Pressure — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Definition of pressure, hydrostatic absolute and gauge pressure formulas, Pascal’s principle, pressure-derived buoyant force, pressure measurement, and problem-solving for static fluid systems including hydraulic applications.

You should already know: Density definition and calculation for uniform fluids, basics of force and area from Newtonian mechanics, core properties of liquid and gas fluids.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Pressure?

Pressure is defined as the magnitude of the normal force exerted per unit area by a fluid on any surface it contacts. Standard notation uses $P$ for pressure, with SI units of Pascals, where $1\text{ Pa}=1{\text{ N/m}}^2$. Common non-SI units you will encounter on the exam include atmospheres ($1\text{ atm}=1.013\times 10^5\text{ Pa}$) and millimeters of mercury ($760\text{ mm Hg}=1\text{ atm}$). Unlike force, pressure is a scalar quantity with no direction, though the force it produces is always perpendicular to the contact surface.

Per the AP Physics 2 Course and Exam Description (CED), Unit 1 Fluids accounts for 10-12% of total exam score, and pressure concepts appear in both multiple-choice (MCQ) and free-response (FRQ) sections. Pressure is almost always a foundational step for larger problems involving buoyancy, fluid flow, or ideal gases, so you can expect 2-4 direct MCQ questions and at least one FRQ part requiring pressure calculations or reasoning. Pressure arises from molecular collisions with a surface, and for static fluids, pressure at any point is equal in all directions.

2. Hydrostatic Pressure in Static Fluids

For a static, incompressible fluid in a uniform gravitational field, pressure increases with depth because fluid at a given depth must support the weight of all fluid above it. To derive the core formula, consider a horizontal cross-section of fluid at depth $h$, with cross-sectional area $A$ and fluid density $\rho$. The weight of the fluid above the cross-section is $W=mg=\rho Vg=\rho Ahg$. Pressure is force per unit area, so the pressure from the overlying fluid is $P=W/A=\rho gh$.

This value is called gauge pressure, the pressure relative to atmospheric pressure at the fluid's surface. To get total absolute pressure, add the atmospheric pressure $P_0$ acting on the fluid surface: $$P_{\text{abs}}=P_0+\rho gh$$ A key result of this formula is that pressure only depends on depth, fluid density, and surface pressure, not the shape of the container. This is the hydrostatic paradox: pressure at the bottom of a wide lake is identical to pressure at the bottom of a narrow tower of water at the same depth, because the extra weight of the wider water is supported by the entire container floor, not just a single point.

Worked Example

A diver descends to 32 m below the surface of the ocean. Seawater has a density of $1025{\text{ kg/m}}^3$, and atmospheric pressure at the surface is $1.01\times 10^5\text{ Pa}$. Calculate (a) the gauge pressure at this depth and (b) the absolute pressure.

1. List given values: $h=32\text{ m}$, $\rho =1025{\text{ kg/m}}^3$, $g=9.8{\text{ m/s}}^2$, $P_0=1.01\times 10^5\text{ Pa}$.
2. Gauge pressure is the pressure from only the overlying water, so use $P_{\text{gauge}}=\rho gh$.
3. Calculate: $P_{\text{gauge}}=(1025)(9.8)(32)=3.21\times 10^5\text{ Pa}$.
4. Add atmospheric pressure to get absolute pressure: $P_{\text{abs}}=P_0+P_{\text{gauge}}=1.01\times 10^5+3.21\times 10^5=4.22\times 10^5\text{ Pa}$.

Exam tip: Always check if the question asks for gauge or absolute pressure. "Pressure from the water" indicates gauge pressure, while "total pressure on the diver" indicates absolute pressure.

3. Pascal's Principle

Pascal's principle states that a change in pressure applied to an enclosed, incompressible static fluid is transmitted undiminished to every portion of the fluid and all walls of the container. This is the working principle behind hydraulic lifts, hydraulic brakes, and other fluid force multipliers that let small input forces produce large output forces.

For a hydraulic lift with two connected pistons, small input piston of area $A_1$ and large output piston of area $A_2$, applying an input force $F_1$ to the small piston creates a pressure change $\Delta P=F_1/A_1$. By Pascal's principle, this pressure change is identical at the large piston, so $\Delta P=F_2/A_2=F_1/A_1$, rearranged to: $$\frac{F_2}{F_1}=\frac{A_2}{A_1}$$ The output force is larger than the input force by the ratio of the piston areas. Energy is still conserved because the large output piston moves a much smaller distance than the small input piston, so work input equals work output (neglecting friction).

