Solving Force Problems in Connected Systems of Objects
by Patrick Ford
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Hey, guys. So by now we've got lots of practice with using f equals M when forces act on individual objects. Now we're gonna start to see what happens when you have forces and connected systems of objects. This happens whenever you have two objects. They're either touching or they could be connected by some rope or cable. Something like that. We're gonna get back to this point. Just a second, 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've got these two masses here three and five. The 5 kg block is pulled by some force. This F is equal to 30. What I wanna do is I want to calculate the acceleration, and I'm not gonna assume there's any friction or anything like that. So I want to calculate acceleration, so I know when you have to use f equals M A. But the first thing I wanna do is I want to draw a free body diagrams. Now there's just multiple objects, so I just draw them for all of the objects. Let's go ahead and get started here. What I'm gonna do is I'm gonna call this object A and B, and so in part, A we're gonna be calculating the acceleration of a and the acceleration of B. Those are my target variables. So I'm gonna draw a free body diagrams here. So this is gonna be my weight force. I'll call this W A. And there's any tensions or applied forces. Well, 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 kg block, so we're drawing a free body diagram for the three. It doesn't actually go there. All right, so then we also have a normal force. And what happens is in this system, these things are only just gonna accelerate along the x axis. So that means that these are the only two forces they have to cancel. All right, let's look at the 5 kg block. It's gonna look similar. We've got the Wake Forest. This is W B. Now we also have an applied force that we just talked about this is our F equals 30 and now is their attention. Will remember that this tension acts throughout the rope. One way to think about this is that B exerts attention on a and so therefore, via action reaction. A also pulls back on be with an equal and opposite force. So it's the exact same tension just points in the opposite direction. And then, of course, we have our normal force. And again, if these all if all the vertical forces are canceling, then they have to be equal to each other this MBG as well. So there's our free body diagram, All right, so now that 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 we're, for instance, we're pulling on this block with 30 Newton's, and if there's no friction or anything like that, and that means it's going to accelerate this way as well. So that's we're gonna choose our direction as positive. It's usually gonna be the same thing as your acceleration, and now we're gonna get into our F equals Emma's. We're gonna write out f equals m A. But we're gonna start off with the simplest object. What that means is the one with the fewest amount of forces on it of use number. So, for example, this one has four forces, but this one only has three. So we're gonna start off with the 3 kg block here. So we want to write our F equals Emma. And now remember, the system is only going to accelerate along the X axis, which means we're just going to be looking at all the ex forces and this equals to mass times acceleration of a. This is what we're trying to solve. So we're just gonna expand out our forces, right? Are some 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 A. And I can just replace the values that I know this is equal to three times a day. And remember, if we're trying to sell for the acceleration, that I'm gonna 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 that 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 move to the Y Axis or anything like that, because we actually have another object. So we're gonna write the the F equals M A for another 5 kg object. And again, we're just looking at all the ex forces. So we want to sulfur this acceleration here. Remember, these are two target variables. So now I just expand all of my forces in the X axis. That's my force and tension. My force points along the direction of positive, and then my tension points against so it picks up a minus sign here. All right, So, actually, now I'm just going to replace the values that I know. I know this force, and I know this mass so I can rewrite this as 30 minus t equals M b times A B actually. Have this five here. So this is five a. Alright, So I've got again want to solve the acceleration. But now I also have two unknowns in this equation, so if you take a look at both of these equations here, I actually have three unknowns between them. I've got the tension and then I've got the two accelerations. So what I'm gonna do is I'm going to label these two equations. I'm gonna call this one number one and this one Number two. And so, unfortunately, what happens here is in these two equations. If there's three unknowns, there's actually nothing I can do about this. I can't solve three unknowns with only two equations that would need three equations. So there's actually something really special about these two accelerations, and that's the main point of this video. If you ever have objects that are attached or connected to each other, they could be touching or they could be connected via some rope. Then that means that they move together. They moved together with the same acceleration and the same velocity. So if you have these two objects, right, and you pull one of them, whatever one of them does, they both have to move together. Because if this one were to accelerate more than this one, then that means the rope would have to magically get longer or stretch, and that doesn't make any sense. So what that means is that instead of having to keep track of all these different accelerations, they're actually all the same. A one is 82 It's all the exact same acceleration, and that's what we can do. You just replace it with a single A. The same thing goes for the velocity. You don't have to keep track of multiple velocities. It's all the same V. So what does that mean for our F equals? M A. Well, that actually means that we're not solving for two different accelerations these days are actually the same. So now we have two equations with two unknowns, and we absolutely can solve that. So that's the fourth step. We're going to solve the acceleration. And to do that, we're going to use one of two methods. They'll both get you the right answer. Those two methods are called equation, addition or equation substitution. Again. They'll both get you the right answer. Your professor may have a preference to just double check with them, make sure you're using the one that they want you to use, so you get the right credit. All right, so this is I'm actually gonna show you how both of them work. Just we can see that we get the same exact answer, so I'll show you how Equation Edition works. And basically, what Equation Edition is it's kind of like the name implies you're just gonna be adding these equations. So here's how you do it we're gonna do is you're gonna line up the equations top to bottom, and then you're just going to add them to eliminate the non target variable. So, for instance, we've got our two equations t equals three a. Now I'm gonna line the second equation of top to bottom, which means I want the tensions to line up and I've got five a and then I've got my f here. Actually, I actually already know this is 30. So remember these are my two equations series, so you line them up top to bottom, and now what you do is you literally just add them straight down. So if you add them straight down, what happens is you have a positive tension and negative tension that cancel out and what you end up with is just 30 equals, and then you get three a and five a, which really is just a A. So you can add those A's together and you'll get an acceleration of 3 75 m per second squared. So that's the answer. All right, so now I'm gonna show you the other method just so we can see that we get the same exact answer. So this is equations substitution, and basically, we're doing here is we're taking one equation, and we're substituting or plugging it into another one to eliminate the target, the non target variable. So the way this works is you're gonna write both equations. This t equals three a. Now we've got this second equation f minus. So this is going to be 30 minus t equals five a. All right, so you've got these two equations again. So now we're gonna do is we're gonna take the simple equation, which is really equation number one the fewest terms, and you're gonna plug it in to the more complicated one to eliminate the non target variable. So, again, we're trying to solve for the acceleration here, so we're gonna want to eliminate the teeth. So what we do here is when we write out the second equation 30 minus. We're not gonna use t anymore because we're basically gonna replace tea with three. Remember, though, t equals three A. So we have 30 minus three equals five A. And so now when you move this to the other side, you get 30 equals a and your acceleration is equal to 3. which is exactly what we got before. It's the same exact number. Just slightly different method. Right? So we got these two answers, and they're the same. All right, so that's it for part A. So now we have to do for part B because we want to solve for the tension force. So how do we do that? Well, basically, that brings us to the last step. If you ever trying to solve for other variables, you're just gonna plug your acceleration now that you've solved it into your equations so that you can solve for other target variables. So which equations were plugging into what's basically just the two that we got stuck on? Because now we actually know what one of these. What this acceleration is we can solve for the other variable the easiest way to do this is just to use the simplest equation. So I'm gonna use equation number one. That's t equals three A. But now we know a it's t equals. Uh, sorry. Three times 3.75. And so you get 11.25 Newtons. And that's your final answer. So what do you want to do? So I want you to pause the video, see if you get the same exact answer. If you actually solve it using equation number two, and you should. That's how you do these kinds of problems. Let me know if you have any questions.