Week 9 Cheatsheet — Pressure Measurement + Submerged Surfaces

How this week breaks down

Two threads, one common technique (reducing distributed pressure to a single force + line of action).

Topic	What you do
Resultant force	$F_R=\rho g{y}_{C}A$. Find the centroid depth, multiply by area, multiply by $\rho g$.
Centre of pressure	$y_R=y_C+I_C/(y_{C}A)$. Use the table for $I_C$.
Vertical dam	Top edge at free surface? Use shortcut: $F_P=\rho g(h/2)(hL)$, acts at $h/3$ above base.
Hinged gate	Find $F_R$ and $y_R$, then $\sum M_{\text{hinge}}=0$ to get the holding force.
Manometer walk	$+\rho gh$ down, $-\rho gh$ up, equal-depth in same fluid means equal pressure.
Absolute vs gauge	Gate exposed to atmosphere on both sides $\Rightarrow$ use gauge. Sealed/evacuated side $\Rightarrow$ use absolute.

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1 — Resultant pressure force on a submerged plane

Definition. The single force equivalent to the linearly varying pressure on the surface:

$F_R=P_C\cdot A=\rho g{y}_C\cdot A$

acting perpendicular to the surface.

• $y_C$ = depth of the centroid of the surface, measured from the free surface.
• $A$ = wetted area of the surface.
• $P_C=\rho g{y}_C$ = pressure at the centroid depth.

Centroid table (depth from top edge / from centre, area)

Shape	$\bar{y}$ from top edge	Area
Rectangle ($b\times h$)	$h/2$	$bh$
General triangle (base on top)	$h/3$	$bh/2$
Triangle (apex on top)	$h/3$ measured down from apex	$bh/2$
Isosceles triangle	$h/3$	$\ell h/2$
Right triangle	$h/3$	$bh/2$
Circle (radius $r$)	$0$ from centre	$\pi r^2$

Vertical dam shortcut (top edge at the free surface)

Plate of height $h$, width $L$:

$F_P=\rho g\left(\frac{h}{2}\right)(h\times L)$

Resultant acts at $h/3$ above the base of the wetted area.

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2 — Centre of pressure

The actual line of action of $F_R$, measured as a depth from the free surface:

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

• $I_C$ = second moment of area about the horizontal axis through the centroid.
• Because $I_C,y_C,A>0$, we always have $y_R>y_C$: the centre of pressure is below the geometric centroid for any plate whose top edge is at or below the free surface.

$I_C$ table (centroidal axis)

Shape	$I_C$
Rectangle ($b\times d$)	$b{d}^3/12$
Triangle (base $b$, height $h$)	$b{h}^3/36$
Circle (radius $r$)	$\pi r^4/4$
Semicircle (radius $r$)	$0.1102r^4$

Quick numerical sanity check

• Square plate side $0.5$ m at $y_C=2.9$ m: $y_R-y_C=0.0052/(2.9\times 0.25)=0.007$ m. Tiny offset for a deep, small plate.
• Lock wall $h=10$ m, $L=20$ m: $y_R-y_C=1667/(5\times 200)=1.67$ m. Large offset because the plate spans from the surface.

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3 — Pressure measurement

Absolute vs gauge

$p_{\text{abs}}=\rho gz+p_0,p_{\text{gauge}}=\rho gz,p_{\text{gauge}}=p_{\text{abs}}-p_{\text{atm}}$

Datum	What it means	When you use it
Absolute	Above perfect vacuum	Closed/evacuated systems, thermodynamics
Gauge	Above atmospheric	Gates and walls with atmosphere on the dry side (atmosphere cancels)

Manometer “walk” rule

Travel a continuous path from point 1 to point 2:

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

Move	Sign
Down through fluid (height $h$)	$+\rho gh$
Up through fluid (height $h$)	$-\rho gh$
Same depth in same continuous fluid	$0$ (no change)

Worked snippets

Setup	Walk	Result
Air duct, Hg manometer, $h=15$ mm, duct side low	$P=P_{\text{atm}}+{\rho }_{Hg}gh$	$P\approx 102$ kPa
Double U-tube freshwater/seawater	$P_1-P_2=g({\rho }_{Hg}{h}_{Hg}-{\rho }_w{h}_w-{\rho }_{sea}{h}_{sea})$	$\approx 3.39$ kPa

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4 — Hinged gate template

Five steps, in this order. Don’t skip.

