Pressure, force and area

Definition

Pressure

The force acting at right angles to a surface on each unit area of that surface.

Definition

Pascal

The unit of pressure, equal to a force of one newton acting at right angles on an area of one square metre.

Pressure describes how concentrated a force is, by comparing the force with the area it acts on.
The equation linking pressure, force and area is one you must recall.
In words, $surface\text{pressure}=\dfrac{\text{force normal to a surface}}{\text{area of that surface}}$ .
In symbols, the equation is $p=FAp=\dfrac{F}{A}$ $p = \frac{F}{A}$ .
1. $p$ is the pressure in pascals, $Pa\text{Pa}$ .
2. $F$ is the force acting at right angles to the surface, in newtons, $N\text{N}$ .
3. $A$ is the area of the surface in square metres, $m2\text{m}^2$ .
The unit of pressure is the pascal, and $Pa1\ \text{Pa}$ is equal to $N/m21\ \text{N/m}^2$ .
Large pressures are often given in kilopascals, where $Pa1\ \text{kPa}=1000\ \text{Pa}$ .
Rearranging the equation gives $F = p A$ to find a force and $A=FpA=\dfrac{F}{p}$ to find an area.

Converting areas

Areas must be in square metres before they are substituted into the pressure equation.
One metre is $100 cm100\ \text{cm}$ $100 cm$ , so one square metre is $100×100=10 000 cm2100\times100=10\,000\ \text{cm}^2$ $100 \times 100 = 10000 cm^{2}$ .
1. Divide an area in $cm2\text{cm}^2$ by $00010\,000$ to convert it into $m2\text{m}^2$ , so $m225\ \text{cm}^2=0.0025\ \text{m}^2$ .
One metre is $1000 mm1000\ \text{mm}$ $1000 mm$ , so one square metre is $1 000 000 mm21\,000\,000\ \text{mm}^2$ $1000000 mm^{2}$ .
1. Divide an area in $mm2\text{mm}^2$ by $0001\,000\,000$ to convert it into $m2\text{m}^2$ , so $m2500\ \text{mm}^2=0.0005\ \text{m}^2$ .
Dividing an area in $cm2\text{cm}^2$ by $100$ instead of $00010\,000$ makes the calculated pressure $100$ times too small.

How contact area affects pressure

For the same force, a smaller area gives a larger pressure, and a larger area gives a smaller pressure.
When the force stays the same, pressure is inversely proportional to area, so halving the contact area doubles the pressure.
Some objects are designed with a small contact area to produce a large pressure.
1. A sharp knife has a very thin edge, so the pressure under it is large enough to cut through food.
2. Drawing pins, nails and needles have fine points, so they push into materials easily.
3. Football boot studs concentrate a player's weight onto small areas, so they sink into the turf and grip.
Other objects are designed with a large contact area to produce a small pressure.
1. Snowshoes spread a walker's weight over a large area, so they do not sink into soft snow.
2. Tractors have wide tyres and some diggers have caterpillar tracks, so they do not sink into muddy ground.
3. The wide straps of a school rucksack spread its weight across your shoulders, so they dig in less than thin straps would.

Hydraulic systems

Definition

Hydraulic system

A system that uses an enclosed liquid to transmit pressure from one piston to another.

Definition

Incompressible

Describes a substance whose volume cannot be reduced by squeezing it.

A hydraulic system uses a liquid, usually oil, trapped in pipes and cylinders to transmit a force from one place to another.
Liquids are almost incompressible, so pushing on the liquid at one point does not squash it but passes the push through it.
A pressure applied to an enclosed liquid is transmitted equally to every part of the liquid.
A simple hydraulic system has a narrow piston where the effort is applied and a wide piston that pushes on the load.
1. A small force $F_1$ on the narrow piston of area $A_1$ produces a pressure $p=F1A1p=\dfrac{F_1}{A_1}$ in the liquid.
2. The same pressure acts on the wide piston of area $A_2$ , producing a force $F_2=pA_2$ .
3. The wide piston has the larger area, so the output force is larger than the input force, and the system is a force multiplier.
The force is multiplied by the ratio of the piston areas, $F2F1=A2A1\dfrac{F_2}{F_1}=\dfrac{A_2}{A_1}$ .
A hydraulic system does not create energy, because the narrow piston must move a long way to push the wide piston a short distance.
Car brakes are hydraulic, because a light push on the brake pedal moves a narrow piston, and the pressure is transmitted through the brake fluid to wider pistons that press the brake pads onto the discs.
A hydraulic car jack, a dentist's chair and the arm of a digger all use the same principle to move heavy loads with a small effort.
Gases are not used in these systems, because a gas is easily compressed, so pushing on it would squash it instead of passing on the pressure.

