Video: Systems of Unbalanced Forces

In this video, we will learn how to apply Newton’s second law of motion and Newton’s third law of motion to analyze systems of forces that produce a net force that is not zero.

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Video Transcript

In this video, we will learn how to use Newton’s second and third laws of motion together in order to analyse systems of unbalanced forces. So let’s start by recalling Newton’s second and third laws.

So let’s start by looking at Newton’s second law. This law tells us that the net force on an object is equal to the mass of the object multiplied by the acceleration experienced by the object. In symbols, we often write this equation as 𝐹 subscript net, the net force on an object, being equal to the mass of the object, 𝑚, multiplied by the acceleration of the object, 𝑎. And it’s worth noting that Newton’s second law doesn’t need refer to the net force on the object, not just any forces acting on the object.

For example, let’s imagine that we’ve got a box here sitting on the floor. Now in this situation, there’s bound to be a few forces acting on the box. Firstly, we know that the box’s weight is going to act in a downward direction. And secondly, we know that there’s going to be an upward contact force from the floor acting onto the box. Now these two vertical forces cancel each other out. So if these are the only two forces acting on the box, then the net force on the box is zero. Even though there are individual forces acting on the box, they are all completely cancelled out by other forces. And hence, because the net force on the box is zero, the acceleration of the box is also going to be zero, regardless of what the mass of the box is.

However, if we were to now add a rightward-acting force, in other words if somebody was to push the box towards the right, then this force we can see is not balanced by another force acting towards the left. And so, in this case, there is a resultant force or unbalanced force acting on the box. And this would even be true if we had a small leftward-acting force on the box. When we say small, we mean smaller than the magnitude of the rightward-acting force. Because now even though there is a leftward-acting force trying to counter the rightward-acting force, the magnitude or size of the leftward-acting force is not large enough. And so, there is a resultant force acting towards the right. And hence, there is a net force acting on the box. And therefore, the box will accelerate in the direction of the net force, which in this case is towards the right. So that is Newton’s second law of motion.

What about Newton’s third law then? Well, Newton’s third law says that when an object — let’s say object 𝐴 — exerts a force on another object — let’s say object 𝐵 — object 𝐵 exerts an equal and opposite force on object 𝐴. Now when we say equal and opposite, we mean equal in magnitude or size but opposite in direction. In other words then, if we think about a box sitting on the floor once again and this time we have a person pushing our box towards the right with let’s say a force which has a magnitude 𝐹. Then because of Newton’s third law of motion, the box will exert an equal in magnitude but opposite in direction force on the person doing the pushing. In other words, on the person’s hands, the box will exert a force of magnitude 𝐹 towards the left in the opposite direction. So that in a nutshell is Newton’s third law of motion.

Now Newton’s second law and Newton’s third law can be used in conjunction with each other. A good example is when we consider the collision between two objects. So let’s say here’s our first object and here’s our second object. Let’s also say that the first object is moving towards the second object, whereas the second object we see to be stationary. Now eventually, the first object arrives at the second object and they collide. And in this situation, let’s say that the second object exerts a force to the left on the first object. And this is the force due to the collision. Let’s call this force 𝐹.

Well then by Newton’s third law of motion, we can say that because this object exerts a force on this object, that means that the first object will exert a force on the second object with the same magnitude but in the opposite direction. In other words, the first object will exert a force towards the right with magnitude 𝐹 on the second object. And since we’re ignoring all other forces, for example, we’re not even considering gravity here, we know that the only force acting on the second object is the force 𝐹. And this means that the force 𝐹 towards the right on the second object is equal to the net force acting on the second object because there are no other forces. And so that overall resultant or net force on the second object is simply the force 𝐹 towards the right.

