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### Course: High school physics > Unit 3

Lesson 7: Inclined planes# Inclined plane force components

Explore the various forces acting on a block sitting on an inclined plane. Learn how to break the force of gravity into two components - one perpendicular to the ramp and one parallel to the ramp. Finally, using geometry and trigonometry, learn how to calculate the magnitude of each component of force that is acting on the block. Created by Sal Khan.

## Want to join the conversation?

- How can something be perpendicular to the surface of the earth which is curved.5:20(58 votes)
- on a small scale it can be neglected.

if you stand on ground, you dont see it curved, you see it plane and straight. Thats because you are tiny comparatively(186 votes)

- Sal said normal force is perpendicular to the surface. But:

1. if normal force is reaction force (weight is 'action force') shouldn't it be positioned on a same line, that is, at same angle as weight of the body on a ramp? Third law states: "if one object exerts a force on another, the other exerts an equal and opposite force on the first." So, why is normal force 'normal' (perpendicular to the surface)?

2. In my experience, if a body is at inclined surface the part of the body closer to the angle of incline plane, (that is, left side of the box) pushes 'harder' to the ground than it's right side. If that is so, normal force shouldn't, in my opinion, be perpendicular to the surface because 'distribution of the pressure' on the inclined plane isn't even.

Thank you. Hope my question is clear...(21 votes)- If there is no friction the surface of on abject can only impart a force on another object perpendicular to surface, any component of the force parallel to the surface requires friction. So the force is broken up into a Normal Force and a Friction Force.

With an incline that is frictionless and you have a block on it the block's weight is directed strait down but the normal force is perpendicular to the incline so when you add the force vectors you end up with a net force parallel to the incline pointing down and this is what causes the block to slide down the incline.

On a frictionless incline the force is the same along the entire surface of the box so there is no portion that pushes harder into the ground than another. Once you introduce friction you have gravity and the normal force essentially acting like it is all at the center of gravity of the box but the friction force is along the surface of contact between the box and the incline. This friction force not in line with the center of gravity of the box produces a torque causing the edge of the box away from the direction of the friction force to put a bit more pressure on the incline that the other.

But regardless of the amount of pressure at each point along the the surface of contact the force without friction is always perpendicular to the surface and is the normal force.(14 votes)

- how do i know what trig-ratio to use??

I'm confused on that..(8 votes)- The way I remember is with SOHCAHTOA. The SOH means that the Sine is Opposite over Hypotenuse, C(osine) = A(djacent) / H(ypotenuse), T(angent) = O(pposite) / A(djacent)(28 votes)

- What are the arrows that Sal is drawing above the F and G?(8 votes)
- Those arrows show that they(F and G) are vector quantities(7 votes)

- Yo chat should i drop out(9 votes)
- what is force(1 vote)
- A force is a push or a pull...

So anytime you move you are pushing on somthing...

Walking- pushing on the ground

Biking- pushing the peddles etc(14 votes)

- i still dont understand why you break up the two forces of gravity. what's the point of doing this?(5 votes)
- We want to see how gravity affects motion on the inclined plane, so we break the force of gravity into components that are parallel and perpendicular to the plane, respectively. That way, we can see how gravity actually affects the sliding motion.(7 votes)

- How would you calculate normal force for a box on an incline? And what about if there is a force being exerted horizontally which keeps the object at rest neglecting friction? How would to force diagram for a scenario like this look and how would you solve for Fnormal and Fapplied?(5 votes)
- Breakout the gravity force vector into components which are parallel -mg*sin(Θ) and perpendicular -mg*cos(Θ) to the incline. The normal force will be equal and opposite to the perpendicular gravity component so N = +mg*cos(Θ). To keep the block from sliding, you would then need to apply a horizontal force equal and opposite to the parallel gravity component so that force would be F = +mg*sin(Θ).(6 votes)

- All of this is really complicated, and I'm having a hard time understanding how to solve this. Is there an equation or something that simplifies this whole process?(6 votes)
- The only quick tip I can give you is that the force of gravity will be multiplied by the sine of the incline and the normal force multiplied by the cosine of the incline.(3 votes)

- How come when we're trying to figure out the force of gravity parallel and force of gravity perpendicular to the inclined plane we use theta as the "focus point"?(3 votes)
- because trigonometry allows us to do that ; use angle as an identity to relate two lines(as in this case with two vectors) in a right angled triangle(4 votes)

## Video transcript

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.