1. Geometric properties. Centroid depth $y_C$ from free surface, area $A$, second moment $I_C$.
2. Resultant force. $F_R=\rho g{y}_{C}A$ (perpendicular to plate).
3. Centre of pressure. $y_R=y_C+I_C/(y_{C}A)$.
4. Moment arm. Distance from the hinge to the line of action of $F_R$ (measured along the gate, not the depth axis).
5. Sum moments. $\sum M_{\text{hinge}}=0$, solve for the holding force.

Worked snippet — Example 1

Gate $2.5$ m × $1.5$ m, hinge at bottom, water $3$ m above top edge, $F$ applied $1.5$ m above hinge.

Step	Calc	Result
$y_C$	$3+1.5/2$	$3.75$ m
$A$	$2.5\times 1.5$	$3.75$ m²
$F_R$	$1000\times 9.81\times 3.75\times 3.75$	$\approx 138$ kN
$I_C$	$2.5\times 1.5^3/12$	$0.70$ m⁴
$y_R$	$3.75+0.70/(3.75\times 3.75)$	$3.80$ m
Arm of $F_R$ above hinge	$(3+1.5)-3.80$	$0.70$ m
$F$ (from $\sum M_H=0$)	$138\times 0.70/1.5$	$\approx 64.4$ kN

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Common mistakes

1. Centroid depth measured from the wrong place. $y_C$ in $F_R=\rho g{y}_{C}A$ is the depth from the free surface — not from the top of the plate, not from the bottom. Add the depth of the top edge to the in-plate centroid offset.
2. Using $b{h}^3/12$ for a triangle. Triangles use $b{h}^3/36$. Rectangles use $b{h}^3/12$. Same shape of formula, different denominator.
3. Putting $y_R$ above $y_C$. The centre of pressure is always below the geometric centroid for plates beneath the free surface. If your $y_R<y_C$, you flipped a sign.
4. Mixing absolute and gauge pressure. For a gate with atmosphere on the dry side, use gauge. Atmospheric pressure cancels on the two faces.
5. Wrong sign on the manometer walk. Down adds, up subtracts. Check by setting all heights to zero and confirming the equation reduces to $P_1=P_2$.
6. Forgetting that the moment arm of $F_R$ goes to $y_R$, not $y_C$. The line of action of the resultant is at the centre of pressure, not the centroid.
7. Inconsistent $g$. Lecture uses $9.81$, tutorial uses $9.8$. Pick one per calculation. Don’t mix.
8. Treating $SG$ as a density. $SG=1.03$ for seawater $\Rightarrow \rho =1030$ kg/m³ (multiply by $1000$).

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Key formulas

$F_R=\rho g{y}_{C}A{y}_R=y_C+\frac{I_C}{y_{C}A}$

$F_P^{\text{vert dam}}=\rho g\left(\frac{h}{2}\right)(hL)\text{acting at}h/3\text{ above base}$

$p_{\text{abs}}=\rho gz+p_0{p}_{\text{gauge}}=\rho gz$

$\text{Manometer:}{P}_1+\sum (\rho gh{)}_{\downarrow }-\sum (\rho gh{)}_{\uparrow }=P_2$

$I_C:\text{rect }\frac{b{d}^3}{12},\text{tri }\frac{b{h}^3}{36},\text{circle }\frac{\pi r^4}{4},\text{semicircle }0.1102r^4$

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