)

Worked example — submerged right triangle (Tutorial Ex 4)

Right-triangular plate, vertical leg $d=2.0$ m and horizontal leg $b=1.0$ m, apex at the top, base at the bottom. Top vertex at depth $a=0.2$ m.

Centroid depth. For a triangle with apex at the top, $\bar{y}=d/3$ measured down from the apex:

$y_C=a+\frac{d}{3}=0.2+\frac{2}{3}\approx 0.866\text{ m}.$

Area.

$A=\frac{bd}{2}=\frac{1\times 2}{2}=1{\text{ m}}^2.$

Resultant.

$F_P=\rho g{y}_{C}A=1000\times 9.8\times 0.866\times 1\approx 8487\text{ N}.$

Second moment of area.

$I_C=\frac{b{d}^3}{36}=\frac{1\times 8}{36}=0.222{\text{ m}}^4.$

Note the 36, not 12 — this is the most reliable place to lose marks on a triangle problem.

Centre of pressure.

$y_R=y_C+\frac{I_C}{y_{C}A}=0.866+\frac{0.222}{0.866\times 1}=1.123\text{ m}.$

The offset $y_R-y_C=0.257$ m is 30% of $y_C$ here — much larger than the rectangular-gate example earlier — because the plate is relatively close to the surface compared to its size. That tracks with the formula: small $y_C$, modest $I_C/A$, so the ratio $I_C/(y_{C}A)$ matters.

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3 — Why the vertical-dam shortcut works

When a plate runs from the free surface (depth $0$) to depth $h$, the wedge of pressure is a perfect triangular prism: pressure goes from $0$ on top to $\rho gh$ on the bottom. The geometry simplifies dramatically.

Quantity	Value	Reason
$y_C$	$h/2$	Centroid of a rectangle is at its mid-height
$A$	$h\times L$	Rectangle
$F_R$	$\rho g(h/2)(hL)$	$\rho g{y}_{C}A$
Line of action	$h/3$ above base	Centroid of a triangular pressure distribution sits $1/3$ of the way from the wide end

You can re-derive the $h/3$ from the centre-of-pressure formula:

$y_R=\frac{h}{2}+\frac{L{h}^3/12}{(h/2)(hL)}=\frac{h}{2}+\frac{h^2/12}{h/2}=\frac{h}{2}+\frac{h}{6}=\frac{2h}{3}.$

So $y_R=2h/3$ below the free surface, i.e. $h-2h/3=h/3$ above the base. The formula and the geometric intuition agree — always a good sign.

Worked example — sea-water lock (Tutorial Ex 3)

Lock wall $20$ m wide retaining sea-water ($SG=1.03$) to depth $10$ m.

Specific gravity. ${\rho }_{\text{sea}}=1.03\times 1000=1030$ kg/m³.

Centroid + area. $y_C=h/2=5$ m; $A=hL=10\times 20=200$ m².

Resultant.

$F_P=\rho g{y}_{C}A=1030\times 9.8\times 5\times 200\approx 1.01\times 10^7\text{ N}=10.1\text{ MN}.$

$I_C$ + centre of pressure.

$I_C=\frac{b{h}^3}{12}=\frac{20\times 1000}{12}\approx 1667{\text{ m}}^4.$

$y_p=5+\frac{1667}{5\times 200}=5+1.67=6.67\text{ m}.$

Sanity check. $10-6.67=3.33$ m $=h/3$ above the base. The dam shortcut and the formula agree.

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4 — Why the manometer walk works

Hydrostatic pressure obeys $dp/dz=\rho g$ for a vertical column of constant-density fluid. Integrate along a vertical path of height $h$:

$\Delta p=p(\text{bottom})-p(\text{top})=\rho gh.$

So down adds $\rho gh$, up subtracts $\rho gh$. Horizontal moves (at constant depth) in the same continuous fluid add zero.

When you walk from $P_1$ to $P_2$ through a manometer, you can break the path into vertical legs (each contributing $\pm \rho gh$ depending on direction and which fluid you’re in) and horizontal legs (zero contribution if the fluid is continuous):

$P_1+{\sum }_{\text{down}}\rho gh-{\sum }_{\text{up}}\rho gh=P_2.$

Crucially: the $\rho$ in each leg is the density of the fluid in that leg, not some fixed reference. That’s why heavier fluids like mercury contribute much more per unit height than water or air.

Worked example — double-U manometer (Tutorial Ex 2)

Freshwater pipe (1) connected to seawater pipe (2) by a mercury manometer with an air gap. ${\rho }_w=1000$, ${\rho }_{sea}=1035$, ${\rho }_{Hg}=13600$ kg/m³. Water leg $h_w=0.6$ m (down from pipe 1), mercury $h_{Hg}=0.1$ m (up to air gap), seawater $h_{sea}=0.4$ m (down to pipe 2). Air column negligible.

Walk equation.

