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Fluid Mechanics · Chapter 3 of 10 · Intermediate

Pressure and Fluid Statics

A fluid at rest still pushes, harder the deeper you go. From that one fact come manometers, the force on a dam, and why steel ships float.

Readiness check

This chapter applies force balance to fluids at rest. Tick only what you can do closed-notes.

Recall pressure as force per area.
Use density and g to get weight per volume.
Find the centroid and moment of inertia of a rectangle.
Balance forces in equilibrium.
Distinguish gauge from absolute pressure.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview density in Chapter 2 and centroids in Mechanics of Materials Chapter 4.

3 or more weak itemsRevisit force balance and area properties first.

The core idea

In a static fluid, pressure increases linearly with depth, P = P₀ + ρgh, and acts equally in all directions. Integrate that pressure to get forces on surfaces and the buoyant force on bodies.

P = P₀ + ρghF = ρg h_c AF_B = ρ_fluid g V_displaced

With no motion there is no shear, so pressure is the only stress, and a force balance on a fluid column gives P = P₀ + ρgh. Pressure at a point is the same in every direction (Pascal), which is what makes manometers work. The resultant force on a submerged surface is the pressure at its centroid times its area, acting below the centroid at the center of pressure. A submerged or floating body feels an upward buoyant force equal to the weight of fluid it displaces.

The skill works when: you track depth carefully through ρgh and place the resultant at the center of pressure, not the centroid.

The skill breaks down when: gauge and absolute pressure are mixed, or the force is placed at the centroid instead of the center of pressure.

The concept. Pressure in a static fluid rises linearly with depth, giving the triangular distribution on a wall. The deeper the point, the harder the fluid pushes, equally in all directions.

The skills, taught in order

Statics is one pressure-depth law applied to instruments, surfaces, and bodies. Five skills cover pressure, manometry, hydrostatic force, the center of pressure, and buoyancy.

3.1 Pressure and depth

Pressure is force per area, the same in all directions at a point. In a static fluid it varies only with depth: P = P₀ + ρgh. Absolute pressure is measured from vacuum; gauge pressure is measured from the local atmosphere (P_abs = P_atm + P_gauge).

Unit	Equivalent
1 atm	101.325 kPa
1 bar	100 kPa
1 atm	760 mmHg

3.2 The manometer

A manometer turns a pressure into a measurable height. Moving down a column adds ρgh, moving up subtracts it; a single fluid of height difference h reads a pressure difference Δp = ρgh. Mercury (SG 13.6) is used for large pressures because it gives a short column.

3.3 Hydrostatic force on a plane surface

The resultant pressure force on a submerged flat surface is the pressure at its centroid times its area: F = P_cA = ρg h_c A, where h_c is the centroid depth. The shape of the surface does not matter for the magnitude, only the centroid depth and area.

3.4 The center of pressure

Because pressure grows with depth, the resultant acts below the centroid, at the center of pressure: y_p = y_c + I_xc/(y_cA). For a vertical rectangle with its top at the surface, this works out to two-thirds of the depth, the classic dam result.

3.5 Buoyancy and stability

Archimedes' principle: a submerged or floating body feels an upward force equal to the weight of displaced fluid, F_B = ρ_fluid g V_displaced. A body floats when it can displace its own weight; stability depends on the metacenter relative to the center of gravity.

Engineering connection: dams, gates, tanks, hydraulic presses, ship hulls, and pressure gauges are all designed with these static relations.

Worked example 1: a mercury manometer

A pressurised tank is connected to a U-tube manometer filled with mercury (SG 13.6). The mercury column reads a height difference of 150 mm. Find the gauge pressure in the tank, and the absolute pressure.

Figure 1. The mercury height difference balances the tank's gauge pressure. Its large density keeps the column short, the reason mercury is used for higher pressures.

ProblemFind the gauge and absolute pressure in the tank in Figure 1.
Given / findSG = 13.6 so ρ = 13 600 kg/m³, h = 0.15 m, P_atm = 101.3 kPa. Find P_gauge and P_abs.
AssumptionsStatic fluid, the gas weight in the connecting line is negligible.
ModelThe mercury column height difference balances the gauge pressure: P = ρgh.
EquationsP_gauge = ρgh P_abs = P_atm + P_gauge
SolveP_gauge = 13 600 × 9.81 × 0.15 = 20.0 kPa. P_abs = 101.3 + 20.0 = 121.3 kPa.
CheckThe same pressure as a water column would need h = P/(ρ_waterg) = 20 000/(998 × 9.81) = 2.04 m, over thirteen times taller. That ratio is exactly the specific gravity, confirming the choice of mercury.
ConclusionA manometer reads pressure as a fluid height through ρgh. The denser the fluid, the shorter and more practical the column.

Result. Gauge pressure 20.0 kPa; absolute pressure 121.3 kPa.

Worked example 2: force on a vertical gate

A vertical rectangular gate 2 m wide and 3 m tall holds back fresh water (ρ = 1000 kg/m³), with its top edge at the water surface. Find the resultant hydrostatic force and the depth of the center of pressure.

Figure 2. The triangular pressure distribution gives a resultant at the centroid depth, but located two-thirds of the way down (the center of pressure), below the area centroid.

