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Systems Connected By Pulleys

Patrick Ford
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Hey, guys. So now that we've gotten some practice with forces and connected systems of objects in this video, I want to show you how to solve problems where you have these things connected by pulleys and basically the differences. Instead of everything moving along the same direction like this, you're gonna have one object that hangs or something like that. So basically, it's gonna go up and down, and the other object is going to a side to side, so these things aren't gonna move in the same direction. We actually saw these problems using the same list of steps that we have before. Let's go ahead and check this out. So we've got this 4 kg block here. It's connected to a 2 kg block and hanging. We want to calculate the acceleration of both blocks. So that's part A. Now, if we want to calculate the acceleration, remember, we're just going to calculate the acceleration of the system just like before. If you have these things connected to each other, then they have the same acceleration and the same velocity. So it's the same principle as we've been dealing with so far. So if we want to calculate the acceleration. We just stick to the steps, and the first thing we have to do is draw the free body diagram for all the objects. So let's go ahead and get started. We've got this object like this. I'm going to call this one a and this one B. And so I got the weight force. So I'll call this a. Now I've got any applied forces or tensions. There's a rope, which means there's gonna be attention that's pulling it to the side like this. And then there's nothing else and there's a normal force. So we've got a normal force like this. And again if this block is only going to slide along the table like this and there's only two vertical forces and they have to cancel, this is R M A G. All right. Now let's look at the hanging block. The hanging block is going to have a weight force. This is gonna be W B. And now we've got attention for us that because of this cable here, so this tension actually points up. So what happens is for massless pulleys. What's important is that the tension on both of these objects points in different directions. And even though it might not look like it's an action reaction pair, it actually is. This hanging block B exerts attention on A that pulls it to the right. And so because of action reaction, A also pulls on B. And it's the same magnitude, all right, so that tension force is actually gonna be the same. All right, so that's it for our free body diagrams. Now we just determine the direction of positive Now again. Before, this was pretty straightforward. If everything was moving to the right like this, then you would just assume the direction of positive is like that. And so now what happens is we have different objects that are moving in different directions. So what happens is to determine the direction of positive. If you have one object that's hanging in the direction of positive is usually the direction to the hanging object will fall. What do I mean? I mean that this object here if we just if we have no friction and we just basically let it go, it's going to start falling like this. So this is going to fall like this. That's gonna be our direction of positive and because of the pulley, what happens is that the the block on the table is going to start accelerating to the right like that. And so this is gonna be our direction of positive. So we have two different directions of positive. But one way you can think about this is that anything that goes to the right and down is going to be a longer direction of positive. All right, so that's our direction of positive. Now we're gonna write f equals m A. We're gonna start out with the simplest object with one with the fewest amount of forces. And that's the 2 kg block. So let's get started here. So we've got our f equals M A. Now again, we're just gonna look at all the forces are acting on the Y axis, so this equals mass times acceleration. Want to figure out the acceleration so we got to expand all of our forces. Now, remember, the rule forces along your direction of positive are written with a plus sign. But remember, our direction of positive here for this hanging object is downwards, which means that MBG actually gets written with a minus sign. So I'm sorry with a plus sign. So MBG is gonna be written with a plus sign and your tension force, which goes against your direction of positive, is written with a minus sign. So this is equal to mass times acceleration. Now, just replace the values that we know. So we know this is gonna be too. This is 9.8 minus. Tension equals to a So what I get is 19.6 minus tension equals to a and I can't go any further because I want to figure out the acceleration. But I don't know what the tension forces. So I've gotten stuck and this is fine. When I get stuck, I just moved to the other object. So I've got this 4 kg block. Now we're going to use f equals m A here. But unlike the, unlike the 2 kg block, this block is just moving horizontally. So we're really just adding up all the, uh, the ex forces and this is equal to mass times acceleration. We only have one X force. It's really just the tension. So our attention is equal to mass times acceleration. And so that means T is equal to four A. So again, I want to figure out the acceleration. But I'm missing the tension, and I've gotten these two equations with two unknowns. Remember, I'm just gonna write them out, so I'm just gonna, like, number them like this, draw a box around them. So I got my two equations with two unknowns here, and that brings us to the fourth step, which is we have to solve our A using equation addition or substitution. I'm gonna use Equation Edition now because I feel like it's the easiest one. So equation addition here. And so what I'm gonna do, remember, I'm just gonna line up these equations top to bottom. So this one, the first one is T equals four a. And then my second equation is 19.6. Minus T equals two A. And so remember, if I line them up like this, then what happens is when I add them straight down, when I add these things straight down, I basically just cancel out the tension force. So I get 19.6 is equal to six a. And so when I solve, I'm gonna get a is equal to 19.6 divided by six, which is equal to 3.27 m per second squared. So that is our acceleration. That's a equals 3.2 27 and that's it for part A. So now what we do is we want to calculate the tension on the string. And now we just basically plug this back into our equations to solve for r the target variables. So if we want tension, we just plug it back into either one of these two equations here to solve for the tension. Now that we know the acceleration, remember, the easiest one is just gonna be the one with the fewest terms. So we'll start off with that one. Tension equals for a tension equals four times three points, 27 and that equals 13.8 Newtons. And that's the answer. So 13.8 again you can pause the video and you can try to solve the tension using this equation, and you're gonna get the exact same number. All right, guys, So let me know if you have any questions. Let's move on