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.
Pressure, force and area are physical properties.
Get your free pressure force area worksheet of 20+ questions and answers. Includes reasoning and applied questions.
Get your free pressure force area worksheet of 20+ questions and answers. Includes reasoning and applied questions.
Area is a measure of the size of space a flat shape takes up. The SI unit for area is the square metre (m^) although we can use other units including square centimetres (cm^), square millimetres (mm^) or square inches (in^) 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 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 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 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 / >. We can also measure pressure using a Bar, where 1 Bar = 100,000 Pa = 100,000 \ N / >.
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 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,
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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.
In order to calculate pressure, force or area using the formula triangle:
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:
A force of 800 \ N acts on an area of 200 \ m^. Calculate the pressure in Pascals.
To calculate the pressure we would need to divide the force by the area.
2 Substitute values for the other two properties and complete the calculation.
Substituting F=800N and A=200m^, we have
3 Write down the solution, including the units.
Pressure = 4 \ N / >=4 \ Pa.
A force of 6.5 \times 10^6 \ N acts on an area of 0.2 \ m^. 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 .
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,
A force of 200 \ N exerts a pressure of 40 \ N/m^. 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 / >, we have
Write down the solution, including the units.
