Unit 8.1 | Density and Pressure
- 18 minutes
- 3 formulas
- 21 questions
- 5 videos
Every fluids problem starts with two ideas: how packed a fluid is and how hard it pushes. This lesson covers density, pressure and how pressure grows with depth.
Unit 8 Breakdown
In this lesson:
- What a fluid is, and what makes a fluid “ideal” on the AP exam
- Density, [katex] \rho = \frac{m}{V} [/katex], and why a piece of an object has the same density as the whole
- Pressure, [katex] P = \frac{F_\perp}{A} [/katex], and how it grows with depth: [katex] P = P_0 + \rho g h [/katex]
- The 4 problem types: density, pressure from a force, pressure at depth, and containers, U-tubes and straws
Introduction
This lesson covers density and pressure, the two ideas every fluids problem is built on. It starts Unit 8, the last unit of AP Physics 1.
A fluid is anything that flows: liquids and gases. A fluid has no fixed shape, so it pushes on every surface it touches. Two numbers describe that push. Density says how much mass is packed into the fluid, and pressure says how hard it pushes on each square meter.
Fluids and the Ideal Fluid
The particles in a fluid move freely and bump into everything around them. Those collisions are what we feel as pressure. In AP Physics 1 you treat every liquid as an ideal fluid:
- Incompressible: its density stays the same everywhere, at any depth and any pressure.
- No viscosity: no internal friction, so no mechanical energy turns into thermal energy as it flows.
Real gases are compressible, which is why air thins out as you climb a mountain. Unless a question says otherwise, treat liquids as ideal.
Density
[katex] \rho = \frac{m}{V} [/katex]
Density [katex] \rho [/katex] (the Greek letter rho) is mass per volume, in kg/m³. Water is about [katex] 1000\,\text{kg/m}^3 [/katex], which is the same as [katex] 1\,\text{g/cm}^3 [/katex]. A substance that is the same all the way through has one density, so any piece of it, big or small, has that same density.
| Substance | Density (kg/m³) |
| Air (sea level) | 1.2 |
| Wood (pine) | about 500 |
| Oil | about 800–900 |
| Water | 1000 |
| Seawater | 1025 |
| Aluminum | 2700 |
| Steel | 7800 |
Pressure
[katex] P = \frac{F_\perp}{A} [/katex]
Pressure is the perpendicular force on a surface divided by its area. Its unit is the pascal: [katex] 1 \text{ Pa} = 1 \text{ N/m}^2 [/katex]. The air around you pushes with about [katex] 1.0 \times 10^5 \text{ Pa} [/katex], called 1 atmosphere ([katex] 1\,\text{atm} [/katex]).
▶ Watch: density and pressure basics (starts at 1:49).
Pressure is a scalar. At any point in a fluid at rest it pushes the same amount in every direction, and the force it makes on a surface always points straight into that surface.
Pressure at a Depth
Go deeper and the fluid above you weighs more, so the pressure grows. Here is where the formula comes from:
- Picture a column of liquid with top area [katex] A [/katex] and height [katex] h [/katex]. Its mass is [katex] m = \rho V = \rho A h [/katex].
- Its weight pushes on the area at the bottom: [katex] \frac{mg}{A} = \frac{\rho A h g}{A} = \rho g h [/katex].
- The air above the surface pushes down too, with [katex] P_0 [/katex]. Add the two.
[katex] P = P_0 + \rho g h [/katex]
[katex] P [/katex] is the absolute pressure, the total. The part from the liquid alone, [katex] P_{gauge} = P – P_0 = \rho g h [/katex], is the gauge pressure: what a tire gauge reads. Only depth matters, not how wide the container is.
Pressure grows linearly with depth
About every [katex] 10\,\text{m} [/katex] of water adds [katex] 1\,\text{atm} [/katex]. At [katex] 10\,\text{m} [/katex] down the absolute pressure is about [katex] 2\,\text{atm} [/katex]; at [katex] 20\,\text{m} [/katex], about [katex] 3\,\text{atm} [/katex].
Every density and pressure question on the AP exam is one of the four types below.
| Problem type | The clue | The setup |
| 1. Density | mass and volume, “cut in half”, a displaced volume | [katex] \rho = \frac{m}{V} [/katex] |
| 2. Pressure from a force | a force pressing on an area: feet, tires, pistons | [katex] P = \frac{F_\perp}{A} [/katex] |
| 3. Pressure at depth | “below the surface”, “absolute”, “gauge” | [katex] P = P_0 + \rho g h [/katex] |
| 4. Containers, U-tubes and straws | odd-shaped containers, two liquids in a U-tube, a straw or barometer | Same fluid, same level, same pressure |
Problem Type 1: Density
How to spot it: A mass and a volume, a shape whose volume you can work out, an object cut into pieces, or a volume measured by how much water rises.
Setup: Find the volume first (a box is [katex] l \times w \times h [/katex], a sphere is [katex] \tfrac{4}{3}\pi r^3 [/katex]), then [katex] \rho = \frac{m}{V} [/katex]. A rise in water level equals the volume of what sank into it.
A solid aluminum block measures [katex] 0.50\,\text{m} [/katex] × [katex] 0.20\,\text{m} [/katex] × [katex] 0.10\,\text{m} [/katex] and has a mass of [katex] 27\,\text{kg} [/katex]. Find its density. The block is then cut in half. What is the density of each half?
