1

concept

## Solving Force Problems in Connected Systems of Objects

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Hey guys. So by now we've got lots of practice with using F. Equals M. A. 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 that are either touching. Or they could be connected by some rope or cable, something like that. So we're gonna get back to this point in 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 five 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 I'm gonna have to use F. Equals M. A. But the first thing I wanna do is I wanna draw 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 wanna 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 we're gonna draw the 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 five kg block. So, when drawing a free body diagram for the three, it doesn't actually go there. Alright, So then we also have a normal force. And what happens is in the 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 five kg block, It's gonna look similar. We've got the weight force. This is W. B. Now we also have an applied force that we just talked about. This is our F. Equals 30. And how is their attention? We'll remember that this tension relax. 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 B. 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 is 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 newtons and if there's no friction or anything like that and that means it's gonna 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 M. A. So 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 a three kg block here. So we're gonna write our F. Equals M. A. And now remember the system is only going to accelerate along the X axis which means we're just gonna 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? 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 A. And I can just replace the values that I know this is equal to three times A. And remember if we're trying to solve 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 I've gotten stuck here and this is fine if we get stuck, we're just gonna 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 five kg object. And again we're just looking at all the X. 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. Alright so actually now I'm just gonna 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. Loops 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 gonna 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 move 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 would 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 a two. 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 A's 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 gonna solve the acceleration. And to do that we're gonna use one of two methods. They'll both get you the right answer. These two methods are called equation addition or equation substitution. Again they'll both get you the right answer. Your professor may have a preference. So just double check with them, make sure you're using the one that they want you to use. So you get the right credit. Alright so this is I'm actually gonna show you how both of them work just so 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. What we're gonna do is you're gonna line up the equations top to bottom and then you're just gonna 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 up 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 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 equation 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 gonna 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 the acceleration here. So we're gonna want to eliminate the T. 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 T with three A. Remember the 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 3.75 which is exactly what we got before. It's the same exact number. Just a 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 sulfur other target variables. So which equations are we plugging into? It's basically just the two that we got stuck on because now we actually know what one of these what this acceleration is. So 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 you get 11 points 25 newtons and that's your final answer. So what I want you to do, so I want you to pause the video, see if you get the same exact answer. If you actually solve it using an equation number two and you should, that's how you do these kinds of problems. Let me know if you have any questions.

