Forces in Connected Systems of Objects: Study with Video Lessons, Practice Problems & Examples

Understanding connected systems of objects involves analyzing forces acting on multiple masses, such as through free body diagrams. When objects are connected, they share the same acceleration, allowing for simplified calculations. For example, using the equation F = m × a, one can combine masses to find acceleration. This approach is effective in pulley systems, where different directions of motion are considered. Ultimately, recognizing the relationship between tension and acceleration is crucial for solving these problems efficiently.

Solving Force Problems in Connected Systems of Objects

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

Three Hanging Blocks

Hey, guys. Let's get to this problem here. We've got these 3 blocks. They're all connected to each other by strings, and we want to figure out the magnitude of the tensions in the topmost string and then the middle string. So what does that mean? So in part a, we're going to be figuring out some tension force here. Let's go ahead and draw the free-body diagrams. That's the first step. So we've got these 3 objects. I'm going to call these blocks a, b, and c. And we've got the free-body diagrams for this 2-kilogram block here is going to be the mass times, 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. This is t. However, there's going to be multiple tensions in this problem. Right? We've got a bunch of cables. So I'm going to call this one t₁. Alright. And then we've got another tension because we also have another cable that's connected to the bottom of the 2-kilogram block. So that means we're going to have another tension force. I'll call that one t₂. But that's it for this one. Right? So there are no friction forces or normals or anything like that. So let's move on to the second block. Now we get the wave force, which is m_bg, and then we've got the tension force that pulls upwards. However, because of the action-reaction, you remember that the tension in this rope is going to be the same throughout the whole thing. So if basically you have this t₂ that points downwards for block 2, it's going to be the same t₂ over here. Right? So it's going to be the same tension, except just like block 2, there's another tension force that's on the bottom of block b, and so I want to call that t₃. Alright. Now for the bottom one here, we've got the weight force, m_cg, and then we also got this tension force, which is t₃ the same exact idea. This is really just an action-reaction pair. Alright? So basically, we've got these 3 tensions t₁, t₂, t₃ and I want to figure out t₁. 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 0. So they're not actually accelerating anywhere, which means we can just stick to our normal rules which are up is positive for all the objects. Alright? So let's now get into our free body diagrams. We're going to 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 topmost string. So it actually makes more sense for us to start with block a. So we've got F = ma. And remember these are all just y forces. So this is all the forces in the y direction. However, we just said was that the acceleration was equal to 0. So that means that this thing's going to be in equilibrium, and that's actually going to be the same for all the objects. They're all in equilibrium. So that means that I've got my t₁ which is upwards, minus t₂ which is down, minus ma_g, and that's equal to 0. Alright. So that's basically, my equation. If I want to figure out this t₁ here, I actually can't solve this because I don't know what the t₂ is. I do know what the mass and gravity are, 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 going to go for block b now. Same exact thing, F = ma. We know the acceleration is 0 because they're all in equilibrium. And so, here what I've got is I've got the t₂ that's upwards minus my t₃ that's downwards minus my m_bg and this is equal to 0. Alright. So, if I want to figure out this T2 and I also have this unknown T3 here and just like this other equation I've also got 2 unknowns in this problem So I'm going to have to go to the 3rd block now So I'm going to have to go to block c and block c is going to be a little bit simpler. So, we know that this is going to be equal to the acceleration zero and we have 2 forces only. So, we have T₃ minus m_cg is equal to 0. Notice how this one has a little bit fewer terms. It doesn't have another tension force. So we get these 3 tensions that are all kind of, like, mashed up in these equations. And remember, we want to find what this tension one is. So how do we solve these kinds of problems when we have systems of equations? Well, Remember, we're just gonna number them. Right? This is going to be number 1, number 2, and then number 3 here. And then we're just going to basically solve by using equation addition or substitution. Now we're not going to be solving 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've got these three equations here. Remember you're just going to line them up top to bottom. So you got T₁ minus T₂ minus m_ag is equal to 0. We want to do is we want to line up this tension 2 in equation number 2 so that it lines up with a T₂ over here. So we've got T₂ minus 3 T₃ minus m_bg is equal to 0. And then finally, we want to line up this t₃ here with this t₃. So our t₃ minus m_cg is equal to 0. So we got our 3 equations here. And remember, what 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 t₂ goes away. Your t₃ is also going to go away. And what's left is you have t₁, which is good. That's what we're trying to solve for. So you've got t₁ minus m_ag. Remember, these are all negative here. So negative negative and negative minus m_bg minus m_cg equals 0 So if you go ahead and move all of this stuff over to the other side, basically, what you get is you get m_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 going to get 2 plus 3 plus 5, which is 10 times 9.8. And so, you get the tension is equal to 98 Newtons. So, again, what we saw here is that this tension force is basically just the combined weight of all of the three blocks here. That's actually important 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. Alright? So that's what we got 98 newtons. Alright. 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 t₂. So what did we just say? Each of the tension tests to support all the weight that is underneath. So a really quick way to do this is you could just look at all the weight that's underneath this, and that's going to be your tension. Right? So this is really just going to be t₂ equals 3 +5 times 9.8, and what you'd get is 78.4 newtons. And this is actually the correct answer. If you really wanted to solve this long way, what you could do is you can go ahead and solve it using this equation here, equation number 1, now that you actually know what t₁

