Lesson Video: Newton’s Third Law of Motion | Nagwa Lesson Video: Newton’s Third Law of Motion | Nagwa

# Lesson Video: Newton’s Third Law of Motion Mathematics

In this video, we will learn how to solve problems on Newton’s third law.

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### Video Transcript

In this video, we’ll learn how to solve problems using Newton’s third law of motion. By this stage, you should’ve encountered and applied Newton’s first and second laws of motion. Newton’s first law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. And Newton’s second law says that the force acting on an object is equal to the mass of that object multiplied by its acceleration, 𝐹 equals 𝑚𝑎. So what does Newton’s third law tell us?

Newton’s third law is most commonly stated as for every action, there’s an equal and opposite reaction. But we’re not talking about the type of reaction you have when, say, your sibling steals your favorite item of clothing. It actually says that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction on object A. As with Newton’s other laws, this is quite intuitive. For example, during a rocket launch, burning fuel releases gas at high velocity. The rocket exerts a downward force on the gas in the combustion chamber. And so the gas exerts an equal and opposite reaction force upward on the rocket. The special name for this is thrust.

Another very important example applies to you at this very second. Whether you’re sitting on a chair or standing up, you are exerting a downward force on the chair or the floor. The chair or the floor is therefore exerting an upwards reaction force on you. This is the very thing that’s stopping you from falling straight through the chair or the ground, like in some sci-fi film. And so we have Newton’s third law of motion and some context behind it. Note that in classical mechanics, this law very rarely stands alone, and for this reason, you must make sure that you’re confident in applying the first two.

Let’s begin then by applying the law in a very simple example.

An elevator is moving vertically upward at a constant speed. A man of mass 150 kilograms is standing inside. Determine the reaction force of the floor on the man. Take 𝑔 to be equal to 9.8 meters per square second.

With questions like these, it can be really helpful to begin by drawing a very simple diagram. Here’s the lift moving upward at a constant speed. Now, if the speed is constant, acceleration must be equal to zero meters per square second. A man of mass 150 kilograms stands inside the lift. And so this means that the man must exert a downward force on the floor of the lift. Now, in fact, to find the value of that downward force, we can use Newton’s second law, force is equal to mass times acceleration. We call the downward force due to the mass “weight”, 𝑤. And we say it’s equal to the mass of the object multiplied by acceleration due to gravity, so 𝑚𝑔. This must mean that since the man has a mass of 150 kilograms, the downwards force of weight is 150𝑔 on the floor of the lift.

Then we recall Newton’s third law of motion. And this says that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction on object A. And so since the man exerts a downwards force on the floor of the elevator, the floor of the elevator must exert an upward force on the man itself. And that’s all the forces we’re interested in for now. We aren’t interested in, for example, tension forces in the cable for the lift. And so we’re going to go back to Newton’s second law of motion, force is equal to mass times acceleration. We’re going to define upwards in this case as being positive, and we’re going to find the net sum of the forces.

In the upwards direction, we have 𝑅, and then downwards, we have 150𝑔. This is acting in the negative direction. And so the net force on our diagram is 𝑅 minus 150𝑔. This is equal to mass times acceleration, 150 times zero. 150 times zero is zero. So our equation becomes 𝑅 minus 150𝑔 equals zero. We can solve for 𝑅 by adding 150𝑔 to both sides, and we find 𝑅 is 150𝑔. But of course we were told 𝑔 is 9.8, so we substitute 𝑔 equals 9.8 into this expression. And we get 𝑅 equals 150 times 9.8, and that’s 1470. We’ve worked with kilograms and meters per square second, so the units for our force are newtons. The reaction force of the floor on the man is 1470 newtons.

Now it’s worth noting that we could have gone straight to this step. We didn’t actually need the first line of working. It’s essentially the reverse of the first law. We’re told that an object will remain at rest or in uniform motion unless acted upon by an external force. In other words, if the sum of the forces acting on the body is zero, then the object will remain at rest or in uniform motion. Now we know that our object remains in uniform motion, and so the sum of the forces did have to be equal to zero.

We’ll see why, though, in our next example, it can be really helpful to use Newton’s second law most of the time.

