UNIT 8

Fluids

Pressure, Density, Buoyancy, and Flow

AP Physics 1

Archimedes · Pascal · Bernoulli · Torricelli

Essential Vocabulary

Terms You Must Know

• Fluid: Any substance that flows (liquids and gases)

• Density (ρ): Mass per unit volume

• Pressure: Force per unit area, acts perpendicular to surfaces

• Buoyancy: Upward force from displaced fluid

• Continuity: Conservation of mass flow in pipes

• Bernoulli's Principle: Conservation of mechanical energy in flowing fluids

• Ideal fluid: Incompressible and nonviscous (assumptions for AP Physics 1)

• Gauge pressure: Pressure relative to atmosphere; Absolute pressure: total pressure from zero

What is a Fluid?

The So Far

Solids hold their shape. Fluids (liquids and gases) flow and reshape themselves.

Why does this matter?

• Water flows downhill

• Air pushes against surfaces

• Boats float and submarines sink

All of these involve forces between fluid particles and objects. Understanding these forces unlocks the physics.

Distinction: Fluids have NO fixed shape. They conform to whatever container holds them.

Density

What is Density?

A measure of how much 'stuff' is packed into a given space.

Density Formula:

ρ = m ÷ V (where m = mass, V = volume)

Everyday Example

• Water: ≈ 1000 kg/m³ (reference standard)

• Air: ≈ 1.2 kg/m³ (about 800× less dense)

• Mercury: ≈ 13,600 kg/m³ (about 13.6× denser than water)

Fact: For an ideal fluid, density is constant regardless of depth or pressure.

Pressure

The Pressure Concept

Imagine pressing a thumbtack into your hand. The same force applied to a wide palm feels different than a point on your finger. Why?

Pressure Formula:

P = F_⊥ ÷ A (where F⊥ = perpendicular force, A = area)

Critical insight: Pressure is a scalar (no direction), but it acts perpendicular to any surface.

Units: Pascals (Pa) = N/m²

• Atmospheric pressure at sea level ≈ 101,325 Pa = 1 atm

Practice Problem 1: Pressure

A circular piston with a diameter of 5.0 cm is used in a water pump. A force of 200 N is applied uniformly to the piston. What pressure does the piston exert on the water?

(Hint: Find the area first, then use P = F/A)

Solution: Pressure Practice Problem 1

Step 1: Find the area

• Diameter = 5.0 cm = 0.05 m, radius = 0.025 m

• A = πr² = π(0.025)² ≈ 0.00196 m²

Step 2: Calculate pressure

• P = 200 N / 0.00196 m² ≈ 102,000 Pa ≈ 1 atm

Note: This is essentially atmospheric pressure. The pump must overcome air pressure to expel water.

Gauge vs. Absolute Pressure (8.2)

Two ways to measure pressure

Absolute Pressure: Total pressure measured from zero (perfect vacuum)

Absolute Pressure:

P_absolute = P₀ + P_gauge

Gauge Pressure: Pressure measured relative to atmospheric pressure (what gauges read)

• When a tire gauge reads 30 psi, absolute pressure = 30 + 14.7 ≈ 45 psi

Why it matters: Fluids always exert pressure in all directions. Submarines must withstand the absolute pressure at depth.

What will happen?

Practice Problem 2: Absolute Pressure

A tank with a rectangular base is filled to a depth of 4 m with water. The tank is open on top. The base of the tank is 3 m long by 2 m wide. Caluculate the absolute pressure of the water at the bottom of the tank.

(Hint: Find the area first, then use P = F/A)

Solution: Pressure Practice Problem 2

P = P₀ + ρgh

P = 101000 + (1000) (10) (4)

P = 141,000 Pa

P = 1.41 X 10⁵

Pressure Increases with Depth (8.2.B)

Hydrostatic Pressure Formula

The deeper you go in a fluid, the more fluid weight is pushing down on you.

Gauge Pressure at depth h:

P_gauge = ρgh (where h = depth below surface)

Why? Think of it as a column of fluid.

