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Lesson 1 of 8

Mechanical Properties of Fluids · Lesson 1 of 8

Fluids, Pressure and Hydrostatic Balance

“Understand fluids, derive pressure isotropy and hydrostatic balance, and distinguish pressure from force.”

Learning Objectives

• Distinguish a fluid’s response to shear from a solid’s response and compare liquid and gas compressibility. • Calculate pressure, density and relative density with correct units. • Explain pressure isotropy using a small fluid wedge. • Derive the hydrostatic pressure law and apply it without confusing force with pressure.

What makes a material a fluid?

Water pours, air escapes from a balloon, and oil takes the shape of its bottle. Liquids and gases are fluids because they can flow. A solid can maintain its shape under a small steady shear force; an ordinary fluid cannot sustain a static shear stress and remain at rest. Continued shear makes neighbouring layers move relative to one another. A fluid can nevertheless resist that relative motion while it flows: that resistance is viscosity, studied later. “Able to flow” does not mean “offering no resistance to motion”.

A liquid has an approximately definite volume under ordinary conditions but takes the shape of its container and can have a free surface. A gas spreads through the available volume. Both can be compressed; the distinction is quantitative. Liquids and solids change volume relatively little under ordinary pressure changes, whereas gases often change volume substantially. Calling water incompressible is a useful approximation, not a claim that its volume can never change. Fluid ideas describe rivers and atmosphere as well as blood, plant sap and many other biological processes.

Pressure describes how a normal force is distributed

A sharp needle produces a much larger local effect than a broad spoon pushed with the same force because the needle shares its load over a smaller contact area. Similarly, a strong board can distribute a load across a larger area. Pressure describes normal force per unit area, rather than the total force alone. For a flat area with uniform pressure, doubling the area doubles the force without changing the pressure.

Average and local pressureLaTeX
F⊥ is the magnitude of the normal force. The limit defines a local continuum quantity; in practice the small area still contains many molecules.

The local definition matters when the pressure varies across a large surface. Divide that surface into sufficiently small patches, find each normal force, and combine the forces with their directions. Do not use one pressure value multiplied by a large area unless the pressure is uniform or the chosen value is its appropriate average.

In a fluid at rest the force on a boundary or imaginary internal surface is perpendicular to that surface. If a tangential component remained, an ordinary fluid would shear and move, contradicting the assumed rest. On an immersed curved body, the normal direction changes from point to point; the pressure can be the same while the force directions differ. Pressure also exists between neighbouring parts of the fluid, even where there is no solid wall.

Pressure acts normally; the pressure itself is scalarNormal forces on an immersed bodyFluid pushes pistonCalibrated spring balances FP = F / A
Normal forces and an ideal pressure probe— Schematic. A calibrated spring balances the force on a piston of known area; orienting the probe changes the force direction, not the local static pressure.

Pressure is a scalar. In F⊥/A the numerator is a normal force magnitude, not an unrestricted force vector. Its unit is the pascal: 1 Pa = 1 N/m², and its dimensions are M L⁻¹ T⁻². A pressure needs no arrow attached to it. A force calculation still needs the orientation of the patch. Do not conclude that equal scalar pressures on two differently oriented surfaces give equal force vectors.

Pressure carried by two thigh bones

Problem
Two femurs, each with cross-sectional area 10 cm², support an upper-body mass of 40 kg. Estimate the average compressive pressure using g = 10 m/s².

  1. 1.The two load-bearing cross-sections act together, so their total area is 20 cm² = 20 × 10⁻⁴ m² = 2.0 × 10⁻³ m².
  2. 2.The total supported force is mg = 40 × 10 = 400 N. It is not 400 N on each bone if the bones share the load equally.
  3. 3.Average pressure = 400/(2.0 × 10⁻³) = 2.0 × 10⁵ Pa. Each bone carries 200 N on 10 cm², which gives the same result. The estimate neglects uneven loading and the bones’ detailed geometry.

