Week 10 In-Depth — Why Fluid Statics Reduces to Three Formulas

Read the cheatsheet first. This note explains why the three formulas ($F_R=P_{C}A$, $y_R=y_C+I_C/(y_{C}A)$, and the sliding/overturning inequalities) work, with one substantial worked example per topic.

⚠️ Confidence note. The official Week 10 lecture PDF was not in the source folder when this note was written. The framing below is reconstructed from Tutorial 10 using standard fluid-statics conventions. Numerical answers are recomputed but should be re-checked against the lecture deck when available.

1 — Why $F_R=P_C\cdot A$ works

Pressure in a static fluid grows linearly with depth:

$$P(y)=\rho gy\text{(gauge, fluid open to atmosphere)}.$$

On a vertical submerged surface, the pressure varies linearly across the surface — high at the bottom, low at the top. The resultant force is the integral of pressure over area:

$$F_R={\int }_{A}P(y)dA={\int }_A\rho gydA=\rho g{\int }_{A}ydA.$$

But by definition of the centroid, ${\int }_{A}ydA=y_C\cdot A$. So

$$F_R=\rho g{y}_{C}A=P_C\cdot A.$$

That is the magic of the formula. The centroid does the integral for you. You never have to integrate pressure across a complicated shape — just look up the centroid depth, multiply by $\rho g$ to get $P_C$, and multiply by area.

Worked example — vertical gate closing a tunnel

A vertical gate $5$ m high and $3$ m wide closes a tunnel running full of water. The pressure at the bottom of the gate is $195kN/{\mathrm{m}}^2$ gauge. Find (a) the resultant force and (b) the centre of pressure.

Set up. The gate is rectangular, so the centroid is at half-height. To find $y_C$ (relative to the free surface) we need the depth of the top of the gate.

Bottom depth. From $P_b=\rho g{y}_b$:

$$y_b=\frac{195,000}{1000\times 9.81}\approx 19.88\mathrm{m}.$$

Top depth. $y_t=y_b-5=14.88\mathrm{m}$.

Centroid depth. $y_C=(y_t+y_b)/2=17.38\mathrm{m}$.

Area. $A=5\times 3=15{\mathrm{m}}^2$.

Resultant.

$$F_R=\rho g{y}_{C}A=1000\times 9.81\times 17.38\times 15\approx 2.56\mathrm{MN}.$$

Centre of pressure. $I_C=b{d}^3/12=3\times 5^3/12=31.25{\mathrm{m}}^4$.

$$y_R=y_C+\frac{I_C}{y_{C}A}=17.38+\frac{31.25}{17.38\times 15}\approx 17.50\mathrm{m}.$$

The centre of pressure sits about $0.12$ m below the centroid — closer to the bottom of the gate, as expected for a deep submerged plate. Notice how small the offset is: when $y_C$ is large, the deep gate “feels” pressure that’s nearly uniform across its face, so the resultant acts almost at the centroid.

The numeric values above were recomputed from the tutorial diagram using ${\rho }_w=1000$ kg/m³ and $g=9.81$ m/s². If the lecture deck uses $g=10$ or a different convention, expect ~2% drift.

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2 — Why the centre of pressure sits below the centroid

The resultant $F_R$ replaces a distributed pressure with a single point force. The depth $y_R$ at which it acts must give the same moment about any chosen axis as the original pressure distribution would.

Take moments of the pressure about the free surface:

$$F_R\cdot y_R={\int }_{A}P(y)\cdot ydA=\rho g{\int }_A{y}^{2}dA=\rho g{I}_{xx},$$

where $I_{xx}=\int y^{2}dA$ is the second moment of area about the free-surface axis.

By the parallel-axis theorem,

$$I_{xx}=I_C+y_C^{2}A,$$

where $I_C$ is the second moment about the surface’s own centroidal axis. Substituting back:

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

Cancel $\rho g$ and divide by $y_{C}A$:

$$y_R=y_C+\frac{I_C}{y_{C}A}.$$

So the "$I_C/(y_{C}A)$" term is exactly the parallel-axis correction made dimensionless. The deeper the surface (larger $y_C$), the smaller that correction — which matches the physics.

The pressure-prism picture

There’s a more visual way to see this. Imagine the pressure distribution as a 3D wedge sitting on the wetted area — the wedge is zero at the free surface and grows linearly with depth. The volume of that wedge is exactly $F_R$. The point at which $F_R$ acts is the centroid of the wedge, projected back onto the surface.

For a rectangular gate with its top at the free surface, the pressure prism is a triangular wedge — its centroid sits two-thirds of the way down, so $y_R=(2/3)H$. Plug $y_C=H/2$, $I_C=b{H}^3/12$, $A=bH$ into the formula:

$$y_R=\frac{H}{2}+\frac{b{H}^3/12}{(H/2)(bH)}=\frac{H}{2}+\frac{H}{6}=\frac{2H}{3}.$$

Same answer. The formula and the prism are the same idea told two ways.

Worked example — triangular gate

A triangular gate (base $b=2.5$ m, height $h=1.5$ m) sits with its top edge $3$ m below the free surface. Find $F_P$ and the holding force $F$ at the top required to keep the gate closed (hinge at the bottom).

Geometry. The triangle’s centroid sits $h/3$ above its base, i.e. $1.0$ m below its top edge. So $y_C=3+1.0=4.0$ m.

Area. $A=bh/2=2.5\times 1.5/2=1.875{\mathrm{m}}^2$.

