CELE Hydraulics & Fluid Mechanics — Hydrostatic Pressure and Forces on SurfacesExam Answer Templates

Marks

1

Topic

Pressure Variation with Depth

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners look for the phrase 'above atmospheric' in the definition and the formula p = γh. You may use p = ρgh but must equate γ = ρg. One sentence plus one formula is all that is needed — do not over-write.

Model Answer

Gauge pressure is the pressure measured above local atmospheric pressure. For a static liquid of unit weight γ at depth h below the free surface: p = γh where p is in Pa (or kPa), γ in N/m³ (or kN/m³), and h in m.

Question Type

very_short_answer

Answer Structure

• One sentence defining gauge pressure as pressure above atmospheric [½ mark]
• State the formula p = γh with correct symbols and units [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing p = ρgh without defining γ = ρg or converting units — risks confusion mark
• Omitting 'above atmospheric' from the definition, making it sound like absolute pressure
• No formula stated — definition alone scores zero for a 'state the formula' question

Key Phrases To Include

• gauge pressure
• above atmospheric
• p = γh
• unit weight γ
• depth h

Marks

1

Topic

Pressure Variation with Depth

Difficulty

easy

Template Id

T2

Examiner Tip

The key discriminator phrase is 'transmitted equally and undiminished in all directions.' Examiners reject answers that only say 'pressure increases with depth' — that is the hydrostatic formula, not Pascal's Law.

Model Answer

Pascal's Law states that pressure applied to a confined fluid is transmitted equally and undiminished in all directions throughout the fluid and acts perpendicular to any surface in contact with it. Application: Hydraulic jack — a small force on a small piston creates the same pressure that acts on a large piston, producing a much larger force.

Question Type

very_short_answer

Answer Structure

• Correct statement of Pascal's Law in one sentence [½ mark]
• One valid engineering application with brief explanation [½ mark]

Scoring Breakdown

Common Mark Deductions

• Confusing Pascal's Law with the hydrostatic pressure variation formula p = γh
• Giving 'water pressure' as an application without explaining the equal-transmission principle

Key Phrases To Include

• transmitted equally
• all directions
• perpendicular to surface
• confined fluid

Marks

2

Topic

Pressure Variation with Depth

Difficulty

easy

Template Id

T3

Examiner Tip

Always show the formula substitution explicitly. If you write only '117.72 kPa' without showing p = γh = 9.81 × 12, you may lose the method mark if the number is slightly off.

Model Answer

Given: h = 12 m, γ = 9.81 kN/m³, patm = 101.325 kPa Step 1 — Gauge pressure: pgauge = γh = 9.81 × 12 = 117.72 kPa ✓ Step 2 — Absolute pressure: pabs = pgauge + patm = 117.72 + 101.325 = 219.05 kPa ✓ Pressure head: h = p/γ = 117.72/9.81 = 12.0 m of water (check) ∴ Gauge pressure = 117.72 kPa; Absolute pressure = 219.05 kPa

Question Type

numerical

Answer Structure

• Line 1: List given data with units [setup — no mark but shows method]
• Line 2: Apply p = γh with substitution → pgauge [1 mark]
• Line 3: Add atmospheric pressure → pabs [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using γ = 9.81 N/m³ instead of 9.81 kN/m³ — answer in Pa instead of kPa, wrong unit
• Forgetting to add patm for absolute pressure — loses 1 mark
• Not labelling which answer is gauge and which is absolute

Key Phrases To Include

• pgauge = γh
• pabs = pgauge + patm
• 117.72 kPa
• 219.05 kPa

Marks

2

Topic

Manometry

Difficulty

medium

Template Id

T4

Examiner Tip

Always state the sign convention (add γh going down, subtract going up) before writing the equation. This earns the method mark even if you make an arithmetic error.