Systems Connected By Pulleys

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 difference is instead of everything moving along the same direction like this, you're going to have one object that hangs or something like that. And so, basically, it's going to go up and down, and the other object is going to go side to side. So these things aren't going to move in the same direction. We actually solve 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 kilogram block here. It's connected to a 2 kilogram block, and it's 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've got I'm going to call this one a and this one b. And so I've got the weight force, so I'll call this w a. Now I've got any applied forces or tensions. Tensions. There's a rope which means this is going to be a tension 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 2 vertical forces, then they have to cancel. So this is our mag. Alright, now let's look at the hanging block. The hanging block is going to have a weight force, so this is going to be wb. And now we've got a tension force 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 a tension 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. Alright. So that tension force is actually going to be the same. Alright. 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, then the direction of positive is usually the direction that the hanging object will fall. What do 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 going to be our direction of positive. And because of the pulley, what happens is that the the the block on the table is going to start accelerating to the right like that. And so this is going to be our direction of positive. So we have 2 different directions of positive. But one way you could think about this is that anything that goes to the right and down is going to be a longer direction of positive. Alright. So that's our direction of positive. Now we're going to write f equals ma. We're going to start out with the simplest object, one with the fewest amount of forces, and that's the 2 kilogram block. So let's get started here. So we've got our f equals ma. Now again, we're just going to look at all the forces that are acting on the y-axis. So this equals mass times acceleration. We 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 m bg gets written with a plus sign. So or sorry with a plus sign. So m bg is going to 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 going to be 2, this is 9.8 minus tension equals 2a. So what I get is 19.6 minus tension equals 2a. And I can't go any further because I want to figure out the acceleration, but I don't know what the tension force is. So I've gotten stuck and this is fine. When I get stuck, I just move to the other object. So I've got this 4 kilogram block. Now we're going to use f equals ma here but unlike this, unlike the 2 kilogram block, this block is just moving horizontally. So we're really just adding up all the x forces. And this is equal to mass times acceleration. We only have one x force. It's It's really just the tension. So our tension is equal to mass times acceleration. And so that means T is equal to 4 a. So again, I wanna figure out the acceleration, but I'm missing the tension. And I've gotten these 2 equations with 2 unknowns. And 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've got my 2 equations with 2 unknowns here, and that brings us to the 4th step, which is we have to solve our a using equation addition or substitution. I'm gonna use equation addition 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 4a, then my second equation is 19.6 minus t equals 2a. 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 6a. And so when I solve, I'm going to get a is equal to 19.6 divided by 6 which is equal to 3.27 meters per second squared. So that is our acceleration. That's a equals 3.27 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 our 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. Now remember, the easiest one is just gonna be the one with the fewest terms. So we'll start off with that 1, tension equals 4a, tension equals 4 times 3.27, and that equals 13.08 newtons, and that's the answer. So 13.08, 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. Alright, guys. So let me know if you have any questions. Let's move on.

