Video: Relating Force, Pressure, and Area

In this video, we will learn how to use the formula for pressure, 𝑃 = 𝐹/𝐴, to calculate pressures that are produced by forces acting on areas.

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Video Transcript

In this video, we will be looking at how to define pressure and how it relates to the concept of force and area. So let’s start by first of all imagining that we have a box. And inside this box, we have a single particle of some gas inside it. Let’s also imagine that this particle is moving around inside the box at a velocity 𝑉 and is basically bouncing around inside the box. Now, let’s also assume that the particle continues to bounce around inside the box with the same speed even though its direction is changing every time it collides with a wall of the box.

And with all of that in mind, let’s consider one particular collision between the particle and the right-hand side wall of the box. So let’s zoom in to this region here, where the particle is colliding with the wall of the box. So here’s the zoomed-in version. Now we said earlier that the particle initially was travelling with a velocity 𝑉 in this direction. And then, after the collision, the particle is travelling with a velocity 𝑉 in this direction.

Now if the magnitude or size of the velocity of the particle before the collision and after the collision are exactly the same or in other words it travels at 𝑉 before the collision and 𝑉 after the collision, then the particle can’t have lost any energy. This is an assumption that we’re making here. We’re assuming that the particle doesn’t lose kinetic energy. So the size or magnitude of the particle’s velocity does not change after the collision, but the direction of the velocity does change. And one thing that we can do is to break up the velocity of the particle into components.

Initially, the particle is travelling at some velocity to the right and some velocity upward. And after the collision, the particle is travelling at some velocity to the left and some velocity upward. Now in a collision with a particle is not losing energy and there’s nothing affecting the upward velocity of the particle, we can say that the upward component of the velocity stays the same. Of course, in the diagram that we’ve drawn, they don’t look exactly identical, but they’re meant to be.

However, what is changing because of the collision is the right left component of the velocity because initially the right left component of the velocity was in this direction and then after the collision, it’s now in this direction. This means that the net result of the collision with the wall is that the particle’s velocity is changing in this direction or we can also say that Δ𝑉 — the change in velocity of the particle — is towards the left. This is because the right left component of the particle’s velocity was such that it was initially travelling towards the right and then after the collision it’s travelling towards the left. So its velocity in the left direction has increased. And we’ve called that change in velocity in the leftward direction Δ𝑉.

Now in addition to this, we can realize that the particle’s collision with the wall takes a certain amount of time. As the particle comes in towards the wall and then bounces off, it spends a certain period of time actually colliding with the wall. Now this period of time might be very very short, but it is a period of time nonetheless. And we can call that period of time when the particle is colliding with the wall Δ𝑡.

Now since the velocity of the particle is changing by Δ𝑉 and it does this over a time interval Δ𝑡, we can therefore work out the rate of change of the particle’s velocity. We can say that the particle changes in velocity by Δ𝑉 over a time interval Δ𝑡. And at this point, we can realize that this quantity that we’re calculating is known as the acceleration of the particle 𝑎. And this is because acceleration is defined as the rate of change of velocity of an object.

Now as well as this, we can say that the particle itself has a mass which we’ll call 𝑚. And then we can recall that when an object with a certain mass undergoes an acceleration, there must have been some net force acting on that object. The reason is because of Newton’s second law of motion: the force on an object 𝐹 is equal to the mass of that object 𝑚 multiplied by its acceleration 𝑎.

And therefore, we can realize that in order for there to be a change in velocity of the particle towards the left, the particle must have experienced the force towards the left exerted by the wall during the collision. We can call this force 𝐹. And we can use Newton’s second law of motion to work out that the force 𝐹 is equal to the mass of the particle which we’ve called 𝑚 multiplied by the acceleration of the particle which was Δ𝑉 divided by Δ𝑡. That’s the change in velocity divided by the time interval over which that velocity changes. So that’s the force exerted on the particle by the wall in the collision.

But then at this point, we can recall Newton’s third law of motion. Newton’s third law of motion tells us that if an object which we’ll call A exerts a force on a second object B, then the second object B exerts an equal and opposite force on object A. Now in our scenario where the particle is colliding with the wall, the wall is exerting a force 𝐹 on the particle. And therefore, by Newton’s third law of motion, the particle must be exerting an equal-in-magnitude force but in the opposite direction on the wall. And so, from this, what we can realize is that the particle colliding with the wall results in a force being exerted on the wall.

Now at this point, we can zoom back out and think about our entire box once again. And instead of just one particle inside the box, let’s think about the box being filled up with lots of particles of gas. Now each one of these particles is moving around inside the box at different velocities and different directions. And they’re all colliding with the walls of the box and bouncing around inside the box. But then, we’ve just seen that every time a particle collides with the wall of the box, it exerts a force on the wall of the box. And furthermore, if we now think about the fact that the box is a three-dimensional object, we can see that each wall of the box has a particular area, which we shall call 𝐴.

Now over time as the particles of gas move around inside the box, lots of them will collide with this particular wall of the box which has an area 𝐴. And each collision is going to exert a particular force on the wall. Now let’s say that we add up all of the forces exerted by the particles on the wall every time they collide with the wall. And let’s call this force 𝐹. Now let’s not get confused here. Earlier, we used 𝐹 to define the force exerted by one particle colliding with the wall. But now we’re adding up all of the forces exerted by the particles colliding with that particular wall of the box and calling that total force 𝐹.

So if we take this total force exerted by all of the particles colliding with the wall and we can do this for any particular instant in time and we can divide that force by the area of the wall itself 𝐴, which will bring us to the definition of a quantity called the pressure. In other words, this force per unit area is known as the pressure exerted by the gas particles on this particular wall which has an area 𝐴. And that is the definition of pressure. It is equal to the force per unit area.

