Two crates connected by a rope lie on a horizontal surface (Fig. ...

Question

Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate A has mass mA, and crate B has mass mB. The coefficient of kinetic friction between each crate and the surface is μk. The crates are pulled to the right at constant velocity by a horizontal force F Draw one or more free-body diagrams to calculate the following in terms of mA, mB, and μk: (a) the magnitude of F

Accepted Answer

Hey everyone in this problem. We have two wooden blocks that are connected using an in elastic wire. The mass of block P. Is M. P. Well, the block Q has a mass M. Q. The blocks are placed on a horizontal bench, you're gonna pull the blocks to the left using a horizontal force F. Which gives the blocks a constant velocity. The blocks experienced the same coefficient of kinetic friction. UK. And were asked to determine the magnitude of F. In terms of M. P, M. Q and mu K. And with the help of some free body diagrams. Alright, so we have our horizontal bench here and we have two blocks and this will be block P. And over here we have block you. Okay? And they're connected using an elastic wire and then we're gonna pull the blocks to the left. Alright, So that's a little diagram of what's going on. Let's go ahead and draw out our free body diagrams. Okay, So for P we have our block P. We always have the normal force pointing straight up, okay, perpendicular to the surface. So this is a normal force and we're going to call it N. Sub P. Okay, So that it's the normal force for the block P. And we always have the weight pointing down as well. The weight for block P. W. P. Okay, similarly, for Q. We're not done with the with the free body diagram for P yet, but we're gonna go to Q. Um And do these forces that we know always act. Okay, So we get in the normal force here for Q Pointing up and the wait for Q. Pointing down. All right now we're pulling this to the left. Okay? And we're starting with block P. So block P. Is going to have a force applied and that's a little bit not horizontal. Okay. It's a horizontal force applied and it's going to be pointing to the left. Okay Because we're pulling to the left and this is what we're trying to find the magnitude of this force. Alright. And then what about to the right? And well we have this rope. Okay so we're gonna have some force dude attention from that wire. What else do we have? We also have friction. Okay friction is going to resist or oppose the motion. So friction is also going to be acting to the right here and it's the blocks are going to be moving. So we know we have kinetic friction. So we're gonna have the force of friction F. And we're gonna say sub K. P. Okay, kinetic friction for block P. Alright, so that's everything happening with P. Now let's go to Q. With Q. Well P. Sorry with P. We had this tension force to the right with Q. We have the kind of equal and opposite force of tension going to the left and then we have our friction going to the right and again kinetic friction because we're moving and we'll call it F. K. Q. Um For block you. Okay. Alright. And this force is not directly applied to Q. It's directly applied to this first block P. And then it's going to transfer to queue through this wire because we don't have the force F directly in queue. Alright, So, let's start with block P. Okay, That's where our force F. Is that we're trying to find. So let's start with P. Okay. And again that force F. Is horizontal. So it's in the X direction. Let's draw coordinate system X. And Y. Okay, so it's in the X direction. So let's start with the forces in the extraction and the sum of the forces in the X direction. According to Newton's second law recall is going to be the mass M. That's going to be the massive P. Because we're talking about block P times the acceleration in the X direction ax. Okay, now the blocks have a constant velocity. If they have a constant velocity, that means the acceleration is zero. Which means that the sum of the forces is just going to equal zero. All right, let's give ourselves more room to work. But we're gonna leave our free body diagrams up so we can see them and refer to them. All right, So, what is the sum of the forces in the X direction here. Okay, We've chosen right to be the positive extraction. Okay, So in the positive X direction we have our force due to tension T. We have the friction force and its kinetic friction of block P. And then in the negative X direction we have the force F. Okay, and this is all equal to zero. And again F. Is what we're trying to solve for. So if we isolate F. We get F. Is equal to the force of tension plus kinetic friction of block P. What is F. K. Recall that the kinetic friction force can be written as mu K. Okay. The coefficient of kinetic friction times the normal end and in this case N. P. For the P. Block. Okay. Alright. So we're trying to find this value of F. We don't know the tension T. And we don't yet know N. P. So we need to find both of those things while we're looking at block P. So let's start with N. P. Let's try to find np first. Okay. So similarly to how we have the sum of the forces in the X. Direction equal to the mass times acceleration, we get the sum of the forces in the Y direction also equal according to Newton's second law, the mass times the acceleration in the Y direction. And our blocks are not moving up and down. The acceleration in the Y direction is zero. And so some of the force is going to be equal to zero. And what forces do we have here? Well, we see that we have the normal force N. P pointing up, we have the weight W. P pointing down. Okay. And so we end up with N. P minus W. P. Is equal to zero. Which tells us that our normal forest P. Is equal to the weight W. P. What is the weight equal to? Well the weight is equal to M times G. So we get the mass M. Of block P times gravitational acceleration G. Alright, so we have N. P. We're gonna need that for our equation for F. Now we also need to find the tension T. And if we go back up to our diagram we see that the tension T. Is part of block P. But that same tension T is also part of block. You working in the opposite direction. And so let's move over to block Q. And again we'll start with some of the forces in the X. Direction because we're trying to find that tension T. Some of the forces in the X. Direction again is equal to the mass and in this case mass block you acceleration in the X. Direction which again is zero because we have constant velocity. Okay, what is the sum of the forces in the extraction? Well we have the force of friction and its kinetic friction because the blocks are moving in the positive extraction. And then we have the tension force and the negative extraction. This is equal to zero. Which tells us that the tension is just equal to the kinetic force of kinetic friction on block you similarly to with block P. the force of kinetic friction is going to be equal to mu k. Okay, it's the same coefficient of kinetic friction for block P and Q. That we're told Okay, times N the normal and in this case en que the normal because of you two. Alright, so we found this equation for the normal or sorry for the tension but it needs the normal of Q. Okay, well, in order to calculate that we're going to need to look in the Y direction. Ok, so we're needing we're having to use kind of all four equations here in this case. So some of the forces in the Y direction. Okay, we're looking at block you is going to be equal to the mass A massive queue. The acceleration in the Y direction. And again we know this is zero because the blocks are not moving up and down, there's no acceleration in the Y direction. Okay, if we look at our diagram, the only forces in the Y direction with the normal force pointing up and the weight pointing down. So we get similarly to the block P we get n q minus W Q is equal to zero and so our normal force and Q is going to be equal to W Q. Okay, And the weight recall is given by the mass times the gravitational acceleration so M Q. G. Alright, so we're going to combine All of these three things. Okay, into our equation here and I'm going to call this equation star Mhm And we're gonna come back to equation star here. Just rewritten the star so you can see where that equation comes from. So we have the force F we're trying to find is equal to the tension T plus mu K times N P. Okay well we know that the tension T is equal to mu K. Times and Q. Then we have the U K N P. What is Np? Np? We know as M P G and N Q is M Q G. Okay, so this tells us that our force F is going to be equal to mu K times M Q. G blast me. Okay times M P G. Alright now we can factor, we have a common factor of mu K time is G. So you get me? Okay, G times M Q. The massive Q plus M P. The massive block P. And so our force that were pulling these blocks with is going to be equal to mu K. The coefficient of kinetic friction times G. The gravitational acceleration times the sum of the masses of the two blocks M Q plus mp. Alright, so if we go back up to our answer choices, we see that we have answer choice. C K F is equal to M P plus M Q. Times mu K. G. Thanks everyone for watching. I hope this video helped see you in the next one

Watch next