If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Breaking down forces for free body diagrams

AP.PHYS:
INT‑3.B (EU)
,
INT‑3.B.2 (EK)
,
INT‑3.B.2.1 (LO)
Sal explains how to draw free body diagrams when forces are applied at an angle. How to find horizontal and vertical components of an angled force using trigonometry.

Want to join the conversation?

  • piceratops seed style avatar for user toukaandkaneki12347890
    At , is it true that, in this example, the normal force is also the "Newton Third Law Partner Force" of the gravity?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user Alexander Wu
      Nope. Action and reaction forces operate on different objects, not the same object. The reaction force to the normal force the surface exerts on the object (up) is the normal force the object exerts on the surface (down). The reaction force of the gravity earth exerts on the object (down) is the gravity the object exerts on earth (up). The reason earth isn’t affected much by this gravity is that its mass is much much greater, so its acceleration is negligible.

      So to sum up, Action and reaction forces operate on different objects and are the same type of force (reaction force of normal force is still normal force, reaction force of gravity is still gravity, reaction force of friction is still friction). In addition to that is Newtons third law: magnitude, direction, orientation, etc.
      (7 votes)
  • aqualine seed style avatar for user HANG
    In this video the 20N force is splitted into 2 forces of 10N and 10√3N respectively, but 10+10√3>20, how is that even possible?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leafers sapling style avatar for user Abosh Upadhyaya
      It is because when you split forces you have to use the Pythagorean theorem, or hypotenuse^2 = side^2 + side^2. In this case, the hypotenuse is 20 N, and the two sides are 10 N and 10sqrt(3).
      If you square the hypotenuse, you get 20^2, which equals 400 N.
      10^2 + (10sqrt(3))^2 = 100 + 300, which equals 400 N, or the hypotenuse squared.

      Therefore, those are the two components of the force, proven by the Pythagorean theorem.

      Think of it this way: if you have a triangle with two sides of length 3 and 4, the hypotenuse does not have to be 3 + 4 = 7; in fact, it will be 5. This is because 3^2 + 4^2 = 5^2 (9 + 16 = 25).
      (7 votes)
  • purple pi pink style avatar for user Neha Bhat
    why isn't there a force to the left as the x-component? kind of like the y component going down?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Ritvik Sharma
      As Sal mentioned at the start of the video, the surface the body is rested on is a flat and FRICTIONLESS surface, and thus, in this case, there won't be any friction working against the motion of the body (or a force to the left of the x-component as you said). Hope this helps, thanks.
      (3 votes)
  • piceratops ultimate style avatar for user mrinmoy
    can include tan theta here? somehow?
    because sinA/cosA=tanA
    (2 votes)
    Default Khan Academy avatar avatar for user
  • mr pink green style avatar for user Ramen23
    are free body diagrams and force diagrams the same thing? Why did we have to use trig when we could just use Pythagorean theorem?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • winston default style avatar for user Samuel Ammonius
    at , why is the force going up 10 + 10*sqrt(3), but the force going down is 2 different forces: 10, and sqrt(3)?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Allen Lin
      The force that is pulling the box down is the force of gravity, 10 N, and the vertical component of the angled force, 10sqrt(3) N. Thus, the total force pulling down on the box is 10 + 10sqrt(3) N. The force going up, the normal force, completely counteracts the force pulling down, so the normal force is also 10 + 10sqrt(3) N. Hope this helps.
      (3 votes)
  • blobby green style avatar for user ravindraravuri800
    why normal force is equal to sum of horizontal component and vertical component of forces in the video at
    (1 vote)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Ritvik Sharma
      Normal force, in this case, is the support force that is acting vertically-upwards on the body. Since it is given that the body has a vertical acceleration = 0m/s2, therefore, the object has net vertical force = 0N. Since there are 2 types of forces being exerted on the body (Weight force, Push force), to have a vertical net force = 0N; there would also be a force acting vertically-upwards that can cancel out these 2 forces. That force is the Normal force (Support force) and to cancel out, the Normal force = Weight force + Push force. Hope this helps, thanks.
      (2 votes)
  • duskpin ultimate style avatar for user Nathan Richardson
    On his free-body diagram at the end, couldn't you have combined the two downward forces to get 10+10sqrt3 N? He did so for the up components.
    (1 vote)
    Default Khan Academy avatar avatar for user
    • starky tree style avatar for user No NamePls
      Technically, you can, but since those 2 values belong to different forces, you have to separate them to show that this value is force of gravity and this value is the angled force's component.