Resultant force.

$$F_P=\rho g{y}_{C}A=1000\times 9.81\times 4.0\times 1.875\approx 73.6\mathrm{kN}.$$

Centre of pressure. $I_C=b{h}^3/36=2.5\times 1.5^3/36\approx 0.234{\mathrm{m}}^4$.

$$y_R=4.0+\frac{0.234}{4.0\times 1.875}\approx 4.031\mathrm{m}.$$

Moment balance. Hinge is at the bottom of the gate (depth $4.5$ m); holding force $F$ is at the top (depth $3$ m). Moments about the hinge:

$$F\cdot (4.5-3.0)=F_P\cdot (4.5-y_R)⟹F=73.6\times \frac{0.469}{1.5}\approx 23\mathrm{kN}.$$

That’s the design tension in the chain/strut holding the gate shut.

The hinge/strut geometry in the tutorial figure may differ from the assumption above (hinge at bottom, strut at top). The method — take moments about the hinge, use the lever arms to $y_R$ — is what matters; re-do the arithmetic if the diagram in your copy has the hinge at a different edge.

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3 — Sliding and overturning: design checks for retaining walls

A retaining wall sees a horizontal water push $F_P$ and resists it through (a) friction at its base, and (b) its own weight creating a restoring moment about the toe.

Sliding

Friction can supply at most $\mu W$, where $W$ is the wall’s weight and $\mu$ the base friction coefficient. For sliding not to occur,

$$\mu W\ge F_P.$$

Per metre of wall length, $W={\rho }_{c}g(t\cdot H\cdot 1)$, so the minimum thickness is

$$t\ge \frac{F_P}{\mu {\rho }_{c}gH}.$$

Overturning

The wall wants to tip about its toe (front-bottom corner, on the dry side). Two moments compete:

• Restoring moment from the wall’s weight, acting through the wall’s centroid at distance $t/2$ from the toe:

$$M_{\text{restoring}}=W\cdot \frac{t}{2}.$$

• Overturning moment from the water force, acting at depth $y_R$, i.e. height $H-y_R$ above the base:

$$M_{\text{overturning}}=F_P\cdot (H-y_R).$$

Safety: $M_{\text{restoring}}>M_{\text{overturning}}$. Their ratio is the safety factor.

Worked example — concrete retaining wall

A $4$ m high concrete wall (${\rho }_c=2550kg/{\mathrm{m}}^3$) retains water $3.5$ m deep. $\mu =0.4$. (a) Find $F_P$ per metre of wall. (b) Find the minimum $t$ to prevent sliding. (c) For $t=1.5$ m, check overturning.

(a) Water force per metre. $y_C=1.75$ m, $A=3.5{\mathrm{m}}^2$:

$$F_P=1000\times 9.81\times 1.75\times 3.5\approx 60.1kN/m.$$

(b) Sliding. $W=2550\times 9.81\times (t\times 4\times 1)\approx 100,062t$ N/m. Solve $0.4W\ge F_P$:

$$t\ge \frac{60,100}{0.4\times 100,062}\approx 1.50\mathrm{m}.$$

(c) Overturning at $t=1.5$ m. $I_C=1\times 3.5^3/12\approx 3.57{\mathrm{m}}^4$:

$$y_R=1.75+\frac{3.57}{1.75\times 3.5}\approx 2.33\mathrm{m}.$$

Height above base: $H-y_R=3.5-2.33=1.17$ m. (Note: $H$ here is the water depth $3.5$ m, not the wall height $4$ m, because $y_R$ was measured from the water surface.)

$$M_{\text{overturning}}=60.1\times 1.17\approx 70.4\mathrm{kN}\cdot m/m.$$ $$W=100,062\times 1.5=150,093N/m,M_{\text{restoring}}=150,093\times 0.75\approx 112.6\mathrm{kN}\cdot m/m.$$

Safety factor $\approx 112.6/70.4\approx 1.6$. The wall passes overturning at the sliding-critical thickness — a typical and reassuring result.

Two subtleties to verify with the lecture deck: (i) whether the water depth equals the wall height (here it doesn’t — wall is $4$ m, water is $3.5$ m), and (ii) whether the lecturer measures the lever arm of $F_P$ as $(H_{\text{wall}}-y_R)$ or $(H_{\text{water}}-y_R)$. The convention here uses the water surface as the reference for $y_R$.

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4 — How the three formulas connect

Three distinct-looking results, one shared idea: integrating a linearly varying pressure over an area reduces to evaluating geometric quantities (centroid, second moment of area).

Result	What pressure-integral it solves
$F_R=\rho g{y}_{C}A$	$\int PdA$ — the total push
$y_R=y_C+I_C/(y_{C}A)$	$\int PydA/F_R$ — the line of action
Sliding + overturning	Apply $F_R$ and $y_R$ to a rigid-body equilibrium problem

If you understand the first integral — pressure varies linearly, so the integral reduces to the centroid times area — the other two are bookkeeping.

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5 — Exam-style template (Model → Visualise → Solve → Assess)

Use this structure on every fluid-statics question in the workshop and exam:

1. Model. Identify the wetted shape. Note its dimensions, fluid, free surface.
2. Visualise. Sketch the surface with the free surface above. Mark $y_C$ and (later) $y_R$.
3. Solve. a. Look up $A$ and $\bar{y}$ from the centroid table; convert to $y_C$ by adding the top-edge depth. b. Compute $F_R=\rho g{y}_{C}A$. c. Look up $I_C$ from the second-moment table. d. Compute $y_R=y_C+I_C/(y_{C}A)$. e. If a wall: compute $W$ per metre, apply sliding and overturning inequalities.
4. Assess. Sanity checks — is $y_R>y_C$? Are units right? Is the safety factor $>1$?

That’s every question on the topic.

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Where to look next

• The cheatsheet for fast revision + the quiz.
• The lecture summary for the source-faithful concept list (still confidence: inferred).
• The study guide for an explicit list of what’s directly supported, inferred, and gap-checked.