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## Physics library

### Course: Physics library>Unit 3

Lesson 7: Treating systems

# Three box system problem

In this video David explains how to easily find the acceleration of a three box system by treating it as a single mass. Created by David SantoPietro.

## Video transcript

- So, check out this problem. You've got a 12 kilogram mass sitting on a table, and on the left hand side it's tied to a rope that passes over a pulley and that rope gets tied to a three kilogram mass. And then on the right side of this 12 kilogram box, you've got another rope and that rope passes over another pulley on the right and is tied to the five kilogram box over here. The question is, what's the acceleration of the 12 kilogram box? Let's make it even harder. Let's say there's a coefficient of kinetic friction between this 12 kilogram box and the table of 0.1. Now you're looking at a really hard problem if you try to solve this the hard way. And by the hard way, I mean using Newton's second law for each box individually and then trying to solve what'd you end up with is at least three equations and three unknowns because you're gonna have three different accelerations. For each of these, you'll have two different tensions cause this left rope is under a different tension from the right rope now. It's only when you have a single rope can you say that it's the same tension. This problem's gonna be hard. There's gonna be tons of Algebra mistakes potentially. And so to avoid that, we can solve this the easy way. And if you remember, the easy way is just by saying, well, let's treat all of these boxes as if they're a single object. And we can do that cause they're all gonna have the same magnitude of acceleration that I'm just calling a system. That's gonna be the magnitude of acceleration of our system. All these boxes will accelerate with the same magnitude. Some may have negative accelerations, some may have positive accelerations like these, but they're gonna all have the same magnitude of acceleration cause we're gonna assume that these ropes don't break. And if they broke, then they'll be different magnitudes or if they stretch, but we're assuming that doesn't happen. And the way we can find this is by just saying, well, if this is just a single object, I don't have to worry about any internal forces, now these tensions become internal forces. And those don't make a system accelerate, only external forces are gonna make a system accelerate. So, all I have to do is find out what are all the external forces that try to make this system go, try to accelerate it, and ones that try to prevent acceleration. I'll call this F external and then I divide by the total mass because this is just simply Newton's second law as if this were one big object. So, what are my external forces? Well, the force that makes it go is gonna be this five kilogram's force of gravity so I'm gonna have a force of gravity over here. That tries to propel the system forward. This is the one that's gonna be driving the system. If I let go of these boxes, it's gonna start shifting in this direction because this five kilogram mass has a larger force of gravity than this three kilogram mass. So, I'm gonna include that as a positive. I'm just gonna define direction of motion as positive, 'cause it's easy. You could do it differently if you wanted to. You could find the other way as positive. So five times 9.8 meters per second squared is how you find this force of gravity. Are there any other forces that are propelling this forward? No, no external forces are. So, are there any forces that are trying to reduce the acceleration? Yeah, there's this force of gravity over here. This force of gravity on the three kilogram mass is trying to prevent the acceleration because it's pointing opposite the direction of motion. The motion of this system is upright and down across this direction but this force is pointing opposite that direction. This force of gravity right here. So, I'm gonna have to subtract three kilograms times 9.8 meters per second squared. Am I gonna have any other forces that try to prevent the system from moving? You might think the force of gravity on this 12 kilogram box, but look, that doesn't really, in and of itself, prevent the system from moving or not moving. That's perpendicular to this direction. I've called the direction of motion, this positive direction. If it were a force this way, if it were a force this way or a force that way it'd try to cause acceleration of the system. This force of gravity just gets negated by the normal force, so I don't even have to worry about that force. So, are there any forces associated with the 12 kilogram box that try to prevent motion? It turns out there is. There is going to be a force of friction between the table because there's this coefficient of kinetic friction. So, I've got a force this way, this kinetic frictional force, that's gonna be, have a size of Mu K times f n. That's how you find the normal force and so this is gonna be minus, the Mu K is 0.1 and the normal force will be the normal force for this 12 kilogram mass. So, I'll use 12 kilograms times 9.8 meters per second squared. You might object, you might say, "Hey, hold on, 12 times 9.8, that's the force of gravity. "Why are you using this force? "I thought you said we didn't use it?" Well, we don't use this force by itself, but it turns out this force of friction depends on this force. So, we're really using a horizontal force, a force that tries to prevent motion, which is why we've got this negative sign here, but it's a horizontal force. It just so happens that this horizontal force depends on a vertical force, which is the normal force. And so that's why we're multiplying by this .1 that turns this vertical force, which is not propelling the system, or trying to stop it, into a horizontal force which is trying to reduce the acceleration of the system. That's why I subtracted and then I divide by the total mass and my total mass is gonna be three plus 12 plus five is gonna be 20 kilograms. Now, I can just solve. If I solve this, I'll get that the acceleration of this system is gonna be 0.392 meters per second squared. So, this is a very fast way. Look it, this is basically a one-liner. If you could put this together right, it's a one-liner. There's much less chance for error than when you're trying to solve three equations with three unknowns. This is beautiful. When you apply this though, be careful. The acceleration of the five kilogram mass would be negative 0.392 because it's accelerating downward. The acceleration of the 12, we'd call positive 0.392 because it's accelerating to the right and we typically call rightward accelerations positive. And then the three kilogram mass also would have positive .392 because it's accelerating upward.