Crates $A$ and $B$ sit at rest side by side on a frictionless horizontal surface. They have masses $m_{A}$ and $m_{B}$ , respectively. When a horizontal force $F$ is applied to crate $A$ , the two crates move off to the right. Draw clearly labeled free-body diagrams for crate $A$ and for crate $B$ . Indicate which pairs of forces, if any, are third-law action–reaction pairs.

Question

Accepted Answer

Hey everybody today we're dealing with the problem about free body diagrams. Now we're being told that a courier service a courier office has two carton boxes, let's draw those in blue and we don't really have much space here. So I'll hold off on drawing it for just a second. But we have two carton boxes, one with a mass. P. And the other with a mass of M. So M. P. And M. Q. Now these two boxes were put together side by side on a horizontal surface with no friction and a quarter boy applies. The force are on to box Q. So that both boxes move in the same direction. So we're being asked to draw a free body diagram of this. So let's um scroll down a bit and let's draw it ourselves. So we have two boxes, let's say this is box Q. Let's say this box Q. And this is box P. And I'm going to put a gap between them for now. But they are indeed uh together. Or rather. Yeah. While the question does say that they are together for the sake of our free body diagram, I'm just going to draw them slightly apart so that we can draw on the relative forces. So assuming that they're pushed together right now before talking about any other forces, we need to recall that uh because we're on a horizontal surface, each box will have a force due to wait. So this is this is weight of Q. Right? And this is the weight of P. Let me draw on our horizontal surface to I think that's important. Let's just say that is a horizontal surface that they're sitting on. And if they have wait, then that means that there is also a normal force acting against the weight, right? Acting in the opposite direction. So this is normal force of Q. And the normal force of peak. And the normal force sort of prevents it from sinking into the floor. It's the action and reaction force that we would see between the floor, the weight and the floor and the weight of the box itself. Recall that normal force is always perpendicular to the surface and in this case it is indeed perpendicular. If we're to draw or if we were to draw a small line all the way down at a former right angle. But anyways, now, what are the other forces acting here? Well, since the two boxes are being pushed on each other, we'll have two different action reaction forces again. Right? So we have a force our that is being applied through. Let me draw this above. We have a force our that is being applied through box Q. Right? And it is pushing both the boxes towards, let's say the right now, as this is happening. If both boxes are to be moving together, then this must mean that box Q is exerting a force on P let's call that force QP. That is Q on P. And similarly P will be forced having an equal and opposite reactionary force P On cue. And this is to prevent the boxes from sort of mushing into another. They're still uh station or they're still standalone boxes, they're just being pushed together. And this explains why the boxes don't sort of meld into each other as the forces applied. Now, with that in mind forces QP. Or force Q. On P. And force P. And Q. Will always be opposite because they are pushing against each other as they're moving to the or moving in the direction of force are. So with this in mind, if we're to draw this without the boxes, right? So for box Q. It would look something like would have the normal force of Q. We would have the, excuse me, we'd have the weight of Q. And we would have the force excuse me of Q. On P. Similarly and I'll do it over here for so this is for Q. Right? For P. Well, we'll have it'll be very similar, will have the weight of P. And the normal force of P. As we describe. We'll also have the reactionary force force P. On cue, but the force our will also be in the positive direction in the right direction. And we don't have force are here because it is in the same direction as uh the force of Q. On P. And the forces going through Q. All the way past P. So the force are would be there and the only answer choice that actually matches this is answer choice B. And we can see that this is right because again we have a few things right? Force P. And Q. And force Q. On P. Going opposite directions. The normal forces and the weights all make sense in uh Other answer choice is the normal force and weight don't make sense right here. Wait is going up in normal forces going down which just makes like which is opposite of what we want. It's the same here. So you can rule those out. And option choice is also incorrect because both Force P. And Q. And force Q. On P. Are going in the same direction. Which doesn't make sense their action erI and reactionary forces so we can rule that out. So I hope this helps. And I look forward to seeing you all in the next one.

Verified video answer for a similar problem:

Key Concepts

Free-Body Diagram

Newton's Third Law of Motion

Frictionless Surface