CELE Hydraulics & Fluid Mechanics — Hydrostatic Pressure and Forces on SurfacesDetailed Explanation
If the summary was not enough, this is the deep dive. Detailed explanations for Hydrostatic Pressure and Forces on Surfaces in the CELE Hydraulics & Fluid Mechanics context, written to turn surface familiarity into genuine understanding. Professional Regulation Commission (PRC) — Board of Civil Engineering's toughest CELE questions on this chapter are answered by the reasoning built here.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Hydraulics & Fluid Mechanics section sits under a "Core" weighting, and Hydrostatic Pressure and Forces on Surfaces is the 2nd chapter in the 10-chapter CELE Hydraulics & Fluid Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Hydraulics & Fluid Mechanics.
Hydrostatic Pressure and Forces on Surfaces - Detailed Explanation
Hydrostatics — the study of fluids at rest — is one of the highest-yield topics in the PRC Civil Engineer Licensure Examination under Hydraulics and Fluid Mechanics. Every dam, retaining wall, tank, sluice gate, and underwater pipeline in the Philippines must be designed to resist hydrostatic loads, making this topic both theoretically fundamental and practically indispensable. This chapter develops the complete framework: how pressure varies with depth (Pascal's law and the hydrostatic equation), how to read manometers, how to compute the total hydrostatic force on plane surfaces and locate its exact line of action (the center of pressure), and how to resolve hydrostatic forces on curved surfaces into manageable horizontal and vertical components. Board-exam problems in this area consistently test three skills: (1) correct identification of centroid depth, (2) proper application of the center-of-pressure formula, and (3) treatment of the vertical force on curved surfaces as the weight of a real or imaginary fluid column. Master these three skills and you will answer virtually every hydrostatics problem the board throws at you.
Concepts
Pressure Variation with Depth and Pascal's Law
In a static, incompressible fluid of unit weight γ (N/m³), the gauge pressure at any point located a vertical depth h below the free surface is: p = γh where γ = ρg. For fresh water, γ = 9.81 kN/m³; for seawater, γ ≈ 10.05 kN/m³. This linear relationship means pressure increases by 9.81 kPa for every 1 m of water depth. The equivalent pressure head h = p/γ lets you express pressure as a height of fluid. Pascal's Law states that pressure applied to an enclosed fluid transmits undiminished in all directions and acts perpendicularly to every surface it contacts. This is why hydrostatic pressure acts normal (perpendicular) to any surface, whether horizontal, vertical, or inclined. Absolute pressure = gauge pressure + atmospheric pressure (p_atm ≈ 101.325 kPa at sea level). In structural and hydraulic calculations, gauge pressure is used unless otherwise specified. For two points 1 and 2 in the same static fluid: p₁ + γz₁ = p₂ + γz₂ (hydrostatic equation) This means all points on the same horizontal plane in a connected fluid have the same pressure — a rule that simplifies manometer analysis enormously.
Examples
This is a direct application of p = γh. Note that the pressure head simply equals the depth for water. For any other fluid, divide gauge pressure by that fluid's unit weight to get head in terms of that fluid.
Scenario
Find the gauge pressure and absolute pressure at the bottom of a 12-m water tank open to the atmosphere.
Solution
p_gauge = γh = 9.81 kN/m³ × 12 m = 117.72 kPa p_abs = p_gauge + p_atm = 117.72 + 101.325 = 219.045 kPa Pressure head = 12 m of water (by definition)
When the surface has a non-zero applied pressure (compressed air, pressurized head), add it directly to the hydrostatic term. This scenario appears frequently in board problems involving pressurized tanks.
Scenario
A closed tank contains oil (s = 0.85) to a depth of 3 m. An air pressure of 20 kPa acts on the oil surface. Find the gauge pressure at the bottom of the tank.
Solution
γ_oil = 0.85 × 9.81 = 8.339 kN/m³ p_bottom = p_air + γ_oil × h = 20 + 8.339 × 3 = 20 + 25.01 = 45.01 kPa
Applications
- Design of water supply pipelines and pressure ratings (working pressure + water hammer allowance)
- Determining minimum wall thickness of cylindrical tanks (thin-wall hoop stress = pr/t)
- Depth-of-submergence calculations for underwater structures and cofferdams
- Pressure-relief valve settings for tanks and boilers
- Altitude corrections for atmospheric pressure in mountain water supply systems in the Cordilleras and Benguet
Misconceptions
- Confusing gauge and absolute pressure — board problems almost always use gauge unless 'absolute' is explicitly stated
- Assuming pressure depends on the shape of the container — it depends ONLY on depth (hydrostatic paradox)
- Using γ = 9.81 kN/m³ for fluids other than fresh water — always check specific gravity
- Forgetting to convert units: 1 kPa = 1 kN/m²; γ in kN/m³ with h in m gives p in kPa directly
Related Concepts
- Manometry
- Force on plane surfaces
- Buoyancy and Archimedes' principle
- Thin-wall pressure vessels
Common Exam Questions
Example
At what depth in seawater (s = 1.025) does gauge pressure equal 200 kPa? → h = p/γ = 200/(1.025 × 9.81) = 19.89 m
Approach
Identify the fluid, find its unit weight (γ = s × 9.81), multiply by depth. If the surface is pressurized, add that pressure.
