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

# Lesson Video: Newton’s Second Law of Motion Physics

In this video, we will learn how to apply Newton’s second law of motion, 𝐹 = 𝑚𝑎, to the find the accelerations caused by forces that act in various directions.

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### Video Transcript

In this video, we will learn about Isaac Newton’s famous second law of motion and apply it to find the accelerations caused by forces that act in various directions. So let’s start by first understanding what a force actually is. Often intuitively thought of as a push or a pull, a force is an interaction that causes a change in the motion of an object. Of course, this assumes that the force in question is not balanced or cancelled out by another force. But we’ll get to that momentarily.

So let’s now see how a force specifically causes a change in the motion of an object. This is when Newton’s second law of motion comes into play. Let’s first consider a block of, say, wood. Let’s say that this block of wood is sitting on the floor and that this block can slide about freely on the floor. Let’s also arbitrarily say that the mass of the block of wood is 𝑚.

Now this block of wood with mass 𝑚 is currently just sitting on the floor basically doing nothing. However, let’s imagine that we now push the block of wood from the left. In other words, we exert a force — let’s call this force 𝐹 — on the block of wood. Well, that force exerted on the block of wood is going to result in an acceleration of the block of wood. In other words, the block is going to stop being stationary and it’s going to start moving in the direction of the force. This is very important. And let’s say that the block of wood will have an acceleration, which we will call 𝑎.

Well, in this situation, Newton’s second law of motion will tell us the relationship between the force 𝐹, the mass 𝑚, and the acceleration 𝑎. The relationship is that the acceleration experienced by the block of wood, 𝑎, multiplied by the mass of that block of wood is equal to the force exerted on the block of wood. In other words, 𝐹 is equal to 𝑚𝑎. And this is Newton’s famous second law of motion. In other words, it gives us the relationship between the force exerted on an object, the mass of the object, and the acceleration experienced by the object due to the force.

However, there is one thing that we need to be extremely careful of here. The force in question is actually the net or overall force on the object. In this scenario where we considered our block of wood sitting on the floor, we were only exerting one force on the object. That force was force 𝐹. And so the net or overall or resultant force on the object was just the force 𝐹.

However, if we now say that the force we exert on the block of wood is called 𝐹 subscript one and we also introduce the concept of friction, in other words, the fact that the floor will exert a force on our wooden block in order to try and oppose the motion of the block. So let’s say that the fact that the block of wood and the floor are in contact with each other and the block of wood is trying to move this way to the right as we’ve drawn it. Then the floor is going to exert another force on the base of the block of wood to try and oppose this motion. Let’s call that force 𝐹 two. And this is a friction force.

Well, in this case, we can see that now we’ve got a large force to the right on the block, which is 𝐹 one, and we’ve got a slightly smaller force, 𝐹 two, to the left. So in this case, what’s the relationship between the mass of the wooden block, the acceleration of the wooden block, and the forces acting on the wooden block, 𝐹 one and 𝐹 two?

Well, Newton’s second law of motion tells us that 𝐹 is equal to 𝑚𝑎. But like we said earlier, 𝐹 has to be the net or overall or resultant force on the object. In other words, let’s say we were trying to calculate the acceleration of the block. What we’d have to do is to rearrange the Newton’s second law equation by dividing both sides by the mass 𝑚, which would give us 𝐹 divided by 𝑚 is equal to 𝑎. In other words, the force — we’ll come back to that in a second — divided by the mass of the block is equal to the acceleration experienced by the block.

So what exactly is the force that we’re talking about? Is it the force 𝐹 one? Or is it the force 𝐹 two? Well, actually, it’s a combination of the two. Like we said earlier, we need to consider the net force on the block. And the way to do this is to add the forces together as vectors. In other words, the force 𝐹 one we can say is acting towards the right on our object, whereas 𝐹 two is acting in the opposite direction.

Well, if this is the case, then the resultant force on the object is going to be something like this, where the size or magnitude of that force is 𝐹 one minus 𝐹 two. This is because 𝐹 two is acting to oppose the force 𝐹 one. And so, overall, the object is going to experience a force 𝐹 one minus 𝐹 two to the right. And that is the force 𝐹 that we’re talking about in this equation. It’s the net or resultant force. And so we’re actually best off placing a subscript net in this equation. This way, we’ll remember that Newton’s second law deals with the net force on an object and that that net force is equal to the mass of the object multiplied by the acceleration experienced by the object.

