https://youtu.be/9HEz-dew_Z0
PHYS 1101: Lecture Fourteen, Part Two
Here’s our brief summary of our previous lecture. Last lecture, we focused on the analysis of and the meaning of Newton’s Third Law. Okay. The f equals ma steps that we’ve been applying to all of these problems, always our starting point to solve these problems, those steps are a reflection of Newton’s Second Law. Rather, these steps help us think through the logical consequence of Newton’s Second Law.
That law only applies to a single object. We do have situations, though, where objects are interacting. We can only apply Newton’s law to each object separately, but then we need to have some information, or what can we say about how multiple objects interact? What are the forces between objects? And information about this comes from Newton’s Third Law.
When do you need this? You’re going to need it if ever you’re asked about the size of the force between objects. For example, the little cartoon on the left. Maybe you have blocks connected with a rope over a pulley, and you’re asked about the tension in that rope. Well, that’s asking about a force that effectively is the action-reaction pair between this object and this object. So you have to consider these two objects separate.
Or further, another example, perhaps you’re asked what’s the force that the bottom box here is providing to hold up Box 2. Again, it’s a force between objects, so you have to treat the objects on both sides of that contact point separately and apply our second law steps separately to each of them.
Here’s other examples. This is a case where you need to consider the small block as a separate object from the bigger block. This problem asks how large would this force p have to be in order for the small block to not slide down, in other words, for it to stay stuck with friction and accelerate straight horizontally, as does the blue block.
You’d have to treat these two separately, because, again, the very force that would be responsible for holding this small purple block up would be a frictional force at that contact point between these two objects. So you’re going to need a magnitude or a value for that force, so you’re going to have to analyze a force at a contact between two objects, so you have to treat the two objects separately, blue box separate from purple box.
And just here’s some other examples we saw in the last lecture. So, when you have to treat two objects separately, Newton’s Third Law gives you some useful information about how the forces on either side of that contact compare, this contact between the two objects. The idea is that the size of the force that each object exerts on each other at that contact point is equal and opposite, meaning the force is in the opposite direction. They always have the same magnitude. That’s regardless of how fast the object’s going, or the objects, if they’re accelerating. Regardless of anything, the size of those two forces have to be the same.
And then you do have to also think through, as you solve these problems, how do the accelerations of the two objects compare, because you’re going to be applying a equals the sum of the forces over m. The second law, to each object, you’re going to want to know how the forces compare between the two objects, so the equation for one object is going to have a same force, opposite direction, compared to this equation for the other object. But then also you can always say something about the accelerations of the two objects. So the left side of this equation for each is going to be related.
Here’s our second quiz question that’s about Newton’s Third Law. We have a cartoon here of two objects. In fact, you could consider it three objects: the ground, the horse, and the cart. You have here descriptions of four different forces that are present in this scene, and you’re asked which two forces form an action-reaction pair.
Well, let me point out the key to this is, for every force, clearly identify what’s the object that the force is applied to. Let me highlight that for you. For each of these, it would be — this is your clue. It says, this is a force on the cart. Number 2 is a force on the horse. Three is a force on the road; four, another force on the horse. Identify the object the force is on, but then for every force, what’s the agent that’s responsible for that?
Well, the agent is in this part of the description. Number 1, for example, is the force of the horse. That’s the agent that’s causing this force on the cart. Number 2, the agent is the cart that’s responsible for this force. Number 3, the agent’s the horse. Number 4, the agent is the road.
If you need to find the action-reaction pair, remember what you need to do is switch the order, or switch the role of object and agent. Let me try to emphasize this by some arrows here. When you switch the object and agent, that is the action-reaction pair, because that will be the force. Those two forces then, with object and agent, and then the switch, agent here and object there, those will be the two forces on either side of that contact point for all contact forces, and so those will be the forces on the two objects due to each other that are equal and opposite.