Question Type
Direct pressure computation
Example
A column of 2 m water sits on 0.5 m oil (s=0.8). Total pressure at base = 9.81×2 + 0.8×9.81×0.5 = 19.62 + 3.924 = 23.54 kPa
Approach
Equate total pressure at the common level: Σ(γ_i × h_i) going down minus Σ(γ_i × h_i) going up.
Question Type
Pressure equivalence / multi-fluid column
Key Points To Remember
- p = γh (gauge pressure); γ_water = 9.81 kN/m³
- Pressure is the same at all points on the same horizontal level in a connected static fluid
- Pressure acts perpendicular (normal) to any surface — this is Pascal's Law
- Absolute pressure = gauge pressure + 101.325 kPa
- Pressure head h = p/γ; converting between pressure units and metres of fluid is a common board-exam step
- Specific gravity s = γ_fluid/γ_water; for mercury s = 13.6, so γ_Hg = 13.6 × 9.81 = 133.4 kN/m³
Manometry
A manometer is a device that measures pressure by balancing fluid columns. The fundamental rule — memorize it exactly — is: Walking along the manometer tube from the known end to the unknown end: • ADD γh when you move DOWNWARD through a fluid • SUBTRACT γh when you move UPWARD through a fluid Set the sum equal to the pressure at the other end. For a simple open piezometer: p_A = γh (h = height of fluid in the tube above point A). For a U-tube manometer with a gage fluid (usually mercury, s = 13.6) connected between two pipes A and B: p_A + γ₁h₁ − γ_m × Δh − γ₂h₂ = p_B where Δh is the manometer deflection (difference in gage-fluid levels), γ_m is the gage-fluid unit weight, and γ₁, γ₂ are the unit weights of the pipe fluids. For a differential manometer measuring p_A − p_B: p_A − p_B = Δh(γ_m − γ_f) [when both pipes carry the same fluid of unit weight γ_f] This form is very common in board exams.
Examples
Starting from the open (atmospheric) end ensures p = 0 at the start. The algebra automatically gives gauge pressure at A. Note: always convert mm to m before multiplying by γ in kN/m³.
Scenario
A U-tube manometer with mercury (s = 13.6) is connected to pipe A carrying water. The mercury in the right (open) arm is 250 mm higher than in the left arm connected to the pipe. The water column above the left mercury level is 400 mm. Find the gauge pressure at A.
Solution
γ_water = 9.81 kN/m³, γ_Hg = 13.6 × 9.81 = 133.42 kN/m³ Starting from the open end (p = 0 gauge): Move down through Hg: +133.42 × 0.250 = +33.35 kPa Move up through water: −9.81 × 0.400 = −3.924 kPa Result at A: p_A = 33.35 − 3.924 = 29.43 kPa
This shortcut formula is valid when both pipes carry the same fluid. It is derived from the walking rule applied to a symmetric differential manometer. Memorize it — it saves time on the board exam.
Scenario
Two water pipes A and B are connected by a differential mercury manometer. The manometer deflection Δh = 300 mm. Find p_A − p_B.
Solution
p_A − p_B = Δh(γ_Hg − γ_water) = 0.300 × (133.42 − 9.81) = 0.300 × 123.61 = 37.08 kPa
Applications
- Measuring pressure drop across filters, valves, and pumps in water treatment plants
- Calibration of pressure gauges in MWSS and local water district facilities
- Laboratory determination of fluid properties
- Pipeline pressure testing before commissioning (required under PD 1067, Water Code of the Philippines)
Misconceptions
- Adding γh when moving UP — the rule is ADD going down, SUBTRACT going up
- Forgetting to account for the pipe-fluid column above the manometer connection point
- Using diameter or cross-sectional area of the tube — manometry depends only on height, not tube geometry
- Confusing 'deflection' Δh with the height of individual arms — Δh is the DIFFERENCE in gage-fluid levels between the two arms
Related Concepts
- Pressure variation with depth
- Specific gravity and unit weight
- Bernoulli equation (pitot-static tube is a form of manometer)
Common Exam Questions
Example
Open manometer, Hg deflection = 150 mm, water above = 200 mm → p = 133.42×0.15 − 9.81×0.20 = 20.01 − 1.96 = 18.05 kPa
Approach
Draw the manometer, label all levels and fluids. Start at the open end (p=0) or the known-pressure end. Walk step by step, adding/subtracting γh. Solve for the unknown.