2

example

## Three Hanging Blocks

7m

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Hey, guys, let's get to this problem here. We've got these three blocks. They're all connected to each other by strings, and we want to figure out the magnitude of the tensions in the top, most string and then the middle string. So what does that mean? So in part A, we're gonna be figuring out some tension force here. Let's go ahead and draw a free body diagrams. That's the first step. So we've got these three objects. I'm gonna call this blocks A B and C, and we've got the free body diagrams for this 2 kg block. Here is gonna be the mass G, which is the weight force. So then we've got another force here that's going to be from this cable or the string that's connected to the ceiling. This is actually what we want to find in this first part Here is this t. However, there's gonna be multiple tensions in this problem, right? We've got a bunch of cables, so I'm gonna call this one t one, all right. And then we've got another tension because we also have another cable that's connected to the bottom of the 2 kg block so that means we're gonna have another tension force. I'll call that one t two, but that's it for this one, right? So there's no friction forces or normals or anything like that. So let's move on to the second block. Now we get the way force, which is MBG, and then we've got the tension force that pulls upwards. However, because of action reaction, remember that the tension in this rope is gonna be the same, uh, throughout the whole thing. So basically, you have this t to that points downwards for block two. It's gonna be the same t two over here, right? So it's gonna be the same tension except just like block to There's another tension force that's on the bottom of block of block B. And so I want to call that t three. Alright, Now for the bottom one here, we've got the weight Force M c G. And then we also got this tension force which is t three against the same exact idea. This is really just an action reaction pair. Alright, so, basically, about these three tensions, t one t two, t three and I want to figure out t one so that brings us to the next step. We just have to figure out the direction of positive. And usually we just choose this to be the direction that the system is going to accelerate. However, we're told in this problem that these blocks are just hanging from the ceiling. What that means is that they're all actually in equilibrium, so the acceleration is equal to zero. So they're not actually accelerating anywhere, which means we can just stick to our normal rules, which is up is positive for all the objects. All right, so let's now get into our free body diagrams. We're gonna start with the one. Usually we would start with the one with the fewest forces, which is going to be this guy down here. However, remember, we're trying to actually look for the tension in the top most strings. So it actually makes more sense for us to start with Block A. So we've got f equals m. I remember these are all just why forces. So this is all the forces in the wider action, however we just said, was that the acceleration was equal to zero. So that means that this thing is gonna be in equilibrium, and that's actually gonna be the same for all the objects. They're all in equilibrium. So that means I've got my t one, which is upwards minus TT, which is down minus my MSG, and that's equal to zero. All right, so that's basically, uh, my equation If I want to figure out this t one here actually can't solve this because I don't know what the t two is. I do know what the mass and gravity is, but I just don't know how those two tensions. So when I get stuck, I just go to another object. Right? So we're gonna go for block B now. Same exact thing. F equals m A. We know the acceleration Zero because they're all in equilibrium. And so he would have got is they've got the t two. That's upwards minus my t three. That's downwards minus my MBG. And this is equal to zero. All right, So if I want to figure out this t two and I also have this unknown t three here And just like this other equation, I've also got two unknowns in this problem, so I'm gonna have to go to the third block now, so I'm gonna have to go to block C and block C is gonna be a little bit simpler, so we know that this is going to be equal to the acceleration zero and we have to. Force is only so we have t three minus M. C g is equal to zero. Notice that this one has a little bit less terms. It doesn't have another tension force. So we get these three tensions that are all kind of like mashed up in these equations. We remember. We want to find what this tension one is. So how do we solve these kinds of problems? We have systems of equations, remember? We're just going to number them right. This is gonna be number one, number two and then number three here and then we're just going to basically solved by using equation, addition or substitution. Now, we're not gonna be selling for a but we're still going to use that step because we're going to see that the tension is going to come out of it. Let's check this out. So we got this. These three equations here remember, you're just gonna line them up top to bottom. So you got t one minus two to minus M. A G is equal to zero. Now we want to do is we want to line up this tension to in equation over Tuesday. It lines up with the T two over here. So you got t two minus three t three minus mbg is equal to zero. And then finally, when I went to line up this t three here with this t three so our t three minus m C G is equal to zero. So we got our three equations here, remember we want to do is you want to line them up and then you want to add them just basically straight down top to bottom so that you cancel out your non target variables. So basically what happens is when you add all these three equations down like this, your t two goes away. Your t three is also going to go away, and what's left is you have t one, which is good. That's what we're trying to solve for So you've got t one minus m a g. These are all negative here. So negative, negative and negative minus m b g minus M C G equals zero. So if you go ahead and move all of this stuff over to the other side, basically what you get is you get an A plus M B plus M c all times gravity. So, really, this is actually just the combined weight of all of the objects that are underneath it. So if you go ahead and solve for this, you're gonna get two plus three plus five, which is 10 times 9.8. And so you get the tension is equal to 98 Newtons. So again, we saw here is that this tension forces basically just the combined weight of all the three blocks here and that's actually important. When when you have a system of objects in which they're all hanging from the ceiling or hanging by multiple ropes or strings, then each one of the tensions in each one of the ropes or cables has to support all the total weight that is underneath it. So, for example, we try to solve for this top tension here. This top tension has to support all of the combined weight that is underneath it. All right, so that's what we got 98 Newtons. All right, so now let's jump into the second part here. We want to figure out the tension in the middle string. Really? This is actually just t two. So what do we just say? The each of the tensions has to support all the weight that is underneath. So really quick Way to do this is you can just look at all the weight that's underneath this, and that's going to be your attention, right? So this is really just gonna be t two equals three plus five times 9.8. And what you get is 78.4 Newton's and this is actually the correct answer. If you really wanted to solve this along with what you could do, you can go ahead and solve it using this equation here, equation number one. Now that you actually know what t one is to, go ahead and pause the video and plug in this t one in for this number. Here, plug in 98 you'll solve for t two, and you'll just get this number over here. All right, so that's it for this one. Guys, let me know if you have any questions.

