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7. Friction, Inclines, Systems

1

concept

6m

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Hey guys. So we've become really familiar with connected systems of objects and also friction. And in this video, I want to talk about a specific type of problem where you have objects that are connected not by ropes or cables, but they're basically just stacked on top of each other. I like to call these problems stacked blocks problems. And this is really interesting things that happened here. So let's go ahead and take a look. Now we're gonna come back to this point in just a second here. We're gonna skip and start with the example, because it's actually very easy to understand what's happening here. The idea is that we have these two blocks. One is 10 and 5 kg. I'll call this one a and this one be. So the idea here is that the floor is frictionless. There's no friction on the bottom surface, But between the two blocs, there is going to be friction. So what happens is I'm gonna pull the bottom block with some force, which means I'm gonna give it some acceleration. Notice how the two blocks stay together. What I want to do is I want to figure out the maximum acceleration that I can give the bottom block so that the two boxes remain moving together. So here's what this means. You pull, there's some acceleration, but the boxes stay together. You pull harder, there's more acceleration. The boxes still stay together. Eventually you can pull hard enough. You can give this an acceleration fast enough that the box is actually starts sliding relative to each other. They are no longer moving together. I want to figure out the maximum acceleration that I can do that with. Right. So basically, this is just gonna be like any other connected objects with system with friction type problem. We're gonna draw the free body diagrams for both. Let's go ahead and get started here. So here I've got A and B the free body diagram for A is gonna look like this. Actually, let me scoot this down a little bit. So I've got my weight force. This is Maggie, and then I have any applied forces or tensions. But there's no applied force that's acting on the top block. Remember, I'm only pulling on the bottom block here, so when I draw this free body diagram, there's no f so I've got a normal force, though, because I've got these two surfaces in contact. One way you can think about this is that a the weight force pushes down on B, so there's a reaction force to that surface push. So I'm gonna call this the normal between the two objects. NBA. Now, finally, what happens is we look at friction, right? So what happens is here you've got this block and it's moving to the rights. So there has to be a force that's acting on the top block to keep it accelerating and moving to the right as well. And that force is the force of friction. Remember that friction tries to stop or prevent the velocity between the two surfaces. So we're pulling on the bottom block, it's moving, and friction actually keeps the top block moving as well. So when you have objects that are stacked on top of each other, the force that acts on the top object that causes it to move is the force of static friction. All right. And so what's interesting about this static friction here? We know this is f s. Is that notice that the velocity of the system is also going to be to the right. So unlike for previous problems, where velocity and friction are always pointing opposite directions, the friction actually acts in the same direction as the direction of motion. All right, the last thing I want to talk about is we have to remember that when we talked about static and kinetic friction, we basically said moving versus not moving We could be a little bit more specific here because now these two surfaces can possibly move relative to each other. So to be more specific, the friction is going to be kinetic. Whatever the relative velocity between the two surfaces is not zero. Anytime you have these two surfaces that are sliding relative to each other, there's gonna be some kinetic friction. Friction is static. On the other hand, whenever the relative velocity between the two surfaces is equal to zero, basically, any time they are moving together like this, that's gonna be static friction. All right, so let's go ahead and get started here. We already know the type of friction that we're dealing with. Um, we actually have to go ahead and draw a free body diagram for B. So let's go and do that real quick. You've got the weight forest that acts on B and now we've got the applied Force. This is my f. We have the normal force that's acting from the floor. This is from the floor on to be so I'm gonna call this n B. But there's also another normal force that's basically part of an action reaction pair be pushes on upwards on a so a pushes downwards on B because of action reaction. So this is a downwards force and I'm going to call this n a B. There's one more action reaction pair, though as well. So I remember we said that there's friction between the two surfaces. There's gonna be a friction force to the right on the top block because of action reaction, there's an equal and opposite force that acts on the bottom blocks on the bottom block. There's another friction force that acts to the left. So really, these two friction forces are actually going to be the same. These are the same fss. Alright, so now we're gonna go ahead and write our F equals m A. Because we're trying to figure out an acceleration so If we're trying to figure out the acceleration, we're gonna start with the simplest object. So basically, we're gonna start off with, uh, the object A So we've got the sum of all forces and really there's only one force to consider, and I'm gonna choose the right direction to be positive. So I got this f s the static friction here, and I'm trying to figure out the maximum acceleration. So if I figure out the maximum acceleration right before the object starts sliding relative to each other, then I'm not just going up against any old friction or static friction. This is actually gonna be f s. Max. What I'm looking for here is the acceleration where the static friction is going to be maximum. So we have an equation for this. So we have, um, UK, or sort of you static times the normal. But what normal are we using in this problem? There's actually three. There's N B A and A B and NB. There's a whole bunch of normals and these problems, which one do we use? Well, basically, the idea is that if you're looking for the friction between the two surfaces Fs Max you're just going to use the normal between the two surfaces. This is going to be the normal force between the two surfaces or between the blocks, right? So really, we're just gonna use em. Use static times N b A, and that's going to equal to mass times a max. So if we can actually go ahead and solve for N B A. Um, if you think about this, these two forces, the weight and the normal have to cancel each other because the top block isn't accelerating vertically. So this NBA here is really just the weight force. So that means we have new static times, M a G equals a times a max. And if you can see here, what happens is that the mass of block A actually cancels out from the equation. And this a max is just gonna equal mu static times gravity. So it's really just going to be 0.7 times 9.8, and the acceleration maximum is going to be 6.86 m per second square. This is the maximum acceleration anything faster than 6 86 and the blocks now starts sliding relative to each other and the friction actually becomes kinetic. So we don't even have to go into the f equals M A for object B and that's it for this one. So hopefully that made sense and thanks for watching.

