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CELE Hydraulics & Fluid Mechanics — Hydrostatic Pressure and Forces on SurfacesMemory Anchors

Summary Concepts Notes Revision Deep Dive Slides Flashcards Mistakes Cheat Mnemonics Templates

Filipino reviewers do well on Hydrostatic Pressure and Forces on Surfaces once they have personal mnemonics — the anchors that make the concept local, memorable, and quick to surface under CELE time pressure. This page gathers the best-working anchors for Professional Regulation Commission (PRC) — Board of Civil Engineering's typical Hydraulics & Fluid Mechanics items on this chapter.

Exam context

For the Civil Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Civil Engineering tests Hydraulics & Fluid Mechanics under a "Core" label, with Hydrostatic Pressure and Forces on Surfaces in the 2nd slot across 10 chapters. CELE candidates must clear the 70% weighted average, no sub-test below 50% cut on the 2026 paper, which draws about a meaningful share of Hydraulics & Fluid Mechanics questions. Date to watch: May and November 2026.

Hydrostatic Pressure and Forces on Surfaces - Memory Anchors

Memory techniques are not shortcuts — they are cognitive scaffolding. Research in educational psychology shows that vivid, emotionally engaging memory anchors (mnemonics, analogies, stories) can increase long-term recall by up to 600% compared to rote repetition. For PRC board exam preparation, where you must retrieve formulas and concepts instantly under pressure, anchoring each key idea to a memorable image, story, or acronym transforms abstract hydraulics into unforgettable mental snapshots. This collection covers every major concept in Hydrostatic Pressure and Forces on Surfaces — from p = γh to curved-surface force components — using techniques proven to stick. Use these anchors during review, then test yourself with the Revision Game. The more vivid and ridiculous the image, the better it adheres to long-term memory.

Anchors

Topic

Pressure Variation with Depth

Concept

Pressure increases linearly with depth: p = γh

Anchor Id

Difficulty

easy

Memory Aid

Imagine you are swimming in Laguna de Bay and a giant stack of PANCAKES is being piled on top of you. Every meter of water depth is one more heavy pancake. The deeper you go, the more pancakes crush down on you — and that crushing force per unit area is exactly the gauge pressure p = γh. The specific weight γ is how heavy each pancake layer is (9.81 kN/m³ for water), and h is how many layers are above you.

Anchor Type

analogy

Why It Works

The pancake stack is a tactile, visual analogy for cumulative weight — it makes the LINEAR increase with depth feel intuitive and physical rather than abstract.

Example Usage

Exam asks: 'Find gauge pressure at 8 m depth in water.' Think: 8 pancakes, each weighing γ = 9.81 kN/m² per meter. p = 9.81 × 8 = 78.48 kPa.

Recall Trigger

Pancake stack in a lake

Topic

Pascal's Law

Concept

Pascal's Law — pressure acts equally in all directions

Anchor Id

Difficulty

easy

Memory Aid

Think of a SIOPAO (steamed bun). When you squeeze it from the top, the filling pushes outward equally in ALL directions — to the left, right, front, back. A static fluid is like the filling inside: pressure at any point spreads equally in every direction and always acts perpendicular to any surface it touches. Blaise Pascal discovered this, so remember: 'Pascal's Siopao — squeeze anywhere, feel it everywhere.'

Anchor Type

analogy

Why It Works

The siopao is culturally familiar to Filipino students, making the abstract concept of omnidirectional pressure tangible and amusing.

Example Usage

When a question states that a dam face is angled, remember that pressure still acts perpendicular to the face at every point — like the siopao filling pushing perpendicularly against its wrapper everywhere.

