Week 9 — Pressure Measurement + Forces on Submerged Surfaces

What this week is about

This week closes the fluid-statics block of EGD102 by stitching together two threads you’ve been carrying since Week 8:

1. How pressure is measured in practice — manometers, the “walk” rule, and the difference between absolute and gauge pressure.
2. How distributed pressure becomes a single force on a submerged plane surface — the resultant $F_R$, its line of action (the centre of pressure), and the moment balance on a hinged gate.

The mental picture you want to leave with: the pressure on a submerged plane is a 3-D pressure prism that varies linearly with depth. We collapse that prism to one equivalent force $F_R=\rho g{y}_{C}A$ acting at $y_R=y_C+I_C/(y_{C}A)$ — always below the geometric centroid because more of the prism’s volume sits at greater depth. Once you have $F_R$ and $y_R$, the gate is just a rigid body and you sum moments about the hinge.

See the in-depth note for why each piece works, or the cheatsheet for the fast revision table + quiz.

Notes in this week

• Lecture summary — the directly-supported reconstruction with every worked example from the lecture and tutorial.
• Cheatsheet — every formula, table and recipe on one scannable page. Includes the quiz (12–18 MCQs, mixed difficulty, reshuffles every visit).
• In-depth analysis — the why: where $y_R>y_C$ comes from, how the pressure prism geometrically becomes $F_R$, the manometer walk as a path integral, and an exam-style worked example end-to-end.
• Study guide — what’s directly supported vs inferred, common mistakes, practice questions, confidence report.
• Workshop prep — 5-minute and 20-minute revision plans for Portfolio 8.

Any notes you drop into this folder will appear here automatically.

What I need to know before the workshop

• Hydrostatic pressure $p=\rho gh$ (Week 8) and the meaning of $g=9.81$ m/s²
• How to find the area and centroid of a rectangle, triangle, circle and semicircle
• Difference between absolute pressure (datum: vacuum) and gauge pressure (datum: atmosphere)
• How to draw a free-body diagram of a hinged gate and sum moments about the hinge
• Second moment of area $I_C$ for the four standard shapes — they live on the formula sheet but you should recognise them on sight

Assessment relevance

• Submerged-surface problems appear on every EGD102 paper — usually as either a rectangular gate (like Tutorial Ex 3) or a triangular plate (like Tutorial Ex 4).
• Manometer problems are common short-answer items; the walk rule ($+\rho gh$ down, $-\rho gh$ up) is the bit that gets marked.
• The L-shaped counter-weighted gate (Example 4) is the highest-difficulty variant — if it appears, it is worth disproportionate marks. Drill it.
• Portfolio tasks expect you to draw the pressure prism and label the line of action of $F_R$, not just compute numbers.

Overview

Week 9 closes the fluid-statics block of EGD102 by combining two threads: (1) how pressure is measured in practice using manometers and the distinction between absolute and gauge pressure (reinforced from Week 8 and used heavily in the tutorial), and (2) how the distributed pressure on a submerged surface is reduced to a single resultant force and a single point of application (the centre of pressure) so that gates, dams and walls can be analysed with ordinary statics.

The big mental picture is that pressure on a submerged plane surface forms a linearly varying pressure prism. We replace that prism with one equivalent force $F_R$ acting at the centroid of the prism — which sits below the geometric centroid of the surface because more of the prism’s volume is at greater depth. Once we have $F_R$ and $y_R$, the gate becomes a rigid body and we just sum moments about the hinge to find the holding force or required counter-weight.

Tutorial 9 includes a Wolfson-style manometer pair (Exercises 1–2) and two submerged-surface problems (Exercises 3–4, rectangle and triangle), so both threads are examinable.