3

concept

## Systems Connected By Pulleys

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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

4

example

## Atwood Machine

6m

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Hey, guys, let's check out this problem here. We have these two blocks that are connected by a cord that goes over a pulley. This is sometimes called an Atwood machine. So we know the masses of the two blocks. One of the larger one is six and smaller ones for We want to figure out the acceleration of this system of connected objects. So we want to figure out what A is equal to. So, luckily for us, we're just gonna go ahead and stick to the steps, right? We can figure out the we're gonna draw the free body diagrams for both objects. Doesn't matter which one I start off with. So if I call this block A and this one block B, then the free body diagrams for this is going to look like the weight we have weight force like this, then the only other force that's acting on this is gonna be the tension from the cord, right? There's no plaque forces and there's no normal or friction or anything like that. Let's move on to the other one, and it's going to look pretty similar. We have a weight force. I'm gonna call this B, which is MBG and then the only other forces acting on this one is also the tension force. And remember that these are actually action reaction pairs. This cord that connects them with a pulls on B B also pulls on a and the tensions actually both act upwards in this case. All right, so that's our free body diagram. And the next thing we have to do is determine the direction of positive. And so here's the rule. Imagine that these two blocks are kind of just hanging there. We know that the right one, the six is heavier. So we know that when you release this thing, one is gonna have to go up and the other one is going to have to go down. So which one is it? We can kind of just guess or predict that the heavier one has more weight pulling on it. So this is the one that's gonna go downwards. So the six is gonna go down, the four is gonna go up, which means that the direction of positive is usually going to be the direction that the heavier object is going to fall. So what that means is that our 6 kg object is going to go down like this, that we're going to choose our direction of positive. But because this pulley goes up and over than if the six goes down, the four is going to go up. So we're gonna choose the upward direction. To be positive for the 4 kg objects or direction of positive is really just the clockwise direction here. That's really important. So now we're gonna write f equals M, and we're gonna start with the simplest object. But what's lucky for us is that we have both objects that are relatively simple. We only have two objects or two forces. So I'm going to start off with object A here, and I want to calculate. I want to use my F equals mass times acceleration. These are all the forces in the Y axis. Now, remember, what we said was that for a anything that's upwards is going to be positive, which means when we expand out our forces, we have the tension that goes in our direction of positive for block four for Block A. So that means that's going to be positive. And then our MSG points downwards against our positive direction. So it's gonna be minus and this is gonna be m a times a. So now we can replace the values that we know. We know 10. So we have tension minus. This is gonna be four times 9.8 and this is going to equal four times a So I could just simplify real quick. I've got four t minus 39.2. Whoops equals for a So I can't go any further because I want to figure out the acceleration. But remember, I don't know what the tension is and it's okay if I get stuck. I'm just gonna go to the other objects. So this is gonna be blocked. Be so I want to write f equals M A for Block B now. So remember now the rule for Block B is that the downward direction is going to be positive. So that means that for Block B, anything that points downwards is going to be our direction of positive. So, for instance, are MBG is actually going to be the positive one here, and our tension points against our direction of positive, So it's gonna be minus, so it's really important that you follow those rules. So we just replace all the values that we know. We know this is six times 98 minus tension equals six times a So this ends up being 58.8 Minus T equals six A. So, again, I can't go any further because I have this acceleration, But I still need the tension force. So predictably, I've gotten two equations with two unknowns. Usually what happens to these problems? So I'm going to box them, and I'm gonna call this one equation number one, and this one equation number two embraces the fourth step. We want to solve this acceleration. We just have to use either equation, addition or substitution. And I'm just gonna pick addition. So basically, what we've got here is we've got one, which is t minus 39.2 equals four A. Now what I want to do is I want to line up the tension right from the equation number two so that we can add them straight down. So this is gonna be 58. 8 looks a little funky, but that's because I'm trying to line up these variables, so I've got these two equations right here. And remember, you're just going to add them straight down. Just add them straight down like this. And what happens is you're gonna cancel out these tensions when you add them. So when you add them straight down, you get 58.8. Remember, this is a minus 39.2, so this is minus 39.2 equals and then four plus six is going to be A. So this turns out to be 19. 6 equals 10 A. And that means your acceleration is 1.96 m per second squared. So let's talk about the direction we got a positive number for acceleration. That just means that our acceleration points in our direction of positive. So that means that a is going to be 1.96 downwards, just exactly like we guessed. All right, that's it for part A. Now we want to move on to Part B, and that's just calculating the tension. And we know if we want to calculate other variables, we're just gonna have to plug our A back into the our equations and then to solve for other targets, right? So we want to figure out we're just going to use one of these equations to solve for T. They both have the same amount of terms. But the thing is that this tea is actually positive in this equation and this one is negative. So this one is slightly more simple. So that means I'm gonna use t minus 39.2 equals four. And then now we know a right. So in T minus 39.2. Actually, I'm just going to do this the other side so t equals four times 1.96. That's what we figured out here, plus 39.2. And if you go ahead and work this sound, you're going to get 47.4 Newton's and that's the answer. That's it for this one. Guys, let me know if you have any questions.