Recall Trigger

Squeezing a siopao

Topic

Force on a Plane Surface

Concept

Total hydrostatic force on a plane surface: F = γ·h̄·A

Anchor Id

Difficulty

easy

Memory Aid

Remember the phrase: 'GAMMA HITS AREA' → γ · h̄ · A = F. Spell it out: G-H-A. Gamma (γ) is the specific weight of the fluid, H-bar (h̄) is the depth to the centroid of the area, and A is the total area. The force F equals the pressure AT THE CENTROID times the entire area. Say it three times fast: 'Gamma-Hbar-Area, Gamma-Hbar-Area, Gamma-Hbar-Area = Force!'

Anchor Type

mnemonic

Why It Works

The verbal rhythm 'GAMMA HITS AREA' creates an auditory hook. Associating the word 'hits' with force reinforces the concept that this computes the resultant force.

Example Usage

Gate is 2 m wide, 3 m tall, top at surface. A = 6 m², h̄ = 1.5 m. Recall 'Gamma Hits Area': F = 9.81 × 1.5 × 6 = 88.29 kN.

Recall Trigger

GAMMA HITS AREA

Topic

Center of Pressure

Concept

Center of pressure is BELOW the centroid

Anchor Id

Difficulty

medium

Memory Aid

Story: Engr. Reyes designs a gate for a dam in Angat Reservoir. She calculates the centroid perfectly at mid-height and hangs a hinge there — but the gate keeps tipping open at the bottom. Why? Because the water pressure is GREATER at the bottom, so the resultant force (center of pressure) acts LOWER than the centroid. Her boss says: 'Reyes, ang pressure ay mas malakas sa ibaba — lagi sa ibaba ang resultant!' (The pressure is always stronger below — the resultant is always lower!). She never forgets: center of pressure is always BELOW the centroid.

Anchor Type

micro_story

Why It Works

The story creates a cause-and-effect narrative with a memorable consequence (gate fails), making the concept emotionally sticky. The Tagalog phrase adds cultural resonance.

Example Usage

When asked where the force acts on a gate, immediately recall the story: it acts at yp, which is BELOW ȳ by the amount Ig/(ȳA). Never say 'at the centroid.'

Recall Trigger

Engr. Reyes's gate tipping open

Topic

Center of Pressure

Concept

Center of pressure formula: yp = ȳ + Ig/(ȳ·A)

Anchor Id

Difficulty

medium

Memory Aid

The formula has TWO parts: the centroid distance ȳ PLUS the 'extra drop' Ig/(ȳA). Remember: 'Y-bar PLUS the IGLOO divided by Y-bar-A.' The IGLOO (Ig) is the centroidal moment of inertia — it's big and heavy, pushing the center of pressure DOWN. Big igloo = bigger drop below centroid. Small igloo (or very deep gate where ȳ is huge) = tiny drop. The deeper the gate, the smaller the extra drop (Ig shrinks relative to ȳA).

Anchor Type

mnemonic

Why It Works

Calling Ig the 'igloo' gives it a memorable visual shape (large, dome-shaped) and the concept of heaviness pulling the CP downward. The humor aids encoding.

Example Usage

yp = ȳ + Ig/(ȳA). For a 2×2 m gate with ȳ = 4 m: Ig = 2(2³)/12 = 1.333 m⁴. Extra drop = 1.333/(4×4) = 0.083 m. yp = 4.083 m below surface.

Recall Trigger

IGLOO pushes yp down

Topic

Center of Pressure — Special Case

Concept

For a surface-piercing vertical rectangle, center of pressure is at 2/3 depth

Anchor Id

Difficulty

easy

Memory Aid

Rhyme: 'When the top is at the surface and the gate goes straight and flat, the force hits at two-thirds down — and that is simply THAT.' For a vertical rectangle with its top edge at the free surface, ȳ = h/2 (centroid at half-depth) and Ig/(ȳA) = h/6, so yp = h/2 + h/6 = 2h/3. This is the classic two-thirds rule. Remember: 'Top-to-surface rectangle? Two-thirds, always, guaranteed.'

Anchor Type

rhyme

Why It Works

The rhyme and the definitive statement 'always, guaranteed' create a confident, retrievable fact. The two-thirds rule is a frequent board-exam answer.

