Density and Pressure in Fluids

🌊 Density: $\rho = m/V$

Part 1 of 7 — Fluids: Density & Pressure

Density is the most fundamental fluid property. Whether an object floats or sinks, why pressure increases with depth, and how hydraulic systems work all start from density.

In this lesson you will learn:

• The definition $\rho = m/V$
• SI units $(kg/m^{3})$ and common conversions ($g/cm^{3}$ ↔ $kg/m^{3}$)
• Densities of common fluids (water, air, mercury)
• How density connects to mass and volume in problem solving

Definition

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

• $m$: mass (kg)
• $V$: volume $(m^{3})$
• $\rho$: density ($kg/m^{3}$, read "rho")

Reference Densities (memorize for AP)

Key Conversion

$1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3$

So water = 1.00 $g/cm^{3}$ = 1000 $kg/m^{3}$.

Why Density Matters

• Floats vs sinks → compare $\rho_{obj}$ to $\rho_{fluid}$
• Hydrostatic pressure $P = \rho g h$ depends on density
• Buoyant force $F_b = \rho g V$ depends on fluid density

Density Concepts 🎯

Density Calculations 🧮

Use SI units throughout.

1. A 6.0 kg block has volume $2.0\times10^{-3}$ $m^{3}$. Density $(kg/m^{3})$?

2. A cube of metal with side length 0.10 m has mass 7.8 kg. Density $(kg/m^{3})$?

3. A 500 mL container is filled with mercury. Mass (kg)? (Hint: 500 mL = $5.0\times10^{-4}$ $m^{3}$, $\rho_{Hg} = 13{,}600$ $kg/m^{3}$)

Density Comparisons 🔍

Exit Quiz — Density ✅

📐 Pressure: $P = F/A$

Part 2 of 7 — Fluids: Density & Pressure

Pressure is force spread over an area. Whether it's the pressure of a finger pushing on a thumbtack or atmospheric pressure pressing on your skin, every fluid problem hinges on this definition.

In this lesson you will learn:

• The definition $P = F/A$
• The pascal (Pa) and useful conversions (kPa, atm, mmHg)
• How pressure differs from force
• Common AP traps with units

Definition of Pressure

$P = \frac{F_\perp}{A}$

• $F_\perp$: force perpendicular (normal) to the surface (N)
• $A$: area over which the force acts $(m^{2})$
• $P$: pressure (Pa = $N/m^{2}$)

Useful Pressure Conversions

Pressure ≠ Force

A nail and a hammer driving on the same hand exert similar forces but very different pressures because the nail's contact area is tiny.

$P_{nail} = \frac{F}{A_{small}} \gg P_{hand} = \frac{F}{A_{large}}$

Pressure in Solids vs Fluids

In a static fluid, pressure acts in all directions at every point — not just downward. This is why a balloon under water gets squeezed evenly from every side.

Pressure Concepts 🎯

Pressure Calculations 🧮

1. A 200 N force pushes on a $4.0\times10^{-2}$ $m^{2}$ surface. Pressure (Pa)?

2. An adult of mass 80 kg stands on one foot of area 0.020 $m^{2}$ ($g = 9.8$). Pressure under the foot (Pa)?

3. A force of 50 N is applied to a tack of contact area $1.0\times10^{-6}$ $m^{2}$. Pressure (Pa)?

Force vs Pressure 🔍

Exit Quiz — Pressure ✅

🌊 Hydrostatic Pressure: $P = \rho g h$

Part 3 of 7 — Fluids: Density & Pressure

When you dive into a pool, you feel pressure on your ears within a few feet. The pressure of a static fluid increases linearly with depth. This is the workhorse equation of AP fluids.

In this lesson you will learn:

• Why pressure increases with depth
• The equation $P = \rho g h$ (gauge form) and $P_{abs} = P_0 + \rho g h$
• That pressure depends only on vertical depth, not container shape
• How to calculate pressure differences between two depths

Why Pressure Increases with Depth

Imagine a horizontal slab of fluid at depth $h$. The fluid above presses down with weight:

$W = mg = \rho V g = \rho (Ah) g$

That weight produces pressure:

$P_{from\ fluid} = \frac{W}{A} = \rho g h$

The Equations

Gauge pressure (pressure due to fluid alone): $P_{gauge} = \rho g h$

Absolute pressure (gauge + atmosphere on top): $P_{abs} = P_0 + \rho g h$

Where:

• $\rho$: fluid density $(kg/m^{3})$
• $g = 9.8$ $m/s^{2}$ (often $g = 10$ on AP estimates)
• $h$: depth below the free surface (m)
• $P_0$: pressure at the surface (often $P_{atm} = 1.013\times10^{5}$ Pa)

The Surprising Truth: Shape Doesn't Matter

Pressure at depth $h$ is the same whether you're at the bottom of a thin tube or a vast lake — only $\rho$, $g$, and $h$ matter.

