Pressure force area

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Pressure Force Area

Here we will learn about pressure, force and area, including what they are and how they are related to one another.

There are also pressure force area worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What are pressure, force and area?

Pressure, force and area are physical properties.

What are pressure, force and area?

Pressure force area worksheet

Get your free pressure force area worksheet of 20+ questions and answers. Includes reasoning and applied questions.

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Pressure force area worksheet

Get your free pressure force area worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE

Area

Area is a measure of the size of space a flat shape takes up. The SI unit for area is the square metre $(m^{2})$ although we can use other units including square centimetres $(cm^{2}),$ square millimetres $(mm^{2})$ or square inches $(in^{2})$ etc.

Area is inversely proportional to the pressure exerted; for a constant force, if the area in which the force is being applied to increases, the pressure exerted across that area decreases. We can therefore state that $A\propto\frac{1}{P}$ for the area $A$ and the pressure $P.$ The constant of proportionality is the force, $F.$

Note: When we consider the area of an object, we are referring to the area of contact of the object with the floor / table / ground / wall / shelf etc.

Force

Force is the energy attributed to a movement or physical action. Force is measured in the standard unit Newtons $(N).$

Newton’s Second Law of Motion states that force is equal to the mass of an object, multiplied by the acceleration, or $F=MA.$ This is why weight and mass are not the same. Mass is a measure of how much matter there is in an object (usually measured in kilograms), whereas weight is the size of the pull of gravity on the object (measured in Newtons).

The acceleration due to gravity on the surface of the Earth is $9.807m/s^2$ and so an average human with a mass of $62kg$ would have a weight of $62 \times 9.807=608N.$

The force acting on an object is directly proportional to the pressure applied over a constant area. The greater the pressure, the greater the force for a constant area and vice versa. We can therefore state that $F \propto P$ for the force F and the pressure P. The constant of proportionality is the area, $A.$

Step-by-step guide: Direct proportion

The forces used for this topic act on an object that is in contact with the floor / table / ground / wall / shelf etc. This means that the force will be perpendicular to the surface for all questions.

Pressure

Pressure is a compound measure. Pressure is defined as the force per unit area. The standard unit of pressure is Pascals $(Pa)$ where $1 \ Pa=1 \ N / {m^{2}}.$ We can also measure pressure using a Bar, where $1$ Bar $= 100,000 Pa = 100,000 \ N / {m^{2}}.$

Imagine an elephant standing on $4$ legs. The pressure applied to the ground is distributed over a wider area than if the elephant was standing on one leg. This is because the force remains constant (the weight of the elephant), but the area in which the force is applied has decreased, so the pressure increases. This is an example of inverse proportion.

As pressure is inversely proportional to the area, we can state that $P\propto{1}{A}$ for the pressure $P$ and the area $A.$ The constant of proportionality is the force, $F.$

Step-by-step guide: Inverse proportion

To calculate either the pressure, force or area of an object, we use the pressure formula,

\text{Pressure}=\frac{\text{Force}}{\text{Area}}

Pressure triangle

This can be written as a formula triangle (sometimes called the pressure triangle).

We can circle what we are trying to find and the formula triangle tells how to calculate the unknown property.

How to calculate pressure, force or area

In order to calculate pressure, force or area using the formula triangle:

Draw the triangle and circle the required property.
Substitute values for the other two properties and complete the calculation.
Write down the solution, including the units.

Explain how to calculate pressure, force or area

Related lessons on compound measures

Pressure force area is part of our series of lessons to support revision on compound measures. You may find it helpful to start with the main compound measures lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

Pressure force area examples

Example 1: calculating pressure (Pa)

A force of $800 \ N$ acts on an area of $200 \ m^{2}.$ Calculate the pressure in Pascals.

Draw the triangle and circle the required property.

To calculate the pressure we would need to divide the force by the area.

2Substitute values for the other two properties and complete the calculation.

Substituting $F=800N$ and $A=200m^{2},$ we have

800\div{200}=4

3Write down the solution, including the units.