Example Usage

3 m tall gate, top at surface: yp = 2/3 × 3 = 2.0 m from surface. Instant answer — no calculation needed once you recall the rhyme.

Recall Trigger

Two-thirds rhyme

Topic

Force on Curved Surfaces — Horizontal Component

Concept

Horizontal component of force on a curved surface = force on vertical projection

Anchor Id

Difficulty

medium

Memory Aid

Imagine a curved BANANA peel lying in water. If you want to know how hard the water pushes it sideways (horizontally), you don't trace the curve — you just FLATTEN the banana into a straight vertical rectangle (its vertical shadow/projection) and calculate the force on THAT flat surface. The water doesn't care about the curve for horizontal pushing; it only sees the silhouette. FH = γ·h̄·Avert is the force on the banana's shadow.

Anchor Type

analogy

Why It Works

The banana analogy replaces the abstract 'vertical projection' with a physical, funny image of flattening a banana — making the replacement step feel logical and visual.

Example Usage

Quarter-circle gate, radius 1.5 m, retains water. Vertical projection = 1.5 m tall × unit width. h̄ of projection = 0.75 m. FH = 9.81 × 0.75 × (1.5 × 1) = 11.04 kN/m.

Recall Trigger

Flatten the banana into its shadow

Topic

Force on Curved Surfaces — Vertical Component

Concept

Vertical component of force on a curved surface = weight of fluid above

Anchor Id

Difficulty

medium

Memory Aid

Story: A scuba diver (named Vergara) is holding up a curved tray above her head in a swimming pool. The weight of ALL the water sitting on that tray pushes DOWN on her arms — that's the vertical hydrostatic force. If the tray is concave (like a bowl facing up), the water above it is real. If the surface curves the other way (concave down, like an inverted bowl), imagine filling that space with imaginary 'ghost water' and compute its weight — the surface actually gets pushed UP by that force. Vergara's rule: 'FV equals the weight of water above me — real or ghost.'

Anchor Type

micro_story

Why It Works

The diver narrative makes the concept of 'weight of fluid above' physical and memorable. The 'ghost water' concept for inverted surfaces adds a creative hook for the harder case.

Example Usage

Quarter-circle gate (concave side facing fluid), radius R = 1.5 m, unit width. FV = γ × Volume above = 9.81 × (π × 1.5²/4 × 1) = 9.81 × 1.767 = 17.33 kN/m (upward if surface curves inward).

Recall Trigger

Vergara the diver holding a tray of water

Topic

Force on Curved Surfaces — Resultant

Concept

Resultant force on curved surface: F = √(FH² + FV²)

Anchor Id

Difficulty

medium

Memory Aid

Picture a RIGHT TRIANGLE on the face of a curved dam. The horizontal leg is FH (pointing sideways), the vertical leg is FV (pointing down or up), and the hypotenuse is the resultant F. This is simply the PYTHAGOREAN THEOREM applied to forces. See the triangle every time you encounter a curved surface problem. For a circular-arc surface, the hypotenuse (resultant) always points toward the CENTER OF CURVATURE — like all spokes of a wheel meeting at the hub.

Anchor Type

visual_association

Why It Works

The right triangle is one of the most ingrained visual memories in any engineering student. Mapping FH and FV to legs of a right triangle makes the combination formula automatic.

Example Usage

FH = 11.04 kN/m, FV = 17.33 kN/m. F = √(11.04² + 17.33²) = √(121.9 + 300.3) = √422.2 = 20.55 kN/m. Angle: θ = arctan(FV/FH).

Recall Trigger

Right triangle on the dam face; spokes meeting at the hub

Topic

Force on Curved Surfaces — Circular Arc

Concept

For a circular-arc gate, the resultant force passes through the center of curvature

Anchor Id

A10

Difficulty

hard

Memory Aid

Think of a bicycle WHEEL submerged in water. All the water pressure forces act radially — like spokes — perpendicular to the curved surface and pointing toward the axle (center of curvature). No matter how you combine them, the resultant must pass through the axle. A circular-arc gate is half a wheel: every pressure force is a spoke, and they all converge at the hub. The resultant is just the strongest spoke, aimed right at the center.

