In this explainer, we will learn how to use the formula for pressure, , to calculate pressures that are produced by forces acting on areas.

Force is a vector quantity, and so a force can be represented by an arrow. Such an arrow may be drawn as a thick or a thin line.

The thickness or thinness of the line has no physical meaning; only the length of the line and its direction correspond to properties of a force (its magnitude and direction).

An ideal line used to represent a force has no thickness, only length. Such a line would of course not be visible in a diagram.

If the arrow representing a force is thought of as having thickness, this leads to the mistaken idea that a force acting on a surface acts on some area, as shown by the yellow region in the following figure.

It is more correct to think of the tip of the arrow acting on the surface, as shown below.

The tip of the arrow is a point; it has an area of zero. A force, therefore, acts at a single point, not on an area.

The idea of a force acting at a point is familiar. Consider a uniform cube at rest on a horizontal surface. It is not unusual to describe the weight of the cube as acting at point at the center of the face of the cube in contact with the surface.

If this model of the weight of the cube as a force was taken literally, however, then at all points on the downward-facing face of the cube other than its center, no force would act.

There would then be no force acting at the point shown by the center of the red cross in the following figure.

An object on the surface at the position of the center of the red cross would however have a force acting on it in reality. An object underneath any point on the downward face of the box would have some force acting on it due to being in contact with the box.

When we wish to model a force acting on an area rather than at a point, we cannot use forces that act at single points. We must instead consider a surface for which forces act at every point on the surface.

When forces act at every point on a surface, a quantity can be defined that is the result of the forces that act. This quantity is called pressure.

The relationship between force, pressure, and area is represented visually by the following figure. Forces act at every point over a rectangular area (not all of these forces are shown). The force at each point is the same.

We can see that the direction in which the forces act is perpendicular to both sides of the rectangle. In order to produce pressure, forces must act perpendicularly to an area. If forces act parallel to an area, then no pressure is produced.

In the following figure, a cross section of an area is shown. The cross-section of an area is a length. Forces act at every point along a length (not all of these forces are shown). The force at each point is the same.

There is a mathematical relationship between the pressure on an area and the force acting perpendicular to all the sides of the area.

### Relationship: The Pressure on an Area and the Force Acting Perpendicularly to the Area

The pressure, , on an area, , is given by where is the force acting perpendicularly to all the sides of the area.

If has the unit newtons and has the unit square metres, then has the unit pascals (Pa). This means that

Pressure is often stated in kilopascals (kPa), where 1 kPa = 1βββ000 Pa.

When we consider the formula we can see that for a fixed value of , the greater the value of is, the smaller the value of must be to produce pressure . Equivalently, a smaller value of corresponds to a greater value of required to produce a fixed value of .

We have seen that the forces due to a pressure have a direction. The area that the pressure acts on can be considered to have a direction compared to the forces.

By showing a rectangular area, we can see that the direction in which the forces act is perpendicular to both sides of the rectangle.

We can then consider both a force and an area to be vector quantities. This is the case even though area is a scalar quantity when it is not related to the direction of a force. Relating an area to a force that is perpendicular to the area changes how the quantity area is used.

When relating force, area, and pressure, we see that multiplying the area perpendicular to the force by the pressure gives the force:

As both force and area are considered as vector quantities when related this way, we see that pressure must be a scalar quantity. This is the case as multiplying a vector quantity by a scalar quantity results in another vector quantity.

We can express the relationship between force, pressure, and perpendicular area then as

A different way of understanding the relationship between force and pressure is to make force the subject of the formula. We can do this by multiplying the formula by :

This gives us

Considering the formula this way, we can say that a pressure on an area is associated with a force acting at every point within that area. The forces all act perpendicularly to the area.

The relationship is represented visually by the following figure, where again a length represents the cross section of an area.

Using the relationship between force, pressure, and area, we can visually represent a change in area for a fixed pressure. This is shown in the following figure in which equal pressures act on two different areas.

We see that the force associated with the pressure is inversely proportional to the area.

Pressure can act on fluids as well as on solid objects. When pressure acts on a fluid, the direction of the forces acting in the fluid can be in various directions.

The following figure shows that increasing the pressure on the top of a container of water can cause the side of the container to rupture.

We can see that the direction of the force on the container wall is not the direction of the force due to the pressure on top of the container. We see then that a pressure does not have to act in the direction of the force that produces the pressure. We cannot then, in general, define the direction of a pressure. This means that pressure is considered a scalar quantity.

Let us now look at an example of determining a pressure.

### Example 1: Determining a Pressure

What pressure is produced by a 100 N
force applied to an area of 2.5 m^{2}?

### Answer

We can determine the pressure using the formula

Substituting the values in the question, we obtain

Let us now look at an example of determining a force using a pressure.

### Example 2: Determining a Force Using a Pressure

A pressure of 400 Pa is applied to an area of
2.5 m^{2}. What force produces this pressure?

### Answer

We can rearrange the formula to make the subject.

We can do this by multiplying the formula by :

This gives us

Substituting the values in the question, we obtain

Let us now look at an example of determining an area using a pressure.

### Example 3: Determining an Area Using a Pressure

A pressure of 75 Pa is produced by a 3βββ000 N force. What area is pressure produced over?

### Answer

We can rearrange the formula to make the subject.

We can do this by first multiplying the formula by :

This gives us

The formula is then divided by :

Substituting the values in the question, we obtain

Let us now look at an example in which a weight is the force acting on an area.

### Example 4: Determining a Pressure given a Weight

A sofa with a mass of 125 kg has a base with an area of
2.5 m^{2}. What pressure does the sofa apply to the ground beneath it?

### Answer

We can determine the pressure using the formula

The force acting is the weight of the sofa. The weight of the sofa is given by where is the mass of the sofa and is the acceleration due to gravity or the gravitational field strength. The value of is given by

Substituting the values in the question, we obtain

Let us now look at an example in which the area a pressure is produced over must be determined to find the weight of an object supported on that area.

### Example 5: Determining a Weight given a Pressure

An empty water tank has a rectangular base with side lengths of 1.2 m and 2.3 m. The weight of the tank applies a pressure of 350 Pa to the surface it is resting on. What is the weight of the tank?

### Answer

The weight of the tank is the force that acts on it. We can represent the force by .

We can rearrange the formula to make the subject.

We can do this by multiplying the formula by :

This gives us

The value of the pressure is stated to be 350 Pa, but the area is not stated.

The tank has a rectangular base, so the area of the base is equal to the product of the lengths of adjacent sides of the rectangle. We have then that

We can now determine , which is given by

Let us now summarize what has been learned in these examples.

### Key Points

- A force acts at a point while a pressure acts on an area.
- The pressure, , on an area, , is given by where is the force acting perpendicularly to the area.
- If a force has the unit newtons and an area has the unit square metres, then a pressure has the unit pascals (Pa), where
- Pressure is often stated in kilopascals (kPa), where 1 kPa = 1βββ000 Pa.
- A component of a force must act perpendicularly to an area to produce a pressure on it.
- Pressure is a scalar quantity.
- For a fixed force, reducing the area that the force acts on increases the pressure produced.