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Current time:0:00Total duration:5:43

- [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.

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