Question Type
Find unknown gauge pressure using U-tube manometer
Example
Δh = 500 mm Hg, water pipes: p_A − p_B = 0.5×(133.42−9.81) = 61.81 kPa
Approach
Use p_A − p_B = Δh(γ_m − γ_f) when both pipes carry the same fluid.
Question Type
Differential manometer — find pressure difference
Key Points To Remember
- Rule: Add γh going down, subtract γh going up — apply at every fluid interface
- At the meniscus (interface between two fluids), pressure is continuous
- All points on the same horizontal level in the SAME fluid have the same pressure — use this to jump across U-tube bottoms
- For differential manometer: p_A − p_B = Δh(γ_gage − γ_pipe_fluid)
- Mercury (s = 13.6) is the most common gage fluid in board problems; sometimes oil (s < 1) is used as a lighter gage fluid for small pressure differences
- Inverted U-tube manometers use a lighter gage fluid (oil or air) and are used when p_A > p_B and both fluids are liquids
Total Hydrostatic Force on a Plane Surface
This is the single most tested hydrostatics concept in the board exam. When a plane surface of area A is submerged in a fluid and its centroid is at depth h̄ below the free surface, the total hydrostatic force is: F = γ h̄ A This is derived by integrating the pressure over the area: F = ∫p dA = ∫γh dA = γ ∫h dA = γ h̄ A. The force F does NOT act at the centroid of the area. It acts at the CENTER OF PRESSURE (CP), which is located BELOW the centroid because pressure increases with depth. The depth of the CP (measured from the free surface along the plane, i.e., along the slant distance ȳ for inclined surfaces) is: y_p = ȳ + I_g / (ȳ A) where: ȳ = slant distance from the surface (or the surface extended) to the centroid, measured along the plane I_g = second moment of area (moment of inertia) of the surface about its own centroidal horizontal axis A = total area of the surface For a VERTICAL surface: ȳ = h̄ (slant distance equals depth). For an INCLINED surface at angle θ: h̄ = ȳ sin θ, so ȳ = h̄ / sin θ. The eccentricity e = I_g / (ȳ A) is always positive, confirming CP is always below the centroid. As the surface goes deeper (ȳ increases), e decreases and CP approaches the centroid. Critical I_g values (memorize for board exam): Rectangle (b × d): I_g = bd³/12 Triangle (base b, height h): I_g = bh³/36 Circle (diameter D): I_g = πD⁴/64 Semicircle (diameter D): I_g = 0.1098 R⁴ The horizontal location of CP coincides with the centroid of the area (pressure is symmetric about the vertical centroidal axis for vertically symmetric surfaces). For asymmetric areas, use I_xy,g / (ȳ A) to find horizontal eccentricity — rarely tested in PRC boards.
Examples
Note that 2.0 m = (2/3) × 3 m — confirming the 'two-thirds rule' for a surface-piercing vertical rectangle. This shortcut is tested directly: for any vertical rectangle with top at the surface, CP is always at two-thirds of the height from the top.
Scenario
A vertical rectangular gate 2 m wide × 3 m tall has its top edge at the water surface. Find F and the location of the center of pressure.
Solution
A = 2 × 3 = 6 m² h̄ = ȳ = 3/2 = 1.5 m (centroid at mid-height for rectangle with top at surface) F = γ h̄ A = 9.81 × 1.5 × 6 = 88.29 kN I_g = bd³/12 = 2(3)³/12 = 54/12 = 4.5 m⁴ e = I_g/(ȳA) = 4.5/(1.5 × 6) = 4.5/9 = 0.5 m y_p = 1.5 + 0.5 = 2.0 m below the surface
When the gate is deeply submerged, CP is very close to the centroid (small eccentricity). The centroid is 3 + 1 = 4 m deep (top edge 3 m + half the gate height 1 m). This is a classic board-exam format.
Scenario
A 2 m × 2 m square vertical gate has its top edge 3 m below the water surface. Find F and y_p.
Solution
h̄ = ȳ = 3 + (2/2) = 3 + 1 = 4 m A = 2 × 2 = 4 m² F = 9.81 × 4 × 4 = 156.96 kN I_g = 2(2)³/12 = 16/12 = 1.333 m⁴ e = 1.333/(4 × 4) = 1.333/16 = 0.0833 m y_p = 4 + 0.0833 = 4.083 m below the surface
For inclined surfaces, ȳ and y_p are measured along the inclined plane (slant distance), NOT as vertical depths. To convert y_p to vertical depth: h_p = y_p × sin θ = 2.667 × sin 60° = 2.31 m. Always clarify whether the answer is asked as slant distance or vertical depth.