Key concepts

• Resultant pressure force on a fully-submerged plane surface equals the pressure at the centroid of the surface times the area of the surface: $F_R=P_C\times A$ with $P_C=\rho g{y}_C$. The force acts perpendicular to the surface.
• Centroid ($\bar{x},\bar{y}$): the geometric centre of a plane shape. Assuming uniform density it is where the area is “concentrated”. For this unit we only use standard shapes (rectangle, general/isosceles/right triangle, circle) from the supplied table; integration definitions $\bar{x}=\frac{1}{A}\int xdA$, $\bar{y}=\frac{1}{A}\int ydA$ are given for reference.
• Vertical dam simplification: when the plate is vertical and extends from the free surface to depth $h$, the centroid is at depth $h/2$ and the area is $h\times L$, giving $F_P=\rho g\frac{h}{2}\cdot hL$. The resultant acts at $h/3$ above the base (centroid of the triangular pressure prism).
• Centre of pressure is the actual line of action of $F_R$. It is located at the centroid of the pressure prism — the 3-D wedge whose height at each point equals the local pressure. Because pressure grows with depth, the centroid of the prism sits deeper than the centroid of the surface, so the centre of pressure is always below the geometric centroid (for a surface whose top edge is at or below the free surface).
• Second moment of area $I_C$ measures a shape’s resistance to bending and appears in the centre-of-pressure formula. Use the table of standard formulas; for this unit you only need rectangle, triangle, circle and semicircle.
• Absolute vs gauge pressure: integrating $dp=\rho gdz$ for a constant density fluid gives absolute pressure $p=\rho gz+p_0$ measured above perfect vacuum. Gauge pressure $p=\rho gz$ uses atmospheric as the datum (so $p_{\text{gauge}}=p_{\text{abs}}-p_{\text{atm}}$).
• Equal-depth rule: in a continuous body of the same fluid, two points at the same depth have the same pressure ($p_A=p_B$). If a continuous path crosses a fluid–fluid interface, you must walk through the manometer adding $+\rho gh$ going down and $-\rho gh$ going up.
• Pressure prism: an imaginary 3-D object whose height represents local pressure across a submerged surface. Larger pressure ordinates sit at greater depth, which is why the resultant acts below the centroid.

Core formulas

Resultant pressure force on a submerged plane surface:

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

Vertical wall/dam from free surface to depth $h$, width $L$:

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

Resultant on a vertical dam acts at $h/3$ above the base of the wetted area.

Depth to centre of pressure (measured from the free surface):

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

where $y_C$ is the depth to the centroid, $I_C$ is the second moment of area about the horizontal axis through the centroid, and $A$ is the wetted area.

Centroid coordinates (integration definition):

$\bar{x}=\frac{1}{A}\int xdA,\bar{y}=\frac{1}{A}\int ydA$

Second moment of area about the centroidal axis ($I_C$):

• Rectangle (width $b$, depth $d$): $I_C=\frac{b{d}^3}{12}$
• Triangle (base $b$, height $h$): $I_C=\frac{b{h}^3}{36}$
• Circle (radius $r$): $I_C=\frac{\pi r^4}{4}$
• Semicircle: $I_C=0.1102r^4$

Centroid depth $\bar{y}$ and area for the standard shapes (from the supplied table):

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

Pressure relations (Tutorial 9):

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

Manometer “walk”: $p_1+\sum (\rho gh{)}_{\text{down}}-\sum (\rho gh{)}_{\text{up}}=p_2$.

Worked examples

Example 1 — Rectangular gate (parts a and b)

A tank gate, hinged at the bottom, is held closed by a horizontal force $F$ applied $1.5$ m above the hinge. The water surface is $3$ m above the top edge of the gate; the gate itself is $1.5$ m tall (vertical) and $2.5$ m wide.

(a) Resultant pressure force on the gate.

1. Geometric properties. The depth to the centroid of the rectangular gate measured from the free surface is the depth to the top edge plus half the gate height (${\bar{y}}_{\text{rect}}=h/2$ from its own top edge):

$y_C=h_C=3+\frac{1.5}{2}=3.75\text{m}$

Area of the submerged rectangle:

$A=b\times h=2.5\times 1.5=3.75{\text{m}}^2$

2. Resultant force:

$F_R=F_P=(\rho g{h}_C)\cdot A=(1000)(9.81)(3.75)(3.75)\approx 137,950\text{N}\approx 138\text{kN}$

(b) Minimum resisting force $F$ to keep the gate closed.