Atwood Machine

Hey, guys. Let's check out this problem here. We've had these 2 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 2 blocks. We know the larger one is 6 and the smaller one is 4. We want to figure out the acceleration of this system and the free body diagram for both objects. It 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 are going to look like the weight. So we have a weight force like this. Then the only other force that is acting on this is going to be the tension from the cord. Right? There's no applied 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 going to call this b, which is m_bg. And then the only other force that is 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. Alright? So that's our free body diagram, and the next thing we have to do is determine a 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 6, is heavier. So we know that when you release this thing, one of them is going to have to go up and the other one's 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 going to go downwards. So the 6 is going to go down. The 4 is going to 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-kilogram object is going to go down like this. That's we're going to choose our direction of positive. But because this pulley goes up and over, then if the 6 goes down, the 4 is going to go up. So we're going to choose the upward direction to be positive for the 4-kilogram objects. So our direction of positive is really just the clockwise direction here. That's really important. So now we're going to write F = ma, and we're going to start with the simplest object. But what's lucky for us is that we have both objects that are relatively simple. We only have 2 objects or 2 forces. So I'm going to start off with object a here, and I want to calculate, I want to use my F = 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 4, for block a. So that means that's going to be positive. And then our MAG points downwards against our positive direction, so it's going to be minus. And this is going to be ma times a. So now we could replace the values that we know. We know tension, so we have tension minus. This is going to be 4 times 9.8, and this is going to equal 4 times a. So I could just simplify real quick. I've got 4, t - 39.2, whoops, equals 4a. 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 going to go to the other objects. So this is going to be block b. So I want to write F = ma 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, our m_bg is actually going to be the positive one here, and our tension points against our direction of positive, so it's going to be minus. So it's really important that you follow those rules. So now we just replace all the values that we know. We know this is 6 times 9.8 minus tension equals 6 times a. So this ends up being 58.8 minus t equals 6a. 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 2 equations with 2 unknowns. It's usually what happens in these problems. So I'm going to box them, and I'm going to call this one equation number 1 and this one equation number 2. And that brings us to the 4th step. We want to solve for this acceleration. We just have to use either equation addition or substitution, and I'm just going to pick addition. So, basically, what we've got here is we've got 1, which is t - 39.2 equals 4a. Now what I want to do is I want to line up the tension, right, from equation number 2 so that we can add them straight down. So this is going to 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 now remember, you're just going to add them straight down. You just add them straight down like this. And what happens is you're going to cancel out these tensions when you add them. So you add them straight down you get 58.8 and remember this is minus 39.2, so this is minus 39.2 equals and then the 4 +6 is going to be 10a. So this turns out to be 19.6 equals 10a, and that means your acceleration is 1.96 meters per second squared. So let's talk about the direction. We got a positive number for our 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. Alright? So 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 going to have to plug our a back into 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 number of terms, but the thing is that this t is actually positive in this equation and this one's negative. So this one is slightly more simple. So that means I'm going to use t - 39.2 equals 4. And then now we know a. Right? So t - 39.2, actually, I'm just going to move this to the other side. So t equals 4 times 1.96, that's what we figured out here, plus 39.2. If you go ahead and work this out, you're going to get 47.04 newtons. That's the answer. That's it for this one, guys. Let me know if you have any questions.

