Treating systems (the hard way) | Forces and Newton's laws of motion | Physics | Khan Academy

All right, this problem is a classic. You're going to see this in basically every single physics textbook. The problem is this: if you've got two masses tied together by a rope and that rope passes over a pulley, what's the acceleration of the masses? In other words, what's the acceleration of the 3 kg mass, and then what's the acceleration of the 5 kg mass?

If you're wondering what the heck a pulley is, so, the pulley is this part right here. This right here is the pulley. So what a pulley does: a pulley is a little piece of plastic or metal that can rotate, and it's usually got a groove in it so that a string or a rope can pass over it. What it does is it rotates freely, so that you can turn what's a horizontal tension on one side into a vertical tension on the other, or vice versa. It turns vertical forces into horizontal forces; it allows you to transfer a force from one direction to another direction. So that's what these pulleys are useful for. If they can spin freely, and if this pulley has basically no mass, if there's no resistance to motion at all, then this tension on this side is going to be equal to the tension on this side. This vertical tension gets transferred fully, undiluted, into a horizontal tension, and these tension values will just be the same if this pulley can spin freely and if its mass is really small, so that there's no inertial reason why it doesn't want to spin.

So that's the problem. Let's say you wanted to figure this out: what is the acceleration of the 3 kg mass, what's the acceleration of the 5 kg mass? Now I got to warn you, there's an easy way to do this and a hard way to do this. Now I'm going to show you the hard way first. Sorry, no one ever ever wants to hear that, but the reason is that the easy way won't make any sense unless I show you the hard way first. It won't make any sense why the easy way works unless I show you the hard way. For two, the hard way isn't really all that hard, so I'm calling it the hard way, but it's not really that bad. And for three, sometimes teachers and professors just want to see you do it the hard way, so you should know how to do this.

So what do we do? We want to find acceleration. Well, you know how to find acceleration; we're going to use Newton's Second Law. So we'll say that the acceleration in a given direction is going to equal the net force in that direction divided by the mass. Now what do we do? What mass are we going to choose? We've got a couple masses here. One thing we could do: let's just pick the 5 kg mass—just pick one of them. So I'm going to say that the acceleration of the 5 kg mass is the net force on the 5 kg mass divided by the mass of the 5 kg mass. And remember, we should always pick a direction as well. So do we want to pick the vertical direction or the horizontal direction? Well, since this box is going to be accelerating horizontally, and that's what we're interested in, I'm going to put one more subscript up here, x, to let us know we're picking the horizontal direction.

So I can fill this out now. I can plug stuff in. The acceleration of the 5 kg mass in the x direction is going to be equal to... all right, what forces do we have to figure out what goes up here? You always draw a force diagram. So what forces do I have on the 5 kg mass? I'm going to have a force of gravity; I'll draw that straight down (Fg), and there's going to be an equal force, normal force, upward. So this normal force up should be equal to the force of gravity in magnitude, because this box is probably not going to be accelerating vertically. There's no real reason why it should be if this table is rigid. And there's one more force on this—there's a force to the right. That's going to be the force of tension; and if there's no friction on this table, then I have no leftward forces here. I'm ignoring air resistance since we usually ignore air resistance. So that's it. The only horizontal force I've got is T (tension), and I divide by the mass of the 5 kg box, which is 5 kg. But we got a problem—look! We don't know the acceleration of the 5 kg mass, and we don't know the tension. I can't solve this.

Normally, what you do in this case is you go to the vertical direction, the other direction, but that's not going to help me either. That's just going to tell me that the normal force is going to be equal to the force of gravity, and we kind of already knew that, so that doesn't help. So what do we do? Well, you might note this is only the equation for the 5 kg mass. So now you have to do this for the 3 kg mass. So let's come over to here; let's say that the acceleration of the 3 kg mass is going to be equal to the net force on the 3 kg mass divided by the mass of the 3 kg mass. And again, which direction should we pick? Well, this acceleration over here is going to be vertical, so let's solve this for the vertical direction. I'm going to add one more subscript, y, to remind myself, and you should do this too, so you remember which direction you're picking.