• More fluid above = more weight = more pressure

At 10 m depth in water: P_gauge = (1000 kg/m³)(10 m/s²)(10 m) = 100,000 Pa ≈ 1 atm

For static fluids, pressure will be the same in all directions, as long as the depth is the same.

Practice Problem 2: Depth and Pressure

A submarine at 400 meters depth must withstand significant pressure. Find the gauge pressure at this depth. (Use ρ = 1025 kg/m³ for seawater.)

(This is approximately 40 times atmospheric pressure—why submarines need such strong hulls!)

Solution: Depth and Pressure

Given: h = 400 m, ρ = 1025 kg/m³, g = 10 m/s² (approximate)

Use: P = ρgh

• P = (1025)(10)(400) = 4,100,000 Pa

• This is approximately 40 atm

At ocean depths, the crushing pressure is why only specially designed submarines can survive. The hull must distribute forces evenly.

Buoyancy:

Archimedes' Principle (circa 250 BCE)

'Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.'

Fluid Displacement: If an object sinks, it displaces a volume of fluid equal to its own volume. If it floats, it displaces a weight of fluid equal to its own weight.

• Fluid particles press harder on the bottom of an object than the top

• The net upward force = weight of fluid that would fit in the object's volume

Buoyant Force:

F_buoy = ρ_fluid × V_displaced × g

Weight of displaced water is less than ball’s weight

Weight of displaced water EQUALS weight of boat

Why Things Float or Sink

Newton's Second Law for Objects in Fluids

Net Force = F_buoy – (m_object g)

Floating (neutral buoyancy):

• F_buoy = Weight of object. Object stays at constant depth.

Sinking:

• Weight > F_buoy. Object accelerates downward.

Rising:

• F_buoy > Weight. Object accelerates upward.

Ships float by displacing enough water to generate a buoyant force equal to their weight.

Practice Problem 3: Buoyancy

A 0.5 m³ block of wood with a mass of 250 kg is fully submerged in water. Calculate the buoyant force and determine whether the wood will float, sink, or stay at depth. (ρ_water = 1000 kg/m³, g = 10 m/s²)

Solution: Buoyancy Practice

Step 1: Find buoyant force

• F _buoy = ρ_water × V × g = 1000 × 0.5 × 10 = 5000 N

Step 2: Find weight of wood

• Weight = 250 × 10 = 2500 N

Step 3: Compare forces

• F_buoy (5000 N) > Weight (2500 N), so the wood will float upward.

Physical result: The wood rises until it partially submerges. At equilibrium, the buoyant force equals its weight.

Fluid Flow: Conservation of Mass (8.4.A)

The Continuity Equation

Imagine water flowing through a pipe. If the pipe narrows, the water speeds up. Why?

Answer: The same amount of mass must flow past any point per unit time.

Continuity Equation:

A₁v₁ = A₂v₂ (for incompressible fluids)

A = cross-sectional area, v = flow velocity

• Narrow section = high velocity

• Wide section = low velocity

Practice Problem 4: Continuity

Water flows through a hose of cross-sectional area 10 cm² at a speed of 2 m/s. The hose narrows to an area of 5 cm². What is the water speed in the narrow section?

Solution: Continuity

Use: A₁v₁ = A₂v₂

• 10 cm² × 2 m/s = 5 cm² × v₂

• 20 = 5v₂

• v₂ = 4 m/s

Result: Water speed doubles as the hose narrows. The same mass flow rate requires higher velocity in the narrower section.

Real-world: This is why a garden hose shoots water farther when you place your thumb over the nozzle (you reduce the area).

Bernoulli's Principle: Energy in Motion

The Energy

A fluid in motion has kinetic energy. A fluid at height has gravitational potential energy. A pressurized fluid has pressure-related energy.

Bernoulli's Equation (derived from conservation of energy):

Bernoulli's Equation:

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

If the flow rate is assumed to be constant (which is what we do), then as fluid flows and speeds up, its pressure decreases (and vice versa).