Anchor Type

analogy

Why It Works

The bicycle wheel is a universally understood object for engineering students. The spoke analogy makes the radial nature of pressure forces on a circular arc geometrically obvious.

Example Usage

When asked 'where does the resultant act on a circular gate?', visualize the wheel — it passes through the center of curvature (the pivot point of the arc's circle).

Recall Trigger

Bicycle wheel axle = center of curvature

Topic

Manometry

Concept

Manometry rule: add γh going DOWN, subtract γh going UP

Anchor Id

A11

Difficulty

medium

Memory Aid

Remember the hiking analogy: 'DESCEND = ADD, ASCEND = SUBTRACT.' Walking DOWN the manometer tube means you are going deeper into fluid — pressure increases, so ADD γh. Walking UP means you are rising — pressure decreases, so SUBTRACT γh. It's like hiking: going downhill (into a valley/deep fluid) adds pressure to your body; going uphill (out of fluid) relieves it. The manometer is just a hiking trail between two pressure points.

Anchor Type

mnemonic

Why It Works

Hiking is a kinesthetic, directional memory. The ADD/SUBTRACT rule becomes a physical direction rule (down/up) rather than an abstract algebraic rule.

Example Usage

U-tube manometer: Start at point A (unknown pressure pA), go DOWN through water (h1 = 0.5 m, add 9.81×0.5), go UP through mercury (h2 = 0.25 m, subtract 9.81×13.6×0.25), end at atmosphere (p = 0). Solve for pA.

Recall Trigger

Hiking: downhill ADD, uphill SUBTRACT

Topic

Pressure Concepts

Concept

Gauge pressure vs. absolute pressure: pabs = pgauge + patm

Anchor Id

A12

Difficulty

easy

Memory Aid

Think of your bank account. Your GAUGE balance is how much extra money you have above zero (reference = zero, just like gauge reference = atmospheric). Your ABSOLUTE balance is your actual total balance including the initial deposit the bank always holds (patm = 101.325 kPa, the 'base deposit' the atmosphere always adds). Most hydraulics problems use GAUGE because we're interested in the extra pressure above atmosphere — just like you care about your spending money above zero, not the bank's minimum balance.

Anchor Type

analogy

Why It Works

Money and bank accounts are instantly relatable for young Filipino graduates. The analogy maps perfectly: gauge = above reference, absolute = total from true zero.

Example Usage

p_abs = p_gauge + 101.325 kPa. If gauge pressure at a gate is 49.05 kPa, then absolute = 49.05 + 101.325 = 150.375 kPa. Board exams almost always want gauge unless stated otherwise.

Recall Trigger

Bank account: gauge = spending money, absolute = total balance

Topic

Hydrostatic Pressure Principles

Concept

Pressure is the same at all points on a horizontal plane in a connected fluid

Anchor Id

A13

Difficulty

easy

Memory Aid

Picture a BILAO (flat round bamboo tray) submerged horizontally in a cauldron of water. Every point on that bilao — left edge, right edge, center — feels EXACTLY the same pressure, because they are all at the same depth. If you tilt the bilao even slightly, the depths change and pressures differ. The bilao must be LEVEL (horizontal) and the fluid must be CONNECTED (same fluid, no walls blocking) for this rule to hold.

Anchor Type

visual_association

Why It Works

The bilao is a distinctly Filipino kitchen item, creating a culturally specific and therefore highly personal memory. The 'tilt breaks equality' addition prevents a common misapplication.

Example Usage

In a manometer problem, when you reach the same horizontal level in the same fluid on both sides of a U-tube, pressures are equal — this is the key equation you write to solve the problem.