Model Answer

Given: Δh_Hg = 0.25 m, s_Hg = 13.6, γwater = 9.81 kN/m³ γHg = 13.6 × 9.81 = 133.416 kN/m³ Manometry equation (start at connection point A, end at open mercury surface): pA + γwater(0) − γHg(0.25) = 0 [right limb open to atmosphere] pA = γHg × 0.25 pA = 133.416 × 0.25 pA = 33.35 kPa (gauge) ✓ ∴ Gauge pressure at connection = 33.35 kPa

Question Type

numerical

Answer Structure

• Line 1: Compute γHg = s × γwater [setup]
• Line 2: Write the manometry pressure equation term by term [1 mark]
• Line 3: Solve for pA with correct unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Δh in mm instead of converting to m — gives answer 1000× too large
• Using γwater instead of γHg for the mercury column height
• Not stating the rule 'add going down, subtract going up' — loses method mark

Key Phrases To Include

• γHg = s × γwater
• manometry equation
• add γh going down, subtract going up
• 33.35 kPa

Marks

2

Topic

Force on a Plane Surface

Difficulty

medium

Template Id

T5

Examiner Tip

The phrase 'always below the centroid' is a key discriminating statement that examiners reward. Back it up with the formula — the term Ig/(ȳA) is always positive, so yp > ȳ always.

Model Answer

Centroid depth (h̄ or ȳ): The depth from the free surface to the centroid (geometric center) of the plane area. It is used to compute the MAGNITUDE of the total hydrostatic force: F = γ h̄ A Center of pressure (yp): The point on the surface where the resultant hydrostatic force effectively acts. Because pressure increases with depth, the pressure distribution is trapezoidal, shifting the resultant below the centroid: yp = ȳ + Ig/(ȳ A) Key distinction: The centroid depth gives the average pressure intensity; the center of pressure is always located BELOW the centroid (since Ig/(ȳA) > 0 for any finite area).

Question Type

short_answer

Answer Structure

• Paragraph 1: Define centroid depth h̄ and its role in force formula F = γh̄A [1 mark]
• Paragraph 2: Define center of pressure yp, give formula yp = ȳ + Ig/(ȳA), and state it is below the centroid [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that the force acts at the centroid — this is the most common conceptual error and loses 1 mark
• Omitting the formula for yp in a 'differentiate' question — loses the second mark
• Confusing ȳ (distance along inclined plane) with h̄ (vertical depth) for an inclined surface

Key Phrases To Include

• centroid depth
• center of pressure
• F = γh̄A
• yp = ȳ + Ig/(ȳA)
• below the centroid
• Ig/(ȳA) > 0

Marks

3

Topic

Force on a Plane Surface

Difficulty

easy

Template Id

T6

Examiner Tip

Always write the three steps as three distinct lines: (1) F, (2) Ig, (3) yp. Examiners mark each step independently. Even if your F is wrong, you still earn the Ig and yp marks if the method is correct.

Model Answer

Given: b = 2 m, d = 3 m, top at surface → h_top = 0 Area: A = 2 × 3 = 6 m² Centroid depth: h̄ = ȳ = 3/2 = 1.5 m (below surface) Step 1 — Total hydrostatic force: F = γ h̄ A = 9.81 × 1.5 × 6 F = 88.29 kN ✓ Step 2 — Centroidal moment of inertia (rectangle): Ig = bh³/12 = 2(3)³/12 = 54/12 = 4.5 m⁴ Step 3 — Center of pressure (measured from water surface along the plane): yp = ȳ + Ig/(ȳA) = 1.5 + 4.5/(1.5 × 6) yp = 1.5 + 4.5/9.0 = 1.5 + 0.5 yp = 2.0 m below the water surface ✓ [Note: yp = 2h/3 = 2(3)/3 = 2.0 m — confirms the two-thirds rule for a surface-piercing rectangle] ∴ F = 88.29 kN acting at 2.0 m below the free surface

Question Type

numerical

Answer Structure

• Line 1–2: Identify A and h̄ [setup]
• Line 3: Apply F = γh̄A → 88.29 kN [1 mark]
• Line 4: Compute Ig = bh³/12 = 4.5 m⁴ [1 mark]
• Line 5: Apply yp = ȳ + Ig/(ȳA) → 2.0 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using h̄ = 3 m (bottom depth) instead of 1.5 m (centroid depth) — loses force mark
• Using Ig = bh³/3 (base moment of inertia) instead of bh³/12 (centroidal) — loses Ig mark
• Reporting yp = 1.5 m (centroid location) instead of 2.0 m — loses location mark

