💧 Density & Pressure — Foundations

Part 1 of 7 — The Language of Fluids

Fluid mechanics is one of the most intuitive and testable topics on the AP Physics 2 exam. You already have everyday experience with fluids — now we'll make it precise.

A fluid is anything that flows: liquids and gases. Both obey the same fundamental laws.

Density

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

Where:

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

Key Densities to Know

Why Density Matters

Density determines whether an object sinks or floats:

• Object denser than fluid → sinks
• Object less dense than fluid → floats
• Object same density as fluid → neutrally buoyant (hovers)

Ice floats because $\rho_{\text{ice}} = 917 < 1000 = \rho_{\text{water}}$.

What Is Pressure?

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

Pressure is the force exerted per unit area, perpendicular to a surface.

• Units: Pascal (Pa) = N/m²
• Type: Scalar — pressure at a point acts equally in all directions
• Other units: 1 atm = 101,325 Pa ≈ $10^5$ Pa

Atmospheric Pressure

The atmosphere presses on everything at Earth's surface:

$P_{\text{atm}} \approx 1.01 \times 10^5 \text{ Pa}$

That's about 10 N per cm² — roughly the weight of a 1 kg mass on your fingertip!

Gauge vs. Absolute Pressure

• Absolute pressure: The total pressure at a point (includes atmospheric)
• Gauge pressure: The pressure above atmospheric: $P_{\text{gauge}} = P_{\text{abs}} - P_{\text{atm}}$
• A tire gauge reading of "32 psi" is gauge pressure. The actual (absolute) pressure inside is $32 + 14.7 \approx 47$ psi.

Concept Check — Density & Pressure Basics

Density Calculation Drill

A solid sphere has radius $r = 0.05$ m and mass $m = 1.5$ kg. (Volume of sphere: $V = \frac{4}{3}\pi r^3$)

1. Volume of the sphere (in m³, round to 3 significant figures)
2. Density of the sphere (in kg/m³, round to nearest integer)
3. Will it sink or float in water? (type "sink" or "float")

Exit Quiz

🏊 Pressure in Fluids — Depth & Pascal's Law

Part 2 of 7 — How Pressure Varies

Pressure in a fluid isn't constant — it increases with depth. This one idea explains why your ears hurt at the bottom of a pool and why dams are thicker at the base.

Pressure vs. Depth

For a static (non-moving) fluid of uniform density:

$P = P_0 + \rho g h$

Where:

• $P$ = absolute pressure at depth $h$
• $P_0$ = pressure at the surface (usually $P_{\text{atm}}$)
• $\rho$ = fluid density (kg/m³)
• $g = 9.8$ m/s² (or 10 for AP estimates)
• $h$ = depth below the surface (m)

Key Insights

1. Pressure depends only on depth — not on the shape of the container
2. At the same depth, pressure is the same everywhere in a connected fluid
3. Pressure increases linearly with depth ($\rho g$ per meter)

How much does pressure increase per meter in water?

$\Delta P = \rho g \Delta h = (1000)(9.8)(1) = 9800 \text{ Pa/m} \approx 10{,}000 \text{ Pa/m}$

Every meter of depth adds roughly 0.1 atm of pressure!

Pascal's Law

A pressure change applied to an enclosed fluid is transmitted undiminished to every point in the fluid and to the walls of the container.

Hydraulic Systems

This principle makes hydraulic lifts possible:

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

A small force on a small piston creates the same pressure as a large force on a large piston.

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

Example

A hydraulic lift has pistons with areas $A_1 = 0.01$ m² and $A_2 = 0.5$ m². You push with 100 N on the small piston.

$F_2 = 100 \times \frac{0.5}{0.01} = 100 \times 50 = 5000 \text{ N}$

You amplified the force by 50×! But there's a trade-off: the small piston moves 50× farther than the large piston (conservation of energy: $W = Fd$).

Concept Check — Depth & Pascal's Law

Depth-Pressure Drill (use $g = 10$ m/s², $P_{\text{atm}} = 10^5$ Pa, $\rho_{\text{water}} = 1000$ kg/m³)

1. Gauge pressure at 5 m depth in water (in Pa)
2. Absolute pressure at 5 m depth (in Pa)
3. Depth at which absolute pressure is 2 atm (in m)

Hydraulic Lift Drill

A car (mass 1500 kg, weight 15,000 N) sits on a hydraulic lift piston with area $A_2 = 0.25$ m².