An elevator was accelerating vertically downward at 1.7 meters per square second. Given that the acceleration due to gravity is 𝑔 equals 9.8 meters per square second, find the reaction force of the floor to a passenger of mass 103 kilograms.

Let’s begin by sketching a diagram. We have an elevator accelerating vertically downward at 1.7 meters per square second. We know that a passenger in the lift has a mass of 103 kilograms. This passenger must exert a downward force on the lift, and we call that downward force weight, where weight is mass times 𝑔 gravity. In this case then, since the passenger has a mass of 103 kilograms, his weight is 103𝑔. Newton’s third law then tells us that if an object exerts a force on a second object, that second object must exert a force of equal magnitude and opposite direction on the first. And so the floor of the lift exerts a reaction force on the passenger itself.

We now have all of the forces that we’re interested in, and so we recall Newton’s second law of motion, force is equal to mass times acceleration. We’re going to find the net sum of the forces in our diagram, and we’re going to set that equal to the mass of the passenger multiplied by acceleration. And so we’re going to need to define a positive direction. Now, this is very much personal preference. I prefer to take the positive direction as the direction at which the object is moving. So in this case, I’m going to assume that downwards is the positive direction. As long as we’re consistent, though, it doesn’t matter which direction we choose.

So forces acting in the positive direction, we have 103𝑔. In the negative direction, we have 𝑅, and so the net sum of the forces in this diagram is 103𝑔 minus 𝑅. That’s equal to the mass of the passenger multiplied by acceleration, so 103 times 1.7. Remember, we’re trying to find the reaction force of the floor to the passenger, so we need to make 𝑅 the subject. Let’s begin by adding 𝑅 to both sides. We could, of course, have multiplied by negative one instead, but we do need 𝑅 to be a positive. Then we’re going to subtract 103 times 1.7 from both sides, and so we get 103𝑔 minus 103 times 1.7 equals 𝑅.

It can be sensible to factor where possible. This makes any mental calculation a little easier. And the left-hand side then becomes 103 times 𝑔 minus 1.7. But, of course, 𝑔 is equal to 9.8, and so since 9.8 minus 1.7 is 8.1, we find 𝑅 must be equal to 103 times 8.1. That’s 834.3. We have been working with kilograms and meters per square second, and so the units for our force are newtons. The reaction force of the floor to a passenger of mass 103 kilograms is 834.3 newtons.

Now, comparing this reaction force to the weight force of the man might remind you of a time where you’ve been in a lift yourself. The weight force 103𝑔 is 103 times 9.8, and that’s 1009.4. 1009.4 is greater than the reaction force, and this actually makes a lot of sense. If you’ve ever been in a lift which accelerates downwards, you almost feel a little bit lighter than normal. Similarly, if you’re in an elevator accelerating upwards, you might feel a little bit heavier than normal. And this is because the reaction force will be greater than the weight force.

In our next example, we’re going to consider how to work with a spring balance.

A body of mass 32 kilograms was suspended from a spring balance fixed to the ceiling of an elevator. Given that the elevator was accelerating upward at 405 centimeters per square second, find the apparent weight of the body. Take 𝑔 to be equal to 9.8 meters per square second.

Let’s begin by sketching a diagram. We have a body of mass 32 kilograms suspended from a spring balance. That spring balance is fixed to the ceiling of an elevator. And so there are a couple of forces at play here. Firstly, there’s the downwards force of the body. That force is the weight, and it’s equal to mass times gravity, so 32 times 𝑔. Now, since the body itself is exerting a downward force on the spring balance, Newton’s third law tells us that the spring balance must also exert an upward force on the body itself. Let’s call that force tension, and actually that tension will give us the reading and the apparent weight of the body.

We know that the elevator itself is accelerating upward at 405 centimeters per square second. But actually, since there are 100 centimeters in a meter, we’re going to divide this value by 100 to get 4.05 meters per square second. And that’s because we’re currently working in kilograms. We’ve got gravity in meters per second, and so our final answer is going to be in newtons. And so we need to be consistent with our units throughout. Now, actually, the spring balance itself will exert a force on the ceiling of the elevator. But we’re not interested in that system of forces. And so we move on, and we use Newton’s second law of motion, force is equal to mass times acceleration.