1. Second moment of area of the rectangle (axis through its centroid, parallel to the water surface):

$I_{C,\text{rect}}=\frac{b{h}^3}{12}=\frac{2.5\times 1.5^3}{12}=0.70{\text{m}}^4$

2. Depth to the centre of pressure from the free surface:

$y_P=y_C+\frac{I_C}{y_{C}A}=3.75+\frac{0.70}{3.75\times 3.75}=3.80\text{m}$

3. Distance from the bottom (hinge) up to the line of action of $F_R$:

$d=(3+1.5)-3.80=0.70\text{m}$

4. Free body diagram of the gate. Taking moments about the hinge $H$ (counter-clockwise positive; $F$ applied at $1.5$ m above the hinge):

$\sum M_H=0:-(F_P)(0.70)+F_g(1.5)=0$

$F_g=\frac{137,950\times 0.70}{1.5}\approx 64,380\text{N}\approx 64.4\text{kN}$

Example 2 — Centre of pressure on a submerged square plate

A square plate with $0.5$ m sides is placed vertically in a water tank so that its centroid is at depth $2.9$ m, top edge parallel to the free surface. Find the depth $y_R$ to the resultant horizontal pressure force.

1. Geometric properties:

$y_C=2.9\text{m},A=0.5\times 0.5=0.25{\text{m}}^2$

$I_{C,\text{sq}}=\frac{b{h}^3}{12}=\frac{0.5\times 0.5^3}{12}=0.0052{\text{m}}^4$

2. Depth to the centre of pressure:

$y_R=y_C+\frac{I_C}{y_{C}A}=2.9+\frac{0.0052}{2.9\times 0.25}=2.9072\text{m}$

So the resultant force acts only about $7$ mm below the centroid — the offset shrinks as $y_C$ grows because $I_C/(y_{C}A)\to 0$ for deep plates.

Example 3 — Rectangular surface, full deliverables

A $2$ m wide, $3$ m tall rectangle is submerged in water with its top edge at depth $2.2$ m below the free surface.

(a) Second moment of area through the centroid.

$I_{C,\text{rect}}=\frac{b{h}^3}{12}=\frac{2\times 3^3}{12}=4.5{\text{m}}^4$

(b) Centre of pressure. First the centroid and area:

$y_C=2.2+\frac{3}{2}=3.7\text{m},A=2\times 3=6{\text{m}}^2$

$y_R=y_P=y_C+\frac{I_C}{y_{C}A}=3.7+\frac{4.5}{3.7\times 6}\approx 3.9\text{m}$

(c) Resultant force.

$F_R=F_P=\rho g{y}_C\cdot A=1000\times 9.81\times 3.7\times 6\approx 217,560\text{N}\approx 218\text{kN}$

The pressure distribution is a trapezoid — non-zero at the top edge ($p=\rho g\cdot 2.2=21.6$ kPa) and larger at the bottom edge ($p=\rho g\cdot 5.2=51.0$ kPa). $F_P$ acts horizontally at $y_R=3.9$ m below the free surface.

Example 4 — L-shaped gate with counter-weight (advanced)

A $5$ m wide L-shaped gate is hinged at $A$, with a $5$ m horizontal arm carrying a weight $W$ at its far end and an $8$ m vertical arm acting as the water-retaining wall. The gate must just open when the water height behind it reaches $6$ m.

Approach: compute the resultant pressure force on the vertical $6$ m of wetted wall, find its line of action (centre of pressure), then take moments about $A$ — the weight $W$ provides the only restoring moment in the absence of water. Setting the two moments equal at the threshold gives the required mass.

Tutorial Exercise 1 — Mercury manometer on an air duct

Mercury ${\rho }_{Hg}=13600$ kg/m${}^3$, manometer height difference $h=15$ mm, atmospheric pressure $100$ kPa.

The duct-side mercury column is lower than the open side, so the air pressure pushes mercury down — duct pressure is above atmospheric.