Density and relative density

Mass density and relative densityLaTeX
Density has SI unit kg/m³ and dimensions M L⁻³. Relative density has no unit; water at 4°C has density approximately 1000 kg/m³.

Mass density tells us the mass contained in a given volume. For example, relative density 2.7 corresponds to 2700 kg/m³ when the reference water density is 1000 kg/m³. Density and pressure are different quantities: a dense fluid is not automatically at high pressure. Both are positive scalars in the ordinary fluid states considered here. Liquid density is approximately constant for this chapter’s pressure calculations; gas density generally depends strongly on pressure and temperature.

FluidRepresentative density (kg/m³)
Water1.00 × 10³
Sea water1.03 × 10³
Mercury13.6 × 10³
Ethyl alcohol0.806 × 10³
Whole blood1.06 × 10³
Air1.29
Oxygen1.43
Hydrogen9.0 × 10⁻²
Interstellar spaceApproximately 10⁻²⁰

These are the chapter’s representative density values. Its STP convention is 0°C and 1 atm; the extremely dilute interstellar entry is an illustrative comparison, not an STP gas sample. Notice the enormous density contrast between ordinary liquids and gases and between mercury and water. Always use the specified density in a numerical problem rather than treating every liquid as water.

Why pressure at a point is the same in every direction

Imagine an extremely small triangular wedge of fluid. Give its inclined face area Aₙ and let that face make angle θ with the horizontal. The vertical face has area Aₓ = Aₙ sin θ, and the horizontal face has area Aᵧ = Aₙ cos θ. The surrounding fluid pushes normally inward on each face. Take the wedge small enough that the weight, which scales with volume, is negligible compared with the face forces, which scale with area.

A small fluid wedge proves pressure isotropyFₓFᵧFₙAᵧ = Aₙ cos θFₓ = Fₙ sin θFᵧ = Fₙ cos θAₓ = Aₙ sin θPₓ = Pᵧ = PₙθWedge weight is negligible in the infinitesimal limit; diagram enlarged.
The small-wedge argument for equal pressures— Cross-section of an infinitesimal triangular prism. Opposite forces on its front and rear faces cancel; the remaining three normal forces balance.
Force balance and area projectionsLaTeX
Dividing matching force and area projections gives Pₓ = Pᵧ = Pₙ. The result is local, not a statement that pressure is identical at all depths.

The angle factors cancel because the same geometry controls force components and projected areas. Rotating the imaginary wedge gives equal static pressure for any surface orientation through that point. Now isolate a horizontal fluid bar. Its end forces must balance, so equal-height points in the same connected resting liquid have equal pressure. Points separated vertically require a different balance because the intervening fluid has weight.

Deriving the pressure increase with depth

Choose a vertical cylinder of resting fluid with area A, height h and uniform density ρ. Label the upper pressure P₁ and lower pressure P₂. The pressure force on the top pushes downward, the force on the bottom pushes upward, and the weight ρAhg acts downward. The fluid is at rest, so the net vertical force must vanish.

Equilibrium of the vertical cylinderLaTeX
h is the positive vertical depth difference; use approximately constant density and g over the column. Side pressure forces have no vertical component for this cylinder.

Area cancels because doubling the cylinder area doubles both its fluid weight and its pressure forces. If the upper point is a free surface at pressure P₀ and the lower point is depth h below it, the result becomes P = P₀ + ρgh. A sealed vessel need not have P₀ equal to atmospheric pressure. With height z measured upward, the same result says pressure decreases as height increases.

Pressure at depth and the vertical gradientLaTeX
The differential form means that a small upward rise dz produces a pressure decrease ρg dz. Constant-density integration gives the preceding finite-depth formula.
Hydrostatic balance and connected vesselsSurface pressure P₀P₁ A ↓↑ P₂ AmghEqual free-surface height → equal base pressureDifferent shapes and fluid volumes
Fluid-column balance and hydrostatic paradox— The column is an imaginary part of the fluid. The connected vessels illustrate equal pressure at equal depth despite different contained volumes; drawings are schematic.