Key Phrases To Include

• F = γh̄A
• Ig = bh³/12
• yp = ȳ + Ig/(ȳA)
• 88.29 kN
• 2.0 m
• center of pressure

Marks

3

Topic

Force on a Plane Surface

Difficulty

medium

Template Id

T7

Examiner Tip

The centroid depth formula h̄ = h_top + (height of gate)/2 is the critical first step. Write it explicitly. Everything else follows from this correct starting point.

Model Answer

Given: b = 2 m, d = 2 m, h_top = 3 m Area: A = 2 × 2 = 4 m² Centroid depth: h̄ = ȳ = 3 + 2/2 = 3 + 1 = 4.0 m Step 1 — Total hydrostatic force: F = γ h̄ A = 9.81 × 4.0 × 4 F = 156.96 kN ✓ Step 2 — Centroidal moment of inertia: Ig = bh³/12 = 2(2)³/12 = 16/12 = 1.333 m⁴ Step 3 — Center of pressure: yp = ȳ + Ig/(ȳA) = 4.0 + 1.333/(4.0 × 4) yp = 4.0 + 1.333/16 = 4.0 + 0.0833 yp = 4.083 m below the free surface ✓ ∴ F = 156.96 kN; Center of pressure = 4.083 m below surface [Observation: As submergence depth increases, yp approaches ȳ — the eccentricity Ig/(ȳA) becomes negligible.]

Question Type

numerical

Answer Structure

• Line 1: Compute h̄ = 3 + d/2 = 4.0 m [setup]
• Line 2: Apply F = γh̄A → 156.96 kN [1 mark]
• Line 3: Compute Ig = bh³/12 = 1.333 m⁴ [1 mark]
• Line 4: Apply yp = ȳ + Ig/(ȳA) → 4.083 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Setting h̄ = 3 m (top edge depth) — loses all three marks since every subsequent calculation is wrong
• Setting h̄ = 5 m (bottom edge depth) — same issue
• Omitting the 0.083 m eccentricity and reporting yp = 4.0 m — loses the location mark

Key Phrases To Include

• h̄ = htop + d/2
• F = γh̄A
• 156.96 kN
• Ig = 1.333 m⁴
• yp = 4.083 m

Marks

3

Topic

Force on a Plane Surface

Difficulty

hard

Template Id

T8

Examiner Tip

For inclined surfaces: ȳ is always measured ALONG the inclined plane, and h̄ = ȳ sinθ converts it to vertical depth. Draw the geometry showing h̄ and ȳ — this prevents the sin/cos confusion that costs most students marks.

Model Answer

Given: b = 2 m, L = 4 m (along incline), θ = 60°, top at water surface Area: A = 2 × 4 = 8 m² Centroid: ȳ = L/2 = 2.0 m (along the inclined plane from surface) Centroid depth: h̄ = ȳ sin θ = 2.0 × sin 60° = 2.0 × 0.866 = 1.732 m Step 1 — Total hydrostatic force: F = γ h̄ A = 9.81 × 1.732 × 8 F = 135.93 kN ✓ Step 2 — Centroidal moment of inertia: Ig = bL³/12 = 2(4)³/12 = 128/12 = 10.667 m⁴ Step 3 — Center of pressure (along the inclined plane from surface): yp = ȳ + Ig/(ȳA) = 2.0 + 10.667/(2.0 × 8) yp = 2.0 + 10.667/16 = 2.0 + 0.667 yp = 2.667 m along the inclined gate from the surface ✓ ∴ F = 135.93 kN; Center of pressure is 2.667 m along gate from top edge