1. Pressure under the car piston (in Pa)
2. Area of small input piston needed if you can push with 300 N (in m²)
3. How far must the small piston move to raise the car 0.10 m? (in m)

Round all answers to 3 significant figures.

Exit Quiz

📏 Manometers & Pressure Measurement

Part 3 of 7 — Measuring Pressure Like a Physicist

How do we actually measure pressure? This part covers the tools and techniques that appear on AP exams — manometers, barometers, and pressure conversions.

The Mercury Barometer

A classic barometer is a tube of mercury inverted in a mercury dish. The atmosphere pushes down on the dish, supporting a column of mercury:

$P_{\text{atm}} = \rho_{\text{Hg}} g h$

At sea level: $h = \frac{P_{\text{atm}}}{\rho_{\text{Hg}} g} = \frac{101{,}325}{(13{,}600)(9.8)} \approx 0.760 \text{ m} = 760 \text{ mm}$

This is why atmospheric pressure is sometimes stated as "760 mmHg" or "760 torr."

Why Mercury?

Mercury is very dense ($\rho = 13{,}600$ kg/m³), so the column is only 76 cm tall. A water barometer would need a column over 10 meters tall!

U-Tube Manometers

A manometer is a U-shaped tube partially filled with liquid, used to measure the pressure of a gas:

Open Manometer

One side is connected to a gas, the other is open to the atmosphere.

• If the gas side is lower, $P_{\text{gas}} > P_{\text{atm}}$
• If the gas side is higher, $P_{\text{gas}} < P_{\text{atm}}$

$P_{\text{gas}} = P_{\text{atm}} + \rho g \Delta h \quad \text{(gas side lower)}$ $P_{\text{gas}} = P_{\text{atm}} - \rho g \Delta h \quad \text{(gas side higher)}$

Where $\Delta h$ is the height difference between the two columns.

Closed Manometer

One side is sealed (vacuum above the liquid). The height difference directly gives the gas pressure:

$P_{\text{gas}} = \rho g h$

Pressure Unit Conversions

You need to be fluent with these on the AP exam:

Quick Conversion Strategy

To convert mmHg → Pa: $P(\text{Pa}) = \frac{P(\text{mmHg})}{760} \times 101{,}325$

Or equivalently: 1 mmHg ≈ 133.3 Pa.

Manometer Reasoning Quiz

Pressure Measurement Drill (use $g = 10$ m/s²)

A closed-end manometer has mercury ($\rho = 13{,}600$ kg/m³) with a height of 0.50 m.

1. Gas pressure in Pa
2. Gas pressure in atm (round to 2 decimals)
3. Gas pressure in mmHg

Round all answers to 3 significant figures.

Exit Quiz

🔧 Pressure Problem-Solving Workshop

Part 4 of 7 — Building Calculation Confidence

You know the formulas. Now let's drill the problem types that appear most often on the AP exam — multi-step depth problems, hydraulics, and unit conversions under time pressure.

Problem-Solving Framework

Every fluid statics problem follows this pattern:

Step 1 — Identify the fluid(s) and their densities

Step 2 — Identify the two points you're comparing pressure between

Step 3 — Apply $P_2 = P_1 + \rho g \Delta h$ moving downward (pressure increases with depth)

Step 4 — Watch units! Convert cm → m, g/cm³ → kg/m³ before plugging in

Common Traps

• Forgetting to add $P_{\text{atm}}$ when absolute pressure is needed
• Using depth from the bottom instead of from the surface
• Mixing up gauge vs. absolute pressure
• Using wrong density when there are multiple fluid layers

Layered Fluids

When two or more immiscible fluids are layered (e.g., oil floating on water), add each layer's contribution separately:

$P_{\text{bottom}} = P_{\text{atm}} + \rho_1 g h_1 + \rho_2 g h_2$

Worked Example

A tank has 0.5 m of oil ($\rho = 800$ kg/m³) floating on 2.0 m of water ($\rho = 1000$ kg/m³). Find the absolute pressure at the bottom.

