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Study Notes

Unit 8 — Fluids

Liquids and gases as systems: density, pressure, buoyancy, and how Newton's laws and the conservation of mass and energy describe fluids at rest and in motion. Each section ends with a short practice set — click "Show answer" to check yourself.

8.1

Internal Structure and Density

A fluid is a substance that can flow and takes the shape of its container: liquids and gases. Its bulk properties come from the arrangement and motion of the particles inside it.

ρ = m / V          units: kg/m³
  • Density (ρ) is mass per unit volume. It is a scalar property of the material, not of the amount.
  • Water: about 1000 kg/m³ (= 1 g/cm³). Air: about 1.2 kg/m³.
  • In a uniform (homogeneous) substance, any sample has the same density. Cutting an object in half changes its mass and volume but not its density.
  • Liquids are nearly incompressible: their density stays almost constant. Gases are compressible, so their density changes with pressure and temperature.
  • Density depends on the mass of the particles and how tightly they are packed, so solids and liquids are far denser than gases.
  • Unit conversion: 1 L = 10⁻³ m³, 1 cm³ = 10⁻⁶ m³.
Tip: convert volume to m³ before using kg/m³.
Practice

1. A block measuring 0.20 m × 0.10 m × 0.050 m has a mass of 2.0 kg. What is its density?

Show answer
V = 0.20 × 0.10 × 0.050 = 1.0 × 10⁻³ m³. ρ = m/V = 2.0/(1.0 × 10⁻³) = 2000 kg/m³.

2. What is the mass of 3.0 L of water?

Show answer
V = 3.0 × 10⁻³ m³. m = ρV = 1000 × 3.0 × 10⁻³ = 3.0 kg.
8.2

Pressure

Pressure is the force applied perpendicular to a surface per unit area. It is a scalar. Its SI unit is the pascal: 1 Pa = 1 N/m².

P = F / A
P = P₀ + ρ·g·h          (absolute pressure at depth h in a static fluid)
Gauge pressure = ρ·g·h
Hydraulic lift:    F₁ / A₁ = F₂ / A₂
  • A fluid exerts pressure in all directions, always perpendicular to any surface it touches.
  • In a static fluid, pressure depends only on depth, not on the container's shape or the amount of fluid above. Points at the same depth have the same pressure.
  • P₀ is the pressure at the fluid surface. For an open container it is atmospheric pressure, about 1.01 × 10⁵ Pa.
  • Pascal's principle: a pressure change applied to an enclosed fluid is transmitted undiminished throughout it. This makes hydraulic lifts and brakes work: a small force on a small piston produces a large force on a large piston.
  • In a hydraulic lift the pressures are equal, but the work is the same: the small piston moves farther than the large one.
Common mix-up: pressure ≠ force. The same force on a smaller area gives a larger pressure (a knife cuts better when sharp).
Practice

1. Find the gauge pressure and the absolute pressure at a depth of 10 m in fresh water. (g = 9.8 m/s², P₀ = 1.01 × 10⁵ Pa)

Show answer
Gauge: ρgh = (1000)(9.8)(10) = 9.8 × 10⁴ Pa. Absolute: P₀ + ρgh ≈ 1.01 × 10⁵ + 0.98 × 10⁵ ≈ 2.0 × 10⁵ Pa.

2. A hydraulic lift has a small piston of area 0.010 m² and a large piston of area 0.50 m². What force on the small piston lifts a load requiring 5000 N on the large one?

Show answer
F₁ = F₂(A₁/A₂) = 5000 × (0.010/0.50) = 100 N.
8.3

Fluids and Newton's Laws

For a fluid at rest, or an object in it, apply Newton's laws with the usual free-body diagram, adding the buoyant force.

F_B = ρ_fluid · V_displaced · g        (Archimedes' principle)
Floating at rest:   F_B = m·g    (V_displaced = submerged volume)
Fraction submerged = ρ_object / ρ_fluid
  • The buoyant force is an upward force equal to the weight of the fluid displaced.
  • It arises because pressure increases with depth: the fluid pushes up on the bottom of the object harder than it pushes down on the top.
  • The buoyant force depends on the fluid's density and the volume submerged, and not on the object's mass or material.
  • If ρobject < ρfluid, the object floats; if greater, it sinks; if equal, it stays at any depth (neutrally buoyant).
  • For a fully submerged object at rest (e.g. held by a string): FT + FB = mg. The scale reading, or apparent weight, is mg − FB.
Key idea: in a floating object the buoyant force equals the object's weight, so it displaces its own weight of fluid, not its own volume.
Practice

1. A 2.0 × 10⁻³ m³ object is completely submerged in water. What buoyant force acts on it? (g = 9.8 m/s²)

Show answer
FB = ρVg = (1000)(2.0 × 10⁻³)(9.8) = 19.6 N, directed upward.

2. A wooden block with density 600 kg/m³ floats in water. What fraction of its volume is submerged?

Show answer
Floating: ρwoodVg = ρwaterVsubg → Vsub/V = 600/1000 = 0.60 (60% submerged).
8.4

Fluids and Conservation Laws

Moving fluids obey conservation of mass and conservation of energy. These are the continuity equation and Bernoulli's equation, for an ideal fluid (incompressible, non-viscous, steady flow).

Continuity (mass):    A₁·v₁ = A₂·v₂        (volume flow rate Q = A·v, in m³/s)
Bernoulli (energy):   P₁ + ½·ρ·v₁² + ρ·g·y₁ = P₂ + ½·ρ·v₂² + ρ·g·y₂
Torricelli:           v = √(2·g·h)        (speed of exit from a small hole at depth h)
  • Continuity: the same volume of fluid passes each point every second, so fluid speeds up in a narrow section of pipe and slows down in a wide one.
  • Bernoulli: along a streamline, the sum of pressure, kinetic energy per volume, and potential energy per volume is constant.
  • In a horizontal pipe (same height), higher speed means lower pressure. This explains lift on a wing and why a shower curtain swings inward.
  • A change in height matters too: fluid pushed to a higher point loses pressure (the ρgy term).
  • Bernoulli's equation is an energy conservation statement. For a fluid at rest (v = 0) it reduces to P = P₀ + ρgh.
Common mix-up: the pressure drop in a fast, narrow section is not caused by the walls squeezing. The energy needed for the higher speed comes from the pressure difference.
Practice

1. Water in a horizontal pipe flows at 2.0 m/s through a section of cross-section 4.0 cm². The pipe narrows to 1.0 cm². Find the new speed and the pressure change. (ρ = 1000 kg/m³)

Show answer
Continuity: v₂ = v₁A₁/A₂ = 2.0 × 4.0/1.0 = 8.0 m/s. Bernoulli (same height): P₁ − P₂ = ½ρ(v₂² − v₁²) = ½(1000)(64 − 4) = 3.0 × 10⁴ Pa. The pressure drops in the narrow section.

2. A large open tank of water has a small hole 1.8 m below the surface. With what speed does the water leave the hole? (g = 9.8 m/s²)

Show answer
v = √(2gh) = √(2 × 9.8 × 1.8) = √35.3 ≈ 5.9 m/s. (Pressure is atmospheric at both the top surface and the hole, and the surface speed is negligible.)