Week 10 Cheatsheet — Forces on Submerged Bodies

How this week breaks down

Three connected questions, all using hydrostatic pressure. Skim this once, then revise from the in-depth note.

Question	Tool
How big is the push?	$F_R=P_C\cdot A$ with $P_C=\rho g{y}_C$
Where does it act?	$y_R=y_C+\frac{I_C}{y_{C}A}$ (always below the centroid)
Will the wall hold?	Sliding: $\mu W\ge F_P$. Overturning: $M_{\text{restoring}}>M_{\text{overturning}}$.

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1 — Resultant pressure force

Definition. $F_R$ is the single equivalent force that replaces the distributed pressure on a submerged surface.

$$F_R=P_C\cdot A{P}_C=\rho g{y}_C\Rightarrow F_R=\rho g{y}_{C}A$$

• $y_C$ is the depth of the centroid of the wetted surface, measured from the free surface.
• $A$ is the wetted area.
• $\rho$ is the fluid density; water $=1000kg/{\mathrm{m}}^3$.

Centroid lookup (standard shapes)

Shape	Area	Centroid $\bar{y}$ (from base)
Rectangle ($b\times h$)	$bh$	$h/2$
Triangle (general, right, isosceles)	$bh/2$	$h/3$ above base
Circle (radius $r$)	$\pi r^2$	At centre
Semicircle (radius $r$)	$\pi r^2/2$	$4r/(3\pi )$ from straight edge

Always add the depth of the shape’s top edge to $\bar{y}$ to get $y_C$ relative to the free surface.

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2 — Centre of pressure

Definition. $y_R$ is the depth at which $F_R$ acts. Below the centroid because pressure grows with depth.

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

The “below-centroid” offset $I_C/(y_{C}A)$ shrinks as the surface gets deeper (because $y_C$ is in the denominator). Deep gates → centre of pressure approaches the centroid.

Second-moment-of-area lookup

Shape	$I_C$ about centroidal horizontal axis
Rectangle ($b\times d$)	$b{d}^3/12$
Triangle ($b\times h$)	$b{h}^3/36$
Circle (radius $r$)	$\pi r^4/4$
Semicircle (radius $r$)	$\approx 0.1102r^4$

Quick exemplars

Setup	$y_C$	$I_C$	$y_R$
Rectangle $3\times 5$, top at free surface	$2.5$ m	$31.25$ m⁴	$3.33$ m (i.e. $2h/3$)
Rectangle (water $3.5$ m deep, per metre width)	$1.75$ m	$3.57$ m⁴	$2.33$ m
Triangle $2.5\times 1.5$, top edge at $3$ m depth	$4.0$ m	$0.234$ m⁴	$4.03$ m

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3 — Sliding and overturning (dam walls)

Work per metre of wall width. Let $H$ be wall height, $t$ thickness, ${\rho }_c$ concrete density, $\mu$ base friction coefficient.

Weight of the wall (per metre width)

$$W={\rho }_{c}g(t\cdot H\cdot 1)$$

Sliding check

Friction must beat the horizontal water push:

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

Overturning check (moments about the toe)

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

Safe when $M_{\text{restoring}}>M_{\text{overturning}}$. The ratio is the safety factor.

Quick exemplar

Wall: $H=4$ m, ${\rho }_c=2550$ kg/m³, $\mu =0.4$, retaining $3.5$ m of water.

Quantity	Value
$F_P$	$60.1$ kN/m
Minimum $t$ for sliding	$1.50$ m
At $t=1.5$ m: $M_{\text{restoring}}$	$112.6$ kN·m/m
At $t=1.5$ m: $M_{\text{overturning}}$	$70.4$ kN·m/m
Safety factor (overturning)	$\approx 1.6$ ✓

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Common mistakes

1. Treating $P_b$ (bottom-edge pressure) as $P_C$. $P_C$ is at the centroid, not the deepest point.
2. Using $\bar{y}$ straight from the lookup. That’s relative to the shape’s base — you need $y_C$ relative to the free surface, so add the depth of the shape’s top edge.
3. Confusing $y_C$ and $y_R$. Centroid is geometric (no fluid involved); centre of pressure is fluid-dependent.
4. Forgetting “per metre” on long walls. Compare like for like.
5. Picking the wrong pivot in the overturning check. Pick the toe of the wall (front bottom corner, downstream of the water) and stick to one sign convention.

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Key formulas

$$F_R=P_{C}A=\rho g{y}_{C}A{y}_R=y_C+\frac{I_C}{y_{C}A}$$ $$\text{Sliding: }\mu {\rho }_{c}gtH\ge F_P\text{Overturning: }W\cdot \frac{t}{2}>F_P\cdot (H-y_R)$$

For the why and full worked examples, see the in-depth note.