Scenario
An inclined rectangular gate (θ = 60° from horizontal) is 1.5 m wide and 4 m long. Its upper edge is at the water surface. Find F and y_p measured along the inclined plane.
Solution
A = 1.5 × 4 = 6 m² ȳ = 4/2 = 2 m (distance along incline to centroid from surface) h̄ = ȳ sin 60° = 2 × 0.866 = 1.732 m F = γ h̄ A = 9.81 × 1.732 × 6 = 101.94 kN I_g = (1.5)(4)³/12 = 32 m⁴ ← wait, I_g = bL³/12 = 1.5(4³)/12 = 1.5(64)/12 = 8 m⁴ e = I_g/(ȳA) = 8/(2 × 6) = 8/12 = 0.667 m (along the incline) y_p = 2 + 0.667 = 2.667 m along the incline from the surface
Applications
- Design of dam faces (gravity dams like Magat, Angat) — overturning and sliding forces
- Hydraulic gate design: sluice gates at NIA irrigation systems
- Tank wall and partition design
- Cofferdam sheet pile design for Metro Manila flood control projects
- Lock gate design for waterway navigation
Misconceptions
- Applying F at the CENTROID instead of the CENTER OF PRESSURE — this is the #1 error in board exams
- Using vertical depth h̄ in the I_g/(ȳA) formula instead of slant distance ȳ for inclined surfaces
- Using I_base (moment of inertia about the base) instead of I_g (about the centroid) — the parallel-axis theorem relationship is I_base = I_g + Aȳ², NOT I_g alone
- Forgetting that the two-thirds rule (y_p = 2H/3) applies ONLY when the top edge is exactly at the free surface
- Computing F as the pressure at the centroid times area but then NOT applying the eccentricity correction for the CP location
Related Concepts
- Moment of inertia (second moment of area) from Engineering Mechanics
- Centroid location of common shapes
- Force on curved surfaces (resolved from the same principles)
- Parallel-axis theorem
- Dam stability analysis (overturning moment)
Common Exam Questions
Example
Circular gate D=1.5 m, center at 4 m depth: A=π(1.5)²/4=1.767 m², F=9.81×4×1.767=69.34 kN, I_g=π(1.5)⁴/64=0.2485 m⁴, e=0.2485/(4×1.767)=0.0352 m, y_p=4.035 m
Approach
Step 1: Find h̄ = depth to centroid. Step 2: Compute F = γ h̄ A. Step 3: Compute I_g for the shape. Step 4: e = I_g/(ȳA). Step 5: y_p = ȳ + e.
Question Type
Find total force and CP on a fully submerged vertical gate
Example
Vertical rectangular gate 2×4 m, hinged at top, water to full height: F=88.29 kN (earlier), y_p=2 m, bottom reaction R=F×(y_p−0)/4... solve by moments
Approach
Take moments about the hinge. F acts at CP. If hinged at top, reaction at bottom R_B = F × (y_p − y_top)/(gate length). If hinged at bottom, R_T = F × (y_bottom − y_p)/(gate length).
Question Type
Find hinge reaction when gate is supported at top or bottom
Example
θ=45°, top at surface, 3m×3m gate: ȳ=1.5 m, h̄=1.5sin45°=1.06 m, F=9.81×1.06×9=93.6 kN
Approach
Use ȳ (along incline) for I_g/(ȳA). Convert depths using h = ȳ sinθ.
Question Type
Inclined gate — force and CP location
Key Points To Remember
- F = γ h̄ A — use CENTROID depth h̄ for force magnitude
- CP is always BELOW the centroid: y_p = ȳ + I_g/(ȳA)
- For vertical surfaces: ȳ = h̄; for inclined surfaces: ȳ = h̄/sinθ
- Memorize I_g for rectangle, triangle, and circle
- As depth increases, CP approaches centroid (eccentricity e = I_g/(ȳA) decreases)
- When the top edge of a vertical rectangle is at the water surface: y_p = 2H/3 (two-thirds depth) — a very common result
- The unit of F is in kN when γ is in kN/m³, h̄ in m, and A in m²
Total Hydrostatic Force on a Curved Surface
Because pressure on a curved surface acts normal to the surface at every point in different directions, direct integration to find a single resultant is complex. The practical approach is to resolve the total hydrostatic force into horizontal and vertical components: F_H (Horizontal Component): F_H equals the hydrostatic force on the VERTICAL PROJECTION of the curved surface. F_H = γ h̄_v A_v where A_v is the area of the vertical projection and h̄_v is the depth to its centroid. The LINE OF ACTION of F_H is found using the center-of-pressure formula applied to the vertical projection. F_V (Vertical Component): F_V equals the WEIGHT of the fluid (real or imaginary) in the volume directly above the curved surface up to the free surface. F_V = γ V where V is the volume of the fluid column above the curved surface. • If the fluid is ABOVE the curved surface (surface curves concave upward relative to the fluid), F_V acts DOWNWARD = weight of fluid directly above. • If the surface curves such that there is NO real fluid above (fluid is below the surface, e.g., the underside of a curved surface), F_V acts UPWARD = weight of the IMAGINARY fluid column that would fill the volume from the surface up to the free surface. Resultant Force: F_R = √(F_H² + F_V²) Direction: tan α = F_V / F_H SPECIAL CASE — Circular Arc Surface: All pressure forces on a circular arc are radial (directed toward or away from the center of curvature). Therefore, the resultant F_R also passes through the center of the circle. This is a powerful geometric property that greatly simplifies problems involving circular gates, cylindrical tanks, and arch dam faces.
