AP Physics 1 · Unit 8: Fluids ·  Lesson 8.3

Deep Dive: Fluids and Newton's Laws

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
Lesson 8.3: Fluids and Newton's Laws
8.3.A.1Concept

Newton's Laws Inside a Fluid

Nothing about being inside a fluid exempts a particle from Newton's laws. Every individual particle in a fluid — whether it's a molecule of water or a molecule of air — still obeys F = ma, still has its motion changed only by a net force, and still exerts equal and opposite forces on whatever it interacts with.

💡This is the same physics you've used since Unit 2, just applied at a scale you can't see directly. Fluids aren't governed by different rules — they're governed by the same rules, applied to an enormous number of particles at once.
8.3.A.2Concept

Microscopic Interactions, Macroscopic Behavior

What you actually observe — water flowing smoothly through a pipe, air pushing on a wing, pressure building with depth — is a macroscopic result. It emerges from two things working together: the internal interactions between the fluid's own particles, and whatever external forcesact on the fluid as a whole (gravity, an applied force, a container wall).

Microscopic — individual particlesEach particle just follows F = ma.Macroscopic — observable flowMillions of particles average into smooth flow.
🔑You'll never track billions of individual particles by hand — and you don't need to. The macroscopic quantities you already know (pressure, density, flow) are exactly the useful summaries that let you skip that step and still get correct answers.
8.3.B.18.3.B.2Concept

The Buoyant Force

The buoyant force is the net upward force a fluid exerts on an object interacting with it. Like fluid pressure itself (Lesson 8.2), it isn't one clean applied force — it's the collective result of countless particle interactions across the object's entire surface.

💡Here's the key insight that makes it upward: pressure increases with depth. That means the fluid pushes harder on the bottom face of a submerged object than on its top face. Everything else — the pressure on the sides — cancels out by symmetry. What's left over is a net force pointing straight up.
8.3.B.3Math

Archimedes' Principle

The magnitude of the buoyant force turns out to be remarkably simple: it's exactly equal to the weight of the fluid the object displaces.

F_buoyant = ρ_fluid · V_displaced · g

Notice what's missing from that equation: nothing about the object's own mass, weight, or material appears anywhere. Only the fluid's density and how much of that fluid the object pushes out of the way.

ExampleGuided Example — Deriving Buoyancy from Pressure

A cube of side length s is fully submerged in a fluid of density ρ, with its top face at depth h below the surface. Derive an expression for the net upward force on the cube from the pressure difference between its top and bottom faces, and show it equals the weight of the fluid displaced.

Step 1 — Write the pressure at the top face
Using the gauge-pressure formula from Lesson 8.2: P_top = P₀ + ρgh

Set the object's volume and mass, and the fluid's density. The bottom-face pressure arrow is always bigger than the top-face arrow — that gap is the buoyant force.

P_bottom (larger)P_top (smaller)F_buoyant
Object Volume V (m³)0.015
Object Mass m (kg)12.000
F_buoyant = ρVg
147.0 N
Weight = mg
117.6 N

F_buoyant (147.0 N) > weight (117.6 N) — the object floats.

Cross-check: computing the net force from (P_bottom − P_top)·A for this cube gives 147.0 N — matching F_buoyant = ρVg above. Same physics, two different starting points.

ExampleWorked Example — Buoyant Force on a Submerged Block

A rectangular block with volume 0.020 m³ is fully submerged in water (ρ_water = 1000 kg/m³). Find the buoyant force on the block. (g = 9.8 m/s²)

← Back to Lesson 8.3Lesson 8.4 — Fluids and Conservation Laws — is next.