Question Type

numerical

Answer Structure

• Line 1: Compute h̄ = ȳ sin θ = 1.732 m — critical inclined-surface step [1 mark]
• Line 2: Apply F = γh̄A → 135.93 kN [1 mark]
• Line 3: Compute Ig and apply yp = ȳ + Ig/(ȳA) → 2.667 m along plane [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using h̄ = ȳ (not multiplying by sinθ) — fundamental error for inclined surfaces
• Reporting yp as a vertical depth instead of distance along the inclined plane
• Using θ from vertical instead of from horizontal — sin and cos swapped

Key Phrases To Include

• h̄ = ȳ sinθ
• ȳ along inclined plane
• F = γh̄A
• Ig = bL³/12
• yp along inclined plane

Marks

2

Topic

Force on a Curved Surface

Difficulty

medium

Template Id

T9

Examiner Tip

The phrase 'real or imaginary fluid above' is crucial. When the fluid is below the surface (e.g., an upward-curved surface), V is the imaginary volume, and FV acts upward. State this clearly for full marks.

Model Answer

For a curved submerged surface, the hydrostatic force cannot be computed by F = γh̄A directly because the pressure vectors are not parallel. Instead, resolve into two orthogonal components: Horizontal component FH: FH = γ h̄_v Av where Av = vertical projection area of the curved surface, and h̄_v = depth to the centroid of that vertical projection. FH acts at the center of pressure of the vertical projection. Vertical component FV: FV = γ V where V = volume of fluid (real or imaginary) directly above the curved surface up to the free surface. FV acts through the centroid of that fluid volume. Resultant: F = √(FH² + FV²) For a circular-arc surface, the resultant passes through the center of curvature because all pressure forces are radial.

Question Type

short_answer

Answer Structure

• Sentence 1: State why direct F = γh̄A cannot be used for curved surfaces [½ mark]
• Line 2: FH = γ h̄_v Av — formula and description [½ mark]
• Line 3: FV = γV — formula and description of V [½ mark]
• Line 4: F = √(FH² + FV²); note on circular arc [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing FV = γh̄A (using plane-surface formula for vertical component) — incorrect
• Saying the vertical force is the weight of the gate, not the fluid above
• Not defining V as 'volume of fluid above the curved surface up to free surface'

Key Phrases To Include

• vertical projection
• FH = γh̄Av
• FV = γV
• weight of fluid above
• resultant F = √(FH² + FV²)
• center of curvature

Marks

5

Topic

Force on a Curved Surface

Difficulty

hard

Template Id

T10

Examiner Tip

Always sketch the geometry first — label the quarter-circle, the rectangular block of fluid above it, and the actual fluid volume = rectangle minus quarter-circle. Examiners reward a clearly labeled diagram. The direction of FV (up or down) depends on whether fluid is above (pushes down) or below (pushes up).

Model Answer

Given: R = 1.5 m, width b = 1 m (per unit width), γ = 9.81 kN/m³ Geometry: Quarter-circle — top at surface (depth = 0), bottom at depth = R = 1.5 m --- Step 1: Horizontal Component FH --- Use the vertical projection of the curved surface: Av = R × b = 1.5 × 1 = 1.5 m² (vertical rectangle, height R = 1.5 m) h̄_v = R/2 = 0.75 m (centroid of vertical projection from surface) FH = γ h̄_v Av = 9.81 × 0.75 × 1.5 FH = 11.036 kN/m ✓ Acts at center of pressure of vertical projection: Ig_vert = (1)(1.5)³/12 = 0.2813 m⁴ yp = 0.75 + 0.2813/(0.75 × 1.5) = 0.75 + 0.25 = 1.0 m below surface --- Step 2: Vertical Component FV --- Volume of fluid above the curved surface (quarter-circle wedge up to free surface): V = (R² − πR²/4) × b [rectangular block minus quarter-circle area] Area above curve = R² − (πR²/4) = (1.5)² − π(1.5)²/4 = 2.25 − π(2.25)/4 = 2.25 − 1.7671 = 0.4829 m² V = 0.4829 × 1 = 0.4829 m³ [Alternatively, the fluid above = full square R×R minus the quarter-circle area] Quarter-circle area = πR²/4 = π(1.5)²/4 = 1.7671 m² Rectangular area = R × R = 2.25 m² Volume of fluid above = (2.25 − 1.7671)(1) = 0.4829 m³ FV = γV = 9.81 × 0.4829 = 4.737 kN/m (↓ downward, fluid pushes down on concave surface) ✓ --- Step 3: Resultant --- F = √(FH² + FV²) = √(11.036² + 4.737²) F = √(121.79 + 22.44) = √144.23 F = 12.01 kN/m ✓ Direction: α = arctan(FV/FH) = arctan(4.737/11.036) = arctan(0.4292) = 23.2° below horizontal ∴ FH = 11.04 kN/m; FV = 4.74 kN/m; F = 12.01 kN/m at 23.2° below horizontal