Examples
The resultant passes through the center of curvature (top-left corner) since this is a circular arc. The vertical force is the weight of the real water volume (quarter-circle prism) above the gate. Note the volume uses the area formula for a quarter-circle: πR²/4.
Scenario
A quarter-circle gate of radius R = 1.5 m and width b = 2 m retains water on its left side. The gate curves from a horizontal floor to a vertical wall with its center of curvature at the top-left corner. The water depth equals R = 1.5 m. Find F_H, F_V, and F_R.
Solution
Vertical projection: A_v = R × b = 1.5 × 2 = 3.0 m² (vertical rectangle, height = R = 1.5 m, top at surface) h̄_v = 1.5/2 = 0.75 m F_H = γ h̄_v A_v = 9.81 × 0.75 × 3.0 = 22.07 kN Volume above curved surface: quarter-circle of radius R, width b V = (πR²/4) × b = (π × 1.5²/4) × 2 = (π × 2.25/4) × 2 = 3.534 m³ F_V = γ V = 9.81 × 3.534 = 34.67 kN (downward, fluid above surface) F_R = √(22.07² + 34.67²) = √(487.1 + 1202) = √1689.1 = 41.10 kN Angle: α = arctan(34.67/22.07) = arctan(1.571) = 57.5° below horizontal
When computing the vertical force on the underside of a curved surface, the imaginary fluid column extends from the surface UP to the free surface. This produces an upward vertical force. The full cylinder's buoyancy equals γ × πR²L, confirming the method is consistent with Archimedes' principle.
Scenario
The underside of a circular cylinder of radius 0.8 m and length 3 m is submerged with the center of the cylinder at 2 m below the water surface. Find the net vertical hydrostatic force on the lower half of the cylinder.
Solution
Volume of semicircle (lower half): V_semi = (πR²/2) × L = (π × 0.64/2) × 3 = 3.016 m³ The lower half: F_V acts upward = weight of imaginary fluid from the lower surface up to free surface. Volume of fluid column = volume of rectangle from axis to surface MINUS volume of semicircle below axis Rectangle volume = R × (depth of center) × L = 0.8 × 2 × 3 = 4.8 m³ ... Actually: F_V (upward) on lower half = weight of fluid in volume = [rectangular prism from lower arc to surface − nothing... ] Correct approach: Volume above lower semicircle up to surface = (area of rectangle 2R wide, height = depth of center + R) × L − (semicircle area) × L is for the full lower half exposed to fluid below. Simplified: F_V upward on bottom half = γ × [(2 × center depth × 2R)/2 × L + upper calculation...] For this scenario using the standard method: F_V_up = γ × V_above_lower_surface = 9.81 × [(π×0.64/2) × 3] = 9.81 × 3.016 = 29.59 kN upward (this equals the buoyant force on the semicircular cross section — consistent with Archimedes' principle)
Applications
- Tainter (radial) gates at dam spillways — the resultant passes through the trunnion pin (center of curvature), so no hydraulic moment is generated about the pin
- Cylindrical tanks and pipelines — hoop stress design
- Arch dam design — distributing hydrostatic force to the abutments
- Ship hull design and submarine pressure hull calculations
- Curved cofferdam cells used in Philippine river channelization projects
Misconceptions
- Using F_V = γ h̄ A_curved — WRONG; F_V is the weight of the fluid volume above, NOT γh̄A
- Thinking F_V always acts downward — it acts downward only if real fluid is above; it acts UPWARD when the fluid is below the surface (imaginary fluid column)
- Forgetting that F_H uses the VERTICAL PROJECTION area, not the actual curved area
- For a circular arc, failing to use the geometric property that F_R passes through the center — missing this leads to unnecessary and incorrect moment calculations
Related Concepts
- Buoyancy and Archimedes' principle (F_V connects directly to buoyancy)
- Force on plane surfaces (F_H is computed using plane-surface method)
- Vector addition and resultant forces
- Tainter gate and radial gate design
Common Exam Questions
Example
Quarter-circle R=2m, width=1m, top at water surface: F_H=γ(R/2)(R×1)=9.81(1)(2)=19.62 kN; F_V=γ(πR²/4)(1)=9.81(π×4/4)=30.82 kN; F_R=√(19.62²+30.82²)=36.54 kN
Approach
Step 1: Identify vertical and horizontal projections. Step 2: F_H = γh̄A_v. Step 3: F_V = γ × volume of fluid prism above. Step 4: Combine vectorially.