Quick Mental Math

• 10 m of water adds about 1 atm (~$10^{5}$ Pa).
• Every 10 m deeper in seawater ≈ 1 additional atm.

Hydrostatic Pressure Concepts 🎯

Hydrostatic Calculations 🧮 (use $g = 9.8$ $m/s^{2}$, $\rho_{water} = 1000$ $kg/m^{3}$, $\rho_{Hg} = 13{,}600$ $kg/m^{3}$)

1. Gauge pressure at the bottom of a 3.0 m deep fresh water pool (Pa)?

2. Absolute pressure at 10 m depth in fresh water with $P_{atm} = 1.0\times10^{5}$ Pa (Pa)?

3. Gauge pressure at the bottom of a 0.76 m mercury column (Pa)? (This is the original definition of 1 atm.)

Hydrostatic Reasoning 🔍

Exit Quiz — Hydrostatic Pressure ✅

📊 Absolute vs Gauge Pressure

Part 4 of 7 — Fluids: Density & Pressure

A tire reads "32 psi" but the absolute pressure inside is closer to 47 psi. The difference is whether you include atmospheric pressure. AP problems mix these freely — knowing which to use is critical.

In this lesson you will learn:

• The definitions of gauge and absolute pressure
• $P_{abs} = P_{atm} + P_{gauge}$
• When AP gives you each
• Common pitfalls (negative gauge ≠ vacuum)

The Two Pressures

Absolute Pressure

Total pressure measured from a perfect vacuum (zero).

$P_{abs} = P_{atm} + \rho g h$

Open container at depth $h$: includes the weight of fluid AND the atmospheric column above.

Gauge Pressure

Pressure relative to the surrounding atmosphere.

$P_{gauge} = P_{abs} - P_{atm}$

A tire gauge reads zero when tire pressure equals atmospheric pressure (a flat tire).

Quick Comparison Table

When to Use Which

• Pressure differences → either works (the $P_{atm}$ cancels)
• Force on a submerged surface with one side exposed to atmosphere → use gauge (atm cancels)
• Force on a fully submerged closed object → atmospheric forces cancel anyway → either works
• Closed container with no atmosphere on the other side → use absolute

Gauge vs Absolute 🎯

Gauge ↔ Absolute 🧮 (use $P_{atm} = 1.0\times10^{5}$ Pa, $g = 10$, $\rho_{water} = 1000$)

1. An open pool is 5 m deep. Absolute pressure at the bottom (Pa)?

2. Same pool — what is the gauge pressure at the bottom (Pa)?

3. A vacuum chamber is at $1.0\times10^{4}$ Pa absolute. Its gauge pressure (Pa) is — give the signed answer.

Choosing Wisely 🔍

Exit Quiz — Gauge vs Absolute ✅

🔧 Pascal's Principle & Hydraulics

Part 5 of 7 — Fluids: Density & Pressure

A small force on a small piston can lift a car. Pascal's Principle is the engineering magic behind hydraulic brakes, jacks, and lifts. The key idea: pressure changes transmit fully through a confined fluid.

In this lesson you will learn:

• Pascal's Principle (statement)
• The hydraulic-press equation $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$
• Why hydraulics multiply force, not energy
• How displacement scales (work conservation)

Pascal's Principle

A pressure change applied to a confined fluid is transmitted undiminished to every part of the fluid and to the walls of the container.

The Hydraulic Press

Apply force $F_1$ on a small piston of area $A_1$. The added pressure $P = F_1/A_1$ travels through the fluid to a large piston of area $A_2$, lifting:

$F_2 = P \cdot A_2 = \frac{A_2}{A_1}\, F_1$

So:

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

If $A_2 = 100\,A_1$, then $F_2 = 100\,F_1$ — a 10 N push lifts 1000 N.

Force Multiplied, Energy NOT Multiplied

Energy is conserved. If $F_2 = 100\,F_1$, the large piston only rises $1/100$ as far as the small piston descends:

$F_1 \cdot d_1 = F_2 \cdot d_2 \quad \Rightarrow \quad \frac{d_2}{d_1} = \frac{A_1}{A_2}$

You trade distance for force, exactly like a lever or pulley system.

