Motion on an Inclined Plane Video Lecture - NEET

Let's say I have some
type of a block here.
And let's say this
block has a mass of m.
So the mass of this
block is equal to m.
And it's sitting on this-- you
could view this is an inclined
plane, or a ramp, or
some type of wedge.
And we want to think about what
might happen to this block.
And we'll start thinking about
the different forces that
might keep it in place
or not keep it in place
and all of the rest.
So the one thing we do know
is if this whole set up
is near the surface
of the Earth--
and we'll assume that it is
for the sake of this video--
that there will be the
force of gravity trying
to bring or attract this
mass towards the center
of the Earth, and vice versa,
the center of the earth
towards this mass.
So we're going to have
some force of gravity.
Let me start right at the center
of this mass right over here.
And so you're going to
have the force of gravity.
The force due to
gravity is going
to be equal to the
gravitational field
near the surface of the Earth.
And so we'll call that g.
We'll call that
g times the mass.
Let me just write it.
The mass times the
gravitational field
near the surface of the Earth.
And it's going to
be downwards, we
know that, or at least towards
the surface of the Earth.
Now, what else is going
to be happening here?
Well, it gets a
little bit confusing,
because you can't really say
that normal force is acting
directly against this
force right over here.
Because remember,
the normal force
acts perpendicular to a surface.
So over here, the surface is
not perpendicular to the force
of gravity.
So we have to think about it a
little bit differently than we
do if this was sitting
on level ground.
Well, the one thing we can do,
and frankly, that we should do,
is maybe we can
break up this force,
the force due to gravity.
We can break it
up into components
that are either
perpendicular to the surface
or that are parallel
to the surface.
And then we can use
those to figure out
what's likely to happen.
What are potentially the netting
forces, or balancing forces,
over here?
So let's see if we can do that.
Let's see if we can
break this force vector,
the force due to gravity,
into a component that
is perpendicular to the
surface of this ramp.
And also another
component that is parallel
to the surface of this ramp.
Let me do that in
a different color.
That is parallel to the
surface of this ramp.
And this is a little bit
unconventional notation,
but I'll call this one over
here the force due to gravity
that is perpendicular
to the ramp.
That little upside down t, I'm
saying that's perpendicular.
Because it shows a line
that's perpendicular to,
I guess, this bottom line, this
horizontal line over there.
And this blue thing
over here, I'm
going to call this
the part of force
due to gravity that is parallel.
I'm just doing these
two upward vertical bars
to show something that is
parallel to the surface.
So this is the
component of force
due to gravity that's
perpendicular, component
of force that is parallel.
So let's see if we
can use a little bit
a geometry and
trigonometry, given
that this wedge is at
a theta degree incline
relative to the horizontal.
If you were to measure
this angle right over here,
you would get theta.
So in future videos we'll
make it more concrete,
like 30 degrees or 45
degrees or whatever.
But let's just keep in general.
If this is theta,
let's figure out
what these components of
the gravitational force
are going to be.
Well, we can break out
our geometry over here.
This, I'm assuming
is a right angle.
And so if this is
a right angle, we
know that the sum of the angles
in a triangle add up to 180.
So if this angle, and this
90 degrees-- right angle
says 90 degrees-- add
up to 180, then that
means that this one and this one
need to add up to 90 degrees.
Or, if this is theta,
this angle right over here
is going to be 90 minus theta.
Now, the other thing
that you may or may not
remember from geometry
class is that if I
have two parallel lines,
and I have a transversal.
So I'm going to assume this
line is parallel to this line.
And then I have a transversal.
So let's say I have a
line that goes like this.
We know from basic
geometry that this angle
is going to be
equal to this angle.
It comes from alternate
interior angles.
And we prove it in
the geometry module,
or in the geometry videos.
But hopefully this makes a
little bit of intuitive sense,
and you could even think
about how these angles would
changes as the transversal
changes, and all of the rest.
But the parallel
lines makes this angle
similar to that angle, or
actually makes it identical,
makes it congruent.
This angle is going to be the
same measure as that angle.
So can we apply that
anywhere over here?
This line is perpendicular
to the surface of the Earth.
Right over here that I'm
kind of shading in blue.
And so is this force vector.
It is also perpendicular to
the surface of the Earth.
So this line over here and
this line over here in magenta
are going to be parallel.
I can even draw that.
That line and that
line are both parallel.
When you look at
it that way, you'll
see that this big line over here
can be viewed as a transversal.
Or you could have this
angle and this angle
are going to be congruent.
They're going to be
alternate interior angles.
So this angle and this angle,
by the exact same idea here.
It just looks a little
bit more confusing here
because I have all
sorts of things.
But this line and this
line are parallel.
You can view this right
over here as a transversal.
So this and this are
congruent angles.
So this is 90 minus
theta degrees.
This too will be 90
minus theta degrees.
90 minus theta degrees.
Now, given that, can we
figure out this angle?
Well one thing, we're assuming
that this yellow force vector
right here is perpendicular
to the surface of this plane
or perpendicular to the
surface of this ramp.