Question Type
Quarter-circle gate — find all force components
Example
A Tainter gate of radius 3 m — the resultant hydrostatic force always passes through the trunnion axis, making the hydraulic moment about the trunnion zero. This is the design advantage of radial gates.
Approach
For any circular arc, state that all pressure vectors are radial, so the resultant F_R passes through the center of curvature (pivot/trunnion).
Question Type
Tainter gate — where does the resultant act?
Key Points To Remember
- F_H = force on the VERTICAL PROJECTION (use plane-surface method on A_v)
- F_V = weight of REAL or IMAGINARY fluid above the curved surface
- F_V acts downward if real fluid is above; acts upward if imaginary fluid (fluid is below surface)
- F_R = √(F_H² + F_V²); angle α = arctan(F_V/F_H)
- For circular-arc surfaces, F_R passes through the CENTER OF THE CIRCLE
- The volume of the fluid prism is often computed as area × length (for 2D problems with unit width or given width)
- F_V for a quarter-circle gate of radius R and width b: V = (πR²/4) × b (quarter-circle area times width)
Practice Problems
The hinge at the top carries a reaction R_T = F − R_B = 50.84 − 33.89 = 16.95 kN. The stop at the bottom carries more than the hinge because the CP (where F acts) is below mid-gate height — it is at the 2/3 point, which is closer to the bottom. This is physically correct: deeper water exerts greater pressure near the bottom, so the bottom reaction is larger.
Problem
Problem 1 (Board-Exam Style — Vertical Gate with Hinge) A vertical rectangular gate 1.8 m wide and 2.4 m high is hinged at its top edge and retains water on one side with water depth equal to the full gate height. A horizontal stop is provided at the bottom edge. Determine: (a) the total hydrostatic force on the gate, (b) the location of the center of pressure, and (c) the reaction at the bottom stop.
Solution
(a) Total Force: A = 1.8 × 2.4 = 4.32 m² h̄ = 2.4/2 = 1.2 m (top edge at surface, centroid at mid-height) F = γ h̄ A = 9.81 × 1.2 × 4.32 = 50.84 kN (b) Center of Pressure: I_g = bh³/12 = 1.8(2.4)³/12 = 1.8(13.824)/12 = 2.074 m⁴ e = I_g/(ȳA) = 2.074/(1.2 × 4.32) = 2.074/5.184 = 0.400 m y_p = 1.2 + 0.400 = 1.6 m below the water surface (Check: y_p = 2H/3 = 2(2.4)/3 = 1.6 m ✓ — confirms two-thirds rule) (c) Reaction at bottom stop (R_B): Take moments about the hinge (top edge, at depth 0): F acts at y_p = 1.6 m from top (along gate = 1.6 m from hinge) Gate height H = 2.4 m R_B × 2.4 = F × 1.6 R_B = 50.84 × 1.6 / 2.4 = 33.89 kN
The key step is establishing the geometry correctly — identifying where each pipe connects and the heights of water and mercury columns. The walking method, applied carefully, automatically accounts for all contributions. A positive result means p_A > p_B, consistent with A being higher and Hg deflecting toward the B side.
Problem
Problem 2 (Board-Exam Style — Differential Manometer) Pipes A and B are 0.5 m apart vertically, with pipe A higher than pipe B. Both carry water. A differential mercury manometer (s_Hg = 13.6) connects them. The mercury levels show a deflection Δh = 180 mm, with the high-mercury level on the side of pipe B. The water column from pipe B down to the lower Hg level is 0.3 m. Find p_A − p_B.
Solution
γ_water = 9.81 kN/m³ γ_Hg = 13.6 × 9.81 = 133.42 kN/m³ Using the walking rule from A to B: Start at A (p_A). Note: pipe A is 0.5 m above pipe B. Δh = 0.180 m. Water above lower Hg level (B side) = 0.3 m. Let the connection from A go down to the Hg level on the A-side. Call the height from A down to the upper Hg level = x. By geometry: x + Δh = distance from A down to lower Hg level. Since A is 0.5 m above B: the lower Hg level is 0.3 m below B, so it is (0.5 + 0.3) = 0.8 m below A. Upper Hg level (A-side) = 0.8 − 0.180 = 0.620 m below A, so x = 0.620 m. Walking from A to B: p_A + γ_w(0.620) − γ_Hg(0.180) − γ_w(0.300) = p_B p_A − p_B = γ_Hg(0.180) − γ_w(0.620) + γ_w(0.300) p_A − p_B = 133.42(0.180) − 9.81(0.620) + 9.81(0.300) p_A − p_B = 24.016 − 6.082 + 2.943 p_A − p_B = 20.88 kPa
The critical distinction for inclined surfaces: use slant distance ȳ in the eccentricity formula, then convert back to vertical depth by multiplying by sinθ. The centroid depth h̄ = 2.061 m was found using the vertical depth of the top plus the vertical drop to the centroid (half the slant length times sinθ).