Why It Works

• Liquids are nearly incompressible.
• Pressure is isotropic in a confined static fluid.
• Closed system → added pressure has nowhere to dissipate.

Pascal's Principle Concepts 🎯

Hydraulic Press Calculations 🧮

1. A hydraulic jack: $A_1 = 4.0\times10^{-4}$ $m^{2}$, $A_2 = 2.0\times10^{-2}$ $m^{2}$. A 200 N force on the small piston lifts how many N?

2. Same jack — to raise the load 0.10 m, how far (m) must the small piston travel?

3. Input piston of area 0.0025 $m^{2}$ needs to lift a 1.5 kN load using a 60 N input. What output area $A_2$ $(m^{2})$ is required?

Hydraulic Reasoning 🔍

Exit Quiz — Pascal's Principle ✅

🛠 Problem-Solving Workshop

Part 6 of 7 — Fluids: Density & Pressure

Time to combine everything: density, hydrostatic pressure, gauge vs absolute, and Pascal's Principle. AP fluids problems often blend two or three of these — let's practice that integration.

Workshop Strategy:

1. Identify the fluid's $\rho$ and the relevant depth $h$.
2. Decide gauge or absolute (does the problem care about $P_{atm}$?).
3. For hydraulic problems, set $P$ equal at both pistons.
4. Always check units: $kg/m^{3}$, m, Pa, N.

Workflow Cheat Sheet

Common AP Combination Problems

• "Find the force on the bottom of a tank" → $F = P_{gauge} A$
• "Find force on a side wall (varying depth)" → gauge pressure grows in proportion to depth, from $0$ at the surface to $\rho g h$ at the bottom, so use the average gauge pressure $P_{avg} = \rho g (h/2)$, then $F = P_{avg} A$
• "Hydraulic lift with depth" → include $\rho g h$ correction if pistons aren't at the same height

Workshop MC 🎯

Workshop Calculations 🧮 (g = 10, $\rho_{water} = 1000$, $P_{atm} = 1.0\times10^{5}$ Pa)

1. A pool 4 m deep, bottom area 8 $m^{2}$. Force from gauge water pressure on the bottom (N)?

2. Hydraulic lift: input piston 0.005 $m^{2}$, output piston 0.10 $m^{2}$, input force 75 N. Output force (N)?

3. A vertical rectangular tank wall 2 m wide and 3 m tall (full to the top). Total horizontal force from water on the wall (N)? Use average-pressure method.

Workshop Reasoning 🔍

Exit Quiz — Workshop ✅

🎯 Synthesis & AP Review — Density & Pressure

Part 7 of 7 — Fluids: Density & Pressure

You've worked through density, pressure definitions, hydrostatic depth dependence, gauge vs absolute, and Pascal's Principle. This final part synthesizes them in AP-style multi-step questions.

Big Ideas Recap:

• $\rho = m/V$ (intensive property — independent of size)
• $P = F/A$ — pressure is force per unit perpendicular area
• $P = \rho g h$ — depth pressure depends only on $\rho$ and $h$
• Pascal: $P_1 = P_2$ throughout a confined fluid; force scales with area

AP Equation Sheet (Fluids — Pressure)

AP-Style Question Patterns

1. "Pressure at depth" → $\rho g h$ (gauge) or $P_0 + \rho g h$ (absolute).
2. "Force on a flat submerged surface" → use $P_{avg} = \rho g h_{centroid}$, then $F = P_{avg} A$.
3. "Compare two open containers" → same $\rho$ + same depth ⇒ same pressure (regardless of width or volume).
4. "Hydraulic lift force/distance" → equal pressure ⇒ $F$ scales with $A$, $d$ scales with $1/A$.
5. "Why doesn't a tall thin tube exert more pressure?" → because $P$ depends only on $\rho$, $g$, $h$.

AP Synthesis MC 🎯

AP Synthesis Calculations 🧮 (g = 10, $\rho_w = 1000$, $P_{atm} = 1.0\times10^{5}$)

1. A 0.50 $m^{3}$ block of metal has mass 1350 kg. Density $(kg/m^{3})$?

2. Diver at 20 m below sea surface (ρ ≈ 1000). Absolute pressure (Pa)?

3. Hydraulic lift: input $A_1 = 0.0010$ $m^{2}$, output $A_2 = 0.040$ $m^{2}$. To lift 4000 N, input force needed (N)?

AP Concept Synthesis 🔍

Exit Quiz — AP Synthesis ✅