Video: Force Pairs and Newton’s Third Law of Motion

In this video we learn about Newton’s Third of Motion, also called the law of action-reaction, and about force pairs acting on an object.

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Video Transcript

In this video, we’re going to learn about force pairs and Newton’s third law of motion. We’ll find what force pairs are, how they relate to the third law, and why the third law of motion does not mean that nothing ever moves. To start out, imagine that you as the owner and operator of a travelling circus are getting set up in a new town for that weekend set of performances. In particular, you want to figure out positioning and timing for the opening act of one of the circuses biggest draws. The entrance by way of being shot from a cannon that’s falling from a parachute of the great acrobat the Tumbling Tuscani. The important thing to keep in mind is that the circus ring will be surrounded by the front row observers. And depending on where the cannon is dropped from, after firing the Tumbling Tuscani out, you want to ensure that the cannon stays clear of the first row as it descends. To figure out how close your canon can be to that first row as it’s descending with the acrobat inside its muzzle, you’ll need to understand a little bit about force pairs and Newton’s third law.

To get an idea for force pairs, imagine you’ve picked up the loose end of a rope which is attached to a wall. And you begin to pull hard on the rope. If we were to draw a line along the midsection of the rope and sketch in the forces acting at that point on the rope, we would have a force, we could call it 𝐹 sub 𝑝, representing the pull that you apply to the rope. And another force, we could call it 𝐹 sub 𝑤 for the force the wall applies on the rope, acts in the opposite direction. These two forces are a force pair. They go together and act in opposite directions as one another. We can see force pairs almost everywhere we look. If a person is walking along a flat surface, their foot in contact with that surface pushes backward, while the frictional force pushes forward on their foot. When a bullet is fired from a gun, the gun exerts a forward direction force on the bullet. And the bullet presses backward on the gun, causing it to recoil. Even in something as simple as a box sitting on a flat surface, there is a force pair involved. The weight force of the box acts down, while the normal force, the force of the table on the box, acts up. These are all examples of pairs of forces which leads us in to Newton’s third law of motion.

It’s possible to put Newton’s third law into words. It says that, for every action, there is an equal and opposite reaction. In other words, we could say that forces come in pairs. If Newton’s third law of motion seems strange, let’s think of a few examples and see if we can find an exception to this law. Let’s say we do something as simple as holding out our hand and swinging it through the air. When we do, our hand exerts force on the air molecules. We can call it 𝐹 sub ℎ. And the air molecules exert an equal and opposite force pushing back on our hand. What about a car driving along the road, to the left as we see it? The car’s tires exert a force on the road. And the road, through friction, exerts force back on the car’s tires, pushing it ahead. Well what about a ball being thrown into a pane of glass and breaking the pane? For the instant in time that the ball is hitting the glass and has yet to break it, the glass forces back on the ball. And the ball pushes on the glass.

When we first learn about Newton’s third law of motion, it may make it seem as though motion would not take place. After all, for every action, there’s an equal and opposite reaction. When we look out at the world of course, we see that motion does happen. And it does happen in the context of Newton’s third law. The motion that we see happening in the world often comes down to a difference in masses involved where there are force pairs. Let’s look at the example of the car driving on the road to see that more clearly. Instead of drawing a close-up view on the car, let’s zoom out to show how it interacts with the mass of the Earth. After all, that’s what it’s riding on.

We’ve said that the car, powered by its engine and mediated by the asphalt of the road, exerts a force on the earth that points opposite the direction of the car’s eventual motion. And we’ve said that there’s a response force through friction on the wheels of the car that push it forward along the road. So the force forward is what powers the car whose motion we’re used to and can experience. And, believe it or not, the force the car exerts on the road and thereby on the Earth has an effect, though infinitesimally small, on the rotation of the Earth. So when we think of equal and opposite forces acting, that doesn’t negate motion from happening. But it does involve motion in ways that we might not expect. Let’s get some practice working with Newton’s third law of motion through a couple of examples.

A sports car of mass 550 kilograms collides with a truck of mass 2200 kilograms. And, during the collision, the net force on each vehicle is the force exerted by the other. The magnitude of the truck’s acceleration is 10 meters per second squared. What is the magnitude of the sports car’s acceleration?

Solving for the magnitude of the sports car’s acceleration, we can call that acceleration 𝑎 sub 𝑐. We’re told the sports car’s mass 550 kilograms, which we can call 𝑚 sub 𝑐. And we’re also told the mass of the truck that the sports car collides with. That mass is 2200 kilograms. And we’ll label it 𝑚 sub 𝑡. We’re also told that the truck is accelerating after the collision at 10 meters per second squared. We’ll call that value 𝑎 sub 𝑡.

Let’s get started on our solution by drawing a sketch of this situation. Well, unfortunately we’ve had an accident. A beautiful high-performing sports car has run straight into a truck coming the other direction. Based on Newton’s third law of motion, which says that for every action there’s an equal and opposite reaction. If we consider the force of the truck on the car and the force of the car on the truck, we can write that these forces are equal and opposite to one another. Even though these two vehicles experience equal and opposite forces, that doesn’t mean that the collision affects the truck the same way it affects the car. If we recall another of Newton’s laws of motion, the second law, we can recall that this law says the net force on an object equals its mass times its acceleration. If we apply this relationship to our force balance equation, and we write the balance equation only in terms of magnitudes. Then we can say that the mass of the truck times the truck’s acceleration is equal to the mass of the car times the car’s acceleration.