Recall Trigger

Horizontal bilao in a cauldron

Topic

Moment of Inertia for Center of Pressure

Concept

Centroidal moments of inertia for common shapes

Anchor Id

A14

Difficulty

medium

Memory Aid

Remember the 'DIRTY DOZEN' fractions for Ig: RECTANGLE: bh³/12 (say 'a dozen, nice and clean'). TRIANGLE: bh³/36 (say 'three dozen — triangle has 3 sides, so multiply denominator by 3'). CIRCLE: πd⁴/64 or πr⁴/4. Trick: Rectangle = /12, Triangle = /36 (12 × 3, because triangle is 1/3 of a rectangle's Ig), Circle = πr⁴/4. Chant: 'Rect twelve, Tri thirty-six, Circle pi-r-four-over-four.'

Anchor Type

chunking

Why It Works

Chunking the three shapes into a short chant with a pattern (the factor of 3 connecting rectangle to triangle) creates a systematic rather than rote memory.

Example Usage

Gate is a 2 m wide, 3 m tall rectangle: Ig = (2)(3³)/12 = 4.5 m⁴. Instantly from the chant: denominator is 12, plug in b and h.

Recall Trigger

Rect-12, Tri-36, Circle-pi-r4-over-4

Topic

Inclined Plane Surfaces

Concept

ȳ vs h̄: equal for vertical surfaces, different for inclined

Anchor Id

A15

Difficulty

hard

Memory Aid

Story: Two engineering reviewees, Ybar and Hbar, are twins. On a VERTICAL gate, they are identical — same depth, same distance from surface. But put the gate on an INCLINE (like a tilted ramp at θ to the horizontal), and Ybar measures along the SLOPE while Hbar measures the VERTICAL depth. They are related by: h̄ = ȳ sin θ. On a vertical gate (θ = 90°), sin 90° = 1, so h̄ = ȳ — the twins are the same. On an inclined gate, Ybar is longer (along the slope) while Hbar is shorter (vertical component). Never confuse the twins on an inclined gate!

Anchor Type

micro_story

Why It Works

Personifying ȳ and h̄ as twins who separate on an incline creates an emotionally engaging narrative that makes the distinction between slope distance and vertical depth unforgettable.

Example Usage

Inclined gate at θ = 60°, centroid at ȳ = 3 m along slope. h̄ = 3 sin 60° = 2.598 m. F = γ h̄ A = 9.81 × 2.598 × A. Center of pressure: yp = ȳ + Ig/(ȳA) — use ȳ (along slope), not h̄.

Recall Trigger

Twins Ybar and Hbar — identical on vertical, different on inclined

Topic

Pressure Head

Concept

Pressure head: h = p/γ — expressing pressure as equivalent fluid depth

Anchor Id

A16

Difficulty

easy

Memory Aid

Pressure head is like converting Philippine Pesos to US Dollars. The pressure (in kPa) is your peso amount. Dividing by γ (the 'exchange rate' of the fluid, 9.81 kN/m³ for water) gives you the equivalent height of water column (metres). A pressure of 98.1 kPa = 98.1/9.81 = 10 m of water head. It's a currency conversion: kPa ↔ metres of water. Mercury has a different 'exchange rate' (γ_Hg = 133.4 kN/m³), so the same pressure converts to a much shorter height in mercury.

Anchor Type

analogy

Why It Works

Currency conversion is an everyday concept for Filipinos. Mapping pressure to its head equivalent via division by γ is structurally identical to currency conversion, making the formula intuitive.

Example Usage

Water pressure p = 49.05 kPa. Head = 49.05/9.81 = 5 m. Mercury equivalent head = 49.05/133.4 = 0.368 m. This is why mercury manometers are compact.

