Pressure: Study Guide (Physics 1 8.2)

This guide is for the algebra-based Physics 1 course. Topic 8.1 described fluids by their density. This topic asks what a fluid, or any object, does to the surfaces it presses on.

Pressure on a surface

Stand on one foot on soft snow and you sink. Lie flat and you do not. Your weight is the same, but it is spread over a different area. Pressure measures how concentrated a force is:

P = F⊥ / A

• F⊥ is the magnitude of the force component perpendicular to the surface, in N.
• A is the area over which that force acts, in m².
• The unit is the pascal: 1 Pa = 1 N/m².

A 600 N person standing on one foot (contact area about 0.015 m²) exerts 600 ÷ 0.015 = 40,000 Pa on the ground. Lying down on 0.50 m², the same person exerts only 1200 Pa: about 33 times less.

Only the perpendicular part counts

If a force pushes at an angle, split it into a component perpendicular to the surface and a component along it. Only the perpendicular component presses into the surface. The parallel component tries to slide things along the surface (that is the job of friction, not pressure).

Worked example 1: pressure under a pushed box

Question. A 15 kg storage box has a base 0.40 m by 0.30 m. (a) Find the pressure it exerts on the floor. (b) A student then pushes on the top of the box with a force of 80 N directed 30° below the horizontal. The box does not move. Find the new pressure on the floor.

1. Contact area: A = 0.40 m × 0.30 m = 0.12 m².
2. (a) The only perpendicular force is the box’s weight: mg = 15 kg × 9.8 m/s² = 147 N. P = 147 N ÷ 0.12 m² = 1225 Pa ≈ 1.2 × 10³ Pa.
3. (b) Split the push. Perpendicular to the floor: 80 sin 30° = 40 N downward. Along the floor: 80 cos 30° = 69 N.
4. The box is at rest vertically, so the floor’s normal force balances both downward forces: 147 N + 40 N = 187 N. By Newton’s third law the box presses on the floor with 187 N.
5. P = 187 N ÷ 0.12 m² = 1558 Pa ≈ 1.6 × 10³ Pa.

Check. The 69 N horizontal component does not appear. Using the whole 80 N would overestimate the pressure. The pressure rose by a factor 187/147 ≈ 1.27, the same factor as the perpendicular force, because the area did not change.

Pressure is a scalar

Force has a direction. Pressure does not: it is a scalar. Think of a tiny pressure sensor held at one point under water. Turn its face up, down or sideways and it reads the same value. What does have a direction is the force that the pressure produces on a particular surface: F = PA, always directed perpendicular to that surface, pushing on it.

So the water in a tank pushes down on the base, sideways on the walls, and even up on the underside of anything submerged. The pressure at a point is one number; the surfaces decide the directions of the forces.

Where fluid pressure comes from

Topic 8.1 pictured a fluid as huge numbers of moving particles. When a particle hits a wall and bounces off, its momentum changes. By Newton’s third law, the wall receives a small impulse pushing outward on it. One collision is tiny, but an enormous number of particles hit every square millimetre each second. Their combined effect is a steady average force on the wall. That force per unit area is the fluid’s pressure.

This picture explains several facts at once:

• Particles move in all directions, so a fluid pushes on every surface it touches, including the top of its container.
• More particles in each cubic metre, or faster particles, mean more collisions each second and so greater pressure.
• For an incompressible liquid, squeezing harder raises the pressure but does not change its volume or density. That is why we can use one value of ρ at every depth below.

Absolute pressure and gauge pressure

The total pressure at a point is the absolute pressure. Often it is easier to measure how much the pressure exceeds some reference pressure P₀. That difference is the gauge pressure:

P = P₀ + P_gauge

Near Earth’s surface the reference is usually atmospheric pressure, P_atm = 1.0 × 10⁵ Pa (the course’s value for 1 atm). A tyre gauge reading 2.2 × 10⁵ Pa means the absolute pressure in the tyre is 2.2 × 10⁵ + 1.0 × 10⁵ = 3.2 × 10⁵ Pa. A flat tyre has gauge pressure zero, not absolute pressure zero: the air inside is still at atmospheric pressure.

Gauge pressure can be negative. Air pumped partly out of a sealed jar has absolute pressure below 1.0 × 10⁵ Pa, so its gauge pressure is below zero.

Pressure under a column of fluid

Why does pressure increase as you go deeper? Picture an imaginary vertical column of liquid inside a still tank, with cross-sectional area A and height h (Figure 2).

The column is at rest, so the vertical forces on it balance (Newton’s first law, Topic 2.4; Topic 8.3 develops this further):

1. Mass of the column: m = ρV = ρAh, so its weight is ρAhg.
2. Balance: P_bottom A = P₀A + ρAhg.
3. Divide by A: P_bottom = P₀ + ρgh.