2

example

4m

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Hey guys, how's it going? So we got this problem here, we've got these two blocks on top of each other, but there's a twist to this one, which is that the block is actually tied to the wall. The idea here is that we're gonna be pulling on this first block, the bottom block with a force of 45 newtons. And we want to figure out the tension that's acting on block A. What is the tension force that's basically keeping block A. To the wall. So let's go ahead and get started here. We know we're gonna have to draw our free body diagrams. So, let me do the first block A. So we've seen this kind of thing before, we have the mass times gravity, that's the weight force of A. Then we have basically attention force. This is basically what we're trying to find here. We have a normal force because these two services are in contact, that normal force is between the two blocks. I'll call it N. B. A. So if these were the only forces that are acting on the object, then obviously the block would have to accelerate to the right. But that doesn't make any sense. It's going to be tied to the wall. There has to be a force that acts to the right, and this is gonna be the force of friction. So, let's go ahead and talk about Block B. Now, Block B is gonna look like this, It has a weight force that acts down, this is MBG. Then we have the applied force. This f. We know that F is 45. And we also have the normal force is there's two of them. There's the normal force on block B from the ground and there's also the action reaction pairs. This is N A B. Right? So these two things are action reaction pairs. There is another action reaction pair. Because we know that if A has a friction force that acts to the right, then B has to have a friction force that acts to the left. So these two things are opposite of each other. So what I'm gonna do is I'm gonna call this F A. B. It's the friction between the two blocks and it has an action reaction pair as well. This is gonna be F A. B as well and these two things have to be equal to each other. However, there's also one more friction force that we have to consider Remember that the coefficients between the two blocks is going to be 0.2. But we're told that the coefficient between all surfaces is 0.2. So that means that there's also some friction because of the bottom surface. So, because this block is being pulled and moves to the rights with some velocity, that means that we also have another friction force that acts to the left. There's two friction forces that actually cause this thing or that actually act backwards. All right. Now we just move on to figuring out what type of friction we're dealing with. And there's two clues that help us out here, we know that we're only given the coefficient of kinetic friction in this problem and we're also told that the block B. Is gonna move to the right. So there's some velocity here, which means that all of these frictions are going to be kinetic friction. So these are these are all kinetic frictions. So I'm gonna call this F A B K. F A B K F B K basically just indicating that they're all kinetic frictions. So let's go ahead and start writing our F. Equals M. A. We would start with the simplest object here, but remember which is actually object A. And also remember that we're trying to figure out the tension force. So we're gonna start with object A. So the sum of all forces equals mass times acceleration. So we just pick a direction of positive. Let's go ahead and choose the rights. Which means that we have F A. B. K. The kinetic friction force between the two blocks is equal to minus the tension is equal to mass times acceleration. Now, what is the acceleration of block A. Well, here's the thing if block A. Has has this tension force that's tying it to the wall then that means that the acceleration of A. Is equal to zero. So there is no acceleration and this thing has to be equilibrium because it's tied to the wall. All the forces have to cancel. So we have our kinetic friction force is equal to the tension. Now this kinetic friction force is kind of weird because we usually have kinetic friction when something is moving. But it really what happens is that the surface, the relative velocity between these two surfaces is moving. That's why we have some kinetic friction even though Block A actually stay still. Okay, so we we expand this force. The friction between the two surfaces is gonna depend on the normal between the two surfaces. So we're gonna use N. B. A. And the Sequels of attention and remember we can solve this N. B. A. And this is really just equal to the weight force. These two forces have to cancel out because the block doesn't accelerate vertically, right? So we have everything, we need to figure this out. This is really just gonna be mass times gravity. This is gonna be the tension force. We don't even have to go into block B. We can just go ahead and sell for this. This is gonna be zero point to the massive A. Is five G. Is 9.8. If you go ahead and solve for this, you're gonna get 9.8 newtons. And that's the answer. That's the tension that's holding this block to the wall. So that's answer choice B. So that's it for this one. Guys. Hopefully that made sense

3

Problem

A 4kg block sits on top of a 6kg block which is on a frictionless surface. The coefficients of friction between the two blocks are μs=0.5 and μk=0.3. Calculate the maximum force you can pull on the bottom block with so that the objects move together.

A

49 N

B

19.6 N

C

3.27 N

D

4.9 N

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