Problem
Problem 3 (Board-Exam Style — Inclined Gate) A rectangular gate 1.2 m wide and 3 m long is inclined at 45° from the horizontal. Its upper edge is 1 m below the water surface (measured vertically). Find: (a) the total hydrostatic force, and (b) the depth of the center of pressure (vertical depth from water surface).
Solution
(a) Total Force: Vertical depth to upper edge: h_top = 1 m Vertical depth to centroid: h̄ = h_top + (L/2)sinθ = 1 + (3/2)sin45° = 1 + 1.5(0.7071) = 1 + 1.0607 = 2.061 m A = 1.2 × 3 = 3.6 m² F = γ h̄ A = 9.81 × 2.061 × 3.6 = 72.77 kN (b) Center of Pressure — Slant Distance: ȳ (slant distance from surface to centroid along incline): ȳ = h̄ / sinθ = 2.061 / sin45° = 2.061 / 0.7071 = 2.914 m I_g = bL³/12 = 1.2(3)³/12 = 1.2(27)/12 = 2.7 m⁴ e = I_g/(ȳA) = 2.7/(2.914 × 3.6) = 2.7/10.490 = 0.2574 m (along incline) y_p (slant) = 2.914 + 0.2574 = 3.171 m from surface along the incline Vertical depth to CP: h_p = y_p × sinθ = 3.171 × sin45° = 3.171 × 0.7071 = 2.242 m below water surface
The quarter-circle volume above the gate equals the quarter-circle area times the gate width. The resultant passes through the pivot (center of curvature) — this is the key geometric property of circular arc gates. In a Tainter (radial) gate design, this means the hydraulic moment about the pivot pin is zero, making operation easier.
Problem
Problem 4 (Board-Exam Style — Quarter Circle Curved Gate) A quarter-circle gate of radius 2 m and width 1.5 m retains water. The gate is positioned with its curved surface on the upstream side, curving from the horizontal channel floor to a vertical wall. The water depth is 2 m (equal to the radius). Determine: (a) F_H, (b) F_V, (c) the resultant force, and (d) its direction.
Solution
(a) Horizontal Component: Vertical projection: rectangle of height R = 2 m, width b = 1.5 m A_v = 2 × 1.5 = 3.0 m² h̄_v = 2/2 = 1.0 m (top of vertical projection at water surface) F_H = γ h̄_v A_v = 9.81 × 1.0 × 3.0 = 29.43 kN (horizontal, acting on the gate) (b) Vertical Component: Volume above the curved surface = quarter-circle cross section × width Area_quarter_circle = πR²/4 = π(2)²/4 = π m² V = π × 1.5 = 1.5π = 4.712 m³ F_V = γ V = 9.81 × 4.712 = 46.22 kN (downward — real fluid above) (c) Resultant: F_R = √(F_H² + F_V²) = √(29.43² + 46.22²) = √(866.1 + 2136) = √3002 = 54.79 kN (d) Direction: tan α = F_V / F_H = 46.22/29.43 = 1.571 α = arctan(1.571) = 57.5° below horizontal The resultant passes through the center of the circular arc (the top-left corner where the wall meets the floor, i.e., the center of curvature).
Pressurized tanks add the air pressure to every point in the fluid. The bottom sees uniform pressure (no distance variation). The side wall sees both the uniform air pressure over the air space AND the variable hydrostatic pressure over the water depth. Splitting the wall into the air strip and water strip, then summing, is the most systematic board-exam approach.
Problem
Problem 5 (Board-Exam Style — Pressurized Tank) A closed rectangular tank 3 m long, 2 m wide, and 2.5 m tall is filled with water to a depth of 2 m. The air space above (0.5 m) is pressurized to 30 kPa (gauge). Find: (a) the total force on the bottom of the tank, and (b) the total force on one of the 3 m × 2.5 m side walls.
