Physics · 1 Motion, forces and energy

1.8 Pressure

Before you start

Force is measured in newtons (N), and the weight of an object is $W = m g$ .
The area of a rectangle is length × width, and $1$ m $= 100$ cm $= 1000$ mm.
Density is mass per unit volume, $ρ = m / V$ , measured in kg/m³.
Rearranging a three-letter equation such as $F = m a$ to make any one letter the subject.
Standard form, e.g. $2.0 \times 10^{5}$ for $200 000$ .

Syllabus objectives covered (4)

Define pressure as force per unit area; recall and use the equation p = F / A
Describe how pressure varies with force and area in the context of everyday examples
Describe, qualitatively, how the pressure beneath the surface of a liquid changes with depth and density of the liquid
Recall and use the equation for the change in pressure beneath the surface of a liquid Δp = ρgΔh

Part 1 of 6

Pressure is force spread over an area

Press the flat end of a pencil into your palm, then press the sharpened tip with the same push. The tip hurts. The force is the same both times. What changed is the area that the force acts on. Pressure is the quantity that measures this difference.
Pressure is defined as force per unit area. 'Per unit area' means 'on each square metre'. If you are asked to Define pressure, write those words, not just the symbols.
$p = \frac{F}{A}$
$p$ is the pressure, in pascals (Pa). $F$ is the force in newtons, acting at right angles to the surface. $A$ is the area the force acts on, in m².
The unit comes straight from the equation: newtons divided by square metres. So $1$ Pa $= 1$ N/m². A pressure of one pascal is a force of one newton on every square metre.
One pascal is a small pressure, so large values are written in kilopascals: $1$ kPa $= 1000$ Pa. Convert kPa to Pa before you put a value into an equation.
Read the equation before you use it. Force is on the top, so for the same area, more force gives more pressure. Area is on the bottom, so for the same force, more area gives less pressure. The rest of this lesson is built on those two sentences.

Part 2 of 6

Getting the area into m² before you divide

The physics in a pressure calculation is one division. The trap here is the area. The equation gives pascals only if the force is in N and the area is in m².
If you are given the sides of a rectangle, convert each side to metres first, then multiply. A face $50$ cm by $40$ cm is $0.50$ m × $0.40$ m $= 0.20$ m².
If you are given an area in cm², you need the conversion factor for area, not for length. A metre is $100$ cm, so a square metre is $100$ cm × $100$ cm.
$1 m^{2} = 100 \times 100 {cm}^{2} = 10 000 {cm}^{2}$
So to change cm² to m², divide by $10 000$ . For example, $50$ cm² $= 50 / 10 000 = 0.0050$ m². If you divide by $100$ instead, your area is $100$ times too big and your pressure is $100$ times too small.
The same idea works for mm². A metre is $1000$ mm, so $1$ m² $= 1000 \times 1000 = 10^{6}$ mm². Divide mm² by $10^{6}$ to get m².
You must also be able to use the equation the other way round. Multiply both sides of $p = F / A$ by $A$ to get the force. Then divide both sides by $p$ to get the area.
$F = p A A = \frac{F}{p}$
There is no formula sheet in Paper 4, so you must recall $p = F / A$ yourself. Practise writing it from memory before every calculation, and write it on the paper as your first line of working.

