## What you need to know

### Pressure Force Area

Area is a measure of the size of a surface and is usually measured in square metres. Force is a measure of how much push or pull is being exerted onto an object and is usually measured in Newtons (N). Pressure is a measure of how much force is applied over a given area of an object, so it is calculated by dividing the amount of force being applied by the area over which its being applied.

The units we used to measure pressure are **compound units **(for more information, see here (https://mathsmadeeasy.co.uk/gcse-maths-revision/conversions-gcse-revision-and-worksheets/)). Another useful topic to revise is rearranging formulae.

### Pressure Force Area Triangle

As we have seen in other topics, such as speed, distance, time, formula triangles can be a useful tool to help determine what order to use a formula. The pressure, force and area triangle works in the same way as others, you cover up the thing you are calculating and the triangle shows you what formula to use.

**Note:** The most common mistakes made on pressure, force and area questions is to mix up the units being used.

### Example 1: Calculating Pressure

A force of 150N is being applied over an area measuring 0.5\text{m}^2 . Calculate the pressure on the object ensuring you give the correct units.

The first step is to select the correct equation to use

Then substitute in the values you are given in the question, to get

\dfrac{150}{0.5}=300\text{ N/m}^2

### Example 2: Calculating Force

A woman is applying 300\text{ N/m}^2 of pressure onto a door with her hand. Her hand has area 0.02\text{ m}^2. Work out the force being applied.

We’re looking for force, so, constructing the triangle and covering up the , we get

Substituting the values into the equation we get

300\times0.02= 6 N

### Example 3: Calculating Area

Pressure of 150\text{ N/m}^2 is experienced when a force of 2 kN is applied. Calculate the area in which the force is applied to obtain the pressure stated.

Now, before we dive into forming our formula triangle, we need to make sure our units are in the same form. In this example we need to convert kN (kilo-newtons) into N (Newtons) bu multiplying by 1000. This gives us 2000 N, now we can look at our formula triangle.

As before, we substitute our values into the equation, giving

\dfrac{2000}{150}=13.3\text{ m}^2

### Example Questions

1) A force of 185.6N is applied to a square of side length 3m. Work out the pressure on the square to 3 significant figures.

We are calculating the pressure, so by covering up p, we can see from the triangle above that we have to divide F by A. Before we can do this calculation however, we need to work out the area of the square. Since the square has a side length of 3m, the area can be calculated as follows:

3\text{m} \times 3 \text{m}=9\text{m²}

Now that we have the area and the force, we can calculate the pressure by substituting the values for the area and the force into the pressure equation as follows:

\text{pressure }=\dfrac{F}{A}=\dfrac{185.6}{9}=20.6222...=20.6\text{ N/m²}\text{ (3 significant figures)}

The units must be N/m² since the units used in the question are N and m².

2) A man with a weight of 740N is standing on one leg. His foot is exerting 2,312.5\text{ N/m²} of pressure onto the ground. What is the surface area of the bottom of his foot?

We are calculating the area, so by covering up A, we can see from the triangle above that we have to divide F by p. So, by substituting the known values for the force and the pressure into the equation, we can calculate the area as follows

\text{area }=\dfrac{F}{p}=\dfrac{2,312.5}{740}=3.125\text{ m}^2

The unit must be m² since the units used in the question are N / m².

3) A shipping container is removed from a ship and placed on the ground. The area of the container in contact with the ground is 16m². The pressure exerted on the floor is 2480 N/m².

What force is being exerted by the shipping container on the ground?

In this question, we are calculating the force, so we will need to rearrange the pressure / force / area equation.

Since

\text{pressure = force} \div \text{ area}

then

\text{force = pressure} \times \text{ area}

By substituting the values for the pressure and the area into the formula above, we can calculate the force as follows:

\text{ force = 16m²} \times 2480 \text{ N/m²} = 39,360 \text{ N}

4) A cylinder is resting on one of its circular faces on the ground. The weight of the cylinder is 4,872N and the cylinder exerts a pressure of 812 N/m² on the ground.

To 2 decimal places, what is the diameter of the cylinder?

This may seem like an impossible question without sufficient information to go by at first glance. We have been given the pressure as well as the weight (force) of the cylinder. Since we know both the force and the pressure, the only thing we are able to calculate at this stage is the area of the circular cylinder face.

Since

\text{pressure = force} \div \text{ area}

then

\text{area = force} \div \text{ pressure}

By substituting the values for the force and the pressure into the formula above, we can calculate the area as follows:

4872 \text{ N} \div 812 \text{ N/m²} = 6\text{ m²}

We now know that the area of the circular face on which the cylinder is resting has an area of 6m², and from this information we need to work out the diameter of the cylinder / circle. In this question, we therefore need to apply some basic knowledge about the circle areas.

The formula for the area of a circle is \pi r².

However, we know the area and want a formula for the radius, so we will need to rearrange this formula.

First of all, we can both sides by \pi:

\dfrac{a}{π} = \text{ r²}

If we now take the square root of both sides, we have a formula for the radius:

\sqrt \dfrac{a}{π} = \text { r}

Since we know the area, all we need to do is divide the area by \pi, and square root the answer.

6 \div \pi = 1.909….

\sqrt 1.909…. = 1.38 \text{ m}

If the radius of the circle is 1.38m, then the diameter will be double this:

2 \times 1.38 = 2.76\text {m}

5) A cube and a square-based pyramid exert the same pressure on the ground.

The square of the pyramid’s base has a side length of 8 metres and has a weight of 440N. The cube has a weight of 110N.

What is the side length of the cube?

This is another of these questions that might seem impossible initially, but there is sufficient information in the question to solve it. Like many maths problems, if you don’t have a clear strategy, just do any calculations you can, as they may lead somewhere useful!

We have been told that the pyramid has a square base which has a side length of 8 metres. From this information, we can calculate the area of the pyramid’s square base. The base of the pyramid has an area of 8 \times 8 = 64 \text{ m².}

For the pyramid, we now have an area and we know its weight (force), meaning we can calculate the pressure it exerts on the ground. Using the standard formula:

\text{pressure = force} \div \text { area}

We can calculate the pressure by substituting the values for force and area into the formula as follows:

\text{ pressure = } 440 \text{ N}\div 64 \text { m²} = 6.875 \text{ N/m²}

We have been told in the question that the cube exerts the same pressure as the square-based pyramid, so the pressure exerted by the cube is also 6.875 N/m². Since we also know the weight (force) of the cube (110 N), we can calculate the area of the cube that is in contact with the floor (this will be the area of a square since this is a cube).

We will now need to rearrange the pressure formula, making area the subject.

If

\text{pressure = force} \div \text { area}

then

\text{area = force} \div \text { pressure}

We can calculate the area by substituting the values for force and pressure into the formula as follows:

\text{ area = } 110 \text{ N}\div 6.875 \text { N/m²} = 16 \text{ m²}

Now that we know the area of the square in contact with the floor, we can work out the side length of the cube by taking the square root of the area:

\sqrt{16} = 4\text{ m}

### Worksheets and Exam Questions

#### (NEW) Pressure Force Area Exam Style Questions - MME

Level 4-5#### Density and Pressure - Drill Questions

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