Pressure $= 4 \ N / {m^{2}}=4 \ Pa.$

Example 2: calculating pressure (Bar)

A force of $6.5 \times 10^6 \ N$ acts on an area of $0.2 \ m^{2}.$ Calculate the pressure in Bar.

Draw the triangle and circle the required property.

To calculate the pressure we would need to divide the force by the area.

Substitute values for the other two properties and complete the calculation.

Converting $6.5 \times 10^6$ to an ordinary number, we get $F=6500000N$ .

$P=6500000\div{0.2}=32500000$

Write down the solution, including the units.

We currently have $32500000 \ Pa.$ As we want the solution in Bar, we need to divide the value by $100,000,$

$32500000\div{100000}=325 \ Bar.$

Example 3: calculating area (m²)

A force of $200 \ N$ exerts a pressure of $40 \ N/m^{2}.$ Calculate the area of the surface that the force is being applied to in square metres.

Draw the triangle and circle the required property.

To calculate the area we would need to divide the force by the pressure.

Substitute values for the other two properties and complete the calculation.

Substituting $F=200N$ and $P=40 \ N / {m^{2}},$ we have

$A=200\div{40}=5$

Write down the solution, including the units.

Area $=5m^{2}.$

Example 4: calculating area (converting units of pressure)

A force of $950 \ N$ exerts a pressure of $53 \ Pa.$ Calculate the area in square metres, correct to $2$ decimal places.

Draw the triangle and circle the required property.

To calculate the area we would need to divide the force by the pressure.

Substitute values for the other two properties and complete the calculation.

As we need the units for the area to be in square metres, we need to convert the pressure from Pascals into $N / {m}^{2}.$ As $1Pa=1N / {m}^{2},$ we can simply state that our pressure $P=53N / {m}^{2}.$

Substituting $F=950N$ and $P=53N / {m}^{2}$ into the pressure triangle, we have

$A=950\div{53}=17.9245283…$

Write down the solution, including the units.

Area $=17.92m^{2}\text{ (2dp)}.$

Example 5: calculating force (N)

A box has a cross sectional area of $1.2 \ m^2$ and exerts a pressure of $60 \ N / {m}^{2}$ on the ground. Calculate the force that the box exerts on the ground, in Newtons.

Draw the triangle and circle the required property.

To calculate the force we would need to multiply the pressure by the area.

Substitute values for the other two properties and complete the calculation.

As $P=60 \ N / {m}^{2}$ and $A=1.2 \ m^2,$ substituting these into the formula triangle we get,

$F=60\times{1.2}=72$

Write down the solution, including the units.

Force $=72 \ N.$

Example 6: calculating force (converting units)

A pressure of $35 \ N/m^2$ is exerted on an area of $40 \ 000 cm^{2}.$ Calculate the force in Newtons.

Draw the triangle and circle the required property.

To calculate the force we would need to multiply the pressure by the area.

Substitute values for the other two properties and complete the calculation.

The area is in square centimetres, so we need to divide by $100^2$ to find the area in square metres.

$40 \ 000 \div{100}^{2}=4$

Substituting $P=35 \ N / {m^{2}}$ and $A=4m^2$ into the pressure triangle, we have

$F=35\times{4}=140$

Write down the solution, including the units.

Force $= 140 \ N.$

Example 7: calculating pressure given the mass

An object has a mass of $2.4kg$ and makes contact with the surface of the Earth with an area of $14.9m^{2}.$ Given that the acceleration due to gravity $g=9.807m/s^{2},$ calculate the pressure exerted on the Earth by the object to $2$ decimal places. State the units in your answer.

Draw the triangle and circle the required property.

Before we use the formula triangle, we need to calculate the force exerted on the Earth by the object. This can be found by multiplying the mass by the acceleration due to gravity $(F=MA).$

$F=2.4\times{9.807}=23.5368 \ N$

As we want to determine the pressure, we need to divide the force by the area.

Substitute values for the other two properties and complete the calculation.