So what forces do I plug in here? You figure that out with a force diagram. I'm going to have a force of gravity on this 3 kg mass, and I'm going to have the same size of friction—or sorry, the same size as tension that I had over here. So the tension on this side of the rope is going to be the same as the tension on this side, assuming this pulley offers no resistance either by its mass or friction. So, assuming that its mass is negligible and there's basically no friction, then I'm going to have a tension, and that tension is going to be the same size, so I'll draw that coming upward. But it's not going to be as big as the force of gravity is on this 3 kg mass. I've got the force of gravity here; this tension's going to be smaller, and the reason is this 3 kg mass is accelerating downward. So these forces can't be balanced; the upward force of tension's got to be smaller than the force of gravity on this 3 kg mass. But this tension here should be the same as this tension here, so I'll plug those in.

So let's plug this in: a of the 3 kg mass in the y direction is going to be equal to... I've got two vertical forces: I've got tension up, so I'll make that positive because we usually treat up as positive; I've got gravity down, and so I'm going to have negative because it's a downward force of 3 kg times the acceleration on Earth (9.8 m/s²). Now what do we do? We divide by 3 kg because that's the mass. But I've still got a problem; I don't know this acceleration or this tension. So what do I do? You might notice, if you're clever, you'll say, wait! I've got my unknown on this side as acceleration and tension. My unknown on this side is acceleration and tension. It seems like I've got two equations, two unknowns; maybe we should combine them, and that's exactly how you do these.

So I've got tension in both of these equations. Let me solve for tension over here where it's kind of simple. I'll just get the tension equals 5 kg times the acceleration of the 5 kg mass in the x direction. So now I know what tension is—tension is equal to this—and that tension over on this side is the same as the tension on this side. So I can take this and I can plug it in for this tension right here. And let's see what we get. We get that the acceleration of the 3 kg mass vertically is going to equal... all right, I'm going to have a big mess on top. What am I going to get? I'm going to get... so T is the same as 5Ax, so I'll plug in 5 kg times the acceleration of the 5 kg mass in the x direction, and then I get all of this stuff over here, so I’ll get the rest of this right here. I just bring that down right there.

All right, now what do I have? I've got 3 kg on the bottom still, so I have to put that here. Are we any better off? Yeah, we're better, because now my only unknowns are acceleration. But these are not the same acceleration—look at this. The acceleration here is the acceleration of the 3 kg mass vertically; this acceleration here is the acceleration of the 5 kg mass horizontally. Now, here's where I'm going to have to make an argument, and some people don't like this, but it's crucial to figuring out this problem. The key idea is this: if this 3 kg mass moves down, let's say 1 m, let's say it moves downward 1 m—well, then this 5 kg mass had better move forward 1 meter. Because if it doesn't, then it didn't provide the 1 meter of rope that this 3 kg mass needed to go downward, which means either the rope broke or the rope stretched. We're going to assume that our rope does not break or stretch. That's kind of a lie; all ropes are going to stretch a little bit under tension. We're going to assume the stretch is negligible.

So the argument is that if this 3 kg mass moves downward a certain amount, this 5 kg mass has to move forward by that same amount in order to feed that amount of rope for this 3 kg mass to go downward by that amount. Otherwise, think about it, if this 5 kg mass just sat here and the 3 kg moved or the 3 kg moved farther than the 5 kg mass, then this rope is stretching or breaking. So if you believe that—if you don't believe it, pause it and think about it, because you got to convince yourself of that. If you believe that, then you can also convince yourself that, well, if the 3 kg mass was moving downward at a certain speed, let's say 2 m/s, then the 5 kg mass had better also be moving forward 2 m/s. Because otherwise, it wouldn't be feeding rope at a rate that this 3 kg needs to move downward at that rate.

And finally, if you believe all that, it's not too much harder to convince yourself that this 3 kg mass—no matter what its acceleration downward—must be... this 5 kg mass had better have the same magnitude of acceleration forward so that it's again feeding the rope. So this rope doesn't break or snap or stretch, because we're going to assume the rope doesn't do that. So what I'm saying is that the acceleration of the 3 kg mass in the y direction had better equal the magnitude... So these magnitudes have to be the same. The size—the sign doesn't have to be the same. So this 3 kg mass has a negative acceleration just because it points down, and we're assuming up is positive, down is negative. This 5 kg mass has a positive acceleration too; it's pointing to the right, and we're assuming rightward is the positive horizontal direction.