Then, we can use Newton’s second law of motion which tells us that the net force on an object is equal to the mass of the object multiplied by its acceleration. And in this particular case for the orange object, we know that the net force is 𝐹. And we know that this is equal to the mass of the orange object which we will call 𝑚 subscript 𝑜 multiplied by the acceleration experienced by the orange object which we will call 𝑎 subscript 𝑜. And so now we know that because the blue object collides with the orange object, the orange object is going to accelerate towards the right with acceleration 𝑎 subscript 𝑜. In other words, the acceleration is going to be in the same direction as the net force on the object. And so, that is how we use Newton’s second and third laws in conjunction to analyse objects and in general systems with unbalanced forces acting. Now the best way to learn about this is to attempt a few example questions. So let’s go about doing that.

Three demolition workers push on a wall. The workers push with forces of 80 newtons, 50 newtons, and 60 newtons on the same side of the wall parallel to each other. What is the total force acting on the wall?

Okay, so in this question, we are considering a wall. So let’s say that this is the wall that we’re talking about and we’re looking at it from above. In other words, we’ve got a bird’s eye perspective. Now, we’ve been told that three demolition workers are pushing on this wall. We’ve been told that the first demolition worker — let’s say they are pushing at this point here — pushes with a force of 80 newtons. The second demolition worker we’ve been told pushes with 50 newtons of force. And the third one we’ve been told pushes with 60 newtons. So those are the three forces acting on the wall.

Now the reason that we can draw them like this is because we’ve been told that the forces are exerted on the same side of the wall and parallel to each other. And because they’re all pushing forces, this means that they must all be pointing in the same direction. What direction that is doesn’t matter as long as we draw them pointing all the same way. In other words, we could have drawn the wall like this for example and said that the workers were pushing in this direction. But as long as we drew all three forces acting in the same direction, then it would have been fine. So anyway, we’ve been asked to find the total force acting on the wall. In other words, we need to combine these three forces acting on the wall to find one overall or resultant or net force acting on the wall.

To do this, we simply add up all of the forces, whilst also remembering that forces are vector quantities. We can recall that a vector quantity is one which has magnitude or size and direction. And as we can see from our three labels, each one of the forces has a certain magnitude: 80 newtons, 50 newtons, and 60 newtons. And each one also has a specific direction. In this particular case, they’re all pointing in the same direction which actually makes life easier for us. Because if they’re all acting in the same direction, then to find the total force acting on the wall, we simply add them up.

If, however, one of the forces was acting in the opposite direction, then we would need to make sure that we label these forces correctly. In other words, if we said that forces acting to the right were positive, then any forces acting towards the left would have to be labelled as negative. But in this case, we don’t have to do that. Instead, we can just say that the total force acting on the wall which we will call 𝐹 subscript total is equal to 80 newtons plus 50 newtons plus 60 newtons. And that ends up being 190 newtons. In other words then, this wall having three demolition workers pushing on it with magnitudes of 80 newtons, 50 newtons, and 60 Newtons, respectively, all in the same direction is equivalent to the wall having one net force of 190 newtons acting in the same direction.

Now in this question, we haven’t been told what that direction is, just that the three forces acting on the wall are acting in the same direction. And so, in order to answer our question, we only need to give the magnitude of the total force. And as we’ve seen already, this is 190 newtons. So we’ve just answered the question where we find the net force or total force acting on an object. Let’s now look at another example question.

A piano of mass 300 kilograms is moved along a horizontal surface by a person pushing from one side and another person pulling from the other. The piano accelerates at 0.25 meters per second squared. The pushing force is 120 newtons. And a friction force of 60 newtons acts in the opposite direction to the piano’s velocity. What is the pulling force? Consider the direction in which the piano moves to be the positive direction.

So let’s say that this is the piano that’s being moved and this is the floor. Let’s also say that the piano is moving in this direction, towards the right. And the way that this is happening is that one person is pushing from one side and another person is pulling from the other side. So in the question, we’ve been told that the force exerted by the person who’s pushing — in other words the force exerted on this side — is equal to 120 newtons. And we’ve also been told that there’s a 60-newton friction force acting in the opposite direction to the piano’s velocity.

