Press Next (or Play) to walk through each story one small step at a time. The numbers come last. After every two stories there is one quick question; if it feels shaky, press Show me another example for one more story. Tap the small round play button next to a caption to hear it read aloud.
1. Ears at the bottom of the pool
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A scuba diver jumps off a boat and starts to swim down. At about 2 m deep she feels a push inside her ears. By 5 m it hurts.
She pinches her nose and blows gently. Her ears "pop" and the pain goes away. Then she swims deeper, and the push comes back.
A submarine has the same problem, much bigger. At 300 m deep, the water pushes on every window with a force of hundreds of thousands of newtons.
The question: what is pressure, why does it grow as you go deeper, and how do you find the force it makes on a surface?
2. Deeper means more water on top
Quick check
Ana swims from 1 m deep down to 3 m deep. The extra pressure from the water on her becomes:
Another example: Three holes in a jug
3. The air is on top too
4. A push on a tiny eardrum
Quick check
What is the absolute pressure 1 m under fresh water? (Air pressure is 101 000 Pa.)
Another example: Ears pop on a mountain road
5. Snowshoes
6. Juice up a straw
Quick check
You stand on soft snow, then lie down flat on it. What happens to the pressure on the snow?
Another example: The thumbtack
7. Check yourself
Think of your answer first, then tap to see it.
a) Two pools are both 2 m deep. One is huge, one is tiny. At which bottom is the water pressure bigger?
Show answer
Neither: it is the same. Liquid pressure depends only on how deep you are (and on the liquid and the air above), not on how wide the pool is. P = P₀ + ρgh has no width in it.
b) Ana pops her ears at 2 m, then swims down to 4 m. Why does the ear pain come back?
Show answer
Deeper water means a bigger push from outside. The air inside her ears is still at the pressure it had at 2 m, so the two pushes no longer match. She has to pop her ears again.
c) What is the gauge pressure 5 m under fresh water? What is the absolute pressure? (Air: 101 000 Pa.)
Three quick questions. Get all three right on the first try and you can skip ahead to the simulation. Not sure? No problem: just read on.
1) A tall, narrow vase and a wide bowl are filled with water to the same depth. Where is the water pressure on the bottom bigger?
2) What is the gauge pressure 2 m under fresh water?
3) A 600 N person stands on 0.04 m² of floor. What pressure does she put on the floor?
The idea in plain words
Picture yourself walking through deep snow in boots. Each step sinks in. Now strap on wide snowshoes. You weigh just the same, but you stay on top.
Your weight did not change. Only the area it spreads over changed. A small area packs the push into a tiny patch, so you sink. A big area spreads the push out, so the snow holds you.
The picture shows the same person twice, with the same weight pushing down. On the left, boots press on a small area: big pressure, and they sink. On the right, snowshoes spread the same weight over a big area: small pressure, and they stay on top.
That is the idea: pressure is force spread over an area. Take the force pushing at right angles to the surface, and divide it by the area:
P = F⊥ / A
The unit is the pascal: 1 Pa = 1 N/m². That is tiny. Air at sea level pushes with about 1.0 × 10⁵ Pa (1 atmosphere). A sharp nail makes a huge pressure with a small force because its area is tiny. Snowshoes make a small pressure because their area is big.
Pressure is a scalar. It has no direction. At a point in a fluid, the fluid pushes the same amount in every direction. The FORCE it makes on a surface points straight into that surface.
Where does fluid pressure come from? A fluid is made of huge numbers of moving particles. They keep hitting every surface they touch. Each hit is a tiny push. All the pushes together, per square metre, are the pressure.
Why deeper means more pressure. The water at depth h holds up all the water above it. A column of water with area A and height h has weight ρ(Ah)g. Spread over the area A, that adds ρgh. Add the air pushing on the surface, P0:
P = P0 + ρgh and Pgauge = ρgh
Here h is the depth, measured down from the surface. P is the absolute pressure (the total). Pgauge is the extra above the surface pressure; it is what a tire gauge reads. ρ is the density of the fluid.
Trap: measuring h up from the bottom of the tank. Instead: h is depth, measured DOWN from the free surface. In a 3.0 m tank, a point 1.0 m above the bottom is at h = 2.0 m.