Of course, if there’s only one force acting on the object in question, then the net force simply is going to be that force. However, if we’re dealing with multiple forces, then we have to add all of those forces acting on the object as vectors, in other words making sure we account for the direction in which they’re acting. And hence, we find the net force on the object, which allows us to use Newton’s second law of motion.

It’s also worth quickly noting by the way that the forces acting on an object don’t necessarily have to be acting on the same line of motion. Because the situation that we’ve drawn here we’ve got a force acting to the right and a force acting to the left.

It is, however, entirely possible that we could place another force acting upwards, for example, if somebody tries to lift the block of wood off the floor. Let’s call that force 𝐹 three. Well, in that case, what we’d have to do is to add all three of the forces as vectors. Let’s simplify our wooden block just to a little blob and think about the force 𝐹 one that’s acting to the right, the force 𝐹 two that’s acting to the left, and the force 𝐹 three that’s acting upwards.

We can start by vectorially adding 𝐹 one and 𝐹 two so that the resultant force is something like this, which is 𝐹 one minus 𝐹 two, to the right. And then we can combine this force, 𝐹 one minus 𝐹 two, with 𝐹 three, the resultant of which is going to look something like this. And to find out the size of that blue vector, we would have to use Pythagoras’s theorem, with 𝐹 one minus 𝐹 two as one of the shorter sides and 𝐹 three as the other shorter side. But the point is that we need to consider all of the forces and all of the directions in which all of those forces are acting in order to find the net force on an object. This is because Newton’s second law of motion deals with the net force on an object.

So now that we’ve looked at Newton’s second law of motion in some detail, let’s get a little bit of practice using this law by looking at a couple of example questions.

How much force is applied to an object of mass five kilograms that is accelerated by that force at a rate of two metres per second squared?

So in this question, we’ve got an object which has a mass — let’s say 𝑚 — of five kilograms and is accelerated — let’s say it’s accelerated to the right — because we can randomly choose a direction. And we’ve been told that this acceleration, which we’ll call 𝑎, is two metres per second squared. We’ve been asked to find the force applied to this object.

Now because the acceleration we’ve said is to the right, this must mean that the force exerted is also to the right, since the force and acceleration must be in the same direction. And we’ll call this force 𝐹.

Now to find the value of this force, we need to recall Newton’s second law of motion. This law tells us that the net force on an object, 𝐹, is equal to the mass of that object multiplied by the acceleration it experiences. So with this equation, we’ll be able to find the net force on the object. And this is exactly what we’ve been asked to find. Even though the question doesn’t necessarily say find the net force, we’ve been asked to find how much force is applied to an object.

Now this may imply that there’s only one force acting on the object, in which case that force is the net force anyway. Or there could be multiple forces acting on the object, but the resultant force or the net force is going to be a combination of all of those forces, whilst accounting for the direction in which they act. And that overall combination can also be thought of as the amount of force applied to an object. And so we can use Newton’s second law of motion to find our answer to this question.

We can therefore say that the force on the object, the net force, is equal to the mass, which is five kilograms, multiplied by the acceleration of the object, which is two metres per second squared. Now taking a quick look at the units, we see that we’ve got kilograms and metres per second squared. And those are the base units of mass and acceleration, respectively. Therefore, when we find our final answer for the force, it’s going to be in its own base unit as well.

Now the base unit of force is the newton, which is equivalent to a kilogram multiplied by metres per second squared. And so when we evaluate our answer — that’s five kilograms multiplied by two metres per second squared — we find that the force is 10 kilograms metres per second squared or 10 newtons. And hence, our answer to this question is that the force applied to this object is 10 newtons.

Okay, so let’s look at another question where, this time, we have to consider multiple forces acting on an object.

A swimmer of mass 45 kilograms uses her legs to push herself away from a swimming pool wall, applying a force of 280 newtons. The water that the swimmer accelerates through applies 160 newtons of force in the opposite direction to that in which she accelerates. What is the swimmer’s acceleration through the water?

Okay, so in this question, we’ve got a swimmer moving through some water. So here’s our swimmer. And we’ve been told that she uses her legs to push herself away from a swimming pool wall. So let’s say that this is the wall that she’s pushing off from. And just for fun, we can draw in the surface of the water.

Anyway, so we’ve been told that the swimmer firstly has a mass of 45 kilograms. Let’s label her mass 𝑚 and say it’s 45 kilograms. We’ve also been told that the swimmer pushes herself off the wall, applying a force of 280 newtons. In other words, the swimmer is exerting a 280-newton force on the wall. But what this means is that we can recall Newton’s third law of motion.