And at this point, we should note that it’s not only gases that can exert a pressure, even solids and liquids can exert pressures. For example, if you were to place your hand on a table and pushed downwards onto the table with a certain force 𝐹, then your hand would be exerting a pressure on the table which would be equivalent to the force 𝐹 divided by the area of contact between your hand and the table. So solids, liquids, and gases can exert pressures. And pressure is defined as the force per unit area.

As well as this, the unit for pressure is the pascal, short form Pa, where one pascal is equivalent to one newton — that’s the unit of force — divided by one meter squared — that’s the unit of area. So now that we’ve had a look at the definition for pressure, let’s get some practice using this with an example question.

The bases of a man’s feet have a total area in contact with the ground of 0.025 meters squared. The man’s weight is 800 newtons. How much is the pressure applied by his feet to the ground in kilopascals?

Okay, so in this question, we have a man who is standing on the ground and his feet have a total area of 0.025 meters squared. In other words, the total area of contact between the man and the ground is 0.025 meters squared. So let’s say that the contact area, which we’ll call 𝐴, is 0.025 meters squared. Now as well as this, we’ve been told that the man’s weight is 800 newtons. Now the man’s weight is going to act in a downward direction. So we can say that the weight, which we’ll call 𝑤, is equal to 800 newtons.

But then, this weight is exactly the force that the man will exert onto the floor. And therefore, if we want to work out the pressure applied by the man’s feet to the ground, then we can recall that the pressure applied to in this case the ground is defined as the force applied by the man in this case to the ground divided by the total area of contact or specifically the area of the floor over which that force is distributed.

Now since we’re working in base units — we’ve been given the area of the man’s feet in meters squared and his weight in newtons — this means that when we calculate the pressure using this equation, we’ll actually find it in its own base unit which is the pascal. However, there is a catch: we need to work out this pressure in kilopascals. We’ll come to that later, but it’s good to be aware of it now.

So we can say that the pressure exerted by the man on the floor is equal to the force that he exerts which is 800 newtons — that’s his weight — divided by the area over which that force is distributed. That area is the same as the contact area between the man’s feet and the floor or in other words the area of the man’s feet which happens to be 0.025 meters squared. Then, when we evaluate the right-hand side of this equation, we find that the pressure ends up being 32000 pascals.

Now we need to convert this pressure into kilopascals. To do this, we can recall that one pascal is equivalent to one thousandth of a kilopascal because the prefix kilo means 1000. So one thousandth of 1000 pascals is the same thing as one pascal. And therefore, what this means is that 32000 pascals is the same thing as 32 kilopascals. Hence, our final answer is that the pressure applied by the man’s feet to the ground is 32 kilopascals.

Okay, let’s take a look at another example question.

A pressure of 20 pascals is applied by an 8000-newton force to a square area. What is the length of a side of the square?

Okay, so in this question, we’ve got a square area, which we can draw something like this if we’re looking at it slightly side-on. And we’ve been told that a force of 8000 newtons is applied to this area. This results in a pressure of 20 pascals. Now at this point, we can recall the relationship between pressure, force, and area.

Pressure is defined as the force per unit area or in other words the force exerted on an area divided by that area. So using this equation, because we know that the pressure exerted on this blue area is 20 pascals and the force exerted is 8000 newtons, we can calculate the area. We can do this by rearranging the equation for which we have to multiply both sides by 𝐴 divided by 𝑃. In this case, the 𝑃s on the left-hand side cancel and the 𝐴s on the right-hand side cancel. What this leaves us with is that area 𝐴 is equal to the force 𝐹 divided by the pressure 𝑃.

Therefore, we can plug in the values for the force which is 8000 newtons and the pressure which is 20 pascals. And since we’re working in base units for both the pressure and the force, our answer ends up being 400 meters squared, where meter squared is the base unit of area. So at this point, we’ve worked out that the total blue area is 400 meters squared.

However, what we’ve been asked to do is to find the length of one of the sides. Luckily though, we’ve been told that this area is actually a square. And hence, we know that the lengths of all of the sides are equal. So we can say that 𝑥 is the length of one of the sides and therefore it’s also the length of all of the sides. Then, we can work out that the area of a square is found by multiplying its length by its width, which because it’s a square, it’s 𝑥 in both cases. Therefore, we can say that the area is equal to the length squared.

And hence, if we want to find the length of one of the sites which is 𝑥, we need to rearrange this equation by taking the square root of both sides. This way, the square root on the right-hand side cancels with the squared. And what we’re left with is that the square root of the area of the square is equal to the length of one of its sides. But then, we’ve just calculated the area to be 400 meters squared. Therefore, the length of one of the sides of the square is the square root of 400 meters squared. And this ends up being 20 meters, which also happens to be the final answer to our question.

Okay, so now that we’ve looked at a couple of examples, let’s summarize what we’ve talked about in this video. Firstly, we saw that pressure is defined as the force exerted on an object per unit area or in an equation, we can say that the pressure 𝑃 is equal to the force 𝐹 divided by the area 𝐴. Secondly, we saw that pressure can be exerted by a solid, liquid, or a gas. There is no restriction to what state of matter can exert a pressure.

And lastly, we saw that pressure has units of Pa which stands for pascals, where one pascal is equivalent to one newton per one meter squared. Or in other words, the unit of pressure is equivalent to the unit of force divided by the unit of area. So this has been an overview on how we can relate pressure to force and area.

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