Trap: mixing up gauge and absolute pressure (forgetting P0, or adding it twice). Instead: absolute P = P0 + ρgh; gauge = ρgh. At 5.0 m of water: gauge 4.9 × 10⁴ Pa, absolute 1.49 × 10⁵ Pa. Read which one the question asks for.
Shape does not matter. Only depth does. In the picture, the three containers are joined at the bottom and filled to the same height. The pressure at the same depth is the same in all of them, even though they hold very different amounts of water.
Trap: "the wide tank holds more water, so its bottom has more pressure". Instead: only depth matters: P = P0 + ρgh. Same liquid, same depth, same pressure, whatever the shape or the amount of water.
Liquids do not squash. Pressure squeezes, but an ideal liquid's volume and density stay the same however big the pressure. That is why ρ in ρgh is the same at the top and the bottom of a tank, and why a hydraulic lift works.
Worked example: the diver's eardrum at 5.0 m
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She is 5.0 m down in a fresh-water lake (ρ = 1000 kg/m³). Air pressure at the surface is 1.0 × 10⁵ Pa. Her eardrum has an area of 0.50 cm². The air inside her middle ear is still at 1.0 × 10⁵ Pa. Depth is measured down from the surface.
Gauge pressure: Pgauge = ρgh = 1000 kg/m³ × 9.8 m/s² × 5.0 m = 4.9 × 10⁴ Pa. Why: this is the extra pressure from the water above her, on top of the air's push.
Absolute pressure: P = P0 + ρgh = 1.0 × 10⁵ Pa + 0.49 × 10⁵ Pa = 1.49 × 10⁵ Pa (about 1.5 atm). Why: the air pushes on the lake's surface, and the water passes that push all the way down.
Area in m²: 0.50 cm² × (10⁻⁴ m² per cm²) = 5.0 × 10⁻⁵ m². Why: 1 cm = 10⁻² m, so 1 cm² = 10⁻⁴ m². Pascals need square metres.
Force from the water side: F = P × A = 1.49 × 10⁵ Pa × 5.0 × 10⁻⁵ m² = 7.5 N, inward. Force from the air inside: 1.0 × 10⁵ × 5.0 × 10⁻⁵ = 5.0 N, outward. Why: the force from a pressure is P × A, pointing into the surface.
Net force on the eardrum: 7.5 N − 5.0 N = 2.5 N inward. Same as Pgauge × A = 4.9 × 10⁴ × 5.0 × 10⁻⁵ = 2.45 N. Why: only the DIFFERENCE in pressure gives a net force. That is why "popping" (letting air in to raise the inside pressure) stops the pain.
Lab 1: the diver's depth gauge
The diver sinks at a steady speed. Depth d is measured down from the surface (the engine's height y = −d, up positive). Change the fluid (fresh water 1000, sea water 1025, oil about 800, mercury 13 600 kg/m³), the pressure on the surface P0 (try 0 for a tank in a vacuum), the speed and the tank depth. The dial reads absolute pressure; the readouts show absolute and gauge pressure in kPa (1 kPa = 1000 Pa).
Pressure vs depth
This graph is against depth, not time. Both lines are straight. The two dots show the diver now.
Slope of both lines = ρg.
Try this: with water and P0 = 1.0 × 10⁵ Pa, find the depth where the absolute pressure is 2 atm (200 kPa). (Answer: about 10.2 m, because ρgh must be 1.0 × 10⁵ Pa.) Then switch to mercury: the pressure doubles in only 0.75 m.
Lab 2: a hydraulic lift
Two cylinders of oil are joined at the bottom. You push the small piston down with force F1. That adds the pressure P = F1/A1 to all of the oil. The big piston, at the same height, feels the same extra pressure, so the oil pushes it up with F2 = P × A2. (We ignore the small height difference of the pistons and their masses.)
Notice: with the lesson values, 200 N on 10 cm² gives F2 = 20 000 N on 1000 cm², more than the 14 700 N weight of a 1500 kg car. But when you push the small piston 30 cm down, the car rises only 3 mm. The oil does not compress, so the same volume moves: A1d1 = A2d2. Work in = 200 N × 0.30 m = 60 J; work done by the oil on the big piston = 20 000 N × 0.003 m = 60 J. No free energy.
Examples
1. Boots or snowshoes? basic
A 600 N hiker stands on snow. Her two boots touch the snow over 0.030 m² in total. Her two snowshoes would touch it over 0.30 m². Find the pressure on the snow each way.
