# Video: Newton’s Third Law of Motion

In this video, we will learn how to apply Newton’s third law of motion to analyze systems of forces.

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### Video Transcript

In this video, we will be learning about Newton’s third law of motion and how to apply it to various different scenarios. So let’s start by looking at a statement of this law. Newton’s third law of motion considers the interaction between two objects. Specifically if we’re talking about an object, let’s say object A, somehow interacting with another object, let’s say object B, in this case, we’re drawing a collision between the two objects. Then, we can say the following. If object A exerts a force on object B, so let’s say that the force exerted by A onto B during this collision is a force towards the right, and we’ll call this force 𝐹. Then, according to Newton’s third law of motion, object B exerts a force that is equal in magnitude or size, but in the opposite direction on object A.

So as we can see in the diagram that we’ve drawn, each object has a force of magnitude 𝐹 acting on it. But the two forces are acting in opposite directions. Now, an often simplified version of this law of motion is stated as every action has an equal and opposite reaction. So in this situation, we’ve got the action force, that’s the force 𝐹, exerted by object A onto object B. And the equal and opposite reaction is the force 𝐹 exerted to the left by object B on object A.

However, this statement of Newton’s third law of motion “every action has an equal and opposite reaction” is not specific enough. For one thing, it makes no mention whatsoever of what the action force and reaction force are actually acting on. And simply memorising this statement often leads to a common misconception. It is very easy to think that if every action has an equal and opposite reaction, then because we’ve got two forces acting in opposite directions and they’re exactly the same in magnitude, that those two forces should cancel each other out.

Now, what that would mean is that either every force ever exerted gets cancelled out by the equal and opposite reaction force. And therefore, nothing in the universe happens. Or our understanding of Newton’s third law of motion is incorrect. And like we said earlier, this statement of Newton’s third law is indeed not specific enough.

However, if we come back to our original statement of Newton’s third law of motion as well as the diagram that we’ve drawn here, we can see that, yes, we are dealing with two equal and opposite forces. But they don’t cancel each other out because each object only has one force acting on it. Object A only has the force 𝐹 acting towards the left on it. And object B only has the force 𝐹 acting towards the right on it.

Therefore, if we were to consider each object by itself, we would see that it only has one force acting on it. And so the forces will not cancel each other out. So let’s ignore this statement here because it’s not specific enough and instead realise that, in Newton’s third law of motion, if an object exerts a force on another object, then the second object exerts a force equal in magnitude but in the opposite direction on the first object.

Now, it’s worth noting that Newton’s third law of motion holds true every time a force is applied. In the diagram that we’ve drawn, there’s a collision between object A and object B. However, it’s not necessary for the force to be a contact force in order for Newton’s third law of motion to apply.

For example, in a situation where an object is falling towards Earth, we often say that the Earth exerts a gravitational force on the object. Specifically, that gravitational force is the weight of the object. But that’s not all that’s going on here. In reality, the Earth exerts a force of 𝑤 on the object. And the object exerts an equal and opposite force 𝑤 onto the Earth. So it’s not only that the Earth is pulling the object towards itself. It’s more that the object and Earth are mutually attracting each other. Now, the reason that we can often ignore the force exerted by the object onto the Earth is seeing if we consider Newton’s second law of motion.

Newton’s second law of motion tells us that the net force on an object is equal to the mass of the object multiplied by the acceleration experienced by that object. And so if we think about the weight of the object, so more specifically, the gravitational force exerted by the Earth onto the object, we see that the weight of the object is equal to the mass of the object multiplied by the acceleration it experiences. But then, more specifically, we can remember that objects falling in Earth’s gravitational field will experience the same acceleration. And that acceleration is what we call 𝑔, the acceleration due to Earth’s gravity or the gravitational field strength of the Earth.

