Boxes A and B are in contact on a horizontal, frictionless surfac...

Question

Boxes A and B are in contact on a horizontal, frictionless surface (Fig. E4.23). Box A has mass 20.0 kg and box B has mass 5.0 kg. A horizontal force of 250 N is exerted on box A. What is the magnitude of the force that box A exerts on box B? (image)

Accepted Answer

Hey everybody today, we're doing a question which are being asked to determine the magnitude of the force that block p. Which weighs 35 kg, applies on block Q. Which weighs 15 kg. Now we know a few things about the system so far. Oops, block P. S. Or blocks P. And Q. Are pushing against each other on a horizontal frictionless surface. And there is a horizontal force of 360 newtons that is being applied on a box p. That pushes both boxes now to determine the magnitude of the force. We should always try and draw a free body diagram first and we could do this both considering the entire system as a whole. Or we could actually separate the boxes and determine the forces acting upon them individually. For this problem, I'd like to do that because it'll make things a lot more clear. So let's draw it out to the side of it. So we have the horizontal surface. Oops we have box P. Oops, Box P. Not the best box. But it'll do. And we have a Box Q. Fox que. Now both boxes Q and P will both have a weight force which is just simply MG. Because all objects have weight if they're on Earth and it will point towards the center of the earth. And similarly they will have a normal force that opposes the weight that prevents the object from sinking into the ground. And it is a byproduct of the electrostatic repulsion between the surface and the object. Now we also have a few other forces were being told that there is a force of newtons being applied to the box p. Which pushes both them and the blocks are therefore being accelerated. Both blocks are therefore being accelerated in the positive X. Direction. However, since these boxes are are together in actuality in order to keep both of them moving together and moving to keep both the moving together without like one stopping the other one pushing into the other and sort of deforming it or anything like that. We also have action eri and reactionary forces. We have a force applied on box P by Q. Which will say F. Q. And similarly, we'll have a reaction force F. F. P. The force of pee on box Q. And for references sake, we'll say that this is the X. Direction and this is the Y direction. So with that in mind, our next step is to sort of think about Newton's second law. The second law, if we can recall states that the net force or the the net force is equal to mass, times acceleration. So let's write this out, this is step one and this is Step two. Now, there's a few things we need to note about the system. We don't have any friction force. So we have no motion in the Y direction. So we can effectively rule out effectively. Not really worry about the normal forces and wait because they don't have any bearing on the motion or the direction of motion that we have here. No, we can add the forces in the X direction for both the blocks. And we can recall that the accelerations here, which is, we can say acceleration to the Q. And acceleration of P. Well oops a F Q will be equal to a F. P. Which will simply just be the acceleration experienced because the blocks are joined and F of Q and F P, as I mentioned earlier, are Newton's third law pairs their reaction and action ery forces. From here. We can determine what the forces are going to be for each block independently. So for block P and I'll write this in red to stick with our uh color color conventions that we've been using so far for block P. We have the sum of the forces in the X direction equals to mass into the acceleration due to in the expression. Now the sum of the forces acting upon P upon box P are going to be f minus f of Excuse me, F F Q. Because FFQ is acting against the direction of the force, the 360 Newton force that is being applied and this will be equal to the massive p into acceleration. So, let us call this um equation one looking at box Q. We can do a very similar convention, we have to find that some of the forces acting in the X direction, multiplied by the mass of Q times acceleration. Now, the only force acting here on uh, Q specifically, simply the force due to P which will just be equal to the massive P. Or sorry, the massive queue, massive Q into acceleration because again, we determined that acceleration is the same for both boxes. So if we go from here and we add our equations right, we can go ahead and add our equations, remembering that F F P is will have the same magnitude as F F Q. Because their action and reaction forces. So if we were to add equation one and we'll call this equation to, we will get let's write this in green, F is equal to M. P. Of a plus M. Q. Of A. Or into A. And we can factor out a give us mp plus massive que massive queue? There we go. Which means therefore the acceleration experience is equal to the force over the sum of masses P and Q. So let's scroll down a little further. We can say that acceleration is equal to, Well, we know the force being applied, it is 360 newtons and we have the masses of P and Q. Right. The massive P. The massive p was 35 kg. And the massive queue was 15 kg Which gives us an answer of 7.2 m/s squared as the acceleration experienced here. Why do we need this? Well, remember the question is asking us, what is the value of the force of P. Or the force experienced by Q. Right? The force experienced by Q. And that is the force that P. Is pushing on to Q. So we can actually substitute our acceleration that we just got back into equation two. And we get that F of P, which is the force we're looking for is equal to 15 kg, which is the massive queue multiplied by the X. New acceleration that we've discovered is 7.2 m per second squared, Which would give us a final answer of 108 newtons or answer choice. Therefore, the force uh Box P exerts on box Q is 108 Newtons. I hope this helps. And I look forward to seeing you all in the next one.

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