Question Type

numerical

Answer Structure

• Step 1: Identify vertical projection Av and centroid depth h̄_v → compute FH = 11.04 kN/m [2 marks]
• Step 2: Identify volume of fluid above curved surface → compute FV = γV = 4.74 kN/m [2 marks]
• Step 3: Compute resultant F = √(FH² + FV²) = 12.01 kN/m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the full rectangle R×R as the fluid volume above — ignoring the quarter-circle cutout
• Computing FV = γ h̄ A (plane-surface method) for the curved surface — wrong approach
• Not specifying the direction (downward or upward) of FV for concave vs convex surfaces
• Omitting the resultant direction angle α — loses presentation mark

Key Phrases To Include

• vertical projection
• FH = γh̄Av
• volume of fluid above
• FV = γV
• F = √(FH² + FV²)
• quarter-circle area = πR²/4

Marks

5

Topic

Force on a Plane Surface

Difficulty

medium

Template Id

T11

Examiner Tip

Part (c) is where most students lose marks. The overturning moment arm is measured from the BASE, not from the surface. Always compute: arm = H − yp. State this clearly as 'distance from base = H − yp' to earn the method mark.

Model Answer

Given: H = 5 m (water height), b = 8 m, γ = 9.81 kN/m³ A = 8 × 5 = 40 m² h̄ = ȳ = H/2 = 2.5 m (centroid of vertical rectangle, top at surface) --- Part (a): Total Hydrostatic Force --- F = γ h̄ A = 9.81 × 2.5 × 40 F = 981.0 kN ✓ --- Part (b): Center of Pressure Depth --- Ig = bH³/12 = 8(5)³/12 = 8(125)/12 = 83.333 m⁴ yp = ȳ + Ig/(ȳ A) = 2.5 + 83.333/(2.5 × 40) yp = 2.5 + 83.333/100 = 2.5 + 0.833 yp = 3.333 m below the free surface ✓ [= 2H/3 = 2(5)/3 = 3.333 m — confirms two-thirds rule] ✓ --- Part (c): Overturning Moment About the Base --- The force F acts at yp = 3.333 m from surface. Distance from base = H − yp = 5 − 3.333 = 1.667 m M_OT = F × (H − yp) = 981.0 × 1.667 M_OT = 1635.0 kN·m ✓ ∴ (a) F = 981.0 kN; (b) yp = 3.333 m; (c) M_OT = 1635.0 kN·m

Question Type

numerical

Answer Structure

• Setup: Identify A = 40 m² and h̄ = 2.5 m
• Part (a): F = γh̄A = 981.0 kN [1 mark]
• Part (b): Ig = bH³/12 = 83.333 m⁴ [1 mark]; yp = ȳ + Ig/(ȳA) = 3.333 m [1 mark]
• Part (c): Lever arm = H − yp = 1.667 m [1 mark]; M_OT = F × arm = 1635.0 kN·m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using yp as the lever arm from base instead of H − yp — measures from surface, not from base
• Not computing overturning moment (stopping after part b) — loses 2 marks
• Using F = γHA (taking h̄ = H instead of H/2) — doubles the force

Key Phrases To Include

• F = γh̄A
• 981.0 kN
• Ig = bH³/12
• yp = 3.333 m = 2H/3
• lever arm = H − yp
• M_OT = 1635 kN·m

Marks

3

Topic

Force on a Plane Surface

Difficulty

medium

Template Id

T12

Examiner Tip

Memorize centroid locations: rectangle = h/2 from top; triangle apex-up = 2h/3 from apex = h/3 from base; semicircle = 4R/3π from diameter. These are tested directly in board exams.

