Hey, guys. So by now, we've got lots of practice with using f=ma when forces act on individual objects. Now we're going to search to see what happens when you have forces in connected systems of objects. This happens whenever you have 2 objects. They're either touching or they could be connected by some rope or cable, something like that. So we're going to get back to this point in just a second. I actually just want to go ahead and start the problem off here because it's very familiar to what we've seen before. Let's check this out. We got these 2 masses here, 35. The 5 kilogram block is pulled by some force. This f is equal to 30. What I want to do is I want to calculate the acceleration, and I'm not going to assume there's any friction or anything like that. So I want to calculate acceleration so I know I'm going to have to use f=ma. But the first thing I want to do is I want to draw free body diagrams. Now there are just multiple objects, so I just draw them for all of the objects. Let's go ahead and get started here. What I'm going to do is I want to call this object A and B. And so in part A, we're going to be calculating the acceleration of A and the acceleration of B. Those are my target variables. So we're going to draw the free body diagrams here. So this is going to be my weight force. I'll call this WA. And is there any tensions or applied forces? I know there's some tension because I have the rope right here. This is T. Now you might be thinking there's an applied force also on this block. But remember that this 30 Newton applied force acts only on the 5 kilogram block. So we're drawing the free body diagram for the 3. It doesn't actually go there. Alright. So then we also know that these are the only two forces they have to cancel. All right. Let's look at the 5 kilogram block. It's going to look similar. We've got the weight force, this is WB. Now we also have an applied force that we just talked about. This is our F equals 30. And now is there tension? Well, remember that this tension acts throughout the rope. One way to think about this is that B exerts a tension on A, and so therefore, via action-reaction, A also pulls back on B with an equal and opposite force. So it's the exact same tension, it just points in the opposite direction. And then of course we have our normal force. And again, if all the vertical forces are canceling then they have to be equal to each other. This is mvg as well. So there's our free body diagram. Alright? So now what we have to do is choose the direction of positive. The easiest way to do this is to think about where the system is actually going to accelerate. So for instance, we're pulling on this block with 30 newtons and if there's no friction or anything like that, then that means it's going to accelerate this way as well. So that's where we're going to choose our direction as positive. It's usually going to be the same thing as your acceleration. And now we're going to get into our f=ma. So we're going to write out f=ma, but we're going to start off with the simplest object. What that means is the one with the fewest amount of forces on it, fewest number. So for example, this one has 4 forces, but this one only has 3. So we're going to start off with a 3 kilogram block here. So we want to write our f=ma. And now remember, this system is only going to accelerate along the x axis which means we're just going to be looking at all the x forces. And this equals to mass times acceleration of A. This is what we're trying to solve so we're just going to expand out our forces. Right? Our sum of all forces. There's really only one, it's the tension force and it follows the same rules. It goes along our direction of positive, so it's a positive sign. So this tension equals mass of A, and I can just replace the values that I know. This is equal to 3 times A. And remember if we're trying to solve for the acceleration then I'm going to actually need the other variable which in this case is tension. But I don't know anything about the tension. Right? It's unknown. So what happens is I've gotten stuck here and this is fine. If we get stuck, we're just going to move to another equation. Now we're not going to do move to the y axis or anything like that because we actually have another object. So we're going to write the f=ma for another 5 kilogram object and again, we're just looking at all the x forces. So we want to solve for this acceleration here. Remember these are our 2 target variables. So now I just expand all of my forces in the x axis. That's my force and tension. My force points a
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6. Intro to Forces (Dynamics)
Forces in Connected Systems of Objects
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