Example

Pressure under a crate

A crate with a weight of $N500\ \text{N}$ rests on the floor on a base with an area of $m20.025\ \text{m}^2$ .

p=\frac{F}{A}=\frac{500\ \text{N}}{0.025\ \text{m}^2}=20\,000\ \text{Pa}

The crate exerts a pressure of $Pa20\,000\ \text{Pa}$ , which is $kPa20\ \text{kPa}$ , on the floor.

Example

Boots and snowshoes

A walker with a weight of $N600\ \text{N}$ stands in boots with a total sole area of $cm2500\ \text{cm}^2$ , then in snowshoes with a total area of $cm23000\ \text{cm}^2$ .

500\ \text{cm}^2=\frac{500}{10\,000}\ \text{m}^2=0.050\ \text{m}^2

p_{\text{boots}}=\frac{600\ \text{N}}{0.050\ \text{m}^2}=12\,000\ \text{Pa}

3000\ \text{cm}^2=\frac{3000}{10\,000}\ \text{m}^2=0.30\ \text{m}^2

p_{\text{snowshoes}}=\frac{600\ \text{N}}{0.30\ \text{m}^2}=2000\ \text{Pa}

The snowshoes have six times the area, so they reduce the pressure on the snow to one sixth.

Example

Force from a tyre pressure

The air in a car tyre is at a pressure of $kPa200\ \text{kPa}$ above the outside air, and the tyre touches the road over an area of $m20.015\ \text{m}^2$ .

F=pA=200\,000\ \text{Pa}\times0.015\ \text{m}^2=3000\ \text{N}

The tyre supports a force of $N3000\ \text{N}$ from the car.

Example

Lifting a car with a hydraulic jack

A mechanic pushes with a force of $N50\ \text{N}$ on a narrow piston with an area of $m20.0010\ \text{m}^2$ , and the wide piston has an area of $m20.050\ \text{m}^2$ .

p=\frac{F_1}{A_1}=\frac{50\ \text{N}}{0.0010\ \text{m}^2}=50\,000\ \text{Pa}

F_2=pA_2=50\,000\ \text{Pa}\times0.050\ \text{m}^2=2500\ \text{N}

The jack multiplies the force by $50$ , which is the ratio of the piston areas, so a $N50\ \text{N}$ push lifts a $N2500\ \text{N}$ load.

Common Mistake

Do not substitute an area in $cm2\text{cm}^2$ into $p=FAp=\dfrac{F}{A}$ , because the area must be in square metres to give pascals.
Do not convert $cm2\text{cm}^2$ into $m2\text{m}^2$ by dividing by $100$ , because the correct factor is $00010\,000$ .
Do not treat pressure and force as the same thing, because a larger force can produce a smaller pressure if it is spread over a larger area.
Do not say that a hydraulic system multiplies energy, because the larger output force moves through a smaller distance.

Exam technique

Pressure calculations

Write $p=FAp=\dfrac{F}{A}$ from memory, and use the force acting at right angles to the surface, which is the weight for an object resting on level ground.
Convert the area into square metres on its own line before substituting, so that the conversion is credited.
Give pressure in pascals, or in kilopascals if the question asks for them, and check that a larger area has given a smaller pressure.
For a hydraulic system, find the pressure at the narrow piston first, then apply the same pressure to the wide piston, and explain that the pressure is transmitted equally through the liquid.

Self review

Write the equation that links pressure, force and area.
Convert an area of $cm240\ \text{cm}^2$ into square metres.
Calculate the pressure produced by a force of $N120\ \text{N}$ acting on an area of $m20.030\ \text{m}^2$ .
Explain why snowshoes stop a walker sinking into soft snow.
Explain how a hydraulic jack lets a small force lift a heavy car.

Pressure, force and area

Definition

Pressure

The force acting at right angles to a surface on each unit area of that surface.

Definition

Pascal

The unit of pressure, equal to a force of one newton acting at right angles on an area of one square metre.

Pressure describes how concentrated a force is, by comparing the force with the area it acts on.
The equation linking pressure, force and area is one you must recall.
In words, $surface\text{pressure}=\dfrac{\text{force normal to a surface}}{\text{area of that surface}}$ .
In symbols, the equation is $p=FAp=\dfrac{F}{A}$ $p = \frac{F}{A}$ .
1. $p$ is the pressure in pascals, $Pa\text{Pa}$ .
2. $F$ is the force acting at right angles to the surface, in newtons, $N\text{N}$ .
3. $A$ is the area of the surface in square metres, $m2\text{m}^2$ .
The unit of pressure is the pascal, and $Pa1\ \text{Pa}$ is equal to $N/m21\ \text{N/m}^2$ .
Large pressures are often given in kilopascals, where $Pa1\ \text{kPa}=1000\ \text{Pa}$ .
Rearranging the equation gives $F = p A$ to find a force and $A=FpA=\dfrac{F}{p}$ to find an area.