Shortcut for Solving Connected Systems of Objects

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 want to do in this video is show you a very powerful shortcut that you can use to solve the accelerations in these kinds of system problems. Let's go ahead and check this out. We had a 3 and a 5 kilogram block being pulled by a force. And to solve this, we had to write f=ma for both, come up with 2 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 going to show you how to do this in 20 seconds. Basically just going to imagine there's a single large object and combine all the masses together. For example, we have these 2, 3 and 5 kilogram blocks, but I'm just going to imagine now there's a big block that's 8. And what happens is when you imagine this, there's only one force that's pulling on this block, which is f=30. So you might be wondering what happens to these tension forces that were connecting the 2 objects, and, basically, they go away. So when you do this, you're going to ignore any tensions or normals between the 2 objects or 3 objects, whatever there is. So these two tensions, it's 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=ma, but now we're going to write f=big ma. So the only force that's acting now is the 30 and this equals 8a. And so a=3.75ms-2, 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 4 kilogram block, the 2 kilogram blocks 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=ma for both, do 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 4. This is 2, which means the large object is actually going to have a mass of 6. 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, it's 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 6 kilogram block, it's as if the weight force is really only still from the 2 kilogram block that's acting. Alright. So then when you do our f=ma, this is f=big ma. So remember this is going to be mbg. 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 going to be positive. So that means our mbg is positive here, and this equals big M times a. So this is going to be 2 times 9.8 equals 6a. This is 19.6 equals 6a. So we get a=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 going to calculate this acceleration of this 3 block system. So if we wanted to do this, we'd have to draw 3 free body diagrams. We have to write f=ma three times. We'd have to get 3 equations, do equation addition, and all that stuff, and instead what you can do is you can just lump these together into a single box of 10. Right? So it's 2 plus 3 plus 5 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 2 and the 3, the 3 and the 5, all of those go away. So now what happens is we just write our f=big ma. There's only just the 40 equals 10a, and so your acceleration is 4. 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 2 weight forces. Remember, this is going to be mbg. Really, this is going to be, like, 6 times 9.8. And then you also have this one which is mag which is 4 times 9.8. And then you had the 2 tension forces. There was a tension up and a tension 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 2 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 had 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 6 times 9.8 is 58.8, which is pulling it down. But then on the other side, you have this upward force which is mag, which is really just 39.2. That's pulling it kind of up. Right? Again, we're just kind of just imagining this is a single object. That's kind of what it works out to. So now when you use your f=big ma, now you have 2 external forces. Remember, anything that's downwards is going to be positive. Anything upwards is going to be negative. So what happens here is you have 58.8 minus 39.2 equals 10a. 19.6 equals 10a, and you get a=1.96, which is exactly what we got when we worked this out the long way. That's it for this one, guys. Let me know if you have any questions.

Pulling Two Connected Blocks Upward

Hey, guys. So we got these 2 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 force is 100 newtons. So we've got the masses of both blocks. What we want to do is we want to calculate the acceleration of the blocks in the first part. So this is really going to be a. And then in the second part, we're going to calculate some tension. So we want to calculate the acceleration. We know we can use this shortcut here. So I've got this 2 kilogram block, this 3. So I'm going to call these 1 a and b. And what I want to do is just really quickly draw free-body diagrams for both of these. So I've got the \(m_ag\), that's the weight force. Then I've got the \(T_{rope}\), that's the tension force in the rope, and I actually know 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 \(m_bg\), and then I've got the upward force of the tension in the connecting string. So we can use the shortcut to find the acceleration. All we're doing really is we're just replacing these 2 blocks with one 5 kilogram block, right, by adding the masses together, and then we're just going to 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 that this one block here is being pulled upwards by \(T_{rope}\), which is 100. And then these weight forces, \(m_ag\) and \(m_bg\), can actually combine. Right? They don't cancel. And so they combine 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 \(F = ma\). So then, actually, we're going to use \(F = m \times a\). We're going to use the same rules, you know, upward is positive, so any forces along and against get positives and negatives. So our \(T_{rope}\) is positive, and then minus our big \(mg\) is negative and this equals big \(ma\). So this is going to be a \(100 - 5 \times 9.8 = 5a\). So we generally getting this \(51 = 5a\), and so \(a = 10.2\) meters per second squared. 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 want to do is we want to find tension in this connecting string over here. So this is actually going to be \(T_s\). So here's the deal. Whenever you are solving for the acceleration in 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, then what you're going to have to do is you're going to have to draw a free-body diagram and write \(F = ma\) 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 going to pick the simplest one, which is really our 3 kilogram block. So if I write for the 3 kilogram block, I write \(F = ma\). Remember, I have to use \(F = ma\) for the little object. Right? So I have to use the little m. So this is going to be \(m_b \times a\). So our forces are going to be \(T_s - m_bg = m_ba\). So when you move this to the other side, the \(m_b\) is actually a common factor, so it can kind of be pulled out of the parenthesis. So we've got \(m_b(a + g\)). And now you can solve this. Right? This is going to be the mass of b which is 3 times a which is 10.2 plus 9.8, that's g. And so if you work this out, what you're going to get is the tension 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.

