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

So in the previous video, we solved this problem the hard way. Maybe you watched it, maybe you didn't. Maybe you just skipped right to here and you're like, "I don't even want to know the hard way. Just show me the easy way, please." Well, that's what we're going to talk about now.

Turns out there's a trick. The trick is, after you solve this problem the hard way with a 5 kg mass and a 3 kg mass, when you find the acceleration, what you get is this: that the acceleration of the 5 kg mass is just ( \frac{29.4}{8} ) kg. But when you do enough of these, you might start realizing, wait a minute, 29.4 newtons? That was just the force of gravity pulling on this 3 kilogram mass.

In other words, the only force that was really propelling this whole entire system forward, at least the only external force propelling it forward, is 8 kg down here. You might be like, "Wait, 8 kg? That's just 5 kg + 3 kg. Is that just a coincidence, or is this telling us something deep and fundamental?" It's not a coincidence.

Turns out you'll always get this: that what you'll end up with after solving this the hard way is that, in the very end, you'll get all the external forces added up here where forces that make it go, like the force of gravity, end up being positive. Forces that try to resist the motion, so if there was friction, that would be an external force that tries to resist motion, would be up top. Then you get the total mass on the bottom.

This makes sense. The acceleration of this entire system, if we think about it as a single object, will only depend on the external forces. In this case, the only external force making it go was this force of gravity right here. You might object; you might be like, "Wait, what about this tension right here? Isn't the tension pulling on this 5 kg mass making the system go?"

It is, but since it's an internal force, now if we're treating this entire system as our one object, since this tension is pulling trying to make it go, you've got another tension over here resisting the motion on this mass trying to make it stop. That's what internal forces do. They're always equal and opposite on one part of the object and the other, so you can't propel yourself forward with an internal force.

So these end up canceling out. Essentially, the only force you have in this case was the force of gravity on top, only external forces, and the total mass on the bottom. And that's the trick. That's the trick to quickly find the acceleration of some system that might be complicated if you had to do it in multiple equations and multiple unknowns, but much, much easier once you realize this.

The trick sometimes is called just treating systems as a single object. So let me just show you really quick. If that made no sense, let me just show you what this means. If we just get rid of this, so what I'm claiming is this: if you ever have a system where multiple objects are required to move with the exact same magnitude of acceleration, right? Because maybe they're tied together by a rope or maybe they're pushing on each other, maybe there are many boxes in a row, and these boxes all have to be pushed at the same acceleration because they can't get pushed through each other.

If there's some condition where multiple objects must have the same magnitude of acceleration, then you can simply find the acceleration of that system as if it were a single object. I'm writing ( S_y ) for system by just using Newton's second law. But instead of looking at an individual object for a given direction, we're just going to do all of the external forces on our system treated as if it were a single object, divided by the total mass of our system.

So when you plug in these external forces, these are forces that are external. External means not internal to the system. So if I think of this 5 kg box and this 3 kg box as a single mass, tension would be an internal force because it's applied internally between these two objects, between objects inside of our system.

But the force of gravity on the 3 kg mass, that's an external force because that's the Earth pulling down on the 3 kg mass, and the Earth is not part of our system. Similarly, the normal force is an external force, but it's exactly canceled by the gravitational force. So even though those are external, they're not going to make it in here. I mean, you could put them in there, but they're just going to cancel anyway.

We only look at forces in the direction of motion. If it's a force that causes motion, we're going to make that a positive force. If it's a force in the direction of motion, like this force of gravity is, we make those positive forces. So forces will be plugged in positively into here if they make the system go. And that might seem weird. You might be like, "Wait, how do I decide if it makes the system go?"

Well, just ask yourself: is that force directed in the same direction as the motion of the system? So we're just saying the system's going to accelerate if there are forces that make it go, and we're going to make negative, we're going to plug in negative forces. The forces that make the system stop or resist the motion of the system.