Show solutionRead the steps as text
The force on the snow is her weight, 600 N, straight down (perpendicular to flat snow). Why: she is at rest, so the snow pushes up 600 N and she pushes down 600 N (third law).
Boots: P = F/A = 600 N / 0.030 m² = 2.0 × 10⁴ Pa. Why: definition of pressure.
Snowshoes: P = 600 N / 0.30 m² = 2.0 × 10³ Pa, ten times less. Why: ten times the area, same force. Low pressure means she does not sink in.
2. A submarine window medium
A submarine is 200 m below the surface of the sea (ρ = 1025 kg/m³). Inside, the air is kept at 1.0 × 10⁵ Pa. A window has an area of 0.10 m². What is the net force on the window?
Show solutionRead the steps as text
Outside absolute pressure: P = P0 + ρgh = 1.0 × 10⁵ + 1025 × 9.8 × 200 = 1.0 × 10⁵ + 2.009 × 10⁶ = 2.11 × 10⁶ Pa. Why: the sea surface has air at 1 atm on it.
Inside pressure: 1.0 × 10⁵ Pa. Difference: 2.11 × 10⁶ − 1.0 × 10⁵ = 2.009 × 10⁶ Pa, which is just ρgh. Why: the 1 atm on each side cancels, so the net push comes from the gauge pressure.
Net force: F = ΔP × A = 2.009 × 10⁶ Pa × 0.10 m² = 2.0 × 10⁵ N, pushing inward. Why: F = PA. That is the weight of about 20 tonnes, on one window.
3. Oil and water in a U-tube AP
A U-tube open at both ends holds water (1000 kg/m³). Oil (800 kg/m³), which does not mix with water, is poured into the left arm until the oil column is 10.0 cm tall. How much higher is the top of the oil than the top of the water in the right arm?
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Pick the level of the oil-water boundary in the left arm. Draw a line across to the right arm at the same height; there, it is in water. Why: in one connected, still fluid (water below this line), points at the same height have the same pressure.
Left: P = Patm + ρoilg(0.100 m). Right: P = Patm + ρwaterg hw, where hw is how far the water surface is above the line. Why: P = P0 + ρgh in each arm, both open to the air.
Set them equal: 800 × 0.100 = 1000 × hw, so hw = 0.080 m = 8.0 cm. Why: Patm and g cancel.
The oil top is 10.0 − 8.0 = 2.0 cm higher than the water top. Why: the less dense oil needs a taller column to make the same pressure.
4. Measuring the air with mercury AP
A barometer is a tube closed at the top, with a vacuum (P = 0) above a column of liquid. The open bottom stands in a dish open to the air at 1.0 × 10⁵ Pa. How tall is the column for mercury (13 600 kg/m³)? For water?
Show solutionRead the steps as text
At the dish surface level, the pressure inside the tube equals the air pressure: 0 + ρgh = Patm. Why: same height in a connected still fluid → same pressure. Above the column P0 = 0.
Mercury: h = Patm/(ρg) = 1.0 × 10⁵ / (13 600 × 9.8) = 1.0 × 10⁵ / 133 280 = 0.75 m. Why: solve for h.
Water: h = 1.0 × 10⁵ / (1000 × 9.8) = 10.2 m. Why: 13.6 times less dense, so 13.6 times taller. That is why barometers use mercury.
AP-style practice
Use g = 9.8 m/s², ρwater = 1000 kg/m³ and Patm = 1.0 × 10⁵ Pa unless told otherwise. Depth is measured down from the surface.
1. Three containers sit on a table. A is a wide tank, B is a narrow tube, C is a cone wide at the top. All three are open to the air and filled with water to the same height, 0.50 m. Which is true about the water pressure on the bottom of each?
Show answer
(D). P = P0 + ρgh depends only on the depth and the fluid, not on the shape or the amount. All three: 1.0 × 10⁵ + 1000 × 9.8 × 0.50 = 1.049 × 10⁵ Pa. (A) and (B) mix up force and pressure: the total force on A's big bottom is larger, but force per area is the same. (C): slanted walls do push on the water, and that is exactly why the extra water above them does not add to the pressure at the bottom.
2. A pressure sensor is lowered into an open lake. Which description best matches the graph of absolute pressure (vertical axis) against depth (horizontal axis)?
