Twenty minutes or less.

Week 9 — Pressure Measurement + Forces on Submerged Surfaces. Pick a mode. Start a timer. That's it.

Pick a mode

The shortest path to walking in prepared.

Timer

5:00

//content

5-minute version

Two threads, one sentence each.

Manometer / pressure measurement — walk the path, down adds $ρ g h$ , up subtracts $ρ g h$ , neglect the air column.
Submerged surface — $F_{R} = ρ g y_{C} A$ , line of action at $y_{R} = y_{C} + I_{C} / (y_{C} A)$ , always below the centroid.

Then for a hinged gate: find $F_{R}$ , find $y_{R}$ , take moments about the hinge.

Open the cheatsheet quiz, do 3 easy questions, close it. You’re prepped.

Time	Action
0–5 min	Skim the cheatsheet tables — especially the centroid table and the $I_{C}$ table.
5–10 min	Re-do Tutorial Exercise 3 (sea-water lock) from scratch — covering pen, write it out. Targets all four steps.
10–15 min	Take the cheatsheet quiz. Don’t worry about the score.
15–20 min	Read the “centre of pressure” derivation in the in-depth note — section 2. The parallel-axis argument is the bit that sticks.

Most students slip on three specific things this week:

Centroid depth measured from the wrong place. $y_{C}$ is from the free surface, not from the top of the plate. Always add the depth of the top edge to the in-plate offset.
$b h^{3} /12$ vs $b h^{3} /36$ . Rectangles use $/12$ . Triangles use $/36$ . Same shape of formula, different denominator.
Manometer walk sign error. Down adds, up subtracts. Sanity check: set all heights to zero, equation should reduce to $P_{1} = P_{2}$ .

If you only fix those three, you’ll pick up the bulk of the available marks.

$F_{R} = ρ g y_{C} A y_{R} = y_{C} + \frac{I _{C}}{y _{C} A}$

$Vertical dam (top edge at free surface): F_{P} = ρ g (\frac{h}{2}) (h L) at h /3 above base$

$p_{abs} = ρ g z + p_{0} p_{gauge} = ρ g z$

$Manometer walk: P_{1} + \sum (ρ g h)_{↓} - \sum (ρ g h)_{↑} = P_{2}$

$I_{C} : rect \frac{b d ^{3}}{12}, tri \frac{b h ^{3}}{36}, circle \frac{π r ^{4}}{4}, semicircle 0.1102 r^{4}$

Task type	What the setup usually looks like
Manometer (single fluid)	One U-tube, height difference $h$ , ambient on one side. Find the gauge or absolute pressure on the other.
Manometer (multi-fluid)	Two pipes connected via Hg + air gap. Walk the path; find $P_{1} - P_{2}$ .
Rectangular submerged plate	Vertical rectangle, top edge at given depth. Find $F_{R}$ , $y_{R}$ , possibly the holding force on a hinge.
Triangular submerged plate	Apex up or apex down; remember $\overset{y}{ˉ} = h /3$ from the apex (apex up) and $I_{C} = b h^{3} /36$ .
Hinged gate	Add a moment balance: $\sum M_{hinge} = 0$ , solve for the unknown force.
Vertical dam (extends to free surface)	Use the shortcut: $F_{P} = ρ g (h /2) (h L)$ , acts at $h /3$ above base.

$y_{C}$ measured from the wrong reference (top of plate, bottom of plate, ground — only the free surface is correct).
Confusing $b h^{3} /12$ (rectangle) with $b h^{3} /36$ (triangle).
Putting the centre of pressure above the centroid.
Mixing absolute and gauge pressure. For a gate open to atmosphere on both sides, use gauge.
Wrong sign on the manometer walk.
Treating $S G = 1.03$ as a density (it’s a ratio — multiply by $1000$ ).
Forgetting the moment arm of $F_{R}$ goes to $y_{R}$ , not $y_{C}$ , when summing moments.
Inconsistent $g$ within a single calculation. Pick $9.81$ or $9.8$ and stick with it.

Try these without notes. Eight minutes total. Use $g = 9.81$ m/s², $ρ_{water} = 1000$ kg/m³.

A vertical rectangular plate $1$ m wide and $2$ m tall has its top edge at the free surface. Find the magnitude and line of action (height above base) of the resultant water force.
A right-triangular plate with vertical leg $d = 3$ m and horizontal leg $b = 1.5$ m has its apex at the water surface (apex up, base at the bottom). Find $y_{C}$ , $A$ , $I_{C}$ , and $y_{R}$ .
A U-tube manometer connects an air tank to atmosphere. Mercury ( $ρ = 13600$ kg/m³) stands $50$ mm higher on the atmosphere side. Atmospheric pressure is $101$ kPa. What is the absolute air-tank pressure?

Answers:

Question	Answer
1	$F_{R} = ρ g (h /2) (h L) = 1000 \times 9.81 \times 1 \times 2 = 19.62$ kN, acts at $h /3 = 0.667$ m above the base.
2	$y_{C} = d /3 = 1$ m; $A = b d /2 = 2.25$ m²; $I_{C} = b d^{3} /36 = 1.5 \times 27/36 = 1.125$ m⁴; $y_{R} = 1 + 1.125/ (1 \times 2.25) = 1.5$ m.
3	Hg higher on atmosphere side $\Rightarrow$ air pressure is below atmospheric. $P_{air} = P_{atm} - ρ_{H g} g h = 101000 - 13600 \times 9.81 \times 0.050 \approx 94.3$ kPa absolute.

Read the cheatsheet tables (centroid + $I_{C}$ ).
Re-did Tutorial Exercise 3 (or Example 1) longhand.
Cheatsheet quiz attempted.
Mini self-test attempted.
Can explain in one sentence why $y_{R} > y_{C}$ (pressure prism wedge → centroid below the surface’s centroid).

That’s it. Close the laptop.