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### Course: Physics archive > Unit 3

Lesson 5: Inclined planes and friction- Inclined plane force components
- Ice accelerating down an incline
- Force of friction keeping the block stationary
- Correction to force of friction keeping the block stationary
- Force of friction keeping velocity constant
- Intuition on static and kinetic friction comparisons
- Static and kinetic friction example
- What is friction?
- What are inclines?

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# Ice accelerating down an incline

Explore the physics of an ice block sliding down an icy incline. Understand the forces at play, including gravity and the normal force, and how they contribute to the block's acceleration. Learn to calculate these forces using trigonometry and Newton's second law. Created by Sal Khan.

## Want to join the conversation?

- Maybe I missed it in the video, but why do the Fg(parallel) and the Fg (perpendicular) not ad up to 98 N? shouldn't they? Because all we did was split up the gravity vector into 2 vectors, so if we add them it should be the same, no?.. Thank you.(24 votes)
- Actually they do add up to 98! :-) These are vectors, so adding them depends on the angle between them, not just their values. If they were pointed the same way, then they would add up to the original value, but they are at 90 degrees to each other.

Explanation:

Look at the screenshot of the video (you don't have to replay it to see). There is a right triangle with the hypotenuse as the force from gravity (98) and the two component values for the perpendicular force and the parallel force. If you use the Pythagorean theorem with the components, you get 98 for the magnitude of the force due to gravity vector.

Details:

Fgparallel = 98 sin (30)

Fgperpendicular = 98 cos (30)

These are the legs of a right triangle.

The hypotenuse then is Ftotal so...

c^2 = a^2 + b^2

Ftotal^2 = Fgparallel ^2 + Fgperpendicular ^2

Ftotal^2 = (98sin(30)) ^2 + (98cos(30)) ^2

Ftotal^2 = (98 * 1/2) ^2 + 98 * (sqrt(3)/2) ^2

Ftotal^2 = 2401 + 7203

Ftotal^2 = 9604 (take the square root of both sides)

Ftotal = 98!(70 votes)

- Hey there Sal, it was a great video and i looove your website but after watching this, I had a question in my mind.

It the force acting on the block parallel to the slope is 49 N and the force due to gravity is 98 N, where does the remaining 49 N go?

Is the remaining 49 N the force that is being counteracted by the normal force on the block?

thx(7 votes)- Hi Tushar, you are on the right lines but remember that forces are vectors so they have direction and cannot be simply added or subtracted like that unless they are acting in the same or opposite directions. What Sal did is to decompose the 98N into two components: the 49N acting parallel to the surface of the slope and the 49(3^0.5)N which is the force that the block exerts perpendicular to the slope. And as you say that component is exactly equal and opposite to the normal force which the slope exerts on the block. But you cannot do 98-49N because they are vectors in different directions. Please remember to be very wary of doing that with vectors.

All the best,

Sheridan(16 votes)

- Hello!

This video has helped me understand the basics of forces on the inclined plane (THANK YOU!) but I have a question:

From3:20to4:10, why is the Cossin calculated on the y component and the sin calculated on the x component? Please explain.

Udyant Aggarwal(5 votes)- Understandably, you seem to have gotten turned around! The cosine is used in this case because what we have is an angle on a right triangle. Based on this angle, we want to know the the side beside it, which we will then compare to the hypotenuse.

Basically, cos and sin should not be thought of as x and y (even though in the unit circle they can be used this way). To keep yourself less confused, the sine of an angle is the ratio of the leg directly opposite to it over the hypotenuse, while the cosine is the leg beside the angle over the hypotenuse. Trying to rotate and reflect everything into terms of x and y will just get you doing extra work.

If you still have questions, I suggest you draw a bunch of triangles and turn them around to see what I mean on your own.(5 votes)

- Does ice itself actually have negligible friction? Isn't it because melting ice creates a surface layer of water that enables it to move smoothly?

(Sorry, this isn't really related to the video in itself, but I've been having this doubt for a while)(6 votes)- the reasons for ice being slippery are not well understood. The explanation that relies on the application of pressure to melt a thin layer of water doesn't hold up to careful experimentation and calculations.(2 votes)

- If a body is held stationary on an incline by more than just friction, would the frictional forcre be regarded as static or dynamic?(3 votes)
- Hi, I know I'm really late :) but it depends on whether the body is accelerating or not. Since it's at rest and not sliding, there is static friction but because of the incline the body is trying to overcome static friction.

