Lesson 11: Fluids

Week 11 — Liquids, gases, and the forces between them

🎯 Learning Objectives

Understand pressure and buoyancy
Apply Pascal's principle and Bernoulli's equation
Analyze fluid flow

🎯 Intuition First: Why Do Ships Float?

Imagine a steel block in water. It sinks. But a steel ship — which is mostly empty space — floats. Why? Because what matters isn't the material, but the average density of the whole object compared to the fluid. A ship displaces a volume of water whose weight equals the ship's weight.

💡 Key Question:

Why does water pressure increase as you dive deeper? Because you're supporting the weight of all the water above you. The deeper you go, the more water is stacking on top.

📋 Pressure

Pressure: P = F/A

Units: Pascal (Pa) = N/m²

Hydrostatic pressure at depth h: P = ρgh + P₀

Atmospheric pressure: P_atm = 1.013 × 10⁵ Pa = 1 atm = 760 mmHg

📝 Pascal's Principle:

Pressure applied to an enclosed fluid is transmitted equally in all directions. This is the basis of hydraulic lifts!

P₁ = P₂ → F₁/A₁ = F₂/A₂

📋 Buoyancy (Archimedes' Principle)

F_b = ρ_fluid × V_displaced × g

The upward buoyant force equals the weight of the fluid displaced by the object.

📝 Worked Example: Floating Object

A block of wood (density 600 kg/m³, volume 0.02 m³) floats in water (density 1000 kg/m³). What fraction of the block is submerged?

Step 1 — Weight of block:

W = ρ_woodVg = 600 × 0.02 × 9.8 = 117.6 N

Step 2 — Buoyant force = weight (floating):

ρ_waterV_subg = 117.6 N

V_sub = 117.6/(1000 × 9.8) = 0.012 m³

Step 3 — Fraction submerged:

f = V_sub/V = 0.012/0.02 = 0.6 = 60%

✅ 60% of the block is underwater. (In general, f = ρ_object/ρ_fluid)

📋 Bernoulli's Equation

P + ½ρv² + ρgh = constant (along a streamline)

This expresses energy conservation for flowing fluids: pressure + kinetic energy density + potential energy density = constant.

💡 Intuition:

Where a fluid flows faster, the pressure is lower. This is why airplane wings generate lift, why a shower curtain gets sucked inward, and why perfume sprays work!

📝 Worked Example: Water in a Pipe

Water flows through a pipe that narrows from 10 cm² to 4 cm². In the wide section, speed is 2 m/s. Find the speed and pressure difference in the narrow section. (Assume horizontal pipe, ρ = 1000 kg/m³)

Step 1 — Continuity equation (Av = constant):

A₁v₁ = A₂v₂

(10)(2) = (4)v₂

v₂ = 20/4 = 5 m/s

Step 2 — Bernoulli (horizontal: h₁ = h₂):

P₁ + ½ρv₁² = P₂ + ½ρv₂²

P₁ − P₂ = ½ρ(v₂² − v₁²) = ½(1000)(25 − 4)

P₁ − P₂ = 500 × 21 = 10,500 Pa

✅ v₂ = 5 m/s, pressure drops by 10,500 Pa in the narrow section

📋 Equation of Continuity

A₁v₁ = A₂v₂ (for incompressible fluids)

What goes in must come out. If the pipe narrows, the fluid must speed up.