Converting areas

Areas must be in square metres before they are substituted into the pressure equation.
One metre is $100 cm100\ \text{cm}$ $100 cm$ , so one square metre is $100×100=10 000 cm2100\times100=10\,000\ \text{cm}^2$ $100 \times 100 = 10000 cm^{2}$ .
1. Divide an area in $cm2\text{cm}^2$ by $00010\,000$ to convert it into $m2\text{m}^2$ , so $m225\ \text{cm}^2=0.0025\ \text{m}^2$ .
One metre is $1000 mm1000\ \text{mm}$ $1000 mm$ , so one square metre is $1 000 000 mm21\,000\,000\ \text{mm}^2$ $1000000 mm^{2}$ .
1. Divide an area in $mm2\text{mm}^2$ by $0001\,000\,000$ to convert it into $m2\text{m}^2$ , so $m2500\ \text{mm}^2=0.0005\ \text{m}^2$ .
Dividing an area in $cm2\text{cm}^2$ by $100$ instead of $00010\,000$ makes the calculated pressure $100$ times too small.

How contact area affects pressure

For the same force, a smaller area gives a larger pressure, and a larger area gives a smaller pressure.
When the force stays the same, pressure is inversely proportional to area, so halving the contact area doubles the pressure.
Some objects are designed with a small contact area to produce a large pressure.
1. A sharp knife has a very thin edge, so the pressure under it is large enough to cut through food.
2. Drawing pins, nails and needles have fine points, so they push into materials easily.
3. Football boot studs concentrate a player's weight onto small areas, so they sink into the turf and grip.
Other objects are designed with a large contact area to produce a small pressure.
1. Snowshoes spread a walker's weight over a large area, so they do not sink into soft snow.
2. Tractors have wide tyres and some diggers have caterpillar tracks, so they do not sink into muddy ground.
3. The wide straps of a school rucksack spread its weight across your shoulders, so they dig in less than thin straps would.

Hydraulic systems

Definition

Hydraulic system

A system that uses an enclosed liquid to transmit pressure from one piston to another.

Definition

Incompressible

Describes a substance whose volume cannot be reduced by squeezing it.

A hydraulic system uses a liquid, usually oil, trapped in pipes and cylinders to transmit a force from one place to another.
Liquids are almost incompressible, so pushing on the liquid at one point does not squash it but passes the push through it.
A pressure applied to an enclosed liquid is transmitted equally to every part of the liquid.
A simple hydraulic system has a narrow piston where the effort is applied and a wide piston that pushes on the load.
1. A small force $F_1$ on the narrow piston of area $A_1$ produces a pressure $p=F1A1p=\dfrac{F_1}{A_1}$ in the liquid.
2. The same pressure acts on the wide piston of area $A_2$ , producing a force $F_2=pA_2$ .
3. The wide piston has the larger area, so the output force is larger than the input force, and the system is a force multiplier.
The force is multiplied by the ratio of the piston areas, $F2F1=A2A1\dfrac{F_2}{F_1}=\dfrac{A_2}{A_1}$ .
A hydraulic system does not create energy, because the narrow piston must move a long way to push the wide piston a short distance.
Car brakes are hydraulic, because a light push on the brake pedal moves a narrow piston, and the pressure is transmitted through the brake fluid to wider pistons that press the brake pads onto the discs.
A hydraulic car jack, a dentist's chair and the arm of a digger all use the same principle to move heavy loads with a small effort.
Gases are not used in these systems, because a gas is easily compressed, so pushing on it would squash it instead of passing on the pressure.

Example

Pressure under a crate

A crate with a weight of $N500\ \text{N}$ rests on the floor on a base with an area of $m20.025\ \text{m}^2$ .

p=\frac{F}{A}=\frac{500\ \text{N}}{0.025\ \text{m}^2}=20\,000\ \text{Pa}

The crate exerts a pressure of $Pa20\,000\ \text{Pa}$ , which is $kPa20\ \text{kPa}$ , on the floor.

Example

Boots and snowshoes

A walker with a weight of $N600\ \text{N}$ stands in boots with a total sole area of $cm2500\ \text{cm}^2$ , then in snowshoes with a total area of $cm23000\ \text{cm}^2$ .