Now, this acceleration is a pretty reasonable acceleration. It’s 9.8 meters per second squared. However, if we then consider the equal and opposite force 𝑤 acting on the Earth, then we can see that, this time, although 𝑤 and 𝑤 are equal in magnitude, we are now considering 𝑤 as being equal to the mass of the Earth multiplied by the acceleration experienced by the Earth. That’s once again using Newton’s second law of motion. But then, at this point, we can see that 𝑤 and 𝑤 are both equal. So we can equate 𝑚𝑔, that’s the mass of the object multiplied by the gravitational field strength of the Earth, to the right-hand side of this equation, which is the mass of the Earth multiplied by the acceleration experienced by the Earth.

Then, if we rearrange to solve for the acceleration of the Earth, we see that it’s equal to the mass of the object multiplied by the gravitational field strength divided by the mass of the Earth. And then, we can realise that the mass of the Earth is absolutely huge. Specifically, the mass of the Earth is about 5.972 times 10 to the power of 24 kilograms. And so because we have a very large number in the denominator of this fraction, the overall value of the fraction becomes very very small. That’s of course assuming that the mass of the object that we’re dealing with isn’t comparable to the mass of the Earth.

In other words, the object could be anything like a tennis ball or a human being or even an aeroplane because the mass of the object is much much smaller than the mass of the Earth. And so once again, this fraction can become very very small for objects that are much smaller than the Earth. And in that situation, we therefore see that the acceleration experienced by the Earth is extremely tiny.

In other words, we can actually ignore the acceleration of the Earth. And that’s why often when we’re dealing with an object falling towards Earth in its gravitational field, we completely ignore the force exerted by the object on the Earth because it has very little impact on the Earth. However, as soon as the object that we’re dealing with becomes remotely comparable in mass to the Earth such as, for example, when we’re thinking about the moon, then we do have to consider the force exerted on the Earth. And then, we also need to think about gravity as the interaction between the object and the Earth, rather than just the Earth pulling the object towards itself.

Now, it’s also worth noting that when we’re with dealing with relatively small objects, that’s small objects compared to the mass of the Earth, then ignoring the value of the force exerted by the object on the Earth is not entirely accurate. But it does simplify our calculations massively because it gives us one listing to think about. And ignoring this force doesn’t result in any major consequences to our calculations either. So now that we’ve had a look at a couple of scenarios where we’d use Newton’s third law of motion, let’s have a go at an example question.

A rock of mass five kilograms is dropped onto Earth and decelerates from a speed of one meter per second to rest in a time of 0.1 seconds. The first part of the question asks us what magnitude force does the deceleration of the rock exert on Earth?

Okay so first of all, we see that, in this question, we’ve got the Earth. And we’ve got a rock of mass five kilograms dropped onto the Earth. We’ve been told that when the rock hits the Earth, it decelerates from a speed of one meter per second to rest. In other words, previously before the rocket hit the Earth, it was travelling downward, with a velocity of one meter per second. Then, as soon as it hit the Earth, it decelerated to a speed of zero metres per second. In other words, it was at rest. And the time taken for the rock to change its velocity from one meter per second to rest was 0.1 seconds. So we can say that the time interval, which we’ll call Δ𝑡, over which the rock lost its velocity was 0.1 seconds.

Then, at this point, we can see that the rock has a changing velocity. And we know the amount of time over which this velocity changed. So we can work out the acceleration of the rock, which we will call 𝑎. And we can recall that the acceleration of any object is given by finding the change in velocity of the object, divided by the amount of time taken for that change in velocity to occur. Now, in this case, we’ve already been given Δ𝑡. That’s 0.1 seconds. And we can work out the change in velocity. To do this, we can find the final velocity minus the initial velocity, which will give us the change in velocity. And this ends up being zero metres per second, the final velocity, minus one meter per second, the initial velocity or, in other words, negative one meter per second.

Now, the reason that the changing velocity is negative is because the rock is losing velocity. In other words, we will see that the value of 𝑎 is going to be negative because Δ𝑣 is negative. And this means that the rock is decelerating. So we can plug in our values for Δ𝑣 and Δ𝑡. And when we do so, we find that the acceleration of the rock is negative 10 metres per second squared. In other words, the deceleration of the rock is 10 metres per second squared.

