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: (b) the tension in the rope connecting the blocks.

Question

Diagram showing two crates P and Q connected by a rope, with a force F pulling them to the right.

Accepted Answer

So for doing anything more, we can determine our acceleration. Our acceleration will be zero because remember we have a constant velocity velocity is not changing, then there's no acceleration either. So let's look at block P first Block P let's do that in blue for block P. When this kinetic friction involved, we need to determine an expression for the normal force. Which means we need to write Newton's second law is equal to M. A. For the Y direction. So let's do that. We have and let me scroll down a bit so I have a little more space to work. That should be good. So we have some of all forces is equal to mass times acceleration in the Y direction. Now, what are the forces in the y direction? Well, sum of all forces will be the normal force to to p minus the weight due to P which will be the mass times acceleration. But we determined that acceleration is zero. So this will simply be zero. Therefore the normal force must be equal to the weight. Just simply equal to MG or mass of P, I should say, and that works fine. That's great. That's everything we need there. However, for the X direction for the X direction we can do the same thing. Its force in the X direction is equal to mass times acceleration in the X direction. So in the X direction, what all forces do we have acting on it? We have T. Of Q on P. Plus the force due to friction because they're acting in the same direction minus the force pulling it will be zero. Recall that the force due to kinetic friction of force due to kinetic friction is equal to the coefficient of kinetic friction multiplied by the normal force. So expanding that we get T. Of Q. On P. Is equal to or sorry. Plus the coefficient of kinetic friction into the normal force of P. As you go to F. And expanding this further we get T. Of Q. On P. Plus coefficient of friction multiplied by massive P times G. Because that is the normal forces we described earlier will be equal to F. Now this is equation one let's say this is equation one but it has that unnecessary F. Term. So we're gonna need to do some more work. So let's also take a look at block. P. Gonna scroll down just a wee bit more for block P. It's very similar. Or sorry, block you my bet for block you. It's very similar. We follow the same steps and for Q. Block you. But again in the Y direction we only have a couple of forces. We have the normal force of Q minus the weight of Q equals zero. Which means that No one forces you must be equal to the weight of Q. Which is simply MG or M. Q. G. I should say. We also have the horizontal direction the X direction is simply the mass times the acceleration in the X direction. Now in the X direction, what all do we have? Well, the forces acting here we have the force of kinetic friction and the tension force acting against it. So this is P on cue. That'll be equal to zero expanding this further. We know that this is just a little closer into normal force of queues equal to T. P. On cue. Two. Oops that's not right, it's minus T. P. On Q. Is equal to zero. Following the same steps as early. We can finally get that T of P. On cue. There's nothing but the force of kinetic friction force of kinetic friction multiplied by the mass of Q times G. So if we have expressed T we have expressed the tension as forth as in the desired terms. And these expressions tell us that the mass or that the tension does not depend on the mass of P. This is really logical because friction on block you is not affected by the mass of block P. And this lines up with answer choice. Excuse me, this lines up with answer choice. E. F. Is equal to the coefficient of kinetic friction multiplied by the mass of Q times G. I hope this helps. And I look forward to seeing you all in the next one

Verified Solution

Key Concepts

Free-Body Diagram

Kinetic Friction

Newton's Second Law