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Masses on incline system problem

In this video David explains how to find the acceleration and tension for a system of masses involving an incline. Created by David SantoPietro.

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  • blobby green style avatar for user jamie_chu78
    Wait, what's an internal force? I presume gravity is an external force, as well as friction, as well the force of large dragons trying to impede your motion. What are forces that come from within? It almost sounds like some sort of chinese proverb.
    (12 votes)
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    • leaf green style avatar for user Mark Zwald
      Internal forces result in conservation of momentum for the defined system, and external forces do not. It depends on what you have defined your system to be. Example, if you are in space floating with a ball and define that as the system. Then when you apply a force to the ball to throw it (and the ball applies a force to you), then the total momentum of the system remains unchanged since all those forces were internal. Now if something from outside your system pulls you (ex. gravity from planet), the system's momentum is no longer conserved because that additional force was external to the system, but if you expand the system to include the planet and take into account its momentum, then the total momentum of the larger system remains conserved. So it depends how you define what your system is, whether a force is internal or external to it.
      (15 votes)
  • aqualine ultimate style avatar for user patri
    At , why is tension considered an internal force? What is the difference between internal and external forces?
    (6 votes)
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    • orange juice squid orange style avatar for user Paige
      My teacher taught me to just draw a big circle around the whole system you're trying to deal with. Anything outside of that circle is external, and anything inside is internal. If you drew a circle around both of the boxes and the string attaching them, the tension force is inside of the circle and thus internal.
      (10 votes)
  • blobby green style avatar for user Slifer123574
    Are the tensions in the system considered Third Law Force Pairs?
    (6 votes)
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  • hopper jumping style avatar for user Sam D
    I've watched all the videos on treating systems as a whole and one thing which I don't get is why don't we consider the coefficient of static friction along with the coefficient of kinetic friction? I mean, before kinetic friction starts acting on the box there's got to be static friction, so what am I missing here?
    (6 votes)
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    • male robot hal style avatar for user safwanashiq123
      CORRECT! But , We're looking at a problem(s) where the beginning of the problem(s) states that the objects have already been in motion before we looked/observed at it, Therefore,We consider Only The Kinetic Friction

      Or if we you are still confused,
      THE OBJECT IS SLIDING NOT ROLLING OR ANYTHING ELSE!
      (6 votes)
  • leafers sapling style avatar for user KuemerleNo
    Are the two tension forces equal? I know at he said that the internal forces cancel, but is that the same thing as saying they are equal in separate directions?
    (5 votes)
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    • piceratops tree style avatar for user AThont
      A pulley is a rotating piece that is meant to convert horizontal tension force into vertical tension force. In these videos, we are assuming there's no resistance from the pulley, so the tension of one string is "converted" into the tension of the other string with no force being subtracted. In short, yes they are equal, but in different directions.
      (6 votes)
  • aqualine tree style avatar for user Hello
    How exactly do we determine which body is more massive? Do we compare the vertical components of the gravitational forces on the two bodies or something?
    (4 votes)
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  • female robot ada style avatar for user thaveeshakithulampitiya
    What if there's a friction in the pulley.. Does it affect the whole system
    (3 votes)
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  • blobby green style avatar for user Jihoolee0716
    Is the tension for 9kg mass the same for the 4kg mass? When David was solving for the tension, why did he only put the acceleration of the system 4.95m/s^2 as negative, but not the acceleration due to gravity 9.8m/s^2?
    (3 votes)
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  • aqualine tree style avatar for user Jurgen Gjonçari
    In this video and in other similar exercises, why don't you consider the static coefficient of friction too?
    (1 vote)
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  • blobby green style avatar for user Olivia Vuong
    I think there's a mistake at minutes, how did he get 4. 75m/s?
    I've been calculating it over and over it it keeps appearing to be 3.7m/s.
    What's going on?
    (1 vote)
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Video transcript