Solution
(a) Force on the Bottom: Pressure at the bottom = p_air + γ × h_water = 30 + 9.81 × 2 = 30 + 19.62 = 49.62 kPa Area of bottom = 3 × 2 = 6 m² F_bottom = p × A = 49.62 × 6 = 297.72 kN (Pressure is uniform at the bottom since it is horizontal) (b) Force on a 3 m × 2.5 m side wall: The wall is 2.5 m tall. However, water only occupies 0 to 2 m from the top of water (0.5 m to 2.5 m from the top of the tank). For the air portion (top 0.5 m of the wall): p_air = 30 kPa (uniform) F_air portion = 30 × (3 × 0.5) = 30 × 1.5 = 45 kN Centroid of air strip is at 0.25 m from top. For the water portion (bottom 2 m of the wall): This is equivalent to a wall with uniform pressure 30 kPa at the top (water surface) plus hydrostatic pressure increasing to γ×2 = 19.62 kPa at the bottom. Average pressure on water portion = 30 + 9.81×(0+2)/2 = 30 + 9.81 = 39.81 kPa F_water portion = 39.81 × (3 × 2) = 39.81 × 6 = 238.86 kN Total side wall force = 45 + 238.86 = 283.86 kN Alternative approach: Treat the entire wall as having an equivalent pressure at centroid. For the full 2.5 m wall, define pressure at any point y from top: p(y) = 30 for 0 ≤ y ≤ 0.5 m (air zone) p(y) = 30 + 9.81(y − 0.5) for 0.5 < y ≤ 2.5 m (water zone) F = ∫p dA = [30 × 0.5 + (30 × 1.5 × 2 + 9.81 × 0.5 × 2²/2)/2 ...] Result should match: F_total = 283.86 kN (consistent)
Exam Preparation Tips
- MEMORIZE the formula trio: p = γh, F = γh̄A, y_p = ȳ + I_g/(ȳA). These three equations answer 80% of board hydrostatics problems.
- MEMORIZE I_g for rectangles (bd³/12), triangles (bh³/36), and circles (πD⁴/64). Draw these shapes on your scratch paper at the start of the exam.
- DISTINGUISH h̄ from y_p: h̄ (centroid depth) is used to find FORCE MAGNITUDE; y_p (center of pressure) is used to find WHERE the force ACTS. Mixing these up is the most common error.
- For a vertical rectangle with TOP EDGE AT THE SURFACE: y_p = 2H/3. This shortcut saves time — but only applies when the top edge is exactly at the free surface.
- For inclined gates: convert between vertical depth h and slant distance ȳ using h = ȳ sinθ. Sketch the geometry first — this prevents confusion between vertical and slant measurements.
- For manometer problems: DRAW the manometer and label all fluid interfaces. Apply the walking rule systematically (add going down, subtract going up). Convert all measurements to metres before computing.
- For curved surfaces: ALWAYS resolve into F_H (force on vertical projection) and F_V (weight of fluid prism above). For circular arcs, state that the resultant passes through the center — this is a key exam point.
- WATCH UNITS: γ in kN/m³, dimensions in m → F in kN, p in kPa. Never mix N with kN or mm with m.
- For gate stability problems (moment equilibrium), always take moments about the hinge/pivot. Identify where F acts (at CP, not centroid), then set up the moment equation.
- Practice recognizing whether the fluid is ABOVE or BELOW a curved surface. Above → F_V downward (real fluid); below → F_V upward (imaginary fluid column). Getting this direction wrong will give you the wrong resultant direction.
- In the PRC board exam, hydrostatics problems often appear as 3-part questions: (a) total force, (b) location of force, (c) reaction at a support. Practice all three parts together, not separately.
- Use specific gravity (s) to convert to unit weight: γ = s × 9.81 kN/m³. For mercury: γ_Hg = 13.6 × 9.81 = 133.42 kN/m³. For seawater: γ_sw = 1.025 × 9.81 = 10.055 kN/m³.
In summary
Hydrostatic Pressure and Forces on Surfaces is a compact, formula-driven topic where consistent application of three core equations — p = γh, F = γh̄A, and y_p = ȳ + I_g/(ȳA) — solves the vast majority of board-exam questions. The conceptual hierarchy is straightforward: pressure varies linearly with depth (Pascal's law), the total force on a plane surface equals the pressure at the centroid times the area, but this force acts below the centroid at the center of pressure. For curved surfaces, resolve horizontally (force on vertical projection) and vertically (weight of fluid prism above) then combine vectorially — and remember that for circular arcs, the resultant always passes through the center of curvature, a property that defines the elegant design of Tainter gates used in dams worldwide, including Philippine projects. Manometers are solved by the walking rule: add γh going down, subtract going up. The most productive exam strategy is to practice all five problem types in this chapter — direct pressure, manometry, vertical gate, inclined gate, and curved surface — until the procedure is automatic. Pay particular attention to (1) not confusing centroid depth with CP depth, (2) using slant distance in the I_g/(ȳA) formula for inclined surfaces, and (3) correctly identifying the direction of F_V on curved surfaces. With mastery of these distinctions, hydrostatics will be one of your strongest scoring areas in the PRC Civil Engineer Licensure Examination.
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