So they can have different signs, but the magnitudes better be the same, so that you're feeding this rope at a rate that the other one needs in order to move. And so we can say that the magnitudes are the same in this case since one is negative of the other. I can say that the acceleration of the 3 kg mass vertically downward is going to be equal to, let's say, the negative of the acceleration of the 5 kg mass in the x direction. I could have written it the other way; I could have wrote that the acceleration of the 5 kg mass in the x direction is the negative of the acceleration of the 3 kg mass in the y direction. They're just different by a negative sign is all that's important here.

Okay, so this is the link we need. This is it. So this allows us to put this final equation here in terms of only one variable, because I know what I've got. A3y on this left-hand side, I know A3y should always be A5x. So if I take this and just plug it in for A3y right here, I'm going to get A5x equals... well, all of this stuff, so I'll just copy this, save some time, copy paste, just equals all of that. All I did was plug in what I know A3y has to be equal to, because now look at—I've got one equation with one unknown. I just need to solve for what A5x is. It's on both sides, but so I'll need to combine these and then isolate it on one side.

So there's going to be a little bit of algebra here. Let's just take this; let's give ourselves some room, move this up just a little bit. Okay, so what do we do? We're going to solve for A5x. Let me just get rid of this denominator; let me multiply both sides by 3 kg. So I'm going to get -3 kg times A5 in the x direction. If I multiply both sides by 3 kg, then I get 5 kg times A5 in the x direction. And I've still got minus... all right, 3 times 9.8 is 29.4 Newtons, so we'll just turn this into what it's supposed to be: 29.4 Newtons. So let's combine our A terms now; let's move this -3A to the right-hand side by adding it to both sides and let's add this 29.4 to both sides.

So I'll get the 29.4 Newtons over here as a positive if I add it to both sides, and it'll disappear on the right-hand side. And then I'll add this term to both sides; I'll add a positive 3 kg times A to both sides, it'll disappear on the left and I'll get 5 kg times A5 in the x direction plus 3 kg times A5 in the x direction. Now we're close—look at that! On the right-hand side, I can combine these terms because 5A plus 3A is the same thing as 8A. So 29.4 Newtons equals 8 kg times... I'll put the parentheses here... times A5 in the x direction.

Now I can divide both sides by eight, and usually, we put the thing we're solving for on the left, so I'm just going to put that over here. I'll get 29.4 Newtons over 8 kg is equal to the acceleration of the 5 kg mass in the x direction. And if we calculate that, so if I put that into my calculator, 29.4 divided by 8, I get 3.67, so we'll just round— we'll just say that's 3.68. 3.68! Whoops! 3.68, and it's positive. That's good! We should get a positive because the 5 kg mass has a positive acceleration. So we get positive 3.68 m/s², but that's just the 5 kg mass. How do we get the acceleration of the 3 kg mass? Well, that's easy; it's got to have the same... it's got to have the same magnitude as the 5 kg mass. All I have to do is take this number—now I know what A5x is, so if I just plug that in right here, well then I know that A3y is just going to be equal to 3.68 m/s².

And I'm done! I did it! We figured out the acceleration of the 3 kg mass. It's negative; no surprise because it's accelerating downward. We figured out the acceleration of the 5 kg mass; it's positive, not a surprise! It was accelerating to the right. The way we did it, recapping really quick, we did Newton's Second Law for the 5 kg mass; that didn't let us solve. We did Newton's Second Law for the 3 kg mass; that didn't let us solve. In fact, it got really bleak because it seemed like we had three unknowns and only two equations. But the link that allowed us to make it so that we only had one equation with one unknown is that we plugged one equation into the other first, and we had to then write the accelerations in terms of each other.

That's because these accelerations are not independent: the accelerations have to have the same magnitude, and in this case, one had the opposite sign. So when we plugged that in, we have one equation with one unknown; we solve, and we get the amount of acceleration. So that's the hard way to do these problems. So in the next video, I'll show you the easy way to do these problems.