$P_1+{\rho }_{w}g{h}_w-{\rho }_{Hg}g{h}_{Hg}-{\rho }_{\text{air}}g{h}_{\text{air}}+{\rho }_{sea}g{h}_{sea}=P_2.$

Drop the air term (see below) and rearrange:

$P_1-P_2=g({\rho }_{Hg}{h}_{Hg}-{\rho }_w{h}_w-{\rho }_{sea}{h}_{sea}).$

Plug in:

$P_1-P_2=9.81[13600\times 0.1-1000\times 0.6-1035\times 0.4]$

$=9.81[1360-600-414]=9.81\times 346\approx 3390\text{ Pa}=3.39\text{ kPa}.$

So freshwater pipe is $3.39$ kPa above the seawater pipe.

Why air drops out

${\rho }_{\text{air}}\approx 1.2$ kg/m³ vs ${\rho }_w=1000$ kg/m³. For a comparable column height, the air contribution is ~$10^{-3}$ times the water contribution. Within the precision of typical manometer problems (3 sig figs), the air term is at the noise floor and is conventionally neglected. You should mention this explicitly in a written solution.

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5 — Absolute vs gauge: when does it matter?

Integrate $dp=\rho gdz$ for a constant-density fluid with surface pressure $p_0$ at $z=0$:

$p(z)=p_0+\rho gz.$

If $p_0=p_{\text{atm}}$ (atmosphere sits above the free surface), $p$ here is the absolute pressure. The corresponding gauge pressure is

$p_{\text{gauge}}(z)=p(z)-p_{\text{atm}}=\rho gz.$

Both are valid descriptions of the same physical state; they just use different datums.

When does the choice matter for a gate?

A gate has water on one side and (usually) air at $p_{\text{atm}}$ on the other. The net force is the difference between the two faces:

$F_{\text{net}}=F_{\text{wet}}-F_{\text{dry}}=(p_{\text{atm}}+\rho g{y}_C)A-p_{\text{atm}}A=\rho g{y}_{C}A.$

The atmospheric term cancels exactly. So in gauge terms, you simply use $\rho g{y}_{C}A$ from the start. The lecture default is gauge pressure for gate problems for this reason.

When you must use absolute:

• Dry side is sealed or evacuated (no $p_{\text{atm}}$ pushing back).
• Compressible fluids and thermodynamic calculations (absolute is intrinsic to the gas laws).

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6 — How these all connect

Three threads, one underlying idea: pressure is a force per unit area that varies linearly with depth in a constant-density fluid. Everything follows.

Thread	Linear pressure law shows up as…
Manometer walk	$\rho gh$ per vertical leg
Resultant on a plate	$\rho g{y}_{C}A$ (centroid-depth pressure × area)
Centre of pressure	$y_R=y_C+I_C/(y_{C}A)$ (parallel-axis on $\int y^{2}dA$)
Vertical-dam shortcut	$h/2$ centroid, $h/3$ line of action — both consequences of triangular pressure distribution

So Week 9 is, fundamentally, the practical exploitation of $dp/dz=\rho g$. Once you internalise that single ODE, every formula on the sheet is a rearrangement.

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7 — Exam-style sample, end-to-end

A $2$ m wide, $3$ m tall rectangular plate is submerged in water with its top edge at depth $2.2$ m below the free surface.

(a) Find $I_C$ about the centroidal horizontal axis. (b) Find the depth to the centre of pressure. (c) Find the resultant pressure force on the plate.

(a) Second moment of area.

$I_C=\frac{b{h}^3}{12}=\frac{2\times 3^3}{12}=\frac{54}{12}=4.5{\text{ m}}^4.$

(b) Centre of pressure. First the centroid and area:

$y_C=2.2+\frac{3}{2}=3.7\text{ m},A=2\times 3=6{\text{ m}}^2.$

Then:

$y_R=y_C+\frac{I_C}{y_{C}A}=3.7+\frac{4.5}{3.7\times 6}=3.7+0.20\approx 3.9\text{ m}.$

(c) Resultant force.

$F_R=\rho g{y}_{C}A=1000\times 9.81\times 3.7\times 6\approx 217560\text{ N}\approx 218\text{ kN}.$

Pressure distribution. Trapezoidal:

• Top edge (depth $2.2$ m): $p=\rho g\cdot 2.2\approx 21.6$ kPa
• Bottom edge (depth $5.2$ m): $p=\rho g\cdot 5.2\approx 51.0$ kPa

$F_R$ acts horizontally (perpendicular to the vertical plate) at depth $y_R\approx 3.9$ m below the free surface. The offset from the centroid is $0.2$ m downward — about $7\%$ of the plate height — consistent with the rule of thumb that the offset shrinks as the plate sits deeper.

That layout — geometric properties → resultant → second moment → centre of pressure → (if a gate) moment balance about hinge — is the template for every submerged-surface problem on the paper.

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Where to look next

• The cheatsheet for fast revision + the quiz.
• The lecture summary for the full reconstruction with all four tutorial exercises worked through.
• The workshop PDFs (cited in frontmatter) for more practice problems.