Recall Trigger

Pressure-to-head is a currency conversion (÷ by γ)

Topic

Center of Pressure Behavior

Concept

The deeper a submerged gate, the closer yp is to ȳ (center of pressure approaches centroid)

Anchor Id

A17

Difficulty

hard

Memory Aid

Imagine you are very deep in the Mariana Trench — so deep that the tiny pressure differences between the top and bottom of a gate are negligible compared to the MASSIVE pressure everywhere. At extreme depth, the pressure distribution across the gate is nearly UNIFORM (almost a flat rectangle), so the center of pressure moves right to the centroid. The extra drop Ig/(ȳA) → 0 as ȳ → ∞. Shallow gate = very non-uniform pressure = big drop. Deep gate = nearly uniform pressure = small drop.

Anchor Type

analogy

Why It Works

The Mariana Trench extreme case creates a memorable limit scenario. Understanding why the formula behaves at extremes prevents misapplication and builds conceptual depth.

Example Usage

Gate with ȳ = 100 m (very deep): extra drop = Ig/(100 × A) — extremely small. Gate at ȳ = 1 m (near surface): extra drop = Ig/(1 × A) — relatively large. Always compute, but know the trend.

Recall Trigger

Mariana Trench: deep gate, pressure nearly uniform, yp ≈ ȳ

Topic

Pressure Direction

Concept

Pressure acts perpendicular to any surface

Anchor Id

A18

Difficulty

easy

Memory Aid

Visualize thousands of tiny ARROWS (like arrows in a medieval battle) all hitting a submerged wall — every single arrow is pointing perfectly PERPENDICULAR (90°) to the wall surface, no matter how the wall is angled. There are no arrows sliding along the wall — that would be friction (shear stress), which is ZERO in a static fluid. The wall can be vertical, tilted at 30°, or curved like a barrel — the arrows always hit at 90°. This is the definition of hydrostatic pressure: pure normal force, no tangential component.

Anchor Type

visual_association

Why It Works

The arrow-shower is a memorable action scene. The contrast with friction (which is absent) reinforces the concept's boundary condition and prevents confusion with moving-fluid shear.

Example Usage

When computing force components on an angled or curved surface, always start from the fact that each elemental pressure force dF = p dA is perpendicular to the local surface. Integration gives the resultant.

Recall Trigger

Perpendicular arrows raining on the wall — no sliding arrows

Topic

Manometry — Mercury

Concept

Mercury specific gravity = 13.6 (key manometer constant)

Anchor Id

A19

Difficulty

easy

Memory Aid

Remember: 13.6 is the ATOMIC NUMBER of... wait, that's aluminum (13). But 13.6 is how much heavier mercury is than water. Chunking trick: '13.6 = a baker's dozen plus 0.6'. Or use: Hg → Hg = 'Heavy Ganda' (Filipino slang mash-up) = 13.6 times denser than water. γ_Hg = 13.6 × 9.81 = 133.4 kN/m³. In manometer problems, whenever you see mercury, multiply height by 13.6 (specific gravity) or by 133.4 kN/m³.

Anchor Type

chunking

Why It Works

The 'baker's dozen plus 0.6' chunking and the playful Filipino phrase create dual encoding — numerical pattern plus language association — for a constant that must be instantly recalled.

Example Usage

U-tube with mercury deflection of 250 mm = 0.25 m. Pressure contribution of mercury column: γ_Hg × h = 133.4 × 0.25 = 33.35 kPa. This equals the gauge pressure at the measurement point.

Recall Trigger

Baker's dozen + 0.6 = 13.6 = SG of mercury

Topic

Common Pitfall — Center of Pressure

Concept

The common board-exam pitfall: F acts at center of pressure, NOT at centroid

Anchor Id

A20

Difficulty

medium

Memory Aid

Story: A review student who always scored 99% in class wrote on the board exam: 'The hydrostatic force acts at the centroid.' He failed the hydraulics portion. His professor said: 'Centroid = location of force MAGNITUDE. Center of pressure = WHERE IT ACTUALLY ACTS.' The centroid is the BILLING ADDRESS; the center of pressure is WHERE THE PACKAGE IS DELIVERED — always further below. After failing, he wrote on a sticky note: 'FORCE IS DELIVERED BELOW THE CENTROID' and passed on his next attempt.

