Density and Pressure

Fluid density, pressure, Pascal's principle, and pressure variation with depth

🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

💧 Density and Pressure

Introduction

Fluid mechanics begins with understanding the fundamental properties of fluids: density and pressure. These concepts form the foundation for analyzing fluid behavior in both static and dynamic situations.

Density

Density is mass per unit volume, measuring how tightly matter is packed:

$\rho = \frac{m}{V}$

where:

$\rho$ (rho) = density (kg/m³)
$m$ = mass (kg)
$V$ = volume (m³)

Common Densities:

| Material | Density (kg/m³) | |----------|----------------| | Water | 1000 | | Ice | 917 | | Air (STP)| 1.29 | | Mercury | 13,600 | | Aluminum | 2700 | | Gold | 19,300 |

Specific Gravity

Specific Gravity (SG) is the ratio of a substance's density to water's density:

$SG = \frac{\rho_{substance}}{\rho_{water}}$

Dimensionless (no units)
SG < 1: floats in water
SG > 1: sinks in water

Pressure

Pressure is force per unit area:

$P = \frac{F}{A}$

where:

$P$ = pressure (Pa or N/m²)
$F$ = force perpendicular to surface (N)
$A$ = area (m²)

Common Units:

Pascal (Pa): 1 Pa = 1 N/m²
Atmosphere (atm): 1 atm = 101,325 Pa ≈ 101 kPa
Bar: 1 bar = 100,000 Pa
mmHg: 760 mmHg = 1 atm
psi: 14.7 psi = 1 atm

Pressure in Fluids

Unlike solids, fluids cannot sustain shear stress - they flow. This creates unique pressure properties:

Pressure acts perpendicular to surfaces - fluid pressure always pushes normal to any surface
Pressure is isotropic - equal in all directions at a point
Pressure increases with depth

Pressure vs. Depth

For a fluid at rest, pressure increases linearly with depth:

$P = P_0 + \rho g h$

where:

$P$ = pressure at depth $h$
$P_0$ = pressure at surface (usually atmospheric)
$\rho$ = fluid density (kg/m³)
$g$ = 9.8 m/s²
$h$ = depth below surface (m)

💡 Key Insight: Pressure depends only on vertical depth, not container shape!

Gauge vs. Absolute Pressure

Absolute pressure: Total pressure including atmosphere $P_{absolute} = P_{atm} + P_{gauge}$
Gauge pressure: Pressure relative to atmosphere $P_{gauge} = \rho g h$

Most pressure gauges (like tire gauges) read gauge pressure.

Pascal's Principle

Pascal's Principle: A change in pressure applied to an enclosed fluid is transmitted undiminished to every point in the fluid.

Hydraulic Systems

This principle enables hydraulic lifts:

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

Force multiplication: $F_2 = F_1 \frac{A_2}{A_1}$

Trade-off: Volume conservation means: $d_1 A_1 = d_2 A_2$

Small piston moves far, large piston moves little distance.

Atmospheric Pressure

At sea level:

$P_{atm} = 101,325 \text{ Pa} = 101.3 \text{ kPa} = 1 \text{ atm}$

This is equivalent to:

10.3 m column of water
760 mm column of mercury

Atmospheric pressure decreases with altitude.

Problem-Solving Strategy

Identify pressure type needed (gauge or absolute)
Choose reference point (usually surface)
Apply $P = P_0 + \rho g h$
Remember $h$ is vertical depth only
Check units (convert to SI)

Common Mistakes

❌ Forgetting atmospheric pressure in absolute calculations ❌ Using horizontal distance for $h$ ❌ Mixing gauge and absolute pressure ❌ Unit confusion (1 atm ≠ 1 Pa) ❌ Thinking pressure depends on container shape

📚 Practice Problems

1Problem 1easy

❓ Question:

A cube of aluminum has sides of length 5.0 cm. The density of aluminum is 2700 kg/m³. What is the mass of the cube?

💡 Show Solution

Given:

Side length: $L = 5.0$ cm $= 0.050$ m
Density: $\rho_{Al} = 2700$ kg/m³

Find: Mass $m$

Solution:

Step 1: Calculate volume of cube. $V = L^3 = (0.050)^3 = 1.25 \times 10^{-4} \text{ m}^3$

Step 2: Use density formula. $m = \rho V = (2700)(1.25 \times 10^{-4}) = 0.338 \text{ kg}$

Answer: 338 g

2Problem 2medium

❓ Question:

A submarine is at a depth of 200 m below the ocean surface. The density of seawater is 1025 kg/m³. What is the absolute pressure at this depth? What is the gauge pressure?

💡 Show Solution

Given:

Depth: $h = 200$ m
Seawater density: $\rho = 1025$ kg/m³
$g = 9.8$ m/s²
$P_{atm} = 1.01 \times 10^5$ Pa

Find: Absolute and gauge pressure

Solution:

Step 1: Calculate gauge pressure (water only). $P_{gauge} = \rho g h = (1025)(9.8)(200)$ $P_{gauge} = 2.01 \times 10^6 \text{ Pa} = 2.01 \text{ MPa}$

Step 2: Calculate absolute pressure (total). $P_{absolute} = P_{atm} + P_{gauge}$ $P_{absolute} = 1.01 \times 10^5 + 2.01 \times 10^6$ $P_{absolute} = 2.11 \times 10^6 \text{ Pa} = 2.11 \text{ MPa}$

Step 3: Convert to atmospheres. $P_{absolute} = \frac{2.11 \times 10^6}{1.01 \times 10^5} \approx 20.9 \text{ atm}$

Answer:

Gauge: 2.01 MPa
Absolute: 2.11 MPa or 20.9 atm

3Problem 3hard

❓ Question:

A hydraulic lift has an input piston (diameter 5.0 cm) and output piston (diameter 30 cm). (a) What force must be applied to lift a 2000 kg car? (b) If the input piston is pushed down 20 cm, how far does the car rise?

💡 Show Solution

Given:

$d_1 = 5.0$ cm $= 0.050$ m, $d_2 = 30$ cm $= 0.30$ m
Car mass: $m = 2000$ kg

Solution:

Part (a): Find input force

Step 1: Calculate areas. $A_1 = \pi r_1^2 = \pi (0.025)^2 = 1.96 \times 10^{-3} \text{ m}^2$ $A_2 = \pi r_2^2 = \pi (0.15)^2 = 7.07 \times 10^{-2} \text{ m}^2$

Step 2: Find output force (car weight). $F_2 = mg = (2000)(9.8) = 19,600 \text{ N}$

Step 3: Apply Pascal's principle. $F_1 = F_2 \frac{A_1}{A_2} = (19,600) \frac{1.96 \times 10^{-3}}{7.07 \times 10^{-2}}$ $F_1 = 543 \text{ N}$

Part (b): Find output displacement

Step 4: Use volume conservation. $A_1 d_1 = A_2 d_2$ $d_2 = d_1 \frac{A_1}{A_2} = (0.20) \frac{1.96 \times 10^{-3}}{7.07 \times 10^{-2}}$ $d_2 = 0.0055 \text{ m} = 5.5 \text{ mm}$

Answer:

(a) Input force: 543 N
(b) Car rises: 5.5 mm

Mechanical advantage = 36, but distance disadvantage = 36. Energy conserved!

Next Topic →

Buoyancy and Archimedes' Principle

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore more AP Physics 2 topics