$P = P_{\text{atm}} + \rho_{\text{oil}} g h_{\text{oil}} + \rho_{\text{water}} g h_{\text{water}}$

$P = 10^5 + (800)(10)(0.5) + (1000)(10)(2.0)$

$P = 100{,}000 + 4{,}000 + 20{,}000 = 124{,}000 \text{ Pa}$

$P = 1.24 \text{ atm}$

Layered Fluid Drill (use $g = 10$ m/s², $P_{\text{atm}} = 10^5$ Pa)

A container has three layers:

• Top: 0.3 m of gasoline ($\rho = 680$ kg/m³)
• Middle: 1.0 m of water ($\rho = 1000$ kg/m³)
• Bottom: 0.1 m of mercury ($\rho = 13{,}600$ kg/m³)

1. Gauge pressure at the bottom of the gasoline layer (in Pa)
2. Gauge pressure at the bottom of the water layer (in Pa)
3. Absolute pressure at the very bottom (in Pa)

Tricky Conceptual Questions — AP exam favorites

AP-Style Problem: The U-Tube

A U-tube contains mercury ($\rho = 13{,}600$ kg/m³). Water ($\rho = 1000$ kg/m³) is poured into the left side to a height of 27.2 cm above the mercury surface.

Question: How far does the mercury level on the right side rise above the mercury level on the left?

Solution

At the mercury-water interface on the left, the pressure from the water column must equal the pressure from the extra mercury column on the right:

$\rho_w g h_w = \rho_{\text{Hg}} g \Delta h$

$h_w \cdot \frac{\rho_w}{\rho_{\text{Hg}}} = \Delta h$

$\Delta h = 0.272 \times \frac{1000}{13{,}600} = 0.020 \text{ m} = 2.0 \text{ cm}$

The mercury on the right rises 2.0 cm above the mercury on the left. The total difference in mercury levels is 2.0 cm (but the left side went down and the right went up, so each moved 1.0 cm from the original level).

Exit Quiz

🎈 Forces on Submerged Surfaces

Part 5 of 7 — Pressure Creates Force

Pressure acts on surfaces. When those surfaces are submerged in a fluid, the pressure creates real forces — forces that can collapse submarines, burst pipes, or hold back an ocean behind a dam.

Force on a Flat Horizontal Surface

For a horizontal surface at depth $h$:

$F = PA = (P_0 + \rho g h) \times A$

The pressure is uniform across the surface, so it's straightforward.

Example: Aquarium Floor

An aquarium (0.5 m × 0.3 m) is filled to a depth of 0.4 m. Force on the bottom (gauge only):

$F = \rho g h \times A = (1000)(10)(0.4)(0.5 \times 0.3) = 600 \text{ N}$

Note: This is just the force from the water pressure — it equals the weight of the water above ($mg = \rho V g = 1000 \times 0.06 \times 10 = 600$ N). Not a coincidence!

Force on Vertical Surfaces

For a vertical surface (like a dam wall), pressure varies with depth. The total force requires integration, but the AP shortcut uses the average pressure:

$F = P_{\text{avg}} \times A = \frac{1}{2}\rho g h_{\text{max}} \times A$

Where:

• $h_{\text{max}}$ = depth of the bottom of the wall
• $A$ = area of the submerged wall surface

The force acts at a depth of $\frac{2}{3}h_{\text{max}}$ from the surface (center of pressure, not center of area).

Why Is the Center of Pressure Below the Centroid?

Because pressure increases with depth. The lower part of the wall experiences more pressure than the upper part, pulling the effective "center" of force downward.

Concept Check — Submerged Forces

Force Calculation Drill (use $g = 10$ m/s²)

A rectangular tank (2.0 m wide × 1.0 m long × 1.5 m deep) is filled completely with water.

1. Force on the horizontal bottom (gauge, in N)
2. Average gauge pressure on one of the 2.0 m wide vertical walls (in Pa)
3. Total force on that vertical wall (in N)

Real-World Applications

Submarine Depth Limits

At 400 m depth, gauge pressure is $\sim 4 \times 10^6$ Pa (≈ 40 atm). The hull must withstand enormous compressive forces. Most military subs max out at ~300-500 m; the deepest dive ever (Mariana Trench, 10,994 m) experienced ~1100 atm.

Blood Pressure

Blood pressure is measured in mmHg. A reading of "120/80" means:

• Systolic (heart pumping): 120 mmHg = 16,000 Pa gauge
• Diastolic (heart resting): 80 mmHg = 10,700 Pa gauge

This is tiny compared to atmospheric pressure — your blood vessels are under less than 0.16 atm of gauge pressure.