Model Answer

Given: Triangular gate — apex at top, base b = 3 m, height h = 2 m Top edge at 1 m depth → apex at depth 1 m, base at depth 3 m Step 1 — Area of triangle: A = (1/2) × base × height = (1/2)(3)(2) = 3 m² Step 2 — Centroid depth of triangle: For a triangle with apex at top: centroid is at 2/3 of height from apex h̄ = 1 + (2/3)(2) = 1 + 1.333 = 2.333 m below surface [Triangle centroid is at h/3 from base = h/3 from bottom, so 2h/3 from top = 2(2)/3 = 1.333 m from apex] Step 3 — Total hydrostatic force: F = γ h̄ A = 9.81 × 2.333 × 3 F = 68.66 kN ✓ ∴ Total hydrostatic force = 68.66 kN

Question Type

numerical

Answer Structure

• Line 1: A = (1/2) × b × h = 3 m² [1 mark]
• Line 2: Centroid depth h̄ = 1 + 2h/3 = 2.333 m (apex-up triangle) [1 mark]
• Line 3: F = γh̄A = 68.66 kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using centroid at h/3 from apex (instead of 2h/3) — common sign confusion for triangles
• Using h̄ = 1 + 1 = 2 m (mid-height) instead of centroid location 2.333 m
• Using rectangle area bh = 6 m² instead of triangle area (1/2)bh = 3 m²

Key Phrases To Include

• A = (1/2)bh
• centroid at 2/3 h from apex
• h̄ = 1 + (2/3)(2)
• F = γh̄A
• 68.66 kN

Marks

3

Topic

Manometry

Difficulty

hard

Template Id

T13

Examiner Tip

Write out the traversal step by step, labeling each term as either '+' (going down) or '−' (going up). Examiners follow your steps — a clearly written equation earns the method mark even if arithmetic is slightly wrong.

Model Answer

Given: Δh_Hg = 0.15 m, hA_above = 0.5 m, hB_above = 0.3 m γwater = 9.81 kN/m³, γHg = 13.6 × 9.81 = 133.416 kN/m³ Rule: Start at point A, walk through the manometer to point B, adding γh going down and subtracting going up. pA + γwater(0.5) − γHg(0.15) − γwater(0.3) = pB [down from A to mercury surface left (+): + γw × 0.5] [up from mercury left to mercury right in Hg column (−): − γHg × 0.15] [up from mercury right to point B (−): − γw × 0.3] pA − pB = − γwater(0.5) + γHg(0.15) + γwater(0.3) pA − pB = −9.81(0.5) + 133.416(0.15) + 9.81(0.3) pA − pB = −4.905 + 20.012 + 2.943 pA − pB = 18.05 kPa ✓ ∴ pA − pB = 18.05 kPa

Question Type

numerical

Answer Structure

• Line 1: State sign convention and compute γHg [setup]
• Line 2: Write manometry equation term by term [1 mark]
• Line 3: Substitute values for each term [1 mark]
• Line 4: Solve pA − pB = 18.05 kPa [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the same fluid (water) for all terms — not switching to γHg for the mercury column
• Adding when should subtract or vice versa — sign error in traversal
• Using mm instead of m for Δh — answer off by factor of 1000

Key Phrases To Include

• add going down, subtract going up
• γHg = 13.6 × γwater
• pA + γw(0.5) − γHg(0.15) − γw(0.3) = pB
• 18.05 kPa

Marks

2

Topic

Force on a Plane Surface

Difficulty

medium

Template Id

T14

Examiner Tip

The pressure prism concept is occasionally tested as a 'show that' or 'alternative method' question. The key equation is F = Vol of prism. For a surface-piercing rectangle: Vol = (1/2)(γH)(H·b) = γ(H/2)(bH) = γh̄A. Write this out step by step.

