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>> Hello class, Professor Anderson here. Let's take a look at another Newton's Third type of problem. We're going to tie two blocks together. And we're going to accelerate the two blocks. But now we're going to add some friction to the problem. Okay, so what we said before was, if block one is M1, and this is M2. And we pull it with T2. Let's ask the following question. What is T1 equal to? But let's also include friction. Okay, this whole thing is going to slide to the right. Which means we're dealing with kinetic friction. And kinetic friction has a coefficient mu sub K. We're going to say the whole thing is moving to the right, okay. So that's what our picture looks like. We're going to ask the question, what is the tension in the rope between the two masses? And let's see if we can figure that out. All right, we go back to our free body. And the free body for M1 looks like this. We've got M1G acting down. We've got N1 acting up. We have Tension 1 acting to the right. What else do we have? What do you guys think? Brandon, what else should I put here? >> Kinetic friction going to your left. >> Kinetic friction going to the left. And we will call that FK1, kinetic friction acting on box one. Perfect. M2. Brandon, what do you think I should draw for M2? >> One more force going up. >> Okay. >> Gravity going down. >> Gravity going down. What else? >> Tension 2 going to your right. >> Tension 2 to the right. >> Tension 1 going to the left. >> Tension 1 going to the left. >> And kinetic friction going to the left. >> Kinetic friction going to the left. And we'll call that one FK2. Perfect. That's the free-body diagram for Mass 2, all right. What are the Newton's third law pairs? It's T1; right? T1 going to the right on Mass 1 means T1 going to the left on Mass 2. Okay, let's see if we can solve this thing now for T1. So we have our free-body diagram. We want to write down the sum of the forces. And we can break it into components. And we'll do the X components on mass M1. What do we have? We have T1 going to the right. We have FK1 going to the left. And that's it. That's equal to the mass times the acceleration in the X direction. What about the sum of the forces in the Y direction? Well, we've got N1 going up. We have M1G going down. That's equal to the mass times the acceleration in the Y direction. But we know that's zero. What about M2? For M2 we're going to have the following. Sum of the forces in the X direction. T2 to the right. T1 to the left. FK2 to the left. And that's equal to the mass times the acceleration in the X direction. And, finally, we have the sum of the forces in the Y direction. And for that we have N2 going up. M2G going down. That's equal to the mass times the acceleration in the Y direction. But again that's zero. We have four equations now. But we need a couple more. Which are the friction equations. Friction is equal to the coefficient of friction times the normal force. So FK1 would be mu KN1. FK2 is mu KN2. And now let's see if we can put all this stuff together and solve for the tension T1. All right, if we're looking for tension T1, we can start with this very first equation. And we can write T1 is equal to M1A sub X plus FK1. All right, what is FK1? FK1 is equal to mu KN1 from our last equation. But from this equation we know that N1 is just M1G. Okay, so this is the tension T1. Now, we don't know exactly what we were given. So let's solve this thing generally for T2 as well, such that we have all the parameters we need. T2 is equal to what? It's equal to M2A sub X plus this stuff right here, T1 plus FK2. M2A sub X plus T1. But I know what T1 is. T1 is right there. M1A sub X plus mu KM1G. And then I still have to add FK2. Which is mu KN2. But I know exactly what that is. So I can combine some stuff here; right? I've got (M1 plus M2)A sub X plus mu K times (M1 plus M2) times G. That's my tension T1. That's my tension T2. Now, you might be given some of these numbers. And so you could manipulate these equations and calculate what you want. But let's go back to our original question. Is T1 bigger than T2? Equal to T2? Or less than T2? Which one of those is true? Okay, [inaudible] what do you think? Which one of these is true? Is T1 bigger than T2? Is T1 equal to T2? Or is T1 less than T2? >> T1 is less than T2. >> T1 is less than T2. Same as we had before; right? In the case of no friction. But even now that we have friction, we still get the same answer. Why do you get the same answer? Because, look, T1 is right here. It's buried in T2; right? All I'm doing is adding some more stuff. I'm adding M2A sub X. I'm adding mu KM2G. If those are positive quantities, then T1 is still less than T2. Good. Daniel. >> What happened to the N2? >> N2 became M2G from this equation. And so it's right there, M2G. And there's my mu K. Other questions about this one? Yeah, David over there. >> Where did you get the second part for the Tension 1? >> Where did I get the second part for Tension 1? >> Yeah, on the last part? >> Down here. >> On the equation for Tension 1 on the equation. >> This one right here? >> Yeah, the second part. >> So that came from this one right here; right? The normal force on 1 was just, since this thing equal to zero, we must have N1 equals M1G. So I just plugged that in right here. N1 equals M1G. Okay? Cool. Any other questions? All right, you guys feel okay with this stuff? Reasonably okay? 50 percent or better? Okay. Let's move onto something else.

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