[katex] V = (0.50)(0.20)(0.10) = 0.010 \text{ m}^3 [/katex], so [katex] \rho = \frac{27}{0.010} = 2700 \text{ kg/m}^3 [/katex]. Each half has half the mass and half the volume: [katex] \frac{13.5}{0.0050} = 2700 \text{ kg/m}^3 [/katex]. Same material, same density.
AP trap: Thinking a bigger or heavier piece is denser. Density belongs to the material, not to the amount of it.
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Problem Type 2: Pressure from a force
How to spot it: A force spread over an area: a person standing, a tire on the road, a nail or a heel, a book on a table.
Setup: Find the force perpendicular to the surface (often a weight, [katex] mg [/katex]), convert the area to m², then [katex] P = \frac{F_\perp}{A} [/katex].
A [katex] 60\,\text{kg} [/katex] skater stands on one foot. Her shoe touches the ice over [katex] 0.015\,\text{m}^2 [/katex]. On her skate the blade touches only [katex] 4.0 \times 10^{-4} \text{ m}^2 [/katex]. Find the pressure on the ice each way.
Her weight is [katex] (60)(9.8) = 588 \text{ N} [/katex]. Shoe: [katex] P = \frac{588}{0.015} \approx 3.9 \times 10^4 \text{ Pa} [/katex]. Blade: [katex] P = \frac{588}{4.0 \times 10^{-4}} \approx 1.5 \times 10^6 \text{ Pa} [/katex]. Same force, about 38 times less area, so about 38 times the pressure.
AP trap: Forgetting to convert cm² to m². [katex] 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 [/katex], not [katex] 10^{-2} [/katex]. Also, double the force and quadruple the area and the pressure is halved, not doubled.
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Problem Type 3: Pressure at depth: absolute and gauge
How to spot it: “At a depth of”, “below the surface”, “absolute pressure”, “gauge pressure”, divers, submarines, dams.
Setup: Measure [katex] h [/katex] down from the free surface. Gauge: [katex] \rho g h [/katex]. Absolute: add [katex] P_0 \approx 1.0 \times 10^5 \text{ Pa} [/katex]. Read which one the question wants.
A diver is [katex] 12\,\text{m} [/katex] below the surface of a lake. Find the gauge pressure and the absolute pressure on her. Use [katex] P_0 = 1.0 \times 10^5 \text{ Pa} [/katex].
Gauge: [katex] \rho g h = (1000)(9.8)(12) \approx 1.18 \times 10^5 \text{ Pa} [/katex]. Absolute: [katex] 1.0 \times 10^5 + 1.18 \times 10^5 \approx 2.2 \times 10^5 \text{ Pa} [/katex], a bit over [katex] 2\,\text{atm} [/katex]. If she dives to [katex] 24\,\text{m} [/katex], the gauge pressure doubles, but the absolute pressure does not.
▶ Watch: pressure at depth, worked from scratch (starts at 5:51).
AP trap: Doubling the absolute pressure when the depth doubles. Only the [katex] \rho g h [/katex] part doubles; [katex] P_0 [/katex] stays the same. And changing the air pressure changes the absolute pressure but not the gauge pressure.
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Problem Type 4: Containers, U-tubes and straws
How to spot it: Containers of different shapes, a U-tube with two liquids, a piston on one arm, a straw, or a barometer.
Setup: In one connected fluid at rest, points at the same level have the same pressure, whatever the shape. Pick a level, write the pressure there from each side, and set them equal.
Oil is poured into the left arm of a U-tube that holds water. The oil column is [katex] 20\,\text{cm} [/katex] tall, and the oil–water interface is [katex] 16\,\text{cm} [/katex] below the water surface in the right arm. Find the density of the oil.
Pick the level of the interface. Left: [katex] P_0 + \rho_o g (0.20) [/katex]. Right: [katex] P_0 + (1000) g (0.16) [/katex]. Set them equal; [katex] P_0 [/katex] and [katex] g [/katex] cancel: [katex] \rho_o = 1000 \cdot \frac{0.16}{0.20} = 800 \text{ kg/m}^3 [/katex]. The lighter oil needs a taller column to make the same pressure.
Straws work the same way. Sucking lowers the pressure in the straw, and the air pushing on the drink outside pushes it up. Even a perfect vacuum can lift water only [katex] h = \frac{P_0}{\rho g} = \frac{1.0 \times 10^5}{(1000)(9.8)} \approx 10 \text{ m} [/katex].
▶ Watch: how air pressure holds up a column of liquid (starts at 2:52).
Pascal’s principle is the same idea in a closed container: a pressure added at one point reaches every point of the fluid. A hydraulic lift uses it, [katex] \frac{F_1}{A_1} = \frac{F_2}{A_2} [/katex], so a small force on a small piston holds up a car on a big one. Pascal’s principle is not in the AP Physics 1 course framework, so it is good to know but not required. If a lift does show up, [katex] P = \frac{F}{A} [/katex] and “same level, same pressure” solve it.
AP trap: Thinking a wider container, or one with more water, has more pressure at the bottom. Pressure depends only on the depth and the liquid. Also, a straw does not “pull” the drink up: the air outside pushes it.
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Lesson 8.2 Preview
In the next lesson the pressure difference between the top and bottom of a submerged object becomes a force: the buoyant force. It explains why ships float, why rocks feel lighter underwater, and how much of an iceberg hides below the surface.
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This article was written by Jason