$P=P_{\text{atm}}+\rho gh=100,000+13,600\times 9.8\times 0.015=102,000\text{Pa}=102\text{kPa}$

Tutorial Exercise 2 — Double U-tube manometer between two pipes

Freshwater pipe (1) connected to seawater pipe (2) by a mercury manometer with an intervening air column. ${\rho }_w=1000$, ${\rho }_{sea}=1035$, ${\rho }_{Hg}=13600$ kg/m${}^3$. Heights along the walk: water column $h_w=0.6$ m (down), mercury $h_{Hg}=0.1$ m (up), seawater $h_{sea}=0.4$ m (down). Air column neglected.

Walking from $P_1$ to $P_2$:

$P_1+{\rho }_{w}g{h}_w-{\rho }_{Hg}g{h}_{Hg}-{\rho }_{air}g{h}_{air}+{\rho }_{sea}g{h}_{sea}=P_2$

Neglecting the air term and rearranging:

$P_1-P_2=g({\rho }_{Hg}{h}_{Hg}-{\rho }_w{h}_w-{\rho }_{sea}{h}_{sea})$

$P_1-P_2=9.81[13,600\times 0.1-1000\times 0.6-1035\times 0.4]=3390\text{Pa}=3.39\text{kPa}$

The freshwater pipe is $3.39$ kPa above the seawater pipe. The air column is negligible because ${\rho }_{\text{air}}\approx 1.2$ kg/m${}^3$ is four orders of magnitude smaller than the other fluids and the column height is small, so $\rho gh$ for air is comparatively zero.

Tutorial Exercise 3 — Sea-water lock (vertical rectangle)

Lock wall $20$ m wide retaining sea-water ($SG=1.03$) to a depth of $10$ m.

(a) $y_C=10/2=5$ m.

(b) $F_P=\rho g{y}_{C}A=(1000\times 1.03)(9.8)(5)(10\times 20)=1.01\times 10^7$ N $=10.1$ MN.

(c) $I_C=\frac{b{h}^3}{12}=\frac{20\times 10^3}{12}=1667$ m${}^4$.

(d) $y_p=y_C+\frac{I_C}{y_{C}A}=5+\frac{1667}{5\times 200}=6.67$ m.

This recovers the classic dam result that the resultant acts at $h/3$ above the base for a wall extending to the free surface: $10-6.67=3.33=h/3$.

Tutorial Exercise 4 — Submerged right triangle

Right-triangular plate, vertical leg $d=2.0$ m and horizontal leg $b=1.0$ m, with its top vertex at depth $a=0.2$ m below the free surface (apex at the top, base at the bottom).

(a) $y_C=\frac{d}{3}+a=\frac{2}{3}+0.2\approx 0.866$ m (using $\bar{y}=h/3$ measured from the top of the right triangle).

(b) Area $A=\frac{bd}{2}=1$ m${}^2$; $F_P=\rho g{y}_{C}A=1000\times 9.8\times 0.866\times 1=8487$ N.

(c) $I_C=\frac{b{d}^3}{36}=\frac{1.0\times 2.0^3}{36}=0.222$ m${}^4$.

(d) $y_R=y_C+\frac{I_C}{y_{C}A}=0.866+\frac{0.222}{0.866\times 1}=1.123$ m.

Things to practise

• Re-do Example 1 without looking, hitting the “find geometric properties → compute $F_R$ → compute $I_C$ → compute $y_P$ → moment about hinge” recipe.
• Run Example 3 end-to-end and draw the trapezoidal pressure distribution with the top and bottom pressures labelled.
• Example 4 (L-shaped gate / counter-weight) is the hardest in the set — set up the free-body diagram of the L, write $\sum M_A=0$, and solve for $W$. Watch the moment arm: $F_P$ acts horizontally at depth $y_R$ below the free surface, so its arm about $A$ is measured along the vertical leg.
• Tutorial 1 and 2 are pure manometer / hydrostatic-pressure practice; drill the manometer walk until you can do it without re-reading the rule.
• Tutorial 4 (triangle) is the most likely exam variant after the rectangle: practise getting the correct $y_C=a+d/3$ (not $a+d/2$) and the correct $I_C=b{d}^3/36$ (not $/12$).
• Memorise the four $I_C$ formulas and the four centroid locations from the Centroid table — they’re given on the lecture slide but not always on formula sheets.