Based on this information as well as the fact that the piano has a mass of 300 kilograms and the fact that it accelerates at 0.25 meters per second squared, we need to find the pulling force. In other words, we can say that the force exerted by the person pulling is 𝑥 and that is what we’re trying to find here. Now to do this, we can start by recalling Newton’s second law of motion. Newton’s second law tells us that the net force on an object is equal to the mass of the object multiplied by the acceleration experienced by the object.

In this case, the object in question is our piano. And luckily, we already know the mass of the piano, 300 kilograms. And we know its acceleration, 0.25 meters per second squared. So using this equation for Newton’s second law, we can calculate the net force on the piano. In other words then, we can say that the net force on the piano is equal to the mass, 300 kilograms, multiplied by the acceleration, 0.25 meters per second squared. And because we’re working in base units for mass and base units for acceleration, our answer will have the base units of force which is newtons. So when we evaluate the right-hand side of this equation, we find that the net force on the piano is 75 newtons. And that net force must be acting towards the right.

Now the reason for this is that we’ve been told that the direction in which the piano moves is the positive direction. In other words then, the direction towards the right is positive. And therefore, anything towards the left is negative. And in our calculation, we took the acceleration of the piano to be positive. We said it was 0.25 metres per second squared and not negative 0.25 meters per second squared. And based on that calculation, we got a net force of 75 newtons, which is also a positive value. And so, the net force 𝐹 subscript net will also be acting towards the right as we’ve drawn this diagram.

Now the net force on the piano is the overall force acting on the piano when we account for all of the individual forces acting on the piano. In other words then, we can say at 𝐹 subscript net, the net force on the piano is equal to firstly the pushing force, which we will call 𝐹 subscript push, plus the pulling force, which we will call 𝐹 subscript pull, plus the frictional force, which we will call 𝐹 subscript friction. But we need to be very careful here when we substitute in all our values because remember the 60-newton friction force acts in the opposite direction to the piano’s velocity.

And so, the value of 𝐹 subscript friction, the frictional force, is actually negative 60 newtons. And hence, we can say that the net force on the piano is equal to 𝐹 subscript push, the pushing force which is 120 newtons, plus 𝐹 subscript pull, the pulling force which is 𝑥 and that’s what we’re trying to find, plus the frictional force, which we’ve said is negative 60 newtons. And so, even though we’re adding the frictional force, we’re actually adding a negative value. And so, we’ve accounted for the directionality of the frictional force. And so, what we find on the right-hand side is 120 newtons minus 60 newtons plus 𝑥 which altogether ends up being 60 newtons plus 𝑥.

But then, this expression on the right-hand side is equal to the net force on the piano which we’ve already found is 75 newtons. And so, we can say that 60 newtons plus 𝑥 is equal to 75 newtons. And when we equate the two of the equation, it looks something like this. Then, we can rearrange simply by subtracting 60 newtons from both sides of the equation because this way the positive 60 newtons cancel with the negative 60 newtons. And we’re left with just 𝑥 on the right-hand side. And on the left, we’re left with 75 newtons minus 60 newtons. At which point, we arrive at our answer. We find that the pulling force, the force exerted by the person pulling on the piano, is 15 newtons. And that is the answer to our question. So now that we’ve had a look at a couple of examples, let’s summarise what we’ve talked about in this lesson.

Firstly, we’ve seen that an unbalanced force acting on an object means that there is a net force on the object. Well, of course, an unbalanced force means that the overall force acting on the object is not zero. And this is when we’ve accounted for all of the forces acting on an object. Now, if we have an unbalanced force acting on an object or a nonzero net force acting on the object, then the object itself will accelerate according to Newton’s second law of motion. This law tells us that the net force on the object is equal to the mass of the object multiplied by its acceleration. And secondly, we saw that we can use Newton’s second and third laws of motion as necessary. So in other words, we don’t always have to use both. Sometimes, we’ll use one, sometimes the other, and sometimes both. And we can use these laws as necessary to analyse systems of unbalanced forces.

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