Part 3 of 6

Changing the force or changing the area: everyday examples

The equation $p = F / A$ gives you two ways to change a pressure. You can change the force, or you can change the area that the force acts on.
For a fixed force, pressure is inversely proportional to area. Double the area and the pressure halves. Halve the area and the pressure doubles.
forces diagram — not drawn yet
The same rectangular block of weight 600 N drawn twice on a floor: once resting on its large 0.50 m × 0.40 m face and once standing on its small 0.40 m × 0.25 m face, with an identical downward weight arrow on each and the contact area shaded.
The weight arrow is the same length in both; only the shaded area changes, so the smaller face gives the greater pressure.
A large area makes the pressure small. Snowshoes and skis spread a person's weight over a large area, so the pressure on the snow is small and the person sinks in less. Wide tyres and caterpillar tracks do the same for heavy vehicles on soft ground. Wide foundations under a building spread its weight so that it does not sink into the ground. A wide strap on a heavy bag presses less into your shoulder than a thin one.
A small area makes the pressure large. A sharp knife has a very thin edge, so a moderate push gives a large pressure that cuts through food. The points of a needle, a nail and a drawing pin work in the same way. A narrow stiletto heel can dent a soft floor that a flat shoe on the same person leaves undamaged.
More force on the same area gives more pressure. This is why you press harder on a blunt knife. It is also why a heavier person on the same snowshoes sinks further than a lighter one.
An Explain answer about an everyday example needs a chain of linked statements. First, name the force and say whether it changes. When an object just stands on something, the force is its weight, and that stays the same. Second, say how the area changes. Third, say how the pressure changes as a result. Then, if the question asks, give the effect: sinks less, cuts more easily, hurts less.
Notice what does not change. Snowshoes do not reduce the force on the snow. The weight is exactly the same. They reduce the pressure, because the same force now acts on a larger area.

Part 4 of 6

Pressure in a liquid increases with depth

A liquid has weight. At any level in a liquid, the liquid above that level presses down on it. The deeper you go, the more liquid there is above you, so the greater the pressure.
You can see this with a tall can of water that has small holes at different heights in its side. Water comes out of the lowest hole fastest, because that hole is deepest below the surface, where the pressure is greatest.
apparatus diagram — not drawn yet
A tall can full of water with three small holes in one side at different heights; a jet of water leaves each hole, with arrows at the holes that are longer for deeper holes to show faster water, and the depth of each hole below the water surface marked.
The deeper the hole is below the surface, the faster the water comes out, because the pressure there is greater.
The same idea explains why a dam is much thicker at the bottom than at the top. The water presses hardest at the greatest depth, so the wall must be strongest there. It also explains why your ears feel the pressure more the deeper you dive in a swimming pool.
At any one point in a liquid at rest, the pressure acts equally in all directions, not only downwards. That is why water comes out sideways from a hole in the side of the can.
Now the awkward case. The pressure depends on the depth, not on the width or shape of the container or on how much liquid there is. A narrow tube and a wide tank, filled to the same depth with the same liquid, have the same pressure at the bottom. The last section of this lesson shows why.

Part 5 of 6

A denser liquid gives more pressure at the same depth

Density is mass per unit volume. Take two columns of liquid with the same height and the same base area, so they have the same volume. The one made of the denser liquid has more mass, so it has more weight pressing on the base.
So at the same depth, the pressure is greater in a denser liquid. Sea water is denser than fresh water, so at the same depth the pressure due to sea water is slightly greater than the pressure due to fresh water.
The objective asks you to describe this qualitatively, which means in words, with no numbers. Two statements carry the description, and each one gives the direction of the change: 'the pressure increases as the depth increases' and 'at the same depth, the pressure is greater in a liquid of greater density'.
If a question changes the liquid, compare the densities first. If the depth stays the same, the less dense liquid gives the smaller pressure, and the denser liquid gives the greater pressure.

