Now that we've discussed the gravitational force,
let's move to the normal force.
If you have books sitting on a table as we've just seen,
they're going to be pulled down
towards the center of the earth from a gravitational force.
Since they're near the surface of the earth,
the magnitude of that force would be equal to m times g,
and it will be pointing downwards.
But this problem, if you look at this diagram,
there's only one force acting on these books that we've shown so far
and if something has a force downwards
and that's the only force acting on that object,
the object should move downwards.
But they're obviously not moving downwards
and nothing on your desk, hopefully, is falling downwards right now,
which means that there must be a corresponding force upwards
keeping them where they are. We call this force the normal force.
We don't call it normal because it's regular or not strange.
We call it normal because in mathematical terms,
that word normal is used to mean perpendicular,
because the normal force is by definition always perpendicular to
whatever surface is causing the force, is acting on an object like these books.
So this normal force will always try to balance
whatever forces are acting into the surface
and, in fact, comes from the electrons in the object, on the surface of that object.
We'll talk about electrons later, but just know that they're coming from the matter,
that the objects don't want to go through each other.
The books don't want to go through the table
and so the table opposes them and acts backward with,
what we call, the normal force.
As we're going through problems what we're going to say
is that for the motion into and out of the surface of this table,
there is no acceleration.
In other words, the books aren't leaving the table of their own accord,
and they're certainly not falling through the table.
So accelerating into and out of the table, the acceleration will be zero.
And for anything on a surface we can always say
that the motion into or out of that surface will have no acceleration
because objects aren't flying off of surfaces or falling into surfaces.
So this argument that the acceleration into or out of surface is zero
will be used many times over and over again.
Now that we've briefly introduced the gravitational and normal forces,
let's move to the inclined plane problem.
So this inclined plane problem is something that's so common and brought up so often.
The gravitational force acting on an object downwards
which is on a slope and then the normal force
trying to keep that object from going into the slope,
that we would like to go over it,
because again, you're gonna see this in many context
whether we're talking about other forces, gravity,
or just equations of motion in an object moving down the slope.
What usually happens in a problem like this is you have a setup exactly as I've shown here.
You have an object that's sitting on a slope and you know the angle,
which I'm calling theta, of the slope relative to the horizontal ground.
What we always do is define first, as with any of these problems, a coordinate system.
We can see here as I've introduced a coordinate system
with the x, the positive x axis pointing down the slope
and then the positive y axis pointing up directly away from the slope
or perpendicular to the slope, and this will be the coordinate system
that we usually use for an inclined plane problem.
The first thing we have to do,
as we've talked about with forces in two dimensions,
is try to get all of our forces acting on our object to be just in our two dimensions,
the x and the y axis. So to do that,
we're going to have to once again do something
that we've introduced a couple of times
which is to take our force that is not in these two directions,
the gravitational force, and try to split it up into two components.
So the gravitational force right now is not just in the x direction
and it's not just in the y direction.
What that mean is you're going to have to try to again break it up into two components,
one on the x and one on the y.
To do this, what I've done is redrawn the figure to your left,
and so you can see the gravitational force acting directly downwards
and we also have our axis, the x and y axis,
and what we wanna do is somehow try to break this factor
pointing straight downwards into two, one on the x and one on the y.
So to do this, we're going to once again need to appeal to trigonometry.
So here's the trigonometric way to think about this problem.
What I've done also here is drawn the horizontal line
which would be like the ground and we have the angle with the ground, theta.
So how can we figure out what right triangle should we use to take the hypotenuse,
which is always the force we're trying to split into two parts,
take the hypotenuse and turn it into two legs of a right triangle?
What you would do is look at this problem and use some trigonometric identities
that you would know from these two triangles, the one on the bottom right,
which I've shown highlighted here in gray and then the one on the left,
which I've shown highlighted in blue
and recognize that the angle at the bottom right of the gray triangle
and the angle at the top of the blue triangle are, in fact, the same angle theta.
This is important because now, if you just look at the blue triangle,
what we have is the hypotenuse
which is representing the gravitational force acting directly downwards,
but then we have the two legs of that blue triangle,
one is purely in the y direction and that is the Fg times the cosine of theta,
and then we have one purely in the x direction,
which is the Fg times the sine of theta
using the same trig identities and trig arguments that we've already brought up before.
This is sort of a long derivation,
but fortunately there's a quick and easy way to remember
and make sure you've gotten them right because at the end of the day you know
that if you have a gravitational force acting directly downwards,
you're going to split it into an x and a y component,
and you also know that this will mean that you have a cosine and a sine of theta.
So all you have to do is make sure you get the cosine and the sine correct
and don't mix them up and put them with the wrong force.
A quick way to always make sure you do this right,
is to once again remember that the sine of zero is zero and the cosine of zero is 1.
In other words, if I took my slope and made it smaller and smaller and smaller
until it was horizontal, I would expect that the gravitational force
would not point along the x direction anymore
because it's something just sitting on a horizontal surface.
The gravitational force can't push it side to side,
it can only push it into and out of the earth.
So, we know that the force in the, what we're calling horizontal direction,
the x direction, if it were flat, must be the Fg times the sine of theta.
And the reason we know that is because if I made my slope go to zero,
that force ought to also go to zero,
and the only force should be a vertical one into and out of the earth.
So that's an easy way to check to make sure you have your trig functions correct.