Deep-Sea Fish

Fish at great depths have no gas-filled cavities that would collapse under pressure. They are adapted to their environment. Bringing them to the surface can be fatal — their internal pressure suddenly exceeds external pressure.

Exit Quiz

🧪 Specific Gravity & Density Applications

Part 6 of 7 — Practical Density Skills

Before we wrap up density and pressure, let's cover specific gravity, density measurement techniques, and the multi-step problems that pull everything together.

Specific Gravity

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

$\text{SG} = \frac{\rho_{\text{substance}}}{\rho_{\text{water}}} = \frac{\rho}{1000\ \text{kg/m}^3}$

Since it's a ratio, SG has no units.

Quick Trick

In CGS units (g/cm³), the numerical value of density equals the specific gravity! Water has $\rho = 1.00$ g/cm³, so SG = density in g/cm³.

Fraction Submerged (Preview of Buoyancy)

When an object floats, the fraction submerged equals the ratio of densities:

$\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}$

Examples

Iceberg insight: "The tip of the iceberg" is only ~8-10% of its total volume. The rest is hidden underwater.

Quick Check — Specific Gravity & Floating

Synthesis Drill (use $g = 10$ m/s², $P_{\text{atm}} = 10^5$ Pa)

A U-tube has water ($\rho = 1000$ kg/m³) in the left arm and oil (SG = 0.80) in the right arm. The water surface is 0.20 m above the oil-water interface.

1. Density of the oil (in kg/m³)
2. Height of oil above the interface (in m)
3. Difference in surface levels — which is higher and by how much? (oil surface height − water surface height, in m)

Round all answers to 3 significant figures.

Exit Quiz

🏆 Density & Pressure — Synthesis & AP Review

Part 7 of 7 — Putting It All Together

Let's tie together everything from Parts 1-6 with cumulative problems, common misconceptions, and AP exam strategies.

Concept Map — Everything Connected

$\text{Density } (\rho = m/V) \longrightarrow \text{Pressure } (P = F/A)$ $\downarrow$ $\text{Depth } (P = P_0 + \rho g h) \longrightarrow \text{Pascal's Law}$ $\downarrow$ $\text{Hydraulics } (F_1/A_1 = F_2/A_2) \longrightarrow \text{Manometers}$ $\downarrow$ $\text{Floating } (\text{fraction} = \rho_{\text{obj}}/\rho_{\text{fluid}}) \longrightarrow \text{Buoyancy (next topic!)}$

The Big Ideas

1. Pressure is a scalar — it acts equally in all directions at a point
2. Pressure increases with depth — linearly, at rate $\rho g$ per meter
3. Shape doesn't matter — only depth determines pressure (hydrostatic paradox)
4. Pascal's Law — pressure changes transmit throughout a fluid
5. Density ratios predict floating behavior

Top 5 AP Mistakes for Density & Pressure

1. Gauge vs. Absolute

• "The pressure at 10 m depth" — is this gauge or absolute? Read carefully!
• Default on AP: usually asking for absolute unless stated otherwise

2. Units

• Density: must be in kg/m³ (not g/cm³) for SI calculations
• Pressure: Pa = N/m² (not kPa, not atm, unless specified)

3. Depth Direction

• $h$ is measured downward from the surface, not upward from the bottom
• In a closed container with pressurized gas above, $P_0 \neq P_{\text{atm}}$

4. The Shape Trap

• Students think wider containers have more pressure at the bottom — they don't!
• Pressure depends only on $\rho$, $g$, and $h$

5. Forgetting the Atmosphere

• Unless the problem says "gauge" or the container is sealed with a specific pressure, assume the surface is at $P_{\text{atm}}$

Synthesis Quiz — Multi-concept questions

Final Mixed Drill (use $g = 10$ m/s², $P_{\text{atm}} = 10^5$ Pa)

1. Pressure at 25 m depth in a lake (absolute, in Pa)
2. A 0.004 m³ block of wood (density 600 kg/m³) floats. Volume above water (in m³)
3. Force needed on a 0.002 m² piston to hold back water at 15 m depth (gauge force, in N)

Round all answers to 3 significant figures.

Final Exam-Style Questions