Free body diagram with angled forces: worked example | AP Physics 1 | Khan Academy

So what we have depicted here, we have a block, and let's say that this block is completely stationary. 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. 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. The first thing, let me do its vertical component, so it would look like that, and its 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. I 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, so toa.

And now with that out of the way, we can draw our free body diagram. So let me draw that free body or let me draw that block, and I'm really just going to focus on the block only. 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's actually a little bit shorter. It's obviously not drawn perfectly to scale, but this would be F sine theta.

And 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. 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 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. So in this situation, the force of friction will act upwards, and so we would have a force of friction just like that. Its magnitude would be F sub lower case 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 its 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; it 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 could say, look, if this box is not accelerating in any way, if 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 F cosine theta.

And we could also say that 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.

And 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.