Topic 8.2: Pressure

What Topic 8.2 requires

Topic 8.2 is the only topic in Unit 8 with two learning objectives, and six essential knowledge statements are split between them.

8.2.A, describe the pressure exerted on a surface by a given force.

• 8.2.A.1 states that pressure is defined as the magnitude of the perpendicular force component exerted per unit area over a given surface area, as described by the equation $P = \dfrac{F_{\perp}}{A}$.
• 8.2.A.2 states that pressure is a scalar quantity.
• 8.2.A.3 states that the volume and density of a given amount of an incompressible fluid are constant regardless of the pressure exerted on that fluid.

8.2.B, describe the pressure exerted by a fluid.

• 8.2.B.1 states that the pressure exerted by a fluid is the result of the entirety of the interactions between the fluid's constituent particles and the surface with which those particles interact.
• 8.2.B.2 states that the absolute pressure of a fluid at a given point is equal to the sum of a reference pressure $P_0$, such as the atmospheric pressure $P_{\text{atm}}$, and the gauge pressure $P_{\text{gauge}}$, and prints the relevant equation $P = P_0 + \rho g h$.
• 8.2.B.3 states that the gauge pressure of a vertical column of fluid is described by the equation $P_{\text{gauge}} = \rho g h$.

The split between the two objectives is worth reading as a structure rather than as a list. 8.2.A is about a surface being pushed on by something, anything, and asks what pressure that push amounts to. 8.2.B is about the fluid doing the pushing, and asks how much pressure it exerts and why. The first is a definition; the second is a model of a fluid. Topic 8.2 prints no boundary statement, so nothing here is explicitly fenced off, and the scope is what the six statements say.

The CED lists four suggested skills for this topic: 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Unit 8 is weighted at 10 to 15 percent of the multiple-choice section, with a suggested 12 to 17 class periods.

Pressure is force per area, and only the perpendicular part counts (8.2.A.1)

Newtons divided by square meters gives pascals. The pascal is one of the eight unit symbols printed in the AP Physics 1 Table of Information, and $1 \, \mathrm{Pa} = 1 \, \mathrm{N/m^2}$ by definition, so the two forms are interchangeable in an answer.

The subscript on $F_{\perp}$ is the part students skip, and the CED spells it out in words: pressure is the magnitude of the perpendicular force component exerted per unit area. If a force arrives at an angle to a surface, resolve it first and use only the component along the surface normal. The component parallel to the surface is a real force that does real things, including sliding the object and generating friction, but it contributes nothing to the pressure on that surface. The first worked example below shows a 24 N push producing the pressure of a 12 N one for exactly this reason.

The area in the denominator is the area over which the force is actually distributed, not the total area of the object. A person standing on one foot puts the same weight through roughly half the area and doubles the pressure on the floor, with nothing about the force changed. That inverse relationship is where most of the qualitative questions on 8.2.A live: at fixed force, halving the contact area doubles the pressure, and a sharp point is just a very small $A$.

Rearranged, $F_{\perp} = PA$ converts a pressure back into a force, and that move makes the rest of the unit work. Multiply a pressure by an area to get the force on a wall, a dam, or a submarine window. Multiply a pressure difference by an area to get a net force, which is exactly where the buoyant force of Topic 8.3 comes from: more pressure on the bottom face of a submerged object than on the top, times the area, gives a net upward push.

Pressure is a scalar, but the force it produces is not (8.2.A.2)

Essential knowledge 8.2.A.2 is one line: pressure is a scalar quantity. It is short, and it is tested, because it collides with a picture most students carry.

A scalar has a magnitude and no direction. A pressure of 200 kPa is not a pressure pointing anywhere. There is no pressure vector, no $\vec{P}$, and no component of pressure along an axis. That is why pressures at a point add as ordinary numbers, which is what lets $P = P_0 + \rho g h$ be a plain sum with no vector arithmetic in it.