Hope this helps.. others as well.(1 vote)

- How does the surface area of an object affect the sliding frictional force present while pulling it across a surface?(5 votes)
- It also depends on what you mean by "surface area." An increase in surface area in the sense that the surface is bumpier or rougher would increase the frictional force (since it increases the coefficient of friction) but just increasing the surface area in the sense that the surface of contact is greater would generally have minimal impact on the frictional force.(1 vote)

- What would happen if kinetic friction was involved in the problem but it was still accelerating (rather than being stationary in the next video)?(4 votes)
- You subtract the force of kinetic friction from the parallel to the surface of the slope component of the force of gravity.(1 vote)

- can you explain why dropping things of different mass reach the ground at the same time but sliding things of different mass down an incline do not?(4 votes)
- Mass changes with the speed of the object ? [Remember m'=m/(1-(v^2/c^2)) ]

Does this make mass a scalar quantity?(2 votes)- Mass is a scalar but not because of general relativity. Even if it did not change with speed it would still be a scalar(4 votes)

- How does normal force relate to liquids?(2 votes)
- It doesn't. Normal forces are forces that solid surfaces exert on one another. Liquids can exert a force that behaves similarly sometimes - buoyant forces.(3 votes)

## Video transcript

Let's say that I have
a ramp made of ice. Looks like maybe a wedge or
some type of an inclined plane made of ice. And we'll make everything
of ice in this video so that we have
negligible friction. So this right here is my ramp. It's made of ice. And this angle right over here,
let's just go with 30 degrees. And let's say on this
ramp made of ice, I have another block of ice. So this is a block of ice. It is a block of ice, it's
shiny like ice is shiny. And it has a mass
of 10 kilograms. And what I want to do
is think about what's going to happen to
this block of ice. So first of all, what are
the forces that we know are acting on it? Well if we're assuming
we're on Earth, and we will, and we're
near the surface, then there is the
force of gravity. There's the force of gravity
acting on this block of ice. And the force of
gravity is going to be equal to-- it's going to
be in the downward direction, and its magnitude is going to
be the mass of the block of ice times the gravitational
field times 9.8 meters per second squared. So it's going to be
98 newtons downward. So this is 98 newtons downward. I just took 10 kilograms. Let me write it out. So the force due
to gravity is going to be equal to 10
kilograms times 9.8 meters per second
squared downward. This 9.8 meters
per second squared downward, that is
the field vector for the gravitational field
of the surface of the earth, I guess is one way
to think about it. Sometimes you'll
see the negative 9.8 meters per second squared. And then that negative is giving
you the direction implicitly because the
convention is normally that positive is upward
and negative is downward. We'll just go with
this right over here. So the magnitude of
this vector is 10 times 9.8, which is 98 kilogram
meters per second squared, which is the same thing as newtons. So the magnitude
here is 98 newtons and it is pointing downwards. Now what we want to do
is break this vector up into the components
that are perpendicular and parallel to the
surface of this ramp. So let's do that. So first, let's think
about perpendicular to the surface of the ramp. So perpendicular to the
surface of the ramp. So this right over
here is a right angle. And we saw in the last
video, that whatever angle this over here
is, that is also going to be this
angle over here. So this angle over here is also
going to be a 30-degree angle. And we can use that
information to figure out the magnitude of this orange
vector right over here. And remember, this orange
vector is the component of the force of gravity that
is perpendicular to the plane. And then there's going to be
some component that is parallel to the plane. I'll draw that in yellow. Some component of
the force of gravity that is parallel to the plane. And clearly this
is a right angle, because this is
perpendicular to the plane. And this is parallel
to the plane. If it's perpendicular
to the plane, it's also perpendicular to
this vector right over here. So we can use some
basic trigonometry, like we did in the last
video, to figure out the magnitude of this orange
and this yellow vector right over here. This orange vector's
magnitude over the hypotenuse is going to be equal
to the cosine of 30. Or you could say that
the magnitude of this is 98 times the cosine
of 30 degrees newtons. 98 times the cosine
of 30 degrees newtons. And if you want the whole
vector, it's in this direction. And the direction going into
the surface of the plane. And, based on the
simple trigonometry-- and we go into this in
a little bit more detail in the last video-- we
know that the component of this vector that is parallel
to the surface of this plane is going to be 98
sine of 30 degrees. Sine of 30 degrees. Sine of 30 degrees. And this comes straight
out of this magnitude, which is opposite to the
angle over the hypotenuse. Opposite over hypotenuse is
equal to sine of an angle. And we did all the
work over here. I don't want to
keep repeating it. But I always want to emphasize
that this is coming straight out of basic