The hydrostatic paradox is that containers with different shapes and water masses can have the same bottom pressure when the liquid, surface pressure and vertical depth are the same. It is pressure, not necessarily total bottom force, that is equal. For equal bottom areas their bottom forces are equal too; forces from sloping walls help balance the different fluid weights. Connected open vessels therefore settle to a common surface level. This equal-level argument requires connected fluid at rest and the same pressure over the free surfaces.

A swimmer 10 m below the surface

Problem
Find the pressure on a swimmer at depth 10 m in fresh water. Use ρ = 1000 kg/m³, g = 10 m/s² and atmospheric pressure 1.01 × 10⁵ Pa.

  1. 1.The water adds ρgh = 1000 × 10 × 10 = 1.00 × 10⁵ Pa to the surface pressure.
  2. 2.Absolute pressure = 1.01 × 10⁵ + 1.00 × 10⁵ = 2.01 × 10⁵ Pa, approximately two atmospheres. The water-induced excess is approximately one atmosphere.
  3. 3.At depth 1 km the corresponding fresh-water excess would be 10⁷ Pa, approximately 100 atmospheres. This illustrates why deep-water structures need substantial strength.

In a standing person, the feet are below the brain. The hydrostatic contribution to blood pressure is therefore larger at the feet. Real circulation includes a beating heart and flowing, viscous blood; the depth law identifies the gravitational contribution rather than a complete circulatory model.

Quick Quiz

Quiz

Quick check

Why can an ordinary fluid not remain at rest under a sustained shear stress?

Explanation

A fluid cannot support a static shear stress. Viscosity resists the relative motion once flow occurs.

Quick check

A normal force of 120 N acts uniformly on 3.0 cm². What is the pressure?

Explanation

3.0 cm² = 3.0 × 10⁻⁴ m²; 120 divided by that area is 4.0 × 10⁵ Pa.

Quick check

Two orientations of a small pressure probe at the same point in resting water measure what?

Explanation

Static pressure is isotropic. Its associated normal force follows the chosen surface orientation.

Quick check

At equal depth in connected open containers of the same liquid, bottom pressure depends on which quantity?

Explanation

For constant density P = P₀ + ρgh. Container shape and total liquid volume do not appear.

Quick check

How much does pressure increase 0.50 m downward in water? Use ρ = 1000 kg/m³ and g = 9.8 m/s².

Explanation

The pressure difference is ρgh = 1000 × 9.8 × 0.50 = 4900 Pa; atmospheric pressure is not added to a difference.

Practice Problems

Practice Problems
  1. A 50 kg student balances on a circular heel of diameter 1.0 cm. Take g = 9.8 m/s². Calculate its contact area and average pressure on the floor; explain the radius-to-diameter step.
  2. A liquid has relative density 0.80. Find the mass of 250 cm³ of it, using water density 1000 kg/m³. State which quantities are dimensionless.
  3. Draw a small fluid wedge. Explain why both force projections and area projections contain the same sine or cosine and why its weight may be neglected in the small-element limit.
  4. Derive P₂ − P₁ = ρgh using a vertical fluid cylinder. Identify every force and explain why the arbitrary cylinder area cancels.
  5. Three connected vessels contain water to depth 0.80 m and have bottom areas 20, 40 and 60 cm². Their surfaces are open at 1.01 × 10⁵ Pa. With g = 10 m/s², calculate bottom pressure in each and the water-induced excess force on each bottom. Explain how different water masses can be consistent with the pressure result.

Key Takeaways

Key Takeaways

• Fluids flow under sustained shear; liquid incompressibility is an approximation. • Pressure is a scalar normal force per area, with unit Pa; density has unit kg/m³. • Static pressure is equal in all directions at a point and at equal heights in a connected homogeneous resting liquid. • Pressure rises downward by ρgh for constant density and g; shape and total volume do not enter.