Skip to main content
Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Anderson Video - Two Blocks on a Table

Professor Anderson
80
Was this helpful?
>> There's block number one and we will use how about pink for the other one. There's block number two. And we'll say that this one has mass M1 and this one has mass M2 and now let's pull on the top block. And we're going to pull on it such that we want to find the maximum F pull that the blocks still don't move. How hard can you pull on that top block such that none of the blocks move? Okay, to do this we're going to probably need a little bit more information than just the masses. What other information are we going to need? What do you guys think? What's holding those blocks in place if I try to pull on them? Static friction. Exactly what we just talked about, right? Static friction is holding them in place and so we're going to need to know the co-efficients of static friction. And this is where it gets a little bit cumbersome because we have two surfaces to worry about. Block M1 is sitting on block M2 and so there is some co-efficient of static friction between block 1 and 2. This is what I mean by all these subscripts. So we don't know exactly what that is but that would be some number and then we can just make something up. How about 0.6? And then there is another co-efficient of friction between M2 and the table so this would be static friction between block 2 and the table and maybe that is 0.5 so these you would be given in the problem okay and if you had numbers for there you would be also given the masses and so forth. So what do we do now? We've got a picture. We need to go to the free body diagram and when we go to the free body diagram we have to treat each one independently so let's see what the free body diagram is going to look like for M1. So M1 becomes a dot. What are the forces that are acting on M1? Just shout them out. What do you guys think? >> Gravity. >> What's the first one we always draw, yeah, gravity. M1 G gravity going down on box M1. What else? >> Normal force. >> Normal force and this is where we need to be specific about which normal force we're talking about normal force on block 1 due to block 2. Okay, normal force on block 1 due to 2. What else? Well, we said we're going to pull it, right? F pull, so that's to the right. >> Static friction. >> Static friction which way? >> Opposite to the pull. >> Yeah, if it's not moving there's got to be something that's balance out F pull and that would be static friction going this way F sub S but we need to put some more subscripts on it because this is static frictional force on block 1 due to block 2, okay? Let's take a look at M2. M2 of course has gravity. What else is acting on M2, okay? >> M1 >> M1 is acting on M2. M1 is sitting on M2. So what is the force that I need to put on this picture? What should I label it? Okay, not M G yet but normal force of 2 onto 1. M1, 2 is up M2, 1 is down. This is block 1 pushing on block 2. It might be that we see that is in fact is equal to M1 times G but let's keep it like this for now to keep it general. Anything else? There's got to be something else if the whole thing is at rest, right? >> N 2 T. >> What's that? >> N 2 T. >> N 2 T. So there must be some normal force on 2 due to the table. The table is holding up block number 2. What else? It's not moving so this could be it. But we know that there's somebody trying to pull on this thing so there's got to be something else holding this whole thing in place which is static friction on 2 due to the table and then there is some force to the right to balance that out which is what? It's F sub S 2, 1. The static friction on 2 due to block 1. So this is what I mean when I say these things get kind of complicated quickly, right? Because look that simple picture and look at this mess that we have over here. It's a nightmare. It's not a nightmare but it's a bad dream. So what are the action/reaction pairs that we have just identified? N12 goes with N21. Block 1 is pushing on 2 but block 2 is pushing on block 1. Is there another action/reaction pair that we've identified? Yes it is F S12 and F S21. The frictional force here is exactly equal and opposite to the frictional force there, okay? And we know that those magnitudes are equal. Magnitude much M1 2 equals the magnitude of M21. Magnitude of F S12 equals the magnitude of F S21. These are the action/reaction pairs that we're talking about in this chapter, okay? So now that we have this free body diagram setup how are we going to deal with this problem? How could we possibly figure out what is the maximum pull that we could have and this thing still doesn't move? Well, we need to do some math. We need to write down Newton's Second Law and then we have to do some math. So let's do that. We have got number 1 sum of the forces in the x direction is equal to the mass times the acceleration in the x direction. F pull minus F S12 is equal to zero. The whole thing is at rest. The vertical components not too bad. Sum of the forces in the y direction equals the mass times the acceleration in the y direction we get N12 minus M 1 G that also equals zero. What about for the second one? For the second one we have sum of the forces in the x direction. What do we have? We have F S21. Minus F S1T. That equals zero and then we have the vertical forces N2T minus M 2 G minus N21 all that equals zero and this was of course the vertical equation. That's writing down Newton's Second Law and now if we're going to figure out the maximum F pull we need to do one more thing. We need to realize that static friction for instance, F S12 is less than or equal to mu S12 times the normal force on 1 due to 2 and F S2T is less than or equal to mu S2 T times the normal force on 2 due to the table. So how do you figure out the max pull? You set these to the maximum. Okay, you set them to equal and then you figure out how to solve for F pull. Okay? Questions about this approach? You can see why Newton just said for every action there's an equal and opposite reaction because otherwise it gets very complicated very quickly.
Divider