Part 6 of 6

Calculating the change in pressure: Δp = ρgΔh

For the Extended paper you must recall and use the equation for the change in pressure beneath the surface of a liquid.
$Δ p = ρ g Δ h$
$Δ p$ is the change in pressure, in Pa. $ρ$ is the density of the liquid, in kg/m³. $g$ is the gravitational field strength, in N/kg. $Δ h$ is the change in depth, in m. The symbol $Δ$ means 'change in'.
Take $g$ as $9.8$ N/kg. The front page of Paper 4 tells you to take the acceleration of free fall as $9.8$ m/s², and $9.8$ m/s² is the same as $9.8$ N/kg.
Here is where the equation comes from. Imagine a column of liquid of height $Δ h$ standing on a base of area $A$ . Its volume is $A Δ h$ , so its mass is $ρ A Δ h$ and its weight is $ρ A Δ h g$ . Pressure is weight divided by the base area $A$ .
$Δ p = \frac{ρ A Δ h g}{A} = ρ g Δ h$
forces diagram — not drawn yet
A vertical column of liquid of height Δh and cross-sectional area A, with a downward arrow labelled 'weight = ρAΔhg' acting on the shaded base of area A.
The area A appears in the weight and in the division, so it cancels: the pressure depends only on ρ, g and Δh.
The objective does not ask you to derive the equation. But the derivation shows you why the width of the container does not matter: the area $A$ cancels. Only the density, $g$ and the depth are left.
$Δ h$ is the vertical distance between the two levels you are comparing. If one of them is the surface, $Δ h$ is just the depth below the surface. If you are given heights above the bottom of a tank, subtract to get the depth. Do not use the height above the bottom.
$Δ p$ is the extra pressure caused by the liquid between the two levels. The air also presses on the surface of an open liquid, so the total pressure at a depth is greater than $ρ g Δ h$ . When a question asks for the change in pressure, or the pressure due to the liquid, $ρ g Δ h$ is the answer.
You can rearrange the equation for any one letter. To find a depth, divide both sides by $ρ g$ , because $ρ g$ is what multiplies $Δ h$ .
$Δ h = \frac{Δ p}{ρ g} ρ = \frac{Δ p}{g Δ h}$
Check the units once, so you can trust them: kg/m³ × N/kg × m gives N/m², which is Pa. If your units do not come out as Pa, one of your values is in the wrong unit.

Worked examples

Example 1. A wooden block measures $0.50$ m × $0.40$ m × $0.25$ m and has a weight of $600$ N. It can rest on the floor on any of its faces. Calculate the least pressure and the greatest pressure that it can exert on the floor.

Show the working

The force on the floor is the weight of the block, $600$ N, whichever face it rests on. Only the area changes. So the least pressure comes from the largest face, and the greatest pressure comes from the smallest face.
The largest face is $0.50$ m by $0.40$ m. The lengths are already in metres, so multiply them directly.
$A = 0.50 \times 0.40 = 0.20 m^{2}$
Write the equation, then substitute.
$p = \frac{F}{A} = \frac{600}{0.20} = 3000 Pa$
The smallest face is $0.40$ m by $0.25$ m.
$A = 0.40 \times 0.25 = 0.10 m^{2}$
Use the same force with the smaller area.
$p = \frac{600}{0.10} = 6000 Pa$
Check that the answer makes sense. The smallest face has half the area of the largest face, so the pressure should be twice as big. $6000$ Pa is twice $3000$ Pa.

Answer: Least pressure $3000$ Pa (on the $0.50$ m × $0.40$ m face); greatest pressure $6000$ Pa (on the $0.40$ m × $0.25$ m face).

Example 2. You push a drawing pin into a board with a force of $20$ N. The head of the pin has an area of $1.0$ cm² and its point has an area of $0.10$ mm². Calculate the pressure on your thumb and the pressure on the board under the point.

Show the working

The pin passes your push straight through. So the same $20$ N acts on your thumb at the head and on the board at the point. The weight of the pin is tiny, so you can ignore it.
Convert the head area to m². $1$ m² $= 10 000$ cm², so divide by $10 000$ .
$A_{head} = \frac{1.0}{10 000} = 1.0 \times 10^{- 4} m^{2}$
Pressure on your thumb:
$p = \frac{20}{1.0 \times 10^{- 4}} = 2.0 \times 10^{5} Pa$
Convert the point area to m². $1$ m² $= 10^{6}$ mm², so divide by $10^{6}$ .
$A_{point} = \frac{0.10}{10^{6}} = 1.0 \times 10^{- 7} m^{2}$
Pressure on the board:
$p = \frac{20}{1.0 \times 10^{- 7}} = 2.0 \times 10^{8} Pa$
Compare the two. The force is the same, but $1.0$ cm² is $100$ mm², so the point has $100 / 0.10 = 1000$ times less area than the head. The pressure at the point is therefore $1000$ times greater. That is why the point goes into the wood and the head does not go into your thumb.

