Three Hanging Blocks

by Patrick Ford
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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.