What Newton’s third law of motion tells us is that if object A exerts a force on object B, then object B exerts an equal and opposite force on object A. Now in our situation here, the swimmer is object A. It’s exerting a 280-newton force on object B, which is the wall. Therefore, Newton’s third law of motion tells us that the wall, object B, will exert an equal, 280 newtons, and opposite, to the right as we’ve drawn it, force on the swimmer, 280 newtons.

Now this is important because, as well as this 280-newton force on the swimmer to the right as we’ve drawn it, we’ve been told that the water exerts a 160-newton force on the swimmer in the opposite direction to that in which she accelerates. Well, the swimmer is accelerating to the right because she’s trying to push away from the wall. And of course, once again, we’ve just arbitrarily drawn the wall to the left of the swimmer. We could’ve equally drawn the swimmer going in the opposite direction and the wall to the right. But the point is that the swimmer is trying to accelerate away from the wall.

And so we can say that her acceleration is to the right as we’ve drawn it. And we’ll call this acceleration 𝑎. But then the water we’ve been told exerts a 160-newton force on the swimmer in the opposite direction to which she’s accelerating. This means that there are now two forces acting on the swimmer: the 280-newton force to the right because of the wall and 160-newton force to the left because of the water.

We can simplify all of this by drawing the swimmer just as a blob and considering that there’s a 280-newton force to the right and a 160-newton force to the left. Then we can account for these two forces and find the resultant or overall force on the swimmer. This overall force is also called the net force. And if we consider arbitrarily that the direction to the right is positive and the direction to the left is negative, then we can add these two forces together.

And the net force, which we’ll call 𝐹 subscript net, is equal to the positive 280 newtons, cause that force is to the right, minus the 160 newtons, which is to the left. And that net force, therefore, ends up being 120 newtons. Therefore, it’s a 120-newton force to the right. And this kind of makes sense. If there’s a 280-newton force to the right and a 160-newton force to the left, then 160 of those 280 newtons to the right are being cancelled by this force to the left. And whatever remains — that’s the 120 newtons — is going to be the resultant force to the right.

If, however, we had arbitrarily chosen the left to be positive and the right to be negative, then we’d have found the net Force, 𝐹 subscript net, to be the positive 160 newtons minus the 280 newtons. And that would’ve given us a net force of negative 120 newtons. So what gives?

Well, the reason for this is because we chose this way, to the left, to be positive. And hence, what we found is that the net force is negative 120 newtons to the left. Or in other words, it’s 120 newtons to the right, exactly as what we found here. So even though we do need to be careful with signs, regardless of which convention we chose, we’d find the net force to be 120 newtons to the right.

So now that we found the net force on our swimmer, we can recall Newton’s second law of motion. This law tells us that the net force on an object is equal to the mass of that object multiplied by the acceleration it experiences. And since we’ve already found the net force and we’ve been given the mass of the swimmer, we can find out the acceleration experienced by the swimmer.

We can do this by rearranging the equation by dividing both sides of the equation by the mass, which results in the mass cancelling out on the right. This way, what we’re left with is that the net force on the swimmer divided by the mass of the swimmer is equal to the acceleration experienced by the swimmer. So we can say that the acceleration, 𝑎, is equal to the 120-newton net force to the right divided by the mass of the swimmer, which is 45 kilograms.

We can also see that we’re working in base units because the base unit of force is the newton and the base unit of mass is the kilogram. And so our answer for the acceleration is going to be in its own base unit, which is metres per second squared. And so when we evaluate this fraction on the right-hand side, we find that the acceleration is equal to 2.6 recurring. And the unit is metres per second squared.

Now because the smallest number of significant figures that we’ve been given in the question is two significant figures, we can give our final answer to two significant figures as well. Now 2.6 recurring is basically two point and then six occurring many times. And the reason we write it like this is so that we can work out that this is the first significant figure and this is the second.

Well, the second significant figure is either going to stay the same or it’s going to round up, depending on what the next significant figure is. The next one is a six. Since six is larger than five, this second significant figure is going to round up. And so our value becomes 2.7 metres per second squared. So at this point, we have our final answer. The swimmer’s acceleration is 2.7 metres per second squared.

All right, so now that we’ve had a look at a couple of examples, let’s summarise what we’ve learnt this lesson. We’ve learnt that the net force on an object is given by multiplying the mass of that object by the acceleration experienced by that object. It’s important that the force we’re discussing is the net force. And it’s also important to remember that because the net force and the acceleration are both vectors, we need to account for the directions. The direction in which the net force on an object acts is the same as the direction in which the object will accelerate. And that is Newton’s second law of motion.