- [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.