5

concept

## Shortcut for Solving Connected Systems of Objects

5m

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Hey guys. So we've already seen how to solve these kinds of problems where you have connected systems of objects. In fact, we've already solved these exact problems before in previous videos, which is why I have them written out already. So what I wanna do in this video, I want to show you a very powerful shortcut that you can use to solve the accelerations in these kind of system problems. Let's go ahead and check this out. So we've already seen this problem again, we had a three and a five kg block being pulled by a force and to solve this, we had to write F equals that. We had to draw free body diagrams. We had to write F equals M. A. For both. Come up with two equations. Then we had to use equation addition or substitution. And then you finally could figure out that the answer was 3.75. I'm gonna show you how to do this in 20 seconds. And basically the shortcut is that we're going to combine all of these smaller masses like M A M. B. And we're basically just gonna imagine there's a single large object and combine all the masses together. For example, we have these 23 and five kg blocks, but I'm just gonna imagine now there's a big block that's eight. And what happens is when you imagine this, there's only one force that's pulling on this block, which is F equals 30. So you might be wondering what happens to these tension forces that were connecting the two objects and basically they go away. So when you do this, you're gonna ignore any tensions or normals between the two objects or three objects, whatever There's So basically any connecting forces. So these two tensions basically is that they don't exist anymore. And the only force that is external, the one that's not between is the 30. So now what we do is write F. Equals M. A. But now we're gonna right F. Equals big M. Times A. So the only force that's acting now is the 30 and this equals A. And so A. Equals 3.75 m per second squared, which is exactly the same answer that we got when we did it the long way. So super powerful shortcut. When you're solving for the acceleration, let's check out the other example here again, we've already done this one before. We have this four kg block the two kg books hanging. And then we want to calculate the acceleration. So again, we had to draw the free body diagrams for both. Come up with the F. Equals M. A. For both through the equation addition for both. And then finally you figure out the answer. So instead what you can do is you can just lump these both together into a single object. This is four. This is two, which means the large object is actually gonna have a massive six. And so now what happens is again the tension forces between those two objects are going to go away. So what's the only external force that's really acting on this? It's actually just the weight force that was on object B. Remember that the weight force of A is canceled out by this normal force. So basically as if you have one block that's being pulled down by a weight force of MBG. So even though we combine the masses together into a single six kg block, it's as if the weight force is really only still from the two kg block, that's acting all right. So then we do our F equals M. A. This is F equals big M times A. So remember this is going to be M. B. G. Now, just like we did before, remember how we considered anything to the right and down as positive. It's the exact same thing here. Anything to the right and down is gonna be positive. So that means our MbG is positive here, and this equals big M times A. So this is gonna be two times 9.8 equals six. A. This is 19.6 equals six A. So we get A. Is equal to 3.27. And that is exactly the same answer that we got before. Let's do these next two examples really quickly. So I can show you again how this works. So now we're gonna calculate this acceleration of this three block system. So we wanted to do this, we have to draw three free body diagrams. We have to write F equals M A. Three times. We have to get three equations do equation edition and all that stuff. And instead what you can do is you can just lump these to get these together into a single box of 10, Right? So it's two plus three plus five is 10. And so now it's as if the only force that's acting is this 40. All the tensions that are between these blocks, right? There's tensions between the two and the three, the three and the five. All of those go away. So now what happens is we just write our F equals big M times A. There's only just the 40 equals 10 A. And so your acceleration is four. That's the answer. Now, this last one here, remember this is like an Atwood kind of machine here. And what happens here is that you have to wait forces. Remember this is gonna be mbG really? This is going to be like six times 9.8. And then you also have this one which is M A G, which is four times 9.8. And then you have the two tension forces. There was attention up and attention up like this. Well, again, if you just combine them into a single object, it's as if you have one block that's actually 10. And what happens is the tension forces will go away. And the only to external forces are really just these weight forces. Remember that the rule that we use is that anything up to the right and down was going to be positive. So you actually have these two weight forces that are kind of like pulling against each other. It's the exact same thing here. Anything up to the right and down is going to be positive except when you replace this into one big object, it's kind of as if you have the weight force which is MbG which we know is six times 9.8 is 58.8, which is pulling it down. But then on the other side you have this upward force which is M. A. G. Which is really just 39.2. That's pulling it kind of up. Right? Again, we're just kind of just imagine this is a single object, that's kind of what it works out to. So now when you use your F equals big M times A. Now you have to external forces, remember anything that's too, that's downwards is going to be positive, anything upwards is gonna be negative. So what happens here is that you have 58.8 minus 39.2 equals 10 A 19. equals 10 A. And you get a equals 1.96 which is exactly what we got when we work this out the long way, That's it for this one guys, let me know if you have any questions