Now we have $F=23.5268 \ N$ and $A=14.9m^{2}.$

Substituting these values into the pressure triangle, we have

$P=23.5368\div{14.9}=1.579651007…$

Write down the solution, including the units.

Pressure $=1.58 \ N / {m^{2}}\text{ (2dp)}.$

Common misconceptions

The area is the total surface area of the object

The area is the amount of surface in which the object is in contact with. This will usually be one face of an object, or given as the cross sectional area.

Mixing up the placement of the letters in the triangle

Formula triangles have limited use and if the letters are incorrectly written in the wrong place, the subsequent calculation will be incorrect. To overcome this, the standard units of pressure are $N/m^2$ which is a force $(N),$ divided by an area $(m^{2})$ and so $P=F \div A.$

Practice pressure force area questions

0.4 \ N / {m^{2}}

40 \ N / {m^{2}}

4 \ N / {m^{2}}

400 \ N / {m^{2}}

Using the pressure triangle,

Pressure force area practice question 1 image 1

and so

Pressure force area practice question 1 image 2

P=800\div{20}=40 \ N / {m^{2}}

2.78 \ Pa

0.361 \ Pa

0.36 \ Pa

2.77 \ Pa

Pressure force area practice question 2 image 3

and so

Pressure force area practice question 2 image 2

P=970\div{350}=2.771428571 \ N / {m^{2}} = 2.77 \ Pa \text{ (3sf)}

0.63m^2

0.16m^2

1.58m^2

227698.93m^2

w=mg=61.1 \times 9.807=599.2077 \ N

Pressure force area practice question 3 image 1

and so

Pressure force area practice question 3 image 2

\begin{aligned} &A=599.2077\div{380}=1.576862368…\\\\ &A=1.58m^{2}\text{ (2dp)} \end{aligned}

1752.0m^2

0.03m^2

30.4m^2

32.9m^2

1Pa=1N / {m}^{2}\text{ and so }7.3Pa=7.3N / {m}^{2}

Using the pressure triangle, we have

Pressure force area practice question 4 image 1

and so

Pressure force area practice question 4 image 2

\begin{aligned} &A=240\div{7.3}=32.87671233…\\\\ &A=32.9m^{2}\text{ (1dp)} \end{aligned}

350 \ N

280 \ N

320 \ N

380 \ N

Pressure force area practice question 5 image 3

and so

Pressure force area practice question 5 image 2

F=40\times{7}=280 \ N

236 \ N

23.1 \ N

23.2 \ N

237 \ N

A = 32000cm^2 = 3.2m^2

Pressure force area practice question 6 image 1

and so

Pressure force area practice question 6 image 2

F=74\times{3.2}=236.8 \ N

Pressure force area GCSE questions

1. Work out the force when the pressure is $48 \ N/m^{2}$ and the area is $6 \ m^{2}.$ Circle your answer.

54 \ N \quad \quad \quad 288 \ N \quad \quad \quad 8 \ N \quad \quad \quad 0.125 \ N

(1 mark)

Show answer

288 \ N

(1)

2. Work out the area when the pressure is $35 \ N/m^2$ and the force is $2100 \ N.$

(2 marks)

Show answer

2100\div 35

(1)

60 \ m^2

(1)

3. A block is resting on the floor.

Pressure force area gcse question 3

The downwards force of the block is $7600 \ N.$

Calculate the pressure exerted on the floor.

(3 marks)

Show answer

\text{Area}=5\times 2=10

(1)

7600\div 10

(1)

760 \ N / {m^{2}}

(1)

Did you know?

Air pressure (atmospheric pressure) can change depending on where we are in the world. If we are up a mountain the air pressure is less than when we are at sea level.
Each time a particle comes into contact with a wall, it exerts a force on the wall. If you increase the number of particles within the container (such as an inflating balloon), the frequency of collisions increases, and so does the pressure of the gas inside. At a high pressure, the balloon may not be able to withstand the increased amount of force on its surface, and so it bursts.

Learning checklist

You have now learned how to:

Convert between related compound units (pressure) in numerical and algebraic contexts

The next lessons are

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