Worked Example

A hydraulic lift has a small piston with radius 5.0 cm and a large piston with radius 50 cm. What input force must be applied to the small piston to lift a 1500 kg car on the large piston?

1. The ratio of areas equals the square of the ratio of radii: $\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2}={\left(\frac{r_2}{r_1}\right)}^2$.
2. Calculate the required output force to lift the car: $F_2=mg=(1500)(9.8)=14700\text{ N}$.
3. Rearrange Pascal's principle to solve for $F_1$: $F_1=F_2\frac{A_1}{A_2}=F_2{\left(\frac{r_1}{r_2}\right)}^2$.
4. Plug in values: $\frac{r_1}{r_2}=0.1$, so $F_1=14700\times (0.1{)}^2=147\text{ N}\approx 150\text{ N}$.

Exam tip: Units cancel in the area ratio, so you do not need to convert length units to SI as long as both radii use the same unit.

4. Pressure and Buoyant Force

Buoyant force, the net upward force on any object submerged in a fluid, arises directly from hydrostatic pressure differences between the top and bottom of the object. Because pressure increases with depth, upward pressure on the bottom of the object is always greater than downward pressure on the top of the object, and the net force from this difference is buoyancy. We can derive Archimedes' principle directly from pressure rules for any submerged object.

For a uniformly shaped rectangular object of height $H$, cross-sectional area $A$, fully submerged in a fluid of density ${\rho }_{\text{fluid}}$, the top of the object is at depth $h$ and the bottom is at depth $h+H$. Downward force on the top is $F_{\text{top}}=(P_0+{\rho }_{\text{fluid}}gh)A$, and upward force on the bottom is $F_{\text{bottom}}=(P_0+{\rho }_{\text{fluid}}g(h+H))A$. The net buoyant force is: $$F_b=F_{\text{bottom}}-F_{\text{top}}={\rho }_{\text{fluid}}gHA={\rho }_{\text{fluid}}{V}_{\text{displaced}}g$$ This matches Archimedes' principle, and it holds for any shape of object because horizontal pressure forces on the sides cancel out, leaving only the net vertical force from the depth difference.

Worked Example

A 10 cm tall solid cube is fully submerged in freshwater (density $1000{\text{ kg/m}}^3$) with its top face 50 cm below the surface. Calculate the buoyant force on the cube using the pressure difference method.

1. Convert to SI units: cube height $H=0.10\text{ m}$, face area $A=(0.10{)}^2=0.01{\text{ m}}^2$, depth of top $h_{\text{top}}=0.50\text{ m}$, depth of bottom $h_{\text{bottom}}=0.60\text{ m}$.
2. Write net force as the difference between upward and downward pressure force: $F_b=P_{\text{bottom}}A-P_{\text{top}}A=(P_{\text{bottom}}-P_{\text{top}})A$.
3. Substitute hydrostatic pressure: $P_{\text{bottom}}-P_{\text{top}}=\rho g(h_{\text{bottom}}-h_{\text{top}})=\rho gH$.
4. Plug in values: $F_b=(1000)(9.8)(0.01)(0.10)=9.8\text{ N}$.

Exam tip: If a question asks you to explain buoyancy using pressure concepts, do not just cite Archimedes' principle — you must explicitly reference the pressure difference between top and bottom to earn full credit.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to add atmospheric pressure when asked for absolute pressure, reporting only gauge pressure. Why: Most problems ask for pressure from the fluid alone, so students form a habit of only calculating $\rho gh$. Correct move: Always explicitly mark whether the question asks for absolute or gauge pressure, and double-check that you added $P_0$ if required.
• Wrong move: Using the height of the submerged object instead of depth from the fluid surface in the hydrostatic pressure formula. Why: Students confuse the size of the object with the depth of the point being measured. Correct move: For any point, depth is always measured straight down from the fluid's open surface to the point, regardless of the object's size.
• Wrong move: In Pascal's principle, using the ratio of radii directly instead of the ratio of areas. Why: Students memorize "force scales with piston size" and forget area scales with the square of linear dimensions. Correct move: Always write the area ratio explicitly before solving for force, and confirm you squared the linear dimension ratio.
• Wrong move: Claiming pressure at the same depth is different in containers of different shapes. Why: Students confuse total weight of fluid with pressure (force per unit area). Correct move: Always apply the rule that pressure at the same depth in a static fluid is identical, regardless of container shape.
• Wrong move: Leaving atmospheric pressure in the final buoyant force calculation. Why: Students forget atmospheric pressure acts on both the top and bottom of the object. Correct move: Always simplify the pressure difference first — atmospheric pressure always cancels out and never affects the buoyant force result.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

