Week 10 — Forces on Submerged Bodies (Fluid Statics)

Overview

Week 10 moves into fluid statics — calculating the forces fluids exert on submerged surfaces (gates, dam walls, tank walls). The week introduces the resultant pressure force, the centre of pressure (where that force effectively acts), and how to combine these with sliding/overturning checks for civil-style problems. The reconstruction below is built from Tutorial 10; the formal Week 10 lecture deck wasn’t in the source folder, so confidence is inferred — verify against the official slides when available.

Key concepts

• Resultant pressure force $F_R$ — single equivalent force that replaces the distributed fluid pressure on a submerged surface. Units: N.
• Pressure at the centroid $P_C$ — gauge (or absolute, depending on convention) pressure of the fluid evaluated at the depth of the surface’s geometric centroid. Units: Pa = N/m². Note the tutorial slide labels its units as “Pa” but it is fundamentally a pressure, so $P_C=\rho g{y}_C$ for a fluid open to atmosphere.
• Submerged surface area $A$ — the wetted area of the gate/wall being analysed. Units: m².
• Centroid (geometric centre) $C$ — the area-weighted centre of a 2D shape. For symmetric standard shapes this sits on the axis of symmetry; the lecturer’s table covers rectangle, general triangle, isosceles triangle, right triangle, and circle.
• Centre of pressure ($y_R$) — the depth at which $F_R$ acts on the submerged surface. Always below the centroid because pressure grows with depth, biasing the resultant towards the deeper edge. Units: m.
• Pressure prism — imaginary 3D wedge representing how pressure varies linearly with depth across the wetted surface; the centre of pressure is the centroid of this prism.
• Second moment of area $I_C$ — measure of how the cross-sectional area is distributed relative to a centroidal axis; quantifies bending/rotational stiffness of a shape. Units: m⁴.
• Sliding failure — wall slips horizontally because hydrostatic push exceeds friction at the base.
• Overturning failure — wall tips about its toe because the overturning moment from $F_P$ exceeds the restoring moment from the wall’s weight.
• Coefficient of friction $\mu$ — dimensionless ratio between maximum static friction and normal force.
• Density $\rho$ — mass per unit volume; concrete in the tutorial is ${\rho }_c=2550kg/{\mathrm{m}}^3$, water ${\rho }_w=1000kg/{\mathrm{m}}^3$.

Core formulas

Resultant pressure force on a submerged surface:

$F_R=P_C\cdot A$

For a fluid open to atmosphere where gauge pressure is used:

$P_C=\rho g{y}_C$

so

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

Depth to the centre of pressure (below the free surface):

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

Standard second moments of area about the centroidal axis (from slide 6):

$I_{c,\text{rect}}=\frac{b{d}^3}{12}{I}_{c,\text{tri}}=\frac{b{h}^3}{36}{I}_{c,\text{circ}}=\frac{\pi r^4}{4}{I}_{c,\text{semicircle}}=0.1102r^4$

Centroid lookup table (slide 4) — for a rectangle: $\bar{x}=b/2,\bar{y}=h/2$, area $=bh$. For a general/right triangle: $\bar{y}=h/3$, area $=bh/2$. Circle: centroid at the centre.

Sliding-resistance (friction) inequality at the base of a wall, per unit width:

$\mu W\ge F_P\text{where }W={\rho }_{c}gtH(1\text{m})$

so the minimum thickness $t$ for sliding is

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

Overturning check (moments about the toe of the wall):

$M_{\text{restoring}}=W\cdot \frac{t}{2}{M}_{\text{overturning}}=F_P\cdot (H-y_R)$

The wall is safe from overturning when $M_{\text{restoring}}>M_{\text{overturning}}$ (a safety factor > 1).

Worked examples

Example 1 — Vertical gate closing a tunnel (slide 7)

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 $195kN/{\mathrm{m}}^2$. Find (a) the force on the gate and (b) the centre of pressure.

Setup. Because the pressure varies linearly from $P_{\text{top}}$ to $P_{\text{bottom}}=195$ kPa, the pressure at the centroid (middle of the gate) is the average. We need the depth $y_C$ to the centroid, which depends on $P_{\text{top}}$; from $P=\rho gh$ we can back out the free-surface depth, then add half the gate height.

Working. Bottom of gate at depth $y_b=P_b/(\rho g)=195,000/(1000\times 9.81)\approx 19.88$ m. Top of gate is 5 m above that, so $y_t\approx 14.88$ m. Centroid depth $y_C=(y_t+y_b)/2\approx 17.38$ m.

$F_R=\rho g{y}_{C}A=1000\times 9.81\times 17.38\times (5\times 3)\approx 2.56\mathrm{MN}$

For the rectangle: $I_C=b{d}^3/12=3\times 5^3/12=31.25{\mathrm{m}}^4$, $A=15{\mathrm{m}}^2$.