- So just to show you how powerful this approach is of treating multiple objects as if they were a single mass let's look at this one, this would be a hard one. We've got a 9kg mass hanging from a rope that rope passes over a pulley then it's connected to a 4kg mass sitting on an incline. And this incline is at 30 degrees, and let's step it up let's make it hard, let's say the coefficient of kinetic friction between the incline and the 4kg mass is 0.2 And that's the coefficient. so there's going to be friction as well. If you tried to solve this the hard way it would be challenging, it's do-able but you're going to have multiple equations with multiple unknowns, if you try to analyze each box separately using Newton's second law. But because these boxes have to accelerate at the same rate well at least the same magnitude of acceleration, then we're just going to be able to find the system's acceleration, at least the magnitude of it, the size of it. This 4 kg mass is going to have acceleration in this way of a certain magnitude, and this 9 kg mass is going to have acceleration this way and because our rope is not going to break or stretch, these accelerations are going to have to be the same. So we get to use this trick where we treat these multiple objects as if they are a single mass. And the acceleration of the single mass only depends on the external forces on that mass. So we're only looking at the external forces, and we're gonna divide by the total mass. So what would that be? If we wanted to find the acceleration of this 4 kg mass, let's say what the magnitude of this acceleration This 9 kg mass is much more massive than the 4 kg mass and so this whole system is going to accelerate in that direction, let's just call that direction positive. So that's one weird part about treating multiple objects as if they're a single mass is defining the direction which is positive is a little bit sketchy to some people. We're just saying the direction of motion this way is what we're calling positive. And that works just fine, so when I plug in and go to solve for what is the acceleration I'm gonna plug in forces which go this way as positive and forces which go the other way as negative. What do I plug in up top? What forces make this go? The force of gravity on this 9 kg mass is driving this system, this is the force which makes the whole system move if I were to just let go of these masses it would start accelerating this way because of this force of gravity right here. So that's going to be 9 kg times 9.8 meters per second squared and that's going to be positive because it's making the system go. There's no other forces that make this system go. So now I'm only going to subtract forces that resist the acceleration, what forces resist the acceleration? The gravity of this 4 kg mass resists acceleration, but not all of the gravity. The gravity of this 4 kg mass points straight down, but it's only this component this way which resists the motion of this system in this direction. What is this component? This is "m" "g" "sin(theta)" so if that doesn't make any sense go back and look at the videos about inclines or the article on inclines and you'll see the component of gravity that points down an incline parallel to the surface is equal to "m" "g" "sin(theta)" so I'm gonna have to subtract 4 kg times 4 kg times 9.8 which is "g" times sin of the angle, which is 30 degrees. We need more room up here because there are more forces that try to prevent the system from moving, there's one more force, the force of friction is going to try to prevent this system from moving and that force of friction is gonna also point in this direction. It's not equal to "m" "g" "sin(theta)" it's equal to the force of kinetic friction "mu" "k" times "Fn" and the "mu" "k" is going to be 0.2, you have to be careful because the "Fn" is not just equal to "m" "g" the reason is that on an incline the normal force points this way so the normal force doesn't have to counteract all of gravity on an incline it just has to counteract that component of gravity that's directed perpendicular to the incline and that happens to be "m" "g" "cos(theta)" for an object on an incline and if that makes no sense go back and look at the video on inclines or look at the article on inclines and you'll see that this component of gravity pointing into the surface is "m" "g" cosine that means that normal force is "m" "g" cosine. Because there's no acceleration in this perpendicular direction and I have to multiply by 0.2 because I'm not really plugging in the normal force up here or the force of gravity in this perpendicular direction. I'm plugging in the kinetic frictional force this 0.2 turns this perpendicular force into this parallel force, so I'm plugging in the force of kinetic friction and it just so happens that it depends on the normal force. That's why I'm plugging that in, I'm gonna need a negative 0.2 times 4 kg times 9.8 meters per second squared. And then I need to multiply by cosine of the angle in this case the angle is 30 degrees. Alright, now finally I divide by my total mass because I have no other forces trying to propel this system or to make it stop and my total mass is going to be 13 kg. You might object and think wait a minute, there's other forces here like this tension going this way, why don't we include that? Well that's internal force and the whole benefit and appeal of treating this two-mass system as if it were a single mass is that we don't have to worry about these internal forces, it's there but that tension is also over here and on this side it's resisting the motion because it's pointing opposite the directional motion. On this side it's helping the motion, it's an internal force the internal force is canceled that's why we don't care about them, that's what this trick allows us to do by treating this two-mass system as a single object we get to neglect any internal forces because internal forces always cancel on that object. So if we just solve this now and calculate, we get 4.75 meters per second squared is the acceleration of this system. So this 4 kg mass will accelerate up the incline parallel to it with an acceleration of 4.75 meters per second squared. This 9 kg mass will accelerate downward with a magnitude of 4.75 meters per second squared. Remember if you're going to then go try to find out what one of these internal forces are, we neglected them because we treated this as a single mass. But you could ask the question, what is the size of this tension? Often that's like a part two because we might want to know what the tension is in this problem, if we do that now we can look at the 9 kg mass individually so I can say for just the 9 kg mass alone, what is the tension on it and what are the force? We can find the forces on it simply by saying the acceleration of the 9 kg mass is the net force on the 9 kg mass divided by the mass of the 9 kg mass. Now this is just for the 9 kg mass since I'm done treating this as a system. This trick of treating this two-mass system as a single object is just a way to quickly get the magnitude of the acceleration. Now that I have that and I want to find an internal force I'm looking at just this 9 kg box. And I can say that my acceleration is not 4.75 but -4.75 if we want to treat downwards as negative and upwards as positive then I have to plug this magnitude of acceleration in as a negative acceleration since the 9 kg mass is accelerating downward and that's going to equal what forces are on the 9 kg mass: I called downward negative so that tension upwards is positive, but minus the force of gravity on the 9 kg mass which is 9 kg times 9.8 meters per second squared divided by 9 kg. I don't divide by the whole mass, because I'm done treating this system as if it were a single mass and I'm now looking at an individual mass only so we go back to our old normal rules for newton's second law where up is positive and down is negative and I only look at forces on this 9 kg mass I don't worry about any of these now because they are not directly exerted on the 9 kg mass and at this point I'm only looking at the 9 kg mass. So if I solve this now I can solve for the tension and the tension I get is 45.5 newtons which is less than 9 times 9.8 it's got to be less because this object is accelerating down so we know the net force has to point down, that means this tension has to be less than the force of gravity on the 9 kg block. So recapping, treating a system of masses as if they were a single object is a great way to quickly get the acceleration of the masses in that system. Once you find that acceleration you can then find any internal force that you want by using Newton's second law for an individual box. You're done treating as a system and you just look at the individual box alone like we did here and that allows you to find an internal force like the force of tension.