The force a pressure exerts on a surface is a vector. Its magnitude is $F_{\perp} = PA$ and its direction is set by the surface, always perpendicular to it and pushing on it. The same pressure at the same point in a fluid produces a downward force on a horizontal surface facing up, a sideways force on a vertical wall, and an upward force on the underside of a submerged block. The pressure did not change direction, because it never had one; the surfaces did.

Two consequences follow, and both are exam-grade.

• A fluid at rest pushes on every surface it touches, whichever way that surface faces. This is the answer to the unit's own essential question about why we do not feel the miles of air above us pushing us down. Air pressure is not a downward quantity. It pushes on you from every side at once, including upward on the soles of your feet, and your body is at equilibrium under the sum.
• You cannot ask which way the pressure points at a depth of 3 m. You can ask which way the force on a given surface at that depth points, and then the answer is perpendicular to it. Questions that seem to be asking the first thing are testing whether you know it is the second.

That puts pressure with mass, work, energy, and density on the scalar side of the ledger, and away from force, velocity, momentum, and torque.

Where a fluid's pressure comes from (8.2.B.1 and 8.2.A.3)

Essential knowledge 8.2.B.1 gives the microscopic account: the pressure exerted by a fluid is the result of the entirety of the interactions between the fluid's constituent particles and the surface with which those particles interact. "The entirety of" is the operative phrase. One particle striking a wall delivers one tiny impulse in a random direction. Pressure is what an enormous number of those interactions look like averaged over a surface and over time: steady, smooth, and with no trace of the individual events left in it.

That is the same move Topic 4.2 makes with impulse, applied to an unimaginable number of collisions instead of one. It explains two things the macroscopic equation does not.

• Why pressure has no direction. The particle interactions arrive from every orientation, so nothing survives the averaging except a magnitude.
• Why pressure exists at all in a fluid at rest. The fluid does not need to be flowing for its particles to be interacting with a surface.

Essential knowledge 8.2.A.3 then pins down the bookkeeping for the fluids this course uses: the volume and density of a given amount of an incompressible fluid are constant regardless of the pressure exerted on that fluid. Push on a sealed syringe full of water and the pressure inside rises immediately, while the volume and the density do not move. This is the property the whole unit rests on, and it is worth noticing which way the dependence runs. Pressure depends on density, through $\rho g h$. Density does not depend on pressure, by 8.2.A.3. That one-way street is what keeps $\rho$ a constant you can carry through a problem, and it is the definition of the ideal fluid from Topic 8.1 doing its work.

Pressure at a depth: P = P0 + rho g h (8.2.B.2)

Declare the geometry before you use it, and keep it for the whole problem: $h$ is measured downward from the level where the pressure equals $P_0$, and it is positive going down. Go deeper and $h$ grows; the pressure grows with it. The AP Physics 1 equation sheet's symbol list defines $h$ as "height", which is a label rather than a direction, so take the direction from essential knowledge 8.2.B.3, which describes a vertical column of fluid, and from the physical setup in front of you.

Read the three symbols on the right.

• $P_0$ is a reference pressure, and the CED's own wording is "a reference pressure $P_0$, such as the atmospheric pressure $P_{\text{atm}}$". The word "such as" matters: atmospheric is the usual choice for an open tank or a lake, but $P_0$ is whatever the pressure is at the level you measured $h$ from. In a sealed, pressurized tank it is the gas pressure above the liquid. When you need a number for atmospheric pressure, the Table of Information gives $1 \, \mathrm{atm} = 1.0 \times 10^5 \, \mathrm{N/m^2} = 1.0 \times 10^5 \, \mathrm{Pa}$.
• $\rho$ is the density of the fluid you are in, not of any object floating in it or dropped through it.
• $g$ is the magnitude of the acceleration due to gravity. The Table of Information prints $g = 9.8 \, \mathrm{m/s^2}$, and every number on this page uses that value. You will also meet $g \approx 10 \, \mathrm{m/s^2}$ in the CED, the value the exam itself uses where a numerical value is required. Either is safe: the Topic 1.3 boundary statement says students will not be penalized for correctly using the more precise commonly accepted values of $9.81 \, \mathrm{m/s^2}$ or $9.8 \, \mathrm{m/s^2}$.