Answer: On your thumb: $2.0 \times 10^{5}$ Pa. On the board: $2.0 \times 10^{8}$ Pa, which is $1000$ times greater.

Example 3. A diver swims down from a depth of $5.0$ m to a depth of $25$ m in sea water of density $1030$ kg/m³. Calculate the increase in pressure on the diver. Take $g = 9.8$ N/kg.

Show the working

$Δ h$ is the vertical distance between the two levels, not either depth on its own.
$Δ h = 25 - 5.0 = 20 m$
Write the equation first, then substitute. Every quantity is already in SI units: kg/m³, N/kg and m.
$Δ p = ρ g Δ h = 1030 \times 9.8 \times 20$
Work it out and keep the unit.
$Δ p = 201 880 Pa \approx 2.0 \times 10^{5} Pa$
The answer is positive because the diver went deeper. Pressure increases with depth, so going down always increases the pressure.

Answer: $2.0 \times 10^{5}$ Pa ( $201 880$ Pa).

Example 4. A sensor at the bottom of a tank of oil measures the pressure due to the oil as $3.6$ kPa. The density of the oil is $800$ kg/m³. Calculate the depth of the oil. Take $g = 9.8$ N/kg.

Show the working

Convert kPa to Pa first: $3.6$ kPa $= 3600$ Pa.
The pressure due to the oil is the change in pressure from the surface down to the bottom. So $Δ p = 3600$ Pa, and $Δ h$ is the depth you want.
Rearrange before you substitute. Divide both sides of $Δ p = ρ g Δ h$ by $ρ g$ , because $ρ g$ is what multiplies $Δ h$ .
$Δ h = \frac{Δ p}{ρ g}$
Substitute and calculate.
$Δ h = \frac{3600}{800 \times 9.8} = \frac{3600}{7840} = 0.459 m$
Round to $2$ significant figures, to match the data.

Answer: $0.46$ m.

Where people go wrong

Mistake: Changing cm² to m² by dividing by $100$ .

Instead: Divide by $10 000$ . For mm² to m², divide by $10^{6}$ .

$1$ m $= 100$ cm, so $1$ m² $= 100 \times 100 = 10 000$ cm². Dividing by $100$ leaves the area $100$ times too big and the pressure $100$ times too small.

Mistake: Saying that snowshoes or wide tyres 'spread the force out so the force is smaller'.

Instead: The force is the same. The area is larger, so the pressure is smaller.

The force on the ground is the weight, and the weight does not depend on what you wear on your feet. Only the area changes, and $p = F / A$ then gives a smaller $p$ . A statement that the force gets smaller is wrong physics, and it contradicts a correct statement that the force is unchanged.

Mistake: Thinking that a wider tank, or more liquid, gives a greater pressure at the bottom.

Instead: For the same liquid, the pressure at a point depends only on its depth below the surface.

In $Δ p = ρ g Δ h$ there is no area and no volume. A wider column has more weight, but that weight is spread over a proportionally larger area, so the pressure is the same.

Mistake: Using the height of a point above the bottom of the tank as $Δ h$ .

Instead: $Δ h$ is the vertical distance between the two levels you compare. For the pressure due to the liquid at a point, that is the depth below the surface.

The pressure comes from the liquid above the point. A tap $0.50$ m above the base of a tank filled to $3.5$ m has $3.0$ m of water above it, not $0.50$ m.

Mistake: Thinking that the pressure in a liquid acts only downwards.