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example

## Pulling Two Connected Blocks Upward

3m

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Hey guys, we got these two blocks and they're connected by a string like this. But this whole system here is actually being pulled upwards by a rope like this and you know, this forces 100 newtons. So we've got the masses of both blocks. But we want to do is we want to calculate the acceleration of the blocks in the first parts. So this is really gonna be A. And in the second part we're gonna calculate some tension. So we want to calculate the acceleration. We know we can use this shortcut here. So I've got this two kg block, this three. So I'm gonna call this one A. And B. And what I want to do is is really really quickly draw free body diagrams for both of these. So I got the M. A G. That's the weight force that I've got the T rope, that's the tension force in the rope. And I actually, no, that's 100. And what I've also got is this tension force in the connecting string and I'll call this T. S. Now for the bottom, for the bottom object. I've got the weight force MBG. And then I've got the upward force of the tension and connecting string. So we can use the shortcut to find the acceleration. All we're doing really is we're just replacing this block with these two blocks with 15 kg block, right? By adding the masses together. And then we're just gonna ignore any of the internal or the connecting forces like the tension in the string. This one doesn't get ignored because it's not actually between the objects. Remember that. Right? So basically what happens is is that this one block here is being pulled upwards by t rope which is 100. And then these weight forces M. A. G. And M. B. G can actually combine, right? They don't cancel. And so they combined to actually produce a total weight which is really just big. M. Times G. So to figure out the acceleration, all we have to do is just use our F. Equals M. A. So then actually we're gonna use F equals big M. Times a day, we're gonna use the same rules, you know, upwards positive. So any forces along and against get positives and negatives. So our T rope is positive and then minus our big mgs negative and this equals big M. A. So this is gonna be 100 times five times 9.8 equals five A. So we ended up getting this 51 equals five A. And so A equals 10.2 m per second square. And that's your answer. So, you know, this system here is going to accelerate upwards because we got a positive number at 10.2. That's as easy as that, right? You can just go ahead and lump all these things together to a single object to solve for the acceleration. So let's move on to part B. Now, now what we wanna do is we want to find the tension in this connecting string over here. So this is actually gonna be T. S. So here's the deal. Whenever you when you whatever you're solving for the acceleration in these uh these kinds of problems remember you can always use the shortcut but if you have to go back and you have to solve for a connecting or internal force, what you're gonna have to do is you're gonna have to draw a free body diagram and write F. Equals M. A. For the simplest object. So for example here we've got both of these objects that involve T. S. Right? They both have T. S. In them. So I'm just gonna pick the simplest one which is really our three kg block. So if I write for the three kg block I write F. Equals M. A. Remember I have to use F. Equals M. A. For the little object. Right? So I have to use the little M. So this is gonna be M. B. Times A. So our forces are gonna be T. S minus M. B. G. Equals M. B. A. So when you move this to the other side, the MB is actually a common factor. So you can kind of be pulled out of the parentheses. So you got M. B. And this is gonna be A. Plus G. And now you can solve this, right? This is gonna be the massive B. Which is three times A. Which is 1 10.2 Plus 9.8, that's g. And so if you work this out, we're going to get his attention is exactly equal to 60 Newtons. Alright, so it's as simple as that. That's the answer. Let me know if you guys have any questions.