      You only combine them when trying to solve something using those values since they're in the same direction. In the FBD though, it's best to leave them be so you can see what forces are acting on the object.
      (1 vote)
  • starky sapling style avatar for user Bom
    What is theta? the zero shape thing...
    (1 vote)
    Default Khan Academy avatar avatar for user
  • starky sapling style avatar for user Bom

    Why does the value of normal force increase instead of the box accelerating downwards?

    Why is the box accelerating to the right? (NO friction?)
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] Let's say we have some type of hard, flat, frictionless surface right over here. That's my drawing of a hard, flat, frictionless surface. And on that, I have a block, and that block is not accelerating in any direction. It is just sitting there. And let's say we know the weight of that block. It is a 10-newton block. So my question to you is, pause this video, and think about what are all of the forces that are acting on this block? All right, now let's work through this together. And to do it, I'm going to draw what's known as a free-body diagram to think about all of the forces. And the reason why it's called a free-body diagram is that we just focus on this one body. And we don't draw everything else around it, and we just draw the forces acting on it. And there's actually two typical ways of drawing a free-body diagram. I'll do them both. So first, I could draw it this way. So this is my block here. Now, I told you that it weighs 10 newtons. The weight of an object, that's the force of gravity acting on that object, and it would be downwards. So we have, from this 10 newtons right over here, we know that there is a downward force, the force of gravity acting on the mass of this object of 10 newtons. It has a magnitude of 10 newtons, and the force is acting downwards. We could say this is the magnitude of the force of gravity. And when you draw a free-body diagram, it's typical to show your vectors originating out of the center of that, of that object in your drawing. Now, my question to you is, is that the only force acting on this? If you think it is, what would happen to the object? Well, it would start or it would be accelerating downwards. But I just said that this is not accelerating in any direction. So there must be something that counteracts this, and there is. There's the normal force of this surface acting on the block. That surface is what's keeping the block from accelerating downwards. And I will do that with this vector right over here. So it's going to be going upwards, and it's going to have the exact same magnitude, just in the opposite direction. So I could say the magnitude of the normal force, the normal force is magnitude right over here, is also going to be equal to 10 newtons, but it's going upwards. And we can see that these two are going to net out, and so you have no net force acting in this vertical dimension. And I have no forces, I haven't thought about any or drawn any in the horizontal direction. And so that's why you have no net force in total, and this thing isn't going to be accelerated. Now, I mentioned that there's other ways to draw a free-body diagram. Another way that you might see it is like this, where you see the body. And from the outside of the body, you see the vectors being drawn. So in this situation, you have ten newtons downward, and you would have ten newtons upward. This is another way that you might see free-body diagrams drawn. Now, what I want to do is do something interesting to this block. Let me redraw it. So I have my surface here, my hard, frictionless surface, and it's flat. And I still have my block here. It's my 10-newton block. But now I'm going to apply a force to it. I am going to apply a force that is in this direction. It's in this direction, and its magnitude, let's say its magnitude is 20 newtons. And just so that we know the direction, this angle right over here, let's say that that is 60 degrees. I'll say theta is equal to 60 degrees. The magnitude of this force is equal to 20 newtons. So what would the free-body diagram now look like? Well, it might be tempting to just draw the force right on one of these free-body diagrams, something like that, something like that, and that would not be inaccurate. But we would have to watch out because this force is acting at an angle. So if we were to break it up into its horizontal and vertical components, some part of that force is acting downward. And so you're actually going to have a larger normal force to counteract that. And some other component is going to be working horizontally. And so what we want to do is actually break things up. 