The equation is a sum of two pressures, not a formula to memorize. Whatever was already pressing on the surface is still pressing at depth, and the column of fluid above the point adds its own contribution on top. That is why the relationship is linear in $h$: doubling the depth doubles only the added part, never the whole.

A fast sanity check. In fresh water, $\rho g = (1000)(9.8) = 9800$ pascals of added pressure per meter of depth, so a few meters is a few tens of kilopascals and ten-something meters is comparable to a whole atmosphere. If your pressure at 2 m came out in megapascals, a unit slipped.

Gauge versus absolute pressure, the trap this topic is built around

This is the distinction Topic 8.2 exists to teach, and the CED separates the two ideas into their own essential knowledge statements rather than leaving one as a rearrangement of the other.

Absolute pressure is the total pressure at a point, counting everything: $P = P_0 + \rho g h$, from 8.2.B.2. It is the number that goes into $F_{\perp} = PA$ when you want the whole force the fluid exerts on a surface. It is never negative.

Gauge pressure is the amount by which the absolute pressure exceeds the reference pressure: $P_{\text{gauge}} = P - P_0$, which for a vertical column of fluid of depth $h$ is $P_{\text{gauge}} = \rho g h$, from 8.2.B.3. It is the number a pressure gauge reads, because a gauge with the atmosphere on one side measures a difference and calls it zero when both sides match. A tire inflated to "220 kPa" is at 220 kPa gauge and about 320 kPa absolute.

Count the equations on the sheet, because the count is the point. The fluids block of the AP Physics 1 equation sheet prints seven equations, in this order: $\rho = m/V$, $P = F_{\perp}/A$, $P = P_0 + \rho g h$, $P_{\text{gauge}} = \rho g h$, $F_b = \rho V g$, $A_1 v_1 = A_2 v_2$, and Bernoulli's equation. Three of the seven belong to Topic 8.2, and two of those three are the pair above. Gauge pressure is printed on its own line rather than left for you to derive under time pressure, which is a fair signal about how often the distinction is asked.

Which one does a question want? Work from the wording.

Two further points that separate a good answer from a nearly right one. First, gauge pressure can be negative. If the absolute pressure at a point is below the reference, as it is inside a straw while you drink, the gauge pressure is a negative number and nothing has gone wrong. Second, in any comparison between two points in the same fluid the reference pressure cancels, so a question asking for a pressure difference never needs $P_0$. Recognizing that saves you from hunting for an atmospheric value the problem never gave you.

Pressure depends on depth, not on the shape or the volume of the container

Look at what is on the right-hand side of $P = P_0 + \rho g h$: a reference pressure, a density, $g$, and a depth. Look at what is not there. No area. No volume. No width. No mention of the container at all.

So the pressure 0.60 m below the surface of a given liquid is the same in a laboratory beaker, in a swimming pool, and in a lake, provided each surface is exposed to the same $P_0$. A tall thin tube holding a cupful of water produces the same pressure at its base as a wide tank holding a hundred liters, if the depths match. Two points at the same depth in the same connected body of still fluid are at the same pressure, whatever path the fluid takes between them, which is why the level is the same in every arm of a set of connected tubes no matter how differently shaped they are.

The result feels wrong, and it is worth seeing why. The instinct says more water means more weight means more pressure. That instinct is thinking of the total weight, but pressure is a force per area, and in a container with straight vertical sides the extra weight and the extra base area grow together, so the ratio is unchanged. Only the height of the column survives.