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

So the centre of pressure sits about 0.12 m below the centroid — closer to the bottom of the gate as expected.

Example 2 — Triangular gate (slides 8–9)

A triangular gate (base 2.5 m, top of the triangle at the water-surface side per the diagram) sits with its top edge 3 m below the surface and is 1.5 m tall. Find $F_P$ and the holding force $F$ required to keep the gate shut.

Setup. Treat the triangle as a standard shape. Centroid of a triangle is at $h/3$ above its base, so $y_C$ = (depth to apex) + $h-h/3$ = $3+1.0=4.0$ m below the free surface. Area $A=bh/2=(2.5)(1.5)/2=1.875{\mathrm{m}}^2$.

Working.

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

For the triangle, $I_C=b{h}^3/36=2.5\times 1.5^3/36\approx 0.234{\mathrm{m}}^4$.

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

Take moments about the hinge at the base of the gate (depth 4.5 m). The holding force $F$ is applied at the top of the gate (depth 3 m). Moment balance:

$F\cdot (4.5-3.0)=F_P\cdot (4.5-y_R)$

$F=F_P\cdot \frac{4.5-4.031}{1.5}\approx 73.6\mathrm{kN}\times 0.313\approx 23\mathrm{kN}$

(The specific moment-arm geometry depends on where the hinge is in the figure — adapt to the actual side-view drawing.)

Example 3 — Concrete retaining wall (slides 10–11)

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

(a) Per metre width, the submerged area $A=3.5\times 1=3.5{\mathrm{m}}^2$, centroid depth $y_C=3.5/2=1.75$ m.

$F_P={\rho }_{w}g{y}_{C}A=1000\times 9.81\times 1.75\times 3.5\approx 60.1kN/m$

(b) Weight per metre width: $W={\rho }_{c}g(t\times 4\times 1)=2550\times 9.81\times 4t\approx 100,062t$ N. Friction must resist sliding: $\mu W\ge F_P\Rightarrow 0.4\times 100,062t\ge 60,100$, so

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

(c) Centre of pressure for a rectangular surface from a free surface: $y_R=y_C+I_C/(y_{C}A)$ with $I_C=1\times 3.5^3/12\approx 3.57{\mathrm{m}}^4$.

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

Moment arm of $F_P$ about the toe of the wall = $3.5-y_R=1.17$ m (the resultant acts $1.17$ m above the base).

$M_{\text{overturning}}=F_P\times 1.17\approx 70.4\mathrm{kN}\cdot m/m$

$M_{\text{restoring}}=W\times t/2=100,062\times 1.5\times 0.75\approx 112.6\mathrm{kN}\cdot m/m$

Restoring > overturning by a factor of ≈ 1.6, so the 1.5 m thick wall is safe from overturning.

Things to practise

The tutorial gives three exercises (all reproduced/worked above as targets):

• Exercise 1 — vertical gate, gauge pressure given at bottom → find $F_R$ and $y_R$.
• Exercise 2 — triangular gate at known depth → find $F_P$ and the holding force at the top.
• Exercise 3 — retaining wall with sliding and overturning checks (per metre of wall width).

Workshop Portfolio 10 is completed in this tutorial (slide 12) — make sure your written solution shows the Model → Visualise → Solve → Assess structure the unit emphasises.

Common pitfalls

• Mixing up $y_C$ and $y_R$: the centroid is geometric only — it does not depend on the fluid; the centre of pressure shifts deeper because pressure grows with depth.
• Forgetting “per metre width”: for long dams/walls, both $F_P$ and $W$ should be computed per metre, otherwise the comparison is meaningless.
• Using $\bar{y}$ from the wrong reference: the centroid table gives $\bar{y}$ relative to the shape; you must add the depth of the shape’s top edge to get $y_C$ relative to the free surface.
• Treating the bottom-pressure $P_b$ as $P_C$: $P_C$ is the pressure at the centroid, not at the deepest point. For a rectangle that’s just $P_b/2$ (if the top sits at the free surface), but the tutorial’s gates do not all start at the free surface.
• Sign of the moment in overturning checks: pick the toe as the pivot and consistently call clockwise positive — $W$ creates a restoring moment about that toe, $F_P$ an overturning one.

Source citations

• Tutorial 10 deck, slides 2–11 (Problem Solving Approach, Resultant Pressure Force, Centroid table, Centre of Pressure, Second Moment of Area, Exercises 1–3).
• Tutorial 10 deck, slide 12 (Learning Activities — Mastering Physics topics: pressure measurement and forces on submerged bodies, fluid statics; Portfolio 10 in workshop).

⚠️ The official Week 10 lecture PDF was missing from the source folder when this note was generated. Confidence is inferred. Treat the numerical answers in the worked examples as guides — re-run them once the lecture’s exact conventions (which way is ”$y$ positive”, whether $P_C$ is gauge or absolute) are confirmed.