ProblemFind the hydrostatic force and center of pressure for the gate in Figure 2.
Given / findb = 2 m, H = 3 m, ρ = 1000 kg/m³, top at surface. Find F and y_p.
AssumptionsStatic fresh water, gate vertical, atmospheric pressure acts on both sides (gauge).
ModelForce is the centroid pressure times area; center of pressure from the parallel-axis offset.
EquationsF = ρg h_c A y_p = h_c + I_xc/(h_cA), I_xc = bH³/12
SolveA = 2 × 3 = 6 m², h_c = 1.5 m, so F = 1000 × 9.81 × 1.5 × 6 = 88.3 kN. I_xc = 2 × 3³/12 = 4.5 m⁴, so y_p = 1.5 + 4.5/(1.5 × 6) = 1.5 + 0.5 = 2.0 m.
CheckThe center of pressure at 2.0 m is exactly 2H/3, the known result for a surface-piercing rectangle. It lies below the centroid (1.5 m) because pressure is higher near the bottom.
ConclusionUse the centroid depth for the force magnitude, but apply it at the center of pressure for moments, the difference that sizes a gate's hinge and latch.

Result. Resultant force 88.3 kN, acting at 2.0 m depth (the center of pressure, 2H/3).

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Gauge vs absolute mixed	Pressure off by one atmosphere	"Measured from vacuum or from atmosphere?"	P_abs = P_atm + P_gauge; state which you mean.
Force at the centroid	Wrong moment, hinge mis-sized	"Where does the resultant act?"	The resultant acts at the center of pressure, below the centroid.
Pressure depends on shape	Container-shape confusion	"Does pressure depend on anything but depth?"	In a static fluid, pressure depends only on depth, not container shape.
Buoyancy uses body weight	Float/sink predicted wrong	"Whose weight is the buoyant force?"	F_B equals the weight of displaced fluid, not the body.

Practice ladder

Level 1 · Direct skill

Find the gauge pressure at 10 m depth in fresh water (ρ = 1000 kg/m³).

Show answer

P = ρgh = 1000 × 9.81 × 10 = 98 100 Pa = 98.1 kPa, close to one atmosphere. Every 10 m of water adds about 1 atm.

Level 2 · Mixed concept

A 0.2 m³ block weighing 1500 N is fully submerged in water. Find the buoyant force and whether it sinks.

Show answer

F_B = ρgV = 1000 × 9.81 × 0.2 = 1962 N. Since the buoyant force (1962 N) exceeds the weight (1500 N), the block rises and will float. It then displaces only 1500/9810 = 0.153 m³.

Level 3 · Independent problem

The gate of Worked Example 2 now has its top edge 1 m below the surface. Find the new resultant force.

Show answer

The centroid is now at h_c = 1 + 1.5 = 2.5 m. F = ρg h_c A = 1000 × 9.81 × 2.5 × 6 = 147.2 kN. Submerging the gate deeper raises the force in proportion to centroid depth.

Level 4 · Transfer to real engineering

Find a real hydrostatic structure (an aquarium window, a lock gate, a retaining wall, a floating dock). Estimate the force on it and where that force acts.

What good work looks like

The centroid depth and area identified, F = ρg h_c A computed, the center of pressure located, and a comment on the resulting moment or buoyancy.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I placed the resultant at the center of pressure, not the centroid."

"Give me five submerged surfaces; I will set up F = ρg h_c A for each."

"Compute the force on the gate." Finding h_c, A, and y_p yourself is the skill.

"Does it float?" Comparing buoyant force to weight is the point.

Portfolio task

Analyse one static-fluid system: read a manometer or compute the hydrostatic force on a surface, locate the center of pressure, and check a buoyancy or moment balance.

Must include: a pressure-depth calculation, a force with its center of pressure, and a buoyancy or equilibrium check.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. How does pressure vary with depth?

Linearly: P = P₀ + ρgh, equal in all directions at a point.

2. How does a manometer read pressure?

As a fluid height: Δp = ρgh along the column.

3. Write the hydrostatic force on a plane surface.

F = ρg h_c A, the centroid pressure times area.

4. Where does that force act?

At the center of pressure, y_p = y_c + I_xc/(y_cA), below the centroid.

5. State Archimedes' principle.

The buoyant force equals the weight of the displaced fluid, F_B = ρgV.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the manometer and gate force from a blank page.

+3 daysOne manometer and one hydrostatic-force problem.

+7 daysMove to fluid kinematics, Chapter 4.

+30 daysConnect pressure to the Bernoulli equation in Chapter 5.

Textbook mapping

Item	Mapping
Primary source	Çengel and Cimbala, Fluid Mechanics: Fundamentals and Applications, Chapter 3 (Pressure and Fluid Statics)
Cross-reference	White, Ch. 2 · Munson, Ch. 2
Core topics	3.1 Pressure and depth · 3.2 Manometry · 3.3 Hydrostatic force · 3.4 Center of pressure · 3.5 Buoyancy
Engineering connection	Dams, gates, tanks, hydraulic presses, and ship hulls.
Read next	Chapter 4: Fluid Kinematics.