Main content

## Physics library

### 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?

© 2022 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Force of friction keeping velocity constant

Calculating the coefficient of kinetic friction (correction made in next video). Created by Sal Khan.

## Video transcript

I