So that's perpendicular.
This right here
is 90 minus theta.
So what is this angle up
here going to be equal to?
This angle, let
me do it in green.
What is this angle up
here going to be equal to?
So this angle plus 90
minus theta plus 90
must be equal to 180, or this
angle plus 90 minus theta must
be equal to-- let
me write this down.
I don't want to do
too much in your head.
So let's call it x.
So x plus 90 minus theta.
Plus this 90 degrees right over
here, plus this 90 degrees,
needs to be equal
to 180 degrees.
Let's see, we can subtract
180 degrees from both sides.
So we subtract 90 twice,
you subtract 180 degrees
and you get x minus
theta is equal to 0,
or x is equal to theta.
So whatever the inclination of
the plane is or of this ramp,
that is also going to be
this angle right over here.
And the value to
that is that now we
can use our basic
trigonometry to figure out
this component
and this component
of the force of gravity.
And to see that a
little bit clearer,
let me shift this force
vector down over here.
The parallel component,
let me shift it over here.
And you can see the
perpendicular component
plus the parallel component
is equal to the total force
due to gravity.
And you should also see that
this is a right triangle
that I have set up over here.
This is parallel to the plane.
This is perpendicular
to the plane.
And so we can use
basic trigonometry
to figure out the magnitudes
of the perpendicular
force due to gravity and the
parallel force due to gravity.
Let's think about
it a little bit.
I'll do it over here.
The magnitude of
the perpendicular
force due to gravity.
Or I should say the
component of gravity
that's perpendicular to
the ramp, the magnitude
of that vector-- a
lot of fancy notation
but it's really just the length
of this vector right over here.
So the magnitude of
this over the hypotenuse
of this right triangle.
Well, what the hypotenuse
of this right triangle?
Well, it's going
to be the magnitude
of the total
gravitational force.
I guess you could say that.
And so you could say that is mg.
We could write it like this.
But that's really-- well,
I could write it like that.
And so this is going
to be equal to what?
We have the, if we're looking
at this angle right here,
we have the adjacent
over the hypotenuse.
Remember.
We can do this in a new color.
We can do this in a new color.
SOH CAH TOA.
Cosine is adjacent
over hypotenuse.
So this is equal to
cosine of the angle.
So cosine of theta is
equal to the adjacent
over the hypotenuse.
So if you multiply both
sides by the magnitude
of the hypotenuse, you get the
component of our vector that
is perpendicular to the
surface of the plane
is equal to the magnitude
of the force due to gravity
times the cosine of theta.
Times the cosine of theta.
We'll apply this
in the next video
just so you can make the
numbers a lot more concrete.
Sometimes just the notation
makes it confusing.
You'll see it's really actually
pretty straightforward.
And then this second thing,
we can use the same logic.
If we think about the parallel
vector right over here,
the magnitude of the
component of the force
due to gravity that is
parallel to the plane
over the magnitude of the
force due to gravity--
which is the
magnitude of mg-- that
is going to be equal to what?
This is the opposite
side to the angle.
So the blue stuff is the
opposite side, or at least
its length, is the
opposite side of the angle.
And then right over
here this magnitude
of mg, that is the hypotenuse.
So you have the opposite
over the hypotenuse.
Opposite over hypotenuse.
Sine of an angle is
opposite over hypotenuse.
So this is going to be
equal to the sine of theta.
This is equal to
the sine of theta.
Or you multiply both sides
times the magnitude of the force
due to gravity and
you get the component
of the force due to gravity
that is parallel to the ramp
is going to be the force
due to gravity total times
sine of theta.
Times sine of theta.
And hopefully you should
see where this came from.
Because if you ever have to
derive this again 30 years
after you took a physics class,
you should be able to do it.
But if you know this right
here, and this right here,
we can all of a sudden start
breaking down the forces
into things that
are useful to us.
Because we could
say, hey, look, this
isn't moving down
into this plane.
So maybe there's
some normal force
that's completely netting
it out in this example.
And maybe if there's
nothing to keep it up,
and there's no friction,
maybe this thing
will start accelerating
due to the parallel force.
And we'll think a
lot more about that.
And if you ever forget these,
think about them intuitively.
You don't have to go through
this whole parallel line
and transversal and all of that.
If this angle went
down to 0, then we'll
be talking about
essentially a flat surface.
There is no inclination there.
And if this angle goes down
to 0, then all of the force
should be acting perpendicular
to the surface of the plane.
So if this going to 0, if
the perpendicular force
should be the same thing as
the total gravitational force.
And that's why it's
cosine of theta.
Because cosine of
0 right now is 1.
And so these would
equal each other.
And if this is equal to 0,
then the parallel component
of gravity should go to 0.
Because gravity
will only be acting
downwards, and once again,
if sine of theta is 0.
So the force of gravity that
is parallel will go to 0.
So if you ever forget, just do
that little intuitive thought
process and you'll
remember which one is sine
and which one is cosine.