So maybe I should say resist the motion of the system. In this case, for this one down here, I don't have any of those. So resist motion of the system, I don't have any of those. I could have if I had a force of friction, then there'd be an external force that resists the motion. I would plug in that external force as a negative because it resists the motion.

So even though this might sound weird, it makes sense if you think about it. The acceleration of our system, treated as a single object, is only going to depend on the forces that try to make the system go and the forces that try to make the system stop or resist the motion. So if we add those accordingly with positives and negatives and we divide by the total mass, which gives a total measure of the inertia of our system, we'll get the acceleration of our system.

It makes sense, and it works. Turns out this always works and it saves a ridiculous amount of time. For instance, if we wanted to do this problem, if you just gave me this problem straight away and you were told to do this however you want, I would use this trick. I would say that the acceleration of this system, which is composed of this 5 kg mass and our 3 kg mass, is just going to be equal to... I'd ask myself what force makes this system go? What force drives this system?

It's this force of gravity on the 3 kg mass that's driving this system, right? If you took this force away, if you eliminated that force, nothing's going to happen here. This is the force making the system go. So I'd put in my 3 kg times 9.8. At this point, you might be like, "Well, okay, that gravity made it go. Should I include this gravity too?"

But no, that gravity is perpendicular to the motion for one. So this gravity isn't making the system go; that's just causing this mass to sit on the table. For two, it's canceled by that normal force, so those cancel anyway. Even though they're external forces, this is it. This is the only one that drives the system. So I'd put that in here and I divide by my total mass because that tells me how much my system resists through inertia changes in velocity.

And this is what I get. I get the same thing I got before; I get back my ( 3.68 , \text{m/s}^2 ) and I get it in one line. I mean, this trick is amazing, and it works! It works in every example where two masses or more masses are forced to move with the same acceleration.

So this is great. This will save you a ton of time. This is supposed to be a three here. And to show you how useful it is, let's say there was friction. Let's say there was a coefficient of friction of 0.3. Well, now I'd have a frictional force, so there'd be an external frictional force here. It'd be applied to this 5 kg mass.

I'd have to subtract it up here so if I get rid of this, so it's not going to be ( 3.68 ) anymore. I'm going to have a force of friction that I have to subtract. So minus ( \mu_k ), so the force of friction... I'll just put "force of friction," and so to solve for the force of friction, the force of friction is going to be equal to... Well, I know ( 3 \times 9.8 ) is... Let me just write this in here: ( 29.4 ) Newtons minus the force of friction is given by... So there's a formula for force of friction. The force of friction is always ( \mu_k F_N ).

So the force of friction on this 5 kilogram mass is going to be ( \mu_k ), which is 0.3, so it's going to be ( 0.3 \times F_N ), not the normal force on our entire system. I don't include this 3 kg mass; it's only the normal force on this 5 kg mass that's contributing to this force of friction here. So even though we're treating the system as a whole, we still have to find individual forces on these individual boxes correctly.

So it won't be the entire mass that goes here. The normal force on the 5 kilogram mass is just going to be ( 5 ) kg times ( 9.8 , \text{m/s}^2 ). I divide by my total mass down here because the entire mass is resisting motion through inertia. And if I solve this for my acceleration of the system, I get ( 1.84 , \text{m/s}^2 ).

So this is less than our ( 3.68 ) and that makes sense. Now there's a resistive force, a resistive external force trying to prevent the system from moving. But you have to be careful. What I'm really finding here, I'm really finding the magnitude of the acceleration. This is just giving me the magnitude. If I'm playing this game where positive forces are ones that make it go and negative forces are ones that resist motion, external forces that is, I'm just getting the magnitude of the acceleration.

Individual boxes will have that magnitude of the acceleration, but they may have positive or negative accelerations. In other words, this 5 kg mass accelerating to the right is going to have a positive acceleration. In other words, the acceleration of the 5 kg mass will be positive ( 1.84 ), and the acceleration of the 3 kg mass, since it's accelerating downward, will be negative ( 1.84 , \text{m/s}^2 ).