500\ \text{cm}^2=\frac{500}{10\,000}\ \text{m}^2=0.050\ \text{m}^2

p_{\text{boots}}=\frac{600\ \text{N}}{0.050\ \text{m}^2}=12\,000\ \text{Pa}

3000\ \text{cm}^2=\frac{3000}{10\,000}\ \text{m}^2=0.30\ \text{m}^2

p_{\text{snowshoes}}=\frac{600\ \text{N}}{0.30\ \text{m}^2}=2000\ \text{Pa}

The snowshoes have six times the area, so they reduce the pressure on the snow to one sixth.

Example

Force from a tyre pressure

The air in a car tyre is at a pressure of $kPa200\ \text{kPa}$ above the outside air, and the tyre touches the road over an area of $m20.015\ \text{m}^2$ .

F=pA=200\,000\ \text{Pa}\times0.015\ \text{m}^2=3000\ \text{N}

The tyre supports a force of $N3000\ \text{N}$ from the car.

Example

Lifting a car with a hydraulic jack

A mechanic pushes with a force of $N50\ \text{N}$ on a narrow piston with an area of $m20.0010\ \text{m}^2$ , and the wide piston has an area of $m20.050\ \text{m}^2$ .

p=\frac{F_1}{A_1}=\frac{50\ \text{N}}{0.0010\ \text{m}^2}=50\,000\ \text{Pa}

F_2=pA_2=50\,000\ \text{Pa}\times0.050\ \text{m}^2=2500\ \text{N}

The jack multiplies the force by $50$ , which is the ratio of the piston areas, so a $N50\ \text{N}$ push lifts a $N2500\ \text{N}$ load.

Common Mistake

Do not substitute an area in $cm2\text{cm}^2$ into $p=FAp=\dfrac{F}{A}$ , because the area must be in square metres to give pascals.
Do not convert $cm2\text{cm}^2$ into $m2\text{m}^2$ by dividing by $100$ , because the correct factor is $00010\,000$ .
Do not treat pressure and force as the same thing, because a larger force can produce a smaller pressure if it is spread over a larger area.
Do not say that a hydraulic system multiplies energy, because the larger output force moves through a smaller distance.

Exam technique

Pressure calculations

Write $p=FAp=\dfrac{F}{A}$ from memory, and use the force acting at right angles to the surface, which is the weight for an object resting on level ground.
Convert the area into square metres on its own line before substituting, so that the conversion is credited.
Give pressure in pascals, or in kilopascals if the question asks for them, and check that a larger area has given a smaller pressure.
For a hydraulic system, find the pressure at the narrow piston first, then apply the same pressure to the wide piston, and explain that the pressure is transmitted equally through the liquid.

Self review

Write the equation that links pressure, force and area.
Convert an area of $cm240\ \text{cm}^2$ into square metres.
Calculate the pressure produced by a force of $N120\ \text{N}$ acting on an area of $m20.030\ \text{m}^2$ .
Explain why snowshoes stop a walker sinking into soft snow.
Explain how a hydraulic jack lets a small force lift a heavy car.

2.3.15 Pressure, force and contact area

Pressure, force and area

Converting areas

How contact area affects pressure

Hydraulic systems

Pressure under a crate

Boots and snowshoes

Force from a tyre pressure

Lifting a car with a hydraulic jack

Pressure calculations

Review 2.3.15 Pressure, force and contact area by teaching Genie

2.3.15 Pressure, force and contact area

Pressure, force and area

Converting areas

How contact area affects pressure

Hydraulic systems

Pressure under a crate

Boots and snowshoes

Force from a tyre pressure

Lifting a car with a hydraulic jack

Pressure calculations

Review 2.3.15 Pressure, force and contact area by teaching Genie

2.3.15 Pressure, force and contact area

Pressure, force and area

Converting areas

How contact area affects pressure

Hydraulic systems

Pressure under a crate

Boots and snowshoes

Force from a tyre pressure

Lifting a car with a hydraulic jack

Pressure calculations

Review 2.3.15 Pressure, force and contact area by teaching Genie

2.3.15 Pressure, force and contact area Revision Guide

2.3.15 Pressure, force and contact area

Pressure, force and area

Converting areas

How contact area affects pressure

Hydraulic systems

Pressure under a crate

Boots and snowshoes

Force from a tyre pressure

Lifting a car with a hydraulic jack

Pressure calculations

Review 2.3.15 Pressure, force and contact area by teaching Genie

2.3.15 Pressure, force and contact area Revision Guide