Now, another way to think about this is that the acceleration of the rock is in the opposite direction to its initial velocity because we’ve said that the initial velocity was in this direction downward towards the Earth. And so for the block to decelerate when it hit the Earth, it would have to accelerate in the opposite direction. So what we figured out is the rock’s acceleration and in which direction the acceleration is occurring.

Now at this point, we can move forward and recall Newton’s second law of motion. This law tells us that the net force exerted on an object, 𝐹, is given by multiplying the mass of the object by the acceleration it experiences. So using this equation, we can work out the net force on the rock when it decelerates after colliding with the Earth. We can say that the force on the rock, 𝐹, is equal to the mass of the rock, which we’ve been told is five kilograms, multiplied by the acceleration experienced by the rock, which is negative 10 meters per second squared.

Now, before we evaluate the right-hand side of this equation, we can see that we’re working in base units, kilograms and metres per second squared. Those are the base units of mass and acceleration, respectively. And so whatever answer we find will be in the base unit of force, which is the newton. Therefore, evaluating the right-hand side gives us a force of negative 50 newtons.

In other words, because we said the rock falling downwards was falling with a positive velocity, positive one metre per second, then this means that the force exerted on the rock must be in the opposite direction because we see that the force is negative. And hence, we can draw an upward pointing arrow, that’s the force 𝐹 exerted on the rock causing it to decelerate. And this force is the force exerted by the Earth on the rock during the collision.

Now, it’s worth noting that what we’ve been asked to find here is the magnitude of the force exerted on the Earth due to the deceleration of the rock. However, what we’ve actually found, the force 𝐹, is the force exerted by the Earth on the rock. So to work out the force that we’re trying to find, we can recall Newton’s third law of motion. This law tells us that if an object A exerts a force on another object B, then B exerts an equal and opposite force on object A. Now, in this situation, the Earth is object A. And the rock is object B because we see that the Earth exerts a force 𝐹 on the rock. And that force is in the upward direction.

So by Newton’s third law of motion, we see that the rock exerts a force 𝐹 in the opposite direction on the Earth. And hence if we want to find the magnitude of the force exerted on the Earth due to the deceleration of the rock, then this force is going to have exactly the same magnitude as the force we’ve already calculated. We don’t even need to worry about directions because all what we’ve been asked to find is the magnitude or size of the force. And hence, we can say that the magnitude of force exerted on the Earth due to the deceleration of the rock is 50 newtons. Okay, so let’s look at the next part of the question.

What magnitude force does the collision with Earth exert on the rock?

So in this part of the question, we’re trying to find the force exerted on the rock that caused the rock to decelerate. However, in trying to find the answer to the first part of the question, we’ve actually already found the answer to this part of the question. Specifically, we found the value of 𝐹, which is the force exerted on the rock in order to decelerate it from one meter per second to zero metres per second, in a time interval of 0.1 seconds. And because we once again only need to give the magnitude of the force, so that’s the size of the force, we don’t need to worry about direction or sign. And hence, we can say that the magnitude of the force the collision with the Earth exerts on the rock is 50 newtons. Moving on to the final part of the question then.

Which of the following statements most correctly describes the motion of Earth due to the collision with the rock? A) The magnitude of the acceleration of Earth due to the collision is equal to the acceleration of the rock but in the opposite direction. B) The magnitude of the acceleration of Earth due to the collision is equal to the mass of the rock divided by Earth’s mass. C) Earth is not accelerated at all by the collision with the rock. D) The magnitude of the acceleration of Earth due to the collision is equal to the force applied in the collision divided by Earth’s mass.