Anchor Type

micro_story

Why It Works

The failure story creates emotional impact (negative consequence = strong memory encoding). The billing address vs. delivery address analogy is modern, relatable, and structurally perfect for the concept.

Example Usage

Every time a question asks 'where does the force act?', immediately recall the delivery address story and write: yp = ȳ + Ig/(ȳA), never just ȳ.

Recall Trigger

Billing address (centroid) vs. delivery address (center of pressure)

Revision Game

F = γ·h̄·A (Total hydrostatic force on a plane surface)

Clue

I am the pressure at the centroid of a gate multiplied by the total gate area. I am the force you compute first before finding where I act. What formula produces me?

Memory Link

A3 — GAMMA HITS AREA mnemonic

Center of pressure (yp), located at yp = ȳ + Ig/(ȳA)

Clue

I am always lower than the centroid on a submerged gate. Engineers who ignore me and put a hinge at the centroid end up with a broken gate. I am the delivery address, not the billing address. What am I?

Memory Link

A5 — IGLOO formula mnemonic and A20 — billing vs. delivery address story

yp = 2 m (= 2/3 × 3 m from surface) — the two-thirds rule for surface-piercing rectangles

Clue

For a vertical rectangular gate whose top is exactly at the free water surface, I am always exactly two-thirds of the gate's height below the surface. Say my value for a 3 m tall gate.

Memory Link

A6 — Two-thirds rhyme

FH = γ·h̄·Avert (force on the vertical projection of the curved surface)

Clue

I am the horizontal component of force on a curved dam face. To find me, you do not trace the curve — you find my shadow on the wall and compute the force on that flat shadow instead. What formula computes me?

Memory Link

A7 — Flatten the banana into its shadow analogy

Mercury (Hg); SG = 13.6; γ_Hg = 133.4 kN/m³

Clue

I am a fluid 13.6 times denser than water. My presence in a manometer means a small height of me balances a large height of water. My specific weight is 133.4 kN/m³. What is my name and specific gravity?

Memory Link

A19 — Baker's dozen + 0.6 chunking

ADD γ·h (going down increases pressure)

Clue

In a manometry problem, you are walking along the tube. Every time you step downward through a fluid layer of height h and specific weight γ, what do you do to your running pressure total?

Memory Link

A11 — Hiking analogy: downhill ADDS, uphill SUBTRACTS

FV = γ·V (weight of real or imaginary fluid above the curved surface)

Clue

I am the vertical component of force on a curved gate. I equal the weight of the fluid sitting directly on top of the curved surface. If the fluid is imaginary (the surface curves downward), I push upward. My formula?

Memory Link

A8 — Vergara the diver holding a tray of water

The center of curvature of the circular arc

Clue

For a circular-arc gate, all pressure forces are perpendicular to the surface and therefore point toward one special location — the center of the circle. Where does the resultant force pass through?

Memory Link

A10 — Bicycle wheel axle = center of curvature analogy

h = pressure head (m of fluid column); p = pressure (kPa); γ = specific weight of fluid (kN/m³). For water: γ = 9.81 kN/m³. For mercury: γ = 133.4 kN/m³. A shorter column of denser fluid (mercury) represents the same pressure as a taller column of water.

F uses centroid depth h̄ — but F acts at center of pressure yp, not at h̄
ȳ (along plane) ≠ h̄ (vertical depth) for inclined surfaces; h̄ = ȳ sin θ
FV on curved surface is weight of fluid ABOVE, not γh̄A of curved area
Gauge vs absolute: almost always use gauge unless stated
Curved-surface resultant passes through center of curvature ONLY for circular arcs

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CELE Hydraulics & Fluid Mechanics — Hydrostatic Pressure and Forces on SurfacesMemory Anchors

Exam context

Hydrostatic Pressure and Forces on Surfaces - Memory Anchors

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