Show answer
(B). P = Patm + ρgh is linear in h with slope ρg = 9800 Pa/m and intercept Patm. (A) is the graph of GAUGE pressure. (C) would need the density to grow with depth, but water is incompressible here. (D) ignores the weight of the water.
3. A diver is 15 m below the surface of the sea (ρ = 1025 kg/m³). What is the absolute pressure on her, closest to?
Show answer
(C). ρgh = 1025 × 9.8 × 15 = 1.51 × 10⁵ Pa (gauge). Add Patm: 1.0 × 10⁵ + 1.51 × 10⁵ = 2.5 × 10⁵ Pa. (B) is the gauge pressure only; (A) forgets the water; (D) adds the atmosphere twice.
4. In an open tank of water, the absolute pressure at depth h is P. What is the absolute pressure at depth 2h?
Show answer
(B). P = Patm + ρgh. Doubling h doubles only the ρgh part: the new pressure is Patm + 2ρgh = P + ρgh, which is less than 2P = 2Patm + 2ρgh. Only the gauge pressure doubles. (A) is the classic trap: the intercept Patm means absolute pressure is not proportional to depth.
5. Liquid X has twice the density of liquid Y. Both are in open containers. Point 1 is 0.30 m deep in X. Point 2 is 0.60 m deep in Y. How do the gauge pressures compare?
Show answer
(C). Pgauge = ρgh. Point 1: (2ρY)g(0.30) = 0.60ρYg. Point 2: ρYg(0.60) = 0.60ρYg. Twice the density at half the depth gives the same pressure.
6. (Ranking) Four open containers. Rank the GAUGE pressure at the marked point, greatest first. Show your numbers.
D > C > B > A. A shortcut: compare ρh only (g is the same): 5000, 5125, 5440, 8000 kg/m².
7. (Hydraulic lift, short free response) A garage lift has a small piston of area 2.0 × 10⁻³ m² and a large piston of area 0.20 m². A 1200 kg car sits on the large piston. Ignore the pistons' masses and the height difference between them.
(a) What force on the small piston holds the car up? (b) The small piston is pushed down 0.50 m. How far does the car rise? (c) Show that the work done on the small piston equals the work done on the car.
Show answer
(a) Weight of car: 1200 × 9.8 = 11 760 N. Pressure needed under the big piston: 11 760 / 0.20 = 58 800 Pa. The same pressure on the small piston: F1 = 58 800 × 2.0 × 10⁻³ = 118 N (about 1.2 × 10² N).
(b) The oil does not compress, so A1d1 = A2d2: d2 = (2.0 × 10⁻³ × 0.50) / 0.20 = 5.0 × 10⁻³ m = 5.0 mm.
(c) W1 = 117.6 N × 0.50 m = 58.8 J. W2 = 11 760 N × 0.0050 m = 58.8 J. Equal: the lift multiplies force, not energy.
8. (Experimental design and analysis) A student lowers a pressure sensor into a tall tank of an unknown liquid. The liquid's surface is open to the air. She records:
Depth (m)
0
0.10
0.20
0.30
0.40
Absolute pressure (kPa)
101.0
102.0
103.0
104.1
105.0
(a) Which quantities should she plot to get a straight line, and what do the slope and intercept mean? (b) Find the density of the liquid. (c) She repeats the experiment on a day when the air pressure is higher. Describe how the new graph compares.
Show answer
(a) Plot absolute pressure (vertical) against depth (horizontal). From P = P0 + ρgh, the slope is ρg and the vertical intercept is the air pressure P0.
(b) Best-fit slope ≈ (105.0 − 101.0) kPa / (0.40 − 0) m = 10.0 kPa/m = 1.00 × 10⁴ Pa/m (the points lie close to this line). ρ = slope / g = 1.00 × 10⁴ / 9.8 ≈ 1.0 × 10³ kg/m³ (about 1020 kg/m³, maybe salt water). The intercept, 101 kPa, is the air pressure that day.
(c) Same slope (same liquid, same g), but the whole line shifts up by the increase in air pressure: a higher intercept, parallel line.
9. (Qualitative/quantitative translation) A cylinder full of water and a cylinder of the same height but twice the diameter, also full of water, stand on a table. (a) Without numbers, compare the pressure on their bottoms and the force of the water on their bottoms. (b) Show it with symbols, using height H and the smaller cylinder's radius r.
Show answer
(a) Same depth and same fluid, so the same pressure. The wide cylinder's bottom has four times the area (area goes with diameter squared), so the force on it is four times bigger. That force is just the weight of the water, and it holds four times as much water.
(b) Gauge pressure on both bottoms: ρgH. Force (gauge part): narrow F = ρgH × πr²; wide F = ρgH × π(2r)² = 4ρgHπr². Ratio 4. Check: weight of water in the narrow one = ρ(πr²H)g, the same expression.
Free response (Translation between representations). A graph shows absolute pressure (vertical) against depth (horizontal) for two open tanks of different
liquids. Line A goes straight from 1.0 × 10⁵ Pa at the surface to 1.49 × 10⁵ Pa at 5.0 m. Line B goes straight from
1.0 × 10⁵ Pa to 1.392 × 10⁵ Pa at 5.0 m.
(a) Find the density of each liquid from the graph.
(b) Describe the graph of GAUGE pressure against depth for liquid A.
(c) The same tanks are taken up a mountain where the air pressure is 0.80 × 10⁵ Pa. Which features of the lines change?
Show answer
(a) The slope is ρg. A: (1.49 − 1.00) × 10⁵ / 5.0 = 9800 Pa/m, so ρ = 9800 / 9.8 = 1000 kg/m³ (water).
B: 3.92 × 10⁴ / 5.0 = 7840 Pa/m, so ρ = 800 kg/m³.
(b) A straight line through the origin with the same slope, 9800 Pa/m, reaching 4.9 × 10⁴ Pa at 5.0 m.
(c) Only the intercept: both lines start at 0.80 × 10⁵ Pa instead of 1.0 × 10⁵ Pa. The slopes stay the same, because ρ and g did not change.
Point guide (4 points): 1 for slope = ρg; 1 for both densities; 1 for the gauge line through the origin; 1 for "intercept changes, slope does not".
Free response (Mathematical routines) A diver is 12 m below the surface of a freshwater lake (ρ = 1000 kg/m³). Air pressure at the surface is 1.01 × 10⁵ Pa.
(a) Calculate the gauge pressure at the diver.
(b) Calculate the absolute pressure at the diver.
(c) Her mask window has an area of 0.020 m². Calculate the force of the water on it.
(d) Calculate the depth where the absolute pressure is three times the surface air pressure.
(d) ρgh = 3P0 − P0 = 2.02 × 10⁵ Pa, so h = 2.02 × 10⁵ / (1000 × 9.8) ≈ 20.6 m.
Point guide (4 points): 1 for ρgh = 1.18 × 10⁵ Pa; 1 for adding P0; 1 for F = PA ≈ 4.4 × 10³ N; 1 for h ≈ 20.6 m.
Common mistakes
The mistake: measuring h from the bottom of the tank.
Why it is wrong: in P = P0 + ρgh, h is the depth below the free surface: how much fluid is above the point.
How to spot it: your pressure should be biggest at the bottom. If it comes out biggest near the top, you used height instead of depth.
The mistake: mixing up absolute and gauge pressure.
Why it is wrong: absolute = P0 + ρgh; gauge = ρgh. They differ by 1.0 × 10⁵ Pa in an open container, which is often more than the answer itself.
How to spot it: read the question for "absolute", "total" or "gauge". For a net force across a wall with air on the other side, the air cancels and you need gauge.
The mistake: "more water above means more pressure", so a wide lake must have more pressure than a narrow well at the same depth.
Why it is wrong: only the depth and the fluid's density matter. The shape and the total amount of fluid do not.
How to spot it: if the width or volume of the container is in your pressure calculation, take it out. (It does matter for the total FORCE on the bottom, F = PA.)
The mistake: giving pressure a direction, like "the pressure points down".
Why it is wrong: pressure is a scalar. At a point, the fluid pushes equally in all directions. The force from pressure on a surface points perpendicular into that surface.
How to spot it: if you drew an arrow labelled P, relabel it F = PA and point it into the surface.
The mistake: forgetting to turn cm² into m².
Why it is wrong: 1 cm² = 10⁻⁴ m², not 10⁻² m². A factor of 100 error.
How to spot it: a force on an eardrum or a coin of thousands of newtons is a warning sign. Square the conversion: (10⁻² m)² = 10⁻⁴ m².