• This explains why airplane wings create lift (fast air = low pressure)

• Why a straw drinks water (your lungs lower pressure)

Practice Problem 5: Bernoulli's Principle

In a horizontal pipe (same height), water at point 1 has pressure 50,000 Pa and flows at 1 m/s. At point 2, the pipe narrows and water flows at 3 m/s. Find the pressure at point 2.

Solution: Bernoulli's Principle

Use Bernoulli with h = constant (horizontal pipe):

P₁ + ½ρv₁² = P₂ + ½ρv₂²

• 50,000 + ½(1000)(1)² = P₂ + ½(1000)(3)²

• 50,000 + 500 = P₂ + 4,500

• P₂ = 46,000 Pa

Higher velocity → lower pressure. The pressure dropped by 4000 Pa.

Torricelli's Theorem: Draining Fluid

A Special Case of Bernoulli

When fluid drains from a container through a hole, Bernoulli's equation simplifies to a beautiful result.

Torricelli's Theorem:

v = √(2gh) where h = height above the opening

Derivation: At the top surface, pressure = atmospheric, v ≈ 0. At the hole, same pressure, but v is what we find.

Exit velocity depends only on depth, not on the hole's size or shape.

• Water draining 5 m high exits at √(2×10×5) ≈ 10 m/s (like a car falling from a bridge)

Practice Problem 6: Torricelli's Theorem

A large water tank has a hole 2 meters below the water surface. At what speed does water exit the hole? (g = 10 m/s²)

Solution: Torricelli's Theorem

Use: v = √(2gh)

• v = √(2 × 10 × 2)

• v = √40 ≈ 6.3 m/s

Physical interpretation: Water exits at about 6.3 m/s (roughly 23 km/h or 14 mph). This is why you cannot hold back water gushing from a dam.

Fascinating fact: This velocity is the same as if you dropped the water from a height h and let it fall freely—the gravitational potential energy converts to kinetic energy.

Bringing It All Together

Fluids

• Density tells us how much 'stuff' is packed into space (8.1)

• Pressure is force distributed over area, and increases with depth (8.2)

• Buoyancy occurs because pressure differences create net upward forces (8.3)

• Moving fluids follow conservation of mass (continuity) and energy (Bernoulli) (8.4)

All fluids follow the same fundamental laws, whether flowing through pipes, supporting ships, or moving through the atmosphere.

Real-World Applications

Engineering & Daily Life

• Pipelines: Continuity equation ensures water reaches cities efficiently

• Submarines & bathyscaphes: Design hulls for crushing pressures at depth

• Aviation: Bernoulli explains wing lift (faster air = lower pressure)

• Medical: IV drips use gravity and pressure to deliver fluids

• Hydraulics: Incompressible fluids transmit force over distance

• Weather: Pressure differences drive wind and storm systems

Avoid These Common Misconceptions

1

"Pressure pushes sideways but not up or down." WRONG: Pressure acts perpendicular to any surface.

2

"Denser objects always sink." WRONG: Density of the fluid matters too (steel sinks in water but floats in mercury).

3

"Buoyant force is constant." WRONG: Buoyancy depends on volume displaced, not the object's weight.

4

"Fast-moving air has high pressure." WRONG: By Bernoulli, fast-moving air has LOW pressure (counterintuitive!).

AP Exam Strategy: Fluids

What to Expect

• Multiple-choice: Conceptual understanding and quick calculations

• Free-response: Multi-step problems combining multiple concepts

Problem-Solving Tips

• Always identify: Is the fluid in motion or static? What is conserved?

• Draw diagrams with pressures, velocities, and heights labeled at different points

• Check units carefully (pressure, velocity, area, depth)

• Use ρ_water = 1000 kg/m³, g ≈ 10 m/s² for quick estimates

All Equations

8.1 Density

ρ = m ÷ V

8.2 Pressure

P = F_⊥ ÷ A and P_gauge = ρgh

8.3 Buoyancy

F_buoy= ρ_fluid × V_displaced × g

8.4 Continuity & Bernoulli

A₁v₁ = A₂v₂ and P + ½ρv² + ρgh = constant

8.4 Torricelli (special case)

v = √(2gh)