The sharper version concerns the force on the base. Multiply the pressure at the base by the base area and you get the force the fluid exerts down on it. For a straight-sided container that force equals the weight of the fluid. For a container that flares outward as it rises, it does not: the fluid weighs more than the downward force it exerts on the base, and the difference is carried by the upward vertical component of the normal forces the slanted walls exert on the fluid. The third worked example below has fluid weighing 94 N and pushing down on the base with 47 N. Nothing is missing from the accounting; the walls hold up the rest, exactly as a free-body diagram of the fluid would show.

One more absence, worth stating because questions probe it: the object being pressed on. The pressure at a depth is set by the fluid and the depth, and it is the same whether there is a rock there, a diver, or nothing at all.

Sketching and reading pressure graphs (skill 1.C)

Skill 1.C, creating qualitative sketches of graphs that represent features of a model or the behavior of a physical system, is listed first among Topic 8.2's suggested skills, and it appears in no other topic in this unit. Expect to draw.

The graph the equation hands you is pressure against depth in a uniform liquid. It is a straight line: intercept $P_0$ at $h = 0$, slope $\rho g$, rising as you go deeper. Everything you might be asked follows from those two features.

• The intercept is the surface pressure. An open tank starts at atmospheric; a sealed pressurized tank starts higher. Plot gauge pressure instead of absolute and the line moves down to pass through the origin, with the same slope. That pair of parallel lines, one offset from the other by $P_0$, is the whole gauge-versus-absolute distinction in one picture.
• **The slope is $\rho g$.** A denser liquid gives a steeper line. Two liquids on the same axes are ranked by slope, not by where they end.
• A layered fluid gives a kinked line. Oil floating on water produces one slope through the oil and a steeper one through the water, joined where the layers meet. The pressure itself is continuous across the boundary; only the slope jumps.

The CED's own sample instructional activity for Topic 8.2 runs this backwards, and it is worth doing once. Students find a published graph of air pressure against elevation and explain why its slope decreases with elevation, the reason being that the air gets less dense higher up, then use the slope at one point to estimate the air density there. Two lessons come out of it. The first is procedural: the magnitude of the slope of a pressure-against-height graph is $\rho g$, so dividing a measured slope by $g$ returns a density, which is a real measurement technique rather than an exercise. The second is about assumptions. The exam conventions printed with the Table of Information say fluids are assumed to be ideal, and an ideal fluid is incompressible, so its pressure graph is straight; the CED's own activity is a case where the fluid is plainly compressible and the graph curves. That is what the convention's closing words, "unless otherwise stated", are for.

How Topic 8.2 is tested, and where it leads

The patterns are narrow, and rehearsing them is most of the preparation.

1. Compute a pressure from a force at an angle, or a force from a pressure and an area (skill 2.B). Resolve the force first.
2. Compute the absolute or the gauge pressure at a stated depth, then convert one to the other (skill 2.B). Read the question twice to see which it wants.
3. Compare pressures at two points, in one fluid or two, or at two depths, or in two containers of different shape (skill 2.C). Most of these need only $\rho g \Delta h$ and no reference pressure.
4. Sketch pressure against depth, or against height, for a described situation, or interpret a given one (skill 1.C).
5. Justify a claim, most often that the container's shape is irrelevant, or that the pressure at equal depths in connected fluid is equal, using the equation as the evidence (skill 3.C).
6. Explain, in words, why a fluid at rest exerts pressure on a surface at all, using the particle picture of 8.2.B.1.

Where it leads is the rest of the unit. Topic 8.3 is this topic turned into a force: because pressure grows with depth, the fluid pushes harder on the bottom of a submerged object than on its top, and that pressure difference times area is the buoyant force $F_b = \rho V g$ that goes on a free-body diagram alongside the weight. Topic 8.4 puts pressure into a conservation law: a difference in pressure between two locations is what makes a fluid flow, and Bernoulli's equation carries a pressure term alongside terms that behave like kinetic and gravitational potential energy per unit volume, the same accounting as the conservation of energy work from Unit 3.

The fluids guide walks all four topics end to end, and Topic 8.1 is worth a second pass if the density in $\rho g h$ still feels like a number rather than a property of the fluid you are standing in.