'Cause if you leave it in this angle, it gets very, very confusing. So what I want to do is I want to break up this new blue force into its horizontal and vertical components. And to do so, we just have to remember a little bit of our basic trigonometry. If I have a force like this, if I have a force like this and it is acting at an angle theta right over here, with the horizontal, and I want to break it up into its horizontal and its vertical components, and its vertical components, if the magnitude of the hypotenuse is capital F, then the magnitude of the adjacent side to this angle, this comes straight out of soh-cah-toa, from our right triangle trigonometry, it would be the magnitude of our hypotenuse times the cosine of this angle. And the magnitude of the vertical component, that would be the magnitude of our hypotenuse times the sine of that angle. And you could think about it the other way as well, if the force was like this, where it's just going in the opposite direction. But once again, you have this angle theta. And now the components would look like this, where the vertical component would have the same magnitude, but now it would be pointing downwards. And the horizontal component would have the same magnitude, but now it is pointing to the left. It is the same idea. If this force has magnitude F that's represented by the hypotenuse of this triangle, then the magnitude of our horizontal component is still going to be F cosine theta. The vector is not going in the other direction. And the magnitude of our vertical component here is going to be F sine theta. And so what about this scenario over here? Well, in this scenario, our vertical component is gonna look like this, and our horizontal component is going to look like this. And so what's the magnitude of our horizontal component? Well, it's going to be the magnitude of our hypotenuse times the cosine of the 60 degrees. So this is going to be 20 newtons times the cosine of 60 degrees. And it's really helpful, both trigonometry and physics classes, to know the values of your sines, your cosines, and your tangents at angles like zero degrees, 30 degrees, 60 degrees, 90 degrees, 45 degrees. You could use a calculator here, but it's useful to know that the cosine of 60 degrees is 1/2. So 20 times 1/2, this is going to be equal to 10 newtons. And if we want to know the magnitude of our vertical component, well, this is going to be 20 newtons times the sine of 60 degrees. Once again, this is useful to know. It is square root of three over two. And so 20 divided by two is 10. So this is going to be 10 square root of three newtons. And so we can use that information. We've broken up our original vector into two component vectors that, if you took their sum, you would get your original one. But what's useful now is that we've broken it up into vectors that are parallel or perpendicular to our surface, which will allow us to think about what nets against these things that I already have in place. So let me draw that. So actually, I'll first draw this type of free-body diagram. So there is my block. And I have the force of gravity acting on it downwards. I will draw it right over here. So that's 10 newtons. That is the force of gravity acting downwards. Now, is that the only thing that I have acting downwards? No, I also have the vertical component of this applied force. And so this is going downwards 10 square roots of three newtons. And these aren't drawn perfectly to scale, but hopefully you get the idea. So this is 10 square roots of three newtons, just like that. And now what is our normal force, assuming that our surface is able to not be compressed in any way, that it is a hard, frictionless surface? Well, now our normal force is going to counteract both of these forces. Our normal force might look something like this. Once again, I haven't drawn it completely to scale. But this would be 10 plus 10 square roots of three newtons. And what about now in the horizontal direction? Well, I have this blue vector right here, and so that is going to the right with a magnitude of 10 newtons, 10 newtons. And so now you can hopefully appreciate why a free-body diagram is really, really, really useful. Just looking at this, I can predict what's going to happen to my block. I would say, look, this upward force is completely netted out by these downward forces, or these downward forces are completely netted out by this upward force. And the only net force that I have is ten newtons to the right. And so that lets me know that, hey, since I have a net force to the right, this block is going to accelerate to the right.