We’re going to work out the net sum of the forces in our system, and then we’re going to set that equal to the mass of the body times the acceleration. And so we need to choose a positive direction. Let’s take upwards to be positive, since that’s the direction in which the body is moving. And since we have 𝑇 tension acting upward and 32𝑔 acting in the negative direction, the net sum on our system is 𝑇 minus 32𝑔. That’s equal to the mass of the body times acceleration. So that’s 32 times 4.05. We’re trying to find 𝑇 since that will tell us the apparent weight of the body. And so we’re going to add 32𝑔 to both sides. 𝑇 is 32 times 4.05 plus 32𝑔, which is equal to 32 times 4.05 plus 32 times 9.8. That gives us a value of 443.2. The apparent weight of the body then is 443.2 newtons.

In our final example, we’ll look at how we can find acceleration by using a system of simultaneous equations.

A body is suspended from a spring balance fixed to the ceiling of an elevator. The reading of the balance was 50 kilogram-weight when the elevator was accelerating upward at 𝑎 meters per square second, and the reading was 10 kilogram-weight when the elevator was accelerating downward at five-thirds 𝑎 meters per square second. Given that the acceleration due to gravity is 𝑔 equals 9.8 meters per square second, determine the value of 𝑎.

Let’s begin by sketching a diagram of each part of this scenario. Initially, the elevator is accelerating upward at 𝑎 meters per square second. The reading of the balance at this point was 50 kilogram-weight. Try not to worry too much about the units here. Kilogram-weight is just another way of measuring force, and so we simply write 50 as the reading on our balance, and that’s essentially the upward force. That upward force is a reaction to the downward force of the weight of the body. Now we don’t know the mass, so we’ll let that to be equal to 𝑚. And the downward force is mass times gravity; it’s 𝑚𝑔. We then have a reading of 10 kilogram-weight when the elevator is accelerating downward at five-thirds 𝑎 meters per square second.

The downwards force of the weight 𝑚𝑔 remains unchanged. And so now we have the relevant forces. Let’s use Newton’s second law, force is equal to mass times acceleration, taking upwards to be positive in our first diagram and the net force is 50 minus 𝑚𝑔. That’s equal to mass times acceleration 𝑚𝑎. Then taking the downwards direction to be positive in our second diagram, the sum of the forces is 𝑚𝑔 minus 10. That’s equal to mass times the new acceleration, 𝑚 times five-thirds 𝑎. Note that we could have chosen upwards to be positive. It really doesn’t matter. We’ll get the same result.

We rewrite the right-hand side as five-thirds 𝑚𝑎, and the question tells us that we’re trying to find the value of 𝑎. So we’re going to make 𝑚 the subject in each of our equations, and then we can set them equal to one another. In our first equation, we add 𝑚𝑔 to both sides, and then we factor. So we get 50 equals 𝑚 times 𝑎 plus 𝑔. Finally, we divide by 𝑎 plus 𝑔, and so 𝑚 is 50 over 𝑎 plus 𝑔. In our second equation, we subtract five-thirds 𝑚𝑎 and add 10. We factor once again, and we’re going to divide by the expression inside our parentheses. 𝑚 is 10 over 𝑔 minus five-thirds 𝑎.

Now that we have two equations for 𝑚, we’re going to set them equal to one another. 50 over 𝑎 plus 𝑔 equals 10 over 𝑔 minus five-thirds 𝑎, and then we divide through by 10. We’re going to multiply by both denominators. That gives us five times 𝑔 minus five-thirds 𝑎 equals 𝑎 plus 𝑔. Let’s distribute the parentheses and then make 𝑎 the subject. Now, of course, 𝑔 itself is a number; it’s 9.8. Adding 25 over three 𝑎 to both sides and subtracting 𝑔, and we get four 𝑔 equals twenty-eight thirds 𝑎. But of course, four 𝑔 is four times 9.8, which is 39.2. Our final step is to divide both sides of this equation by twenty-eight thirds. 39.2 divided by twenty-eight thirds is 4.2, and so the acceleration 𝑎 is 4.2 meters per square second.

Let’s now recap the key points from this lesson. In this video, we learned that Newton’s third law of motion says that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction on object A. We also saw that this law is very rarely used alone. It will be used quite commonly alongside the first and second laws for motion.