Here’s what students ask on this topic:

How do you calculate the acceleration of connected objects in a frictionless system?

To calculate the acceleration of connected objects in a frictionless system, follow these steps:

1. Draw free body diagrams for each object to identify all forces acting on them.

2. Write the equations of motion using Newton's second law, $F=ma$, for each object. Consider only the forces in the direction of motion.

3. Recognize that the connected objects share the same acceleration, so set their accelerations equal.

4. Solve the system of equations to find the acceleration. You can use methods like equation addition or substitution to eliminate unknowns.

For example, if two blocks of masses $m_1$ and $m_2$ are connected by a rope and pulled by a force $F$, the combined mass is $m_1+m_2$. The acceleration is $a=\frac{F}{m_1+m_2}$.

What is the role of tension in connected systems of objects?

Tension plays a crucial role in connected systems of objects as it transmits the force between the objects. When two objects are connected by a rope or cable, the tension in the rope ensures that both objects accelerate together. According to Newton's third law, the tension force exerted by one object on the rope is equal and opposite to the tension force exerted by the rope on the other object. This means that the tension is the same throughout the rope, assuming it is massless and inextensible. In equations of motion, tension appears as a force that either aids or opposes the applied forces, depending on the direction of motion.

How do you solve problems involving pulleys in connected systems?

To solve problems involving pulleys in connected systems, follow these steps:

1. Draw free body diagrams for each object, identifying all forces, including tension and weight.

2. Write the equations of motion using Newton's second law, $F=ma$, for each object. Consider the direction of motion for each object.

3. Recognize that the objects connected by the pulley share the same magnitude of acceleration but may move in different directions.

4. Solve the system of equations to find the acceleration and tension. Use methods like equation addition or substitution to eliminate unknowns.

For example, if a 4 kg block on a table is connected to a 2 kg hanging block by a pulley, the acceleration can be found by combining the masses and considering the net force due to gravity on the hanging block.

What is the shortcut method for finding acceleration in connected systems?

The shortcut method for finding acceleration in connected systems involves combining the masses of the connected objects into a single equivalent mass. This simplifies the problem by eliminating internal forces like tension. Follow these steps:

1. Combine the masses of all connected objects into a single mass, $M$.

2. Identify the net external force acting on the system.

3. Use Newton's second law, $F=Ma$, to solve for the acceleration.

For example, if two blocks of masses 3 kg and 5 kg are pulled by a force of 30 N, the combined mass is 8 kg. The acceleration is $a=\frac{F}{M}=\frac{30}{8}=3.75$ m/s².

How do you determine the direction of positive in pulley systems?

In pulley systems, the direction of positive is typically determined by the expected motion of the objects. Follow these guidelines:

1. For a hanging object, the direction of positive is usually the direction in which the object will fall (downwards).

2. For an object on a table connected to a hanging object, the direction of positive is the direction in which the object will move (sideways).

3. Consistently apply the chosen direction of positive to all forces and accelerations in your equations of motion.

For example, if a 2 kg block hangs and a 4 kg block is on a table, the direction of positive for the hanging block is downwards, and for the block on the table, it is to the right. This ensures consistency in solving the equations of motion.