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Free body diagram with angled forces: worked example

Sal draws a free body diagram for a box held stationary against a wall with a force at an angle theta.

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  • duskpin tree style avatar for user Vaishnavi  Dumbali
    I don’t get it. Friction is against the direction of motion. So why is the frictional force(Ff) pointing upwards in the direction of motion instead of downwards?
    (14 votes)
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    • female robot grace style avatar for user Afrika.Hooks
      So in this scenario, the vertical component of the angled force is less than gravity. Therefore, if there was no friction, then the block would accelerate downwards. Since friction always counteracts the motion, the friction force would point upwards.

      In order for the friction force to point downwards, the block would have to be trying to accelerate upwards. That would happen if the horizontal component of the angled force were greater than gravity.

      Hope this helps!
      (10 votes)
  • mr pink green style avatar for user Ramen23
    whats the purpose of doing math without numbers?
    (0 votes)
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  • female robot amelia style avatar for user I love Khan Academy
    What about the equal and opposite force exerted by block which is pulling the earth?
    (1 vote)
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    • eggleston blue style avatar for user IsotonicFlaccidCell21
      You are right in saying that the block is pulling up the earth, but it is really small. THis is seen through f=ma. THe forces have to be the same according to newton's 3rd law, but the mass would be far far far far far larger - the earth is very heavy, thus the acceleration is really really really really small as mass is indirectly proportional to acceleration when the force is the same. Because it is too small, it is mostly disregarded.
      (4 votes)
  • duskpin ultimate style avatar for user BeyMaster82
    At , Sal says that he doesn't draw the arrow atop F sub g because he's talking about the magnitude of the vector. Do you always need to draw the arrow? It seems like a bit of extra work for the same purpose. Thanks!
    (1 vote)
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    • starky tree style avatar for user Dragon
      F sub g refers to the magnitude of the vector, while F sub g with an arrow refers to the whole vector (both magnitude and direction). Here the downward arrow already tells us the direction so it is fine to write only F sub g, which tells us the magnitude of the vector Fsub g. We need to draw the arrow if we want to refer to the “whole” vector and not just its magnitude.
      (4 votes)
  • hopper happy style avatar for user Averell Chen
    Non of those forces have to equal our right or is it just this particular set of question?
    (2 votes)
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    • female robot amelia style avatar for user Johanna
      (I know this is a year late, but I’ll answer the question anyway just in case someone else has the same one.)

      The forces don’t always have to equal (even) out or add up to a net force of zero when we do these kinds of problems, but they do cancel out here.

      In this problem, we’re told that the box is “held stationary”, meaning it’s at rest. Staying stationary/at rest means that the box has 0 velocity and 0 acceleration. Newton’s 2nd law says that the sum of the forces equals mass times acceleration ΣF=ma

      We know the acceleration is 0m/s^2, so we can substitute it to get ΣF=m(0), which multiplies to ΣF=0.

      ΣF=0 means that the sum of the forces equals 0, or that all of the forces have to add up to zero (cancel out). Even though the forces won’t always cancel out, that’s why they do here.
      (1 vote)
  • male robot donald style avatar for user Sivakumar Kunapuli
    Why is the normal force going sideways?
    (2 votes)
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    • piceratops ultimate style avatar for user Charles Araya
      According to Newton's 3rd law of motion, every force action (the horizontal component Fcosθ) exerted on an object (the box on the wall), has an opposite force reaction (the wall on the box) equal in magnitude and opposite in direction. This force is called the normal Force.

      Hope it helps!
      (1 vote)
  • mr pink red style avatar for user ilesha253
    If Fsinθ + Ff= Fg then wouldn't there be no movement because the forces are equal?
    (1 vote)
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  • duskpin tree style avatar for user Jacob A
    so what do i do if force isn't given in newtons? one of my homework questions has the magnitude of an angled force given in mg (milligrams) and no matter where i look, i can't find any help on how to deal with that.
    edit: turns out the force was 1/2 of the mass * gravity, which had been written as 0.500mg. :( lost ten marks because the textbook couldn't write out full words...
    (1 vote)
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    • eggleston blue style avatar for user IsotonicFlaccidCell21
      If you have mg (milligrams), then you know that is the mass. To find the downwards force, you need to find out weight, which is w=mg, weight = mass x gravitational field strength. So if the object is on earth, you would do mass x 9.81 or mass x 10. Remember weight is a force, and is measured in newtons.
      (2 votes)
  • winston baby style avatar for user Yash Singh
    If we did not have the assumption that F*sin(theta) < F_g that he shows at what would be the default assumption?
    (1 vote)
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  • leaf green style avatar for user annie.w0803
    In the practice, I don't know why acceleration becomes 0 when solving for Fn, but stays the same when solving for mass.
    (1 vote)
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Video transcript

