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 mbg, 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, mcg, 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 mag, 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 mbg 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 mcg 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 mag 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 mbg is equal to 0. And then finally, we want to line up this t3 here with this t3. So our t3 minus mcg 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 mag. Remember, these are all negative here. So negative negative and negative minus mbg minus mcg 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 ma plus mb plus mc 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
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6. Intro to Forces (Dynamics)
Forces in Connected Systems of Objects
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