Okay, so there’re four different options here. And each one of them is relatively complicated in terms of what it’s trying to say. But what we’re trying to find is the statement that most correctly describes the motion of Earth due to the collision with the rock. And specifically, each one of these statements refers to the magnitude of the acceleration of Earth. So which one of the statements A, B, C, or D gives us a correct description of how to find the acceleration of Earth due to this collision?

Well, first of all, we’ve seen already that there is a force 𝐹 exerted on the Earth. And this force was found due to Newton’s third law of motion. Because the Earth exerted a force 𝐹 upward onto the rock, the rock therefore exerted a force 𝐹 downward onto the Earth. And then, we can see, using Newton’s second law of motion, that if an object experiences a force and it has a certain mass, then it must experience a certain acceleration. And because the force exerted on the Earth is nonzero, the acceleration of the Earth can also not be zero. Therefore, we can immediately rule out option C, which says that the Earth is not accelerated at all by the collision with the rock.

Then, we can see that the force 𝐹 exerted on Earth, using Newton’s second law of motion, must be equal to the mass of the Earth, which we’ll call 𝑚 subscript 𝐸, multiplied by the acceleration experienced by the earth, which we’ll call 𝑎 subscript 𝐸. And therefore, if we want to find the acceleration experienced by Earth, we divide both sides of the equation by the mass of Earth. This way, 𝑚 subscript 𝐸 cancels on the right-hand side. And so what we’re left with is 𝐹 divided by 𝑚 subscript 𝐸 is equal to 𝑎 subscript 𝐸.

In other words, the acceleration experienced by Earth is equal to the force applied on the Earth divided by the mass of the Earth. And of course, the force applied on the Earth during this collision is the same in magnitude as the force applied on the rock during this collision, but in the opposite direction. However, because each one of the statements only refers to the magnitude of the acceleration, we therefore don’t need to worry about the directions in which these forces or accelerations are acting. All we need to do is to find the statement that tells us that the acceleration experienced by Earth is equal to the force applied in the collision divided by the mass of Earth.

So looking at statement A, this one says that the magnitude of the acceleration of Earth due to the collision is equal to the acceleration of the rock, but in the opposite direction. So what this one is saying is that the acceleration of Earth is equal to the acceleration experienced by the rock, but in the opposite direction. Now, of course, the rock is falling in the Earth’s gravitational field. So the acceleration that the rock experiences is 𝑔. That’s the acceleration due to gravity. And what this statement is saying is that the acceleration of Earth is negative 𝑔 because it’s in the opposite direction.

Well, this doesn’t match what we’ve worked out here. And also as we said earlier, because we’re only thinking about magnitudes, we don’t need to worry about directions. Therefore, statement A is not the answer to our question. Statement B then says that the magnitude of the acceleration of Earth due to the collision is equal to the mass of the rock divided by Earth’s mass. In other words, this one is saying that the acceleration of Earth is equal to the mass of the rock 𝑚 divided by the mass of the Earth 𝑚𝐸. But once again, this is not what we found here. So option B is not the right answer.

Therefore, it looks like option D is the answer to our question. This one says that the magnitude of the acceleration of Earth due to the collision is equal to the force applied in the collision divided by the Earth’s mass. And that’s exactly what we found over here. It’s the force applied in the collision divided by the Earth’s mass. Therefore, option D is the answer to our question.

Okay, so now that we’ve had a look at an example, let’s quickly summarise what we’ve talked about in this lesson.

Firstly, in this video, we saw a statement of Newton’s third law of motion, which says that if an object A exerts a force on object B, then B exerts an equal and opposite force on object A. Secondly, we saw that the two forces mentioned in the above statement do not cancel each other out as they act on different objects. Therefore, each object only has one force acting on it. And finally, we saw that Newton’s Third Law also applies when considering gravity. But we often ignore the force on the much much larger body as its effects are negligible. Such as, for example, when we’re considering a ball falling towards Earth, we often consider the weight of the ball. But we tend to ignore the force exerted by the ball on the Earth because the acceleration that this force will cause on the Earth is minuscule. So this is an overview of Newton’s third law of motion.