- [Instructor] So what we have depicted here. We have a block and let's say that this block is completely stationary and it's being pushed up against this non-frictionless wall. So this wall does have friction with the block. It's being pushed by this force of magnitude F and its direction makes an angle of theta with the horizontal. What I want you to do is pause this video and construct a free body diagram for this block and include the vertical and horizontal components of this force right over here, but also include other things. Include the force, include the force of gravity acting on this block. Include the force of friction acting on this block and include the normal force of the wall acting on the block as well. Pause the video and try to have a go at it. So before I even start to draw the free body diagram, let's break down this force into its vertical and horizontal components. So the first thing, let me do its vertical component. So it would look like that, and it's horizontal component would look like this. And so what's the magnitude of the vertical component? Well it is opposite this angle that we know. This is the angle that is theta and so this is going to be, the vertical component is going to be F sine of theta. We've seen that in previous videos and it comes straight out of right triangle trigonometry. Encourage you to review that if this looks unfamiliar. And the magnitude of the horizontal component, that is going to be F cosine theta. This side right over here is adjacent to the angle, SOH CAH TOA. And now with that out of the way we can draw our free body diagram. So let me draw that free body, let me draw that block and I'm really just gonna focus on the block only. And we know a couple of things that are going on. We know that we have this horizontal force, F cosine theta, so let me draw that. So the magnitude there is F cosine theta. We know we have this vertical force, F sine theta, so let me draw that. So this would be F and that one is actually a little bit shorter, it's obviously not drawn perfectly to scale, but this would be F sine theta. Now let's think about these characters right over here. We have the force of gravity and so that's going to be acting downward. So it would look something like this. So we have, and it would have magnitude F sub g. I'm not drawing the arrow now because I'm just talking about the magnitude of this vector. Here I'm referring to the entire vector, I'm referring to its magnitude and direction. Now what about the force of friction? Let's assume that we're in a situation where the magnitude of the vertical component of this applied force, F sine theta is less than the magnitude of the force of gravity. Well in that situation if there was no friction, the block would start accelerating downward, because you would have a net force downward. We haven't talked about the forces to the left and right yet. But as we mentioned, this thing is stationary and the force of friction is going to act against the direction of motion. And so in this situation, the force of friction will act upwards, so we would have a force of friction just like that and its magnitude would be F sub lowercase f. And then last but not least what about the normal force? Well if this block is not accelerating in any direction that means that the normal force must perfectly counteract this force to the right which is the horizontal component of this applied force. And so our normal force is going to go to the left and it would look like this. So it's magnitude is F sub capital N. So there you have it, we have drawn a free body diagram for this scenario right over here. If these two were equal, then you would have no force of friction or the force of friction would be zero. There would be nothing for it to be counteracting. These two would perfectly cancel out. And if the magnitude of the vertical component of the applied force were greater than the force of gravity without friction, it would start to accelerate upwards and so the force of friction would act against that motion, but let's just go with this scenario right over here and start to appreciate why free body diagrams are so useful. Because we can start to set up equations that relate these magnitudes. We can say look if this box is not accelerating in any way, there's zero net forces in the horizontal and the vertical directions, well then we could say that F sub N completely counteracts F times cosine of theta. So we could say F sub N is equal to, is equal to F cosine theta, and we could also say that F sine theta, F sine theta plus the magnitude of the force of friction. Plus the magnitude of the force of friction that these would completely counteract the magnitude of the force of gravity, because it's going in the opposite direction. So these would be equal to F sub g. So once you set up equations like this, if you know all but one of these variables, you can figure out the other ones which is very useful in physics.