trigonometry, straight out of basic trigonometry. So once you do that, we know
the different components. We can calculate them. Cosine of 30 degrees is
square root of 3 over 2. Sine of 30 degrees is 1/2. That's just one of those
things that you learn and you can derive it yourself
using 30-60-90 triangles, or actually even
equilateral triangles. Or you could use a calculator. But it's also one
of those things that you memorize when
you take trigonometry. So no kind of magical
trick I did here. And so if you evaluate this,
98 times the square root of 3 over 2 newtons,
tells us that-- let me write it in that
same orange color-- the force, the
component of gravity that is perpendicular
to the plane. And this kind of implicitly
gives us this direction, it's perpendicular to the plane. But the force
component of gravity that's perpendicular
to the plane is equal to 98 times
square root of 3 over 2. 98 divided by 2 is 49. So it's equal to 49 times
the square root of 3 newtons. And its direction is into
the surface of the plane, or downward or, let me just
write, into surface of plane. Surface of the plane, or
the surface of the ramp. And it's in this
direction over here. And I have to do this
because it's a vector. I have to tell you what
direction it's going in. And the component of the force
of gravity that is parallel. The component of the force
of gravity that is parallel, I drew it down here, but I
could shift it up over here. It's the same exact vector. The component of
gravity that is parallel to the surface of the plane
is 98 times sine of 30. That's 98 times 1/2,
which is 49 newtons. And it's going in that
direction, or parallel to the surface of the plane. Parallel, I always have
trouble spelling parallel. Parallel to-- don't even
know if I spelled it right-- surface of the plane. So what's going to happen here? Well, if these were the
only forces acting on it. So if we had a net force
going into the surface of the plane of 49 square
roots of 3 newtons. If this was the only force
acting in this dimension or in the dimension that is
perpendicular to the surface of the plane, what would happen? Well, then the block
would just accelerate. At least just due to this force
it would accelerate downward. It would accelerate into
the surface of the plane. But we know it's not
going to accelerate. We know that there's this
big wedge of ice here that is keeping it from
accelerating in that direction. So at least in this dimension,
there will be no acceleration. When I talk about
this dimension, I'm talking about in
the direction that is perpendicular to the
surface of the plane. There will be no acceleration
because this wedge is here. So the wedge is exerting a force
that completely counteracts the force, the perpendicular
component of gravity. And that force. You might guess
what it's called. So the wedge is
exerting a force, just like that, that's going
to be 98 newtons upward. The wedge is going to
be exerting a force that is 49 square roots of 3,
because this right here is 49 square roots
of 3 newtons into. And so this is 49 square
roots of 3 newtons out of the surface,
out of the surface. And this is the normal force. It is the force
perpendicular to the surface that essentially, you could
kind of view as the contact force that the, in this case,
that the surface is exerting to keep this block of ice from
accelerating in that direction. We're not talking about
accelerating straight towards the center of the earth. We're talking about
accelerating in that direction. We broke up the force into kind
of the perpendicular direction and the parallel direction. So you have this
counteracting normal force. And that's why you don't
have the block plummeting or accelerating into the plane. Now what other
forces do we have? Well, we have the force that's
parallel to the surface. And if we assume that
there's no friction-- and I can assume that there's
no friction in this video because we are assuming
that it is ice on ice-- what is going to happen? There's no counteracting
force to this 49 newtons. 49 newtons parallel downwards,
I should say parallel downwards, to the surface of the plane. So what's going to happen? Well, it's going to
accelerate in that direction. You have force is equal to
mass times acceleration. Force is equal to mass
times acceleration. Or you divide both sides by
mass, you get force over mass is equal to acceleration. Over here, our
force is 49 newtons in that direction,
parallel downwards to the surface of the plane. And so if you
divide both by mass, if you divide both
of these by mass. So that's the same thing as
dividing it by 10 kilograms, dividing by 10 kilograms, that
will give you acceleration. That will give you
our acceleration. So acceleration is
49 newtons divided by 10 kilograms
in that direction, in this direction
right over there. And 49 divided by
10 is 4.9, and then newtons divided by kilograms
is meters per second squared. So then you get
your acceleration. Your acceleration is going to be
4.9 meters per second squared. And maybe I could say parallel. That's two bars. Or maybe I'll write parallel. Parallel downwards
to the surface. Now I'm going to
leave you there, and I'll let you think
about another thing that I'll address in
the next video is, what if you had this
just standing still? If it wasn't
accelerating downwards, if it wasn't accelerating
and sliding down, what would be the
force that's keeping it in a kind of a static state? We'll think about that
in the next video.