An open-ended U-tube contains two immiscible liquids at rest. Liquid A has density $2\rho$ and Liquid B has density $\rho$. The height of Liquid A on the left side above the interface is $h$. What is the height of Liquid B on the right side above the interface? A) $h/2$ B) $h$ C) $\sqrt{2}h$ D) $2h$

Worked Solution: For static fluids, pressure at the same elevation (the interface elevation) must be equal on both sides of the U-tube. Pressure from the left side at the interface is $P_0+2\rho gh$, and pressure from the right side is $P_0+\rho g{h}_B$, where $h_B$ is the height of Liquid B. Atmospheric pressure $P_0$ cancels out from both sides. Dividing by $\rho g$ gives $2h=h_B$. The correct answer is D.

Question 2 (Free Response)

A student builds a barometer to measure atmospheric pressure using a closed-end tube filled with glycerin, inverted into an open dish of glycerin. Glycerin has a density of $1260{\text{ kg/m}}^3$, and atmospheric pressure on the experiment day is $1.00\times 10^5\text{ Pa}$. (a) Calculate the height of the glycerin column in the closed tube above the dish surface, assuming pressure at the top of the tube is zero. (b) Mercury has a density of $13600{\text{ kg/m}}^3$. Explain why mercury barometers are much shorter than glycerin barometers for the same atmospheric pressure. (c) If some air leaks into the top of the closed glycerin tube, will the measured height of the glycerin column be higher, lower, or the same as the true atmospheric pressure value? Justify your answer.

Worked Solution: (a) Atmospheric pressure at the dish surface equals the hydrostatic pressure from the glycerin column: $P_0=\rho gh$. Solving for $h$: $h=\frac{P_0}{\rho g}=\frac{1.00\times 10^5}{(1260)(9.8)}\approx 8.1\text{ m}$. (b) For a fixed pressure, hydrostatic pressure equals $\rho gh$, so height is inversely proportional to density. Mercury is more than 10 times denser than glycerin, so its column height is less than 1/10 that of glycerin, resulting in a much shorter barometer. (c) The measured height will be lower than the true value. Leaked air adds positive pressure to the top of the column, giving the relationship $P_0=P_{\text{air}}+\rho gh$. Rearranged, $h=\frac{P_0-P_{\text{air}}}{\rho g}$, which is smaller than the true $h=P_0/(\rho g)$ when $P_{\text{air}}=0$.

Question 3 (Application / Real-World Style)

Bottle-nosed dolphins can dive to depths of up to 300 m to hunt. Estimate the total absolute pressure on a dolphin's body at this depth, in atmospheres. Use a seawater density of $1025{\text{ kg/m}}^3$ and $1\text{ atm}=1.013\times 10^5\text{ Pa}$.

Worked Solution: First calculate the gauge pressure from the overlying seawater: $P_{\text{gauge}}=\rho gh=(1025)(9.8)(300)=3.01\times 10^6\text{ Pa}$. Convert gauge pressure to atmospheres: $P_{\text{gauge}}=\frac{3.01\times 10^6}{1.013\times 10^5}\approx 29.7\text{ atm}$. Add the 1 atm of atmospheric pressure at the surface to get total pressure: $P_{\text{abs}}\approx 31\text{ atm}$. This means the total pressure on the dolphin at 300 m is ~30 times higher than standard sea-level atmospheric pressure, requiring special adaptations to avoid collapse of air-filled body structures.

7. Quick Reference Cheatsheet

8. What's Next

Pressure is the foundational concept for all other fluid topics in AP Physics 2 Unit 1. Next, you will build on pressure differences to fully master buoyancy and fluid dynamics, where Bernoulli's principle describes how pressure changes with fluid speed. Without a solid understanding of static pressure, you will not be able to correctly interpret pressure changes in moving fluids or connect buoyant force to its physical origin, making many FRQ parts impossible to earn full credit. Pressure also connects to Unit 2 Thermal Physics, where the ideal gas law relates pressure to volume and temperature of gases, and to real-world applications of fluid flow in biological systems like the human circulatory system.

• Buoyancy
• Fluid Dynamics
• Ideal Gas Law
• Thermal Properties of Matter