Since we want to solve for the acceleration of the car due to the collision, we divide both sides of the equation by the mass of the car and find that the car’s acceleration is equal to the ratio of the truck’s mass to the car’s mass times the truck’s acceleration, 𝑎 sub 𝑡. Since we’re given 𝑚 sub 𝑐, 𝑚 sub 𝑡, and 𝑎 sub 𝑡 in the problem statement, we can plug in and solve for 𝑎 sub 𝑐. When we calculate this fraction, we find that, to two significant figures, 𝑎 sub 𝑐 is 40 meters per second squared. So even though the force magnitude on the car is the same as it was on the truck, the car’s acceleration is four times greater than the truck’s acceleration, thanks to the fact that the car’s mass is one-fourth the truck’s mass. Let’s look at another example of Newton’s third law.

When starting a race, a 55.0-kilogram-mass sprinter exerts an average force of 635 newtons backward on the ground for 0.560 seconds. What is the sprinter’s speed after the acceleration? How far does the sprinter move during the acceleration?

We can start by recording some of the given data about this example. Our sprinter’s mass is 55.0 kilograms. We’ll label that 𝑚. The sprinter exerts an average force, 635 newtons backward. We’ll call that 𝐹 sub 𝑠. And the duration of this force is 0.560 seconds. We’ll call that value 𝑡. We want to solve first for the sprinter’s speed after the acceleration is complete. And second, we want to solve for the distance. How far does the sprinter move during the acceleration. We’ll label these values 𝑣 sub 𝑓 and 𝑑, respectively.

Let’s draw a sketch of this sprinter exerting this force on the ground. We’re told that as our sprinter runs along, the sprinter exerts a force, we’ve called it 𝐹 sub 𝑠, backward on the ground. And the question is, is this the only horizontal force involved in this picture. Well, we know the sprinter is moving forward. What force causes that? And what’s its magnitude? We can recall Newton’s third law of motion. This third law says that every action has an equal and opposite reaction. In the case of our sprinter, this equal and opposite reaction is the force of friction that presses forward on the sprinter’s foot as the sprinter presses down and back on the ground. Also by the third law, we know that the magnitudes of these two forces are equal.

At this point, we can recall Newton’s second law of motion, which says that the net force acting on an object equals that object’s mass times its acceleration. Looking back at our diagram, we notice that 𝐹 sub 𝑓, the force of friction, acts on the sprinter while 𝐹 sub 𝑠, the force of the sprinter, acts on the Earth. In the horizontal direction, the net force, the only force acting on the sprinter, is the force of friction 𝐹 sub 𝑓. By the second law, we can write that that force of friction is equal to the sprinter’s mass times their acceleration. We can rearrange this expression to solve for acceleration and realize that 𝑚, the sprinter’s mass, is given to us. And the frictional force we know is equal in magnitude to the force of the sprinter on the ground, which is also given. We could solve then for 𝑎. But what we want to solve for in part one is the final speed of the sprinter after accelerating for 0.560 seconds.

Since the acceleration of the sprinter over that time is a constant value, that means that the kinematic equations of motion fit to describe the motion of the sprinter. Looking over these four equations of motion, we want to pick the one that helps us solve for the variable we’re interested in, in this case 𝑣 sub 𝑓, based on the information we have. The first kinematic equation written down can help us. In our case, this equation says that the sprinter’s final speed, 𝑣 sub 𝑓, is equal to initial speed plus acceleration times time. If we assume that the sprinter started from rest, 𝑣 sub zero goes to zero. And the equation simplifies to 𝑣 sub 𝑓 is equal to 𝑎 times 𝑡. We know the acceleration 𝑎 in terms of the frictional force and the sprinter’s mass. And we’re given the time 𝑡 over which this acceleration occurs.

First, what we do is substitute in 𝐹 sub 𝑓 over 𝑚 for 𝑎 in the equation. Next, we recognize that the magnitude of 𝐹 sub 𝑓 is equal to the magnitude of 𝐹 sub 𝑠, the force the sprinter applies to the ground, which is given to us as 635 newtons. With that value plugged in, we then enter the values for mass, 55.0 kilograms, and time, 0.560 seconds. When we calculate this product, we find that, to three significant figures, 𝑣 sub 𝑓 is 6.47 meters per second. That’s the final speed of the sprinter after accelerating.

Next, we want to figure out the distance traveled by the sprinter while this acceleration is taking place. We look again to the list of kinematic equations to help us. And scanning the list, we see that the second equation written is a fit for what we want to solve for. It says that the sprinter’s final speed squared, which we’ve just solved for in part one, is equal to the sprinter’s initial speed squared, which is equal to zero, plus two times the sprinter’s acceleration times 𝑑, which is what we want to solve for. If we rearrange this equation, dividing both sides by two 𝑎, we see that 𝑑 equals 𝑣 sub 𝑓 squared over two times the acceleration 𝑎. And we can recall that 𝑎, the sprinter’s acceleration, we found earlier is equal to 635 newtons divided by the sprinter’s mass, 55.0 kilograms. Similarly, with 𝑣 sub 𝑓, we solved for that value in part one and can plug that number in. When we enter these values on our calculator, we find that, to three significant figures, 𝑑 is 1.81 meters. That’s the distance the sprinter travels while accelerating.

Now, let’s summarize what we’ve learned about force pairs and Newton’s third law of motion. We’ve seen that Newton’s third law of motion, which is also called the law of action-reaction, says for every action, there’s an equal and opposite reaction. This is to say that forces come in pairs. They oppose one another to an equal extent. Even though force pairs balance one another out, when they act on unequal masses, those different masses experience unequal acceleration by Newton’s second law. This is the cause of differences in object motion.

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