Instead: At any point in a liquid at rest, the pressure acts equally in all directions.

Water comes out sideways from a hole in the side of a container, and a dam is pushed sideways by the water behind it. A pressure that acted only downwards could do neither.

Check yourself

Define pressure and give its unit.
Answer
Pressure is force per unit area, $p = F / A$ . Its unit is the pascal (Pa), and $1$ Pa $= 1$ N/m².
A force of $50$ N acts on an area of $250$ cm². Calculate the pressure in Pa.
Answer
$A = 250 / 10 000 = 0.025$ m². $p = F / A = 50 / 0.025 = 2000$ Pa.
A pressure of $2.5 \times 10^{5}$ Pa acts on an area of $0.040$ m². Calculate the force.
Answer
$F = p A = 2.5 \times 10^{5} \times 0.040 = 1.0 \times 10^{4}$ N ( $10 000$ N).
A tractor has very wide rear tyres. Explain, in terms of pressure, how this helps it to cross soft, muddy ground.
Answer
The weight of the tractor is the same, but the wide tyres give a larger area of contact with the ground. So the pressure on the ground is smaller, and the tractor sinks in less.
A crate stands on the floor. It is turned onto a face with half the area. Its weight does not change. What happens to the pressure on the floor?
Answer
The pressure doubles. The force is the same and the area is halved, and $p = F / A$ .
Tank X contains water and tank Y contains oil, which is less dense than water. Both are filled to the same depth. At the bottom of which tank is the pressure due to the liquid greater? Give a reason.
Answer
Tank X (water). The depth is the same, and at the same depth the pressure is greater in the liquid of greater density.
Calculate the increase in pressure from the surface to a depth of $2.0$ m in a swimming pool. The density of water is $1000$ kg/m³. Take $g = 9.8$ N/kg.
Answer
$Δ p = ρ g Δ h = 1000 \times 9.8 \times 2.0 = 19 600$ Pa ( $2.0 \times 10^{4}$ Pa).
In a liquid of density $1200$ kg/m³, the pressure increases by $5880$ Pa between the surface and a point P. Calculate the depth of P. Take $g = 9.8$ N/kg.
Answer
$Δ h = Δ p / (ρ g) = 5880 / (1200 \times 9.8) = 5880 / 11 760 = 0.50$ m.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A walker of weight $600$ N stands still on soft snow in boots. The total area of her two boot soles in contact with the snow is $0.040$ m².

(a) Define pressure. 1 mark No calculator

Mark scheme

B1 force per unit area (accept force ÷ area); force × area scores zero

(b) Calculate the pressure she exerts on the snow. 2 marks

Mark scheme

C1 p = F / A or 600 / 0.040 seen
A1 15 000 Pa (accept 1.5 × 10⁴ Pa, 15 000 N/m² or 15 kPa); unit required

(c) She puts on a pair of snowshoes. The total area of the snowshoes in contact with the snow is $2400$ cm². Calculate the pressure she now exerts on the snow. 3 marks

Mark scheme

C1 area converted to m²: 2400 / 10 000 = 0.24 m²
C1 p = F / A used with F = 600 N
A1 2500 Pa (accept 2500 N/m² or 2.5 kPa); unit required

(d) Explain, in terms of force, area and pressure, why she sinks less deeply into the snow when she wears the snowshoes. 2 marks No calculator

Mark scheme

B1 the force on the snow (her weight) is the same / unchanged
B1 the area in contact is larger, so the pressure on the snow is smaller, ORA

Question 2. An open storage tank contains water of density $1000$ kg/m³ to a depth of $3.5$ m. Three taps, P, Q and R, are fitted to one side of the tank at heights of $3.0$ m, $2.0$ m and $0.50$ m above the base. Take $g = 9.8$ N/kg.