Common pitfalls

• Forgetting where the centroid is measured from. $y_C$ in $F_R=\rho g{y}_{C}A$ is depth from the free surface, not from the top of the plate. Always add the depth of the top edge to the in-plate centroid offset.
• Using $b{h}^3/12$ for a triangle. The centroidal $I_C$ of a triangle about a horizontal axis is $b{h}^3/36$. Same shape, different number.
• Putting $y_R$ above $y_C$. The centre of pressure is always below the centroid for a vertical surface beneath the free surface (because $I_C/(y_{C}A)>0$).
• Mixing absolute and gauge pressure. When computing $F_R$ on a gate exposed to atmosphere on the dry side, use gauge pressure $\rho g{y}_C$ — atmospheric pressure cancels on the two faces. Only switch to absolute when the dry side is sealed or evacuated.
• Wrong sign in the manometer walk. Going down through a fluid you add $\rho gh$; going up you subtract. Verify by checking that the final equation reduces sensibly when all heights are zero.
• Specific gravity vs density. $SG=1.03$ for seawater means ${\rho }_{\text{sea}}=1.03\times 1000=1030$ kg/m${}^3$ (Tutorial uses 1035 for the manometer problem because it specifies a slightly different mix — always read the question).
• Moment arm for the gate force. In Example 1, $F$ is applied $1.5$ m above the hinge, but the moment arm of $F_P$ is the distance from the hinge up to the centre of pressure, which is not the same as the distance to the centroid. Compute $y_R$ first.
• Forgetting to use $g=9.81$ m/s${}^2$ (or $9.8$). The lecture notes mix both values; the tutorial solutions use $9.8$. Be consistent within a single calculation.

Source citations

• Lecture slides: EGD102-Physics/Lecture9_CTP1.pdf — title slide (p. 1), $F_R=P_C\times A$ definition (p. 2), centroid definitions and standard shapes (pp. 3–5), Example 1 setup (p. 6), vertical dam simplification (p. 7), centre of pressure formula and pressure prism (pp. 8–9), $I_C$ for standard shapes (p. 10), Example 2 (p. 11), Example 1 continued (p. 12), Example 3 (p. 13), L-shaped gate Example 4 (p. 15), learning activities (p. 17).
• Lecturer’s worked notes: EGD102-Physics/EGD102_Week9_Lecture.pdf — Example 1 numerics $y_C=3.75$ m and $F_R\approx 138$ kN (p. 1), Example 2 centre of pressure $y_R=2.9072$ m (p. 2), Example 1 continued with $y_P=3.8$ m and $d=0.7$ m from the bottom (p. 3), FBD and moment balance giving $F_g=64380$ N (p. 4), Example 3 worked end-to-end giving $I_C=4.5$ m${}^4$, $y_R\approx 3.9$ m, $F_P\approx 217560$ N (p. 5).
• Tutorial: EGD102-Physics/Tutorial 9.pdf — problem-solving approach (p. 2), absolute vs gauge pressure (p. 3), pressure difference between two points and equal-depth rule (p. 4), Exercise 1 mercury manometer (p. 5), Exercise 2 double U-tube (p. 6), repeats of $F_R$, centroid, centre of pressure and $I_C$ slides (pp. 7–10), Exercise 3 sea-water lock (p. 11), Exercise 4 submerged triangle (p. 12).
• Tutorial solutions: EGD102-Physics/Tutorial 9_Solutions.pdf — Exercise 1 $P=102$ kPa (p. 1), Exercise 2 manometer walk yielding $P_1-P_2=3.39$ kPa (p. 2), Exercise 3 $F_P=10.1$ MN, $I_C=1667$ m${}^4$, $y_P=6.67$ m (p. 3), Exercise 4 $y_C=0.866$ m, $F_P=8487$ N, $I_C=0.222$ m${}^4$, $y_R=1.123$ m (p. 4).