So the gauge pressure at a depth h below the surface is

P_gauge = ρgh

Three things to notice:

• h is the vertical depth below the surface, not the distance along a sloping wall or tube.
• A cancelled. The pressure at a given depth does not depend on how wide the column is. The same liquid at the same depth gives the same pressure in a narrow tube, a wide tank or an odd-shaped vase, as long as the liquid is connected and at rest.
• Every 10 m of water adds about 1000 × 9.8 × 10 = 98,000 Pa ≈ 1 atm of gauge pressure. A diver at 10 m feels roughly twice the absolute pressure felt at the surface.

Graphs of pressure against depth

For an incompressible liquid, P = P₀ + ρgh is a straight line when P is plotted against depth h:

• Vertical intercept = P₀, the pressure at the surface.
• Slope = ρg. A denser liquid gives a steeper line.
• A graph of gauge pressure against depth has the same slope but starts at zero.

Background: air is different. Air is compressible, so it is denser near the ground and thinner higher up. A graph of air pressure against height above the ground still falls as you climb, but it curves: it gets less steep with height because each metre of thinner air weighs less. You do not need a formula for this curve. Over a small height change the air density is nearly constant, so the steepness of the graph at a point is about ρg for the air there.

Worked example 2: a viewing window in the floor of a tank

Question. An aquarium tank holds fresh water 6.0 m deep. Its floor has a horizontal glass window 0.20 m by 0.30 m. Below the window is a viewing room at atmospheric pressure. Find (a) the gauge and absolute pressure at the window and (b) the net force of the water and air on the window.

1. (a) Gauge pressure: P_gauge = ρgh = 1000 kg/m³ × 9.8 m/s² × 6.0 m = 58,800 Pa ≈ 5.9 × 10⁴ Pa.
2. Absolute pressure: P = P_atm + P_gauge = 1.0 × 10⁵ + 0.588 × 10⁵ = 1.588 × 10⁵ Pa ≈ 1.6 × 10⁵ Pa.
3. (b) Window area: A = 0.20 × 0.30 = 0.060 m².
4. Water (with the air above it) pushes down: 1.588 × 10⁵ Pa × 0.060 m² = 9528 N.
5. Room air pushes up: 1.0 × 10⁵ Pa × 0.060 m² = 6000 N.
6. Net force = 9528 − 6000 = 3528 N ≈ 3.5 × 10³ N, downward.

Check. Atmospheric pressure acts on both sides, so it cancels. The net force is simply gauge pressure × area: 58,800 × 0.060 = 3528 N. That shortcut works whenever the same reference pressure acts on the other side.

Worked example 3: identifying a liquid from a pressure reading

Question. A gauge at the bottom of a beaker reads a gauge pressure of 6.2 kPa when the beaker holds 0.50 m of an unknown liquid. Find the liquid’s density and compare it with water.

1. Rearrange P_gauge = ρgh: ρ = P_gauge / (gh).
2. ρ = 6200 Pa ÷ (9.8 m/s² × 0.50 m) = 1265 kg/m³ ≈ 1.3 × 10³ kg/m³.
3. Water at the same depth would give 1000 × 9.8 × 0.50 = 4900 Pa = 4.9 kPa. The reading is higher, so the liquid is denser than water, by a factor 6.2 / 4.9 ≈ 1.3.

Interpretation. Glycerol has a density of about 1260 kg/m³, so the reading is consistent with glycerol. The beaker’s width is not needed: only the depth matters.

Comparing scenarios

• Double the depth: gauge pressure doubles (ρgh ∝ h), but absolute pressure does not. In water, P at 3.0 m is 1.29 × 10⁵ Pa and at 6.0 m is 1.59 × 10⁵ Pa: a factor of only 1.23.
• Same depth, denser liquid: gauge pressure rises in proportion to ρ.
• Same depth, wider container: no change in pressure. More liquid means a larger total force on a larger base, but the force per area is the same.

Common misconceptions

• “Pressure is a force.” Pressure is force per area, a scalar, in Pa. The force a pressure makes depends on the area and acts perpendicular to the surface.
• “Pressure only pushes down.” Fluid pressure pushes on every surface in every orientation, including upward on the bottom of a submerged object.
• “A bigger container has more pressure at the bottom.” At the same depth, pressure depends only on ρ, g and h (plus P₀), not on the amount of liquid.
• “h is the length of the tube.” h is the vertical depth below the free surface. For a slanted tube, use the vertical drop only.
• Mixing gauge and absolute pressure. Read the question: a tyre gauge or a pressure-difference question wants gauge; “total pressure” wants absolute.
• Using the whole force at an angle. Only F⊥ counts in P = F⊥ / A.
• “Doubling depth doubles the pressure.” True for gauge pressure only.
• Forgetting units. Depth in m, area in m², density in kg/m³. A depth in cm gives an answer 100 times too big.

Where this leads

Topic 8.3 (Fluids and Newton’s Laws) uses the pressure difference between the top and bottom of a submerged object to explain the buoyant force and why some objects float. Read the Topic 8.3 study guide next. First, try the practice questions, then use the revision notes and the checklist to consolidate. You can also return to the course roadmap.