Model Answer

The pressure prism is a geometric solid whose base is the submerged plane area A and whose height at any point equals the gauge pressure p = γh at that depth. For a vertical rectangular surface of width b and height H with top at the free surface: - The pressure varies linearly from 0 at the top to γH at the bottom. - The pressure prism is a triangular prism (wedge). - Total force F = Volume of pressure prism = (1/2)(γH)(bH) = γ(H/2)(bH) = γ h̄ A ✓ The resultant force acts through the centroid of the pressure prism volume, which for a triangular prism is H/3 from the base (or 2H/3 from the top) — confirming yp = 2H/3 for a surface-piercing rectangle. For a submerged gate (pressure non-zero at top), the prism becomes a trapezoidal solid, and F equals its volume.

Question Type

short_answer

Answer Structure

• Sentence 1: Define pressure prism — base = area, height = γh [½ mark]
• Sentence 2: F = volume of pressure prism — show equivalence with γh̄A [½ mark]
• Sentence 3: Location of force at centroid of prism volume — confirms yp [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing the pressure prism as a 2D diagram rather than a 3D solid — loses concept mark
• Not connecting volume of prism to F = γh̄A — the answer must show they are equivalent
• Saying the force acts at the base of the prism instead of the centroid of the prism

Key Phrases To Include

• pressure prism
• volume of pressure prism
• triangular prism
• centroid of prism volume
• F = γh̄A
• 2H/3 from top

Marks

5

Topic

Force on a Plane Surface

Difficulty

hard

Template Id

T15

Examiner Tip

Gate problems always require moment equilibrium. Draw the FBD showing three forces: total hydrostatic force F at yp, stop force Fs at the stop location, and hinge reaction at the hinge. Take moments about the hinge to eliminate the hinge reaction from the equation — this is the fastest path to Fs.

Model Answer

Given: b = 3 m, H = 4 m, top of gate at water surface; hinged at top, stop at bottom. A = 3 × 4 = 12 m², h̄ = H/2 = 2.0 m --- Step 1: Total Hydrostatic Force --- F = γh̄A = 9.81 × 2.0 × 12 = 235.44 kN --- Step 2: Center of Pressure (location of F) --- Ig = bH³/12 = 3(4)³/12 = 192/12 = 16 m⁴ yp = ȳ + Ig/(ȳA) = 2.0 + 16/(2.0 × 12) yp = 2.0 + 16/24 = 2.0 + 0.667 = 2.667 m from top [= 2H/3] --- Step 3: Moment Equilibrium (Free Body Diagram of gate) --- Take moments about the hinge (top): Clockwise moment by F: F × yp = 235.44 × 2.667 = 627.7 kN·m Counter-clockwise moment by stop force Fs: Fs × H = Fs × 4 ΣM_hinge = 0: Fs × 4 = 235.44 × 2.667 Fs = 627.7 / 4 Fs = 156.93 kN ✓ --- Check: ΣFx = 0 --- Hinge reaction RH = F − Fs = 235.44 − 156.93 = 78.51 kN (→, toward water) ∴ Force on bottom stop = 156.93 kN (away from water, outward)

Question Type

numerical

Answer Structure

• Step 1: F = γh̄A = 235.44 kN [1 mark]
• Step 2: Ig = 16 m⁴; yp = 2.667 m from top [1 mark]
• Step 3: Draw FBD of gate showing F at yp, Fs at bottom, hinge at top [1 mark]
• Step 4: ΣM_hinge = 0 → Fs × 4 = F × yp [1 mark]
• Step 5: Fs = 156.93 kN with correct unit and direction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Taking moments about the wrong point (e.g., about bottom instead of hinge) — sets up wrong equation
• Using F acting at H/2 = 2.0 m instead of yp = 2.667 m — wrong moment arm
• Not drawing or describing the FBD — loses the method mark
• Forgetting to state the direction of Fs (outward/away from water)

Key Phrases To Include

• F = γh̄A
• yp = 2.667 m = 2H/3
• ΣM_hinge = 0
• Fs × H = F × yp
• 156.93 kN
• FBD of gate