(a) State which tap is at the point of greatest water pressure, and explain why. 2 marks No calculator

Mark scheme

B1 R
B1 R is at the greatest depth below the surface and the pressure increases with depth, owtte

(b) Calculate the increase in pressure from the surface of the water down to the level of tap R. 3 marks

Mark scheme

C1 Δh = 3.5 − 0.50 = 3.0 m
C1 Δp = ρgΔh, or 1000 × 9.8 × Δh, seen
A1 29 400 Pa (accept 2.9 × 10⁴ Pa, 29 400 N/m² or 29.4 kPa); unit required

(c) The tank is drained and refilled to the same depth with oil, which is less dense than water. Explain how the pressure at tap R due to the liquid changes. 2 marks No calculator

Mark scheme

B1 the pressure at R decreases / is smaller
B1 the depth of R is unchanged and the density is lower; at the same depth a less dense liquid gives a smaller pressure, or Δp = ρgΔh with the same Δh and a smaller ρ

(d) The oil has a density of $800$ kg/m³. Some oil is drained off. The pressure due to the oil at the level of tap R is now $9800$ Pa. Calculate the depth of tap R below the surface of the oil. 3 marks

Mark scheme

C1 Δp = ρgΔh recalled, or 9800 = 800 × 9.8 × Δh
C1 Δh = 9800 / (800 × 9.8)
A1 1.25 m (accept 1.3 m); unit required

Question 3. A cook finds it hard to slice a ripe tomato with a kitchen knife.

(a) Explain two different changes that would increase the pressure that the blade exerts on the tomato. 2 marks No calculator

Mark scheme

B1 push down with a larger force; for the same area the pressure is greater
B1 sharpen the blade so the edge has a smaller area of contact; for the same force the pressure is greater

How this is examined

Calculating a pressure with $p = F / A$ when the area is not given in m²: a two- or three-mark Calculate in which the area is given in cm² or mm², or as the two sides of a rectangle that you multiply first $p = F / A$ written, or the numbers substituted into it (this can earn a C mark even if the final answer is wrong); the area changed to m² (cm² ÷ $10 000$ ); the final value with the unit Pa or N/m². A correct number with no unit does not earn the A mark.
Finding a change in liquid pressure with $Δ p = ρ g Δ h$ , including working back to a depth: a two- or three-mark Calculate; the depth may have to be found from two heights first, or $Δ h$ or $ρ$ may be the unknown The equation recalled and written down (there is no formula sheet); $Δ h$ as the vertical distance between the two levels, in m; $g = 9.8$ N/kg; kPa changed to Pa; the answer with its unit.
Why a change in area or in force makes something sink in, cut or hurt: a two-mark Explain about an everyday object, such as snowshoes, wide tyres, a knife, a drawing pin or the strap of a heavy bag Two linked statements: the force (the weight, for an object standing still) is the same; the area is larger or smaller, so the pressure is smaller or larger. 'The force is spread out so it is smaller' is wrong physics and contradicts the first statement.
How depth and density change the pressure in a liquid: a one- or two-mark Describe, Explain or Predict: which hole the water comes out of fastest, why a dam is thicker at the bottom, or what happens to the pressure when the liquid is changed 'The pressure increases with depth' and 'at the same depth, the pressure is greater in a denser liquid', each with the direction of the change. For a dam: 'the pressure is greater at greater depth, so the wall must be stronger at the bottom'.
The definition of pressure in words: a one-mark Define, before a calculation 'Force per unit area' or 'force divided by area'. 'Force times area' does not earn it.

Words to know

Pressure: Force per unit area, $p = F / A$ , where the force acts at right angles to the surface. Measured in pascals.
Pascal (Pa): The unit of pressure: $1$ Pa $= 1$ N/m². $1$ kPa $= 1000$ Pa.
Density: Mass per unit volume, $ρ = m / V$ , measured in kg/m³.
Depth: The vertical distance of a point below the surface of a liquid.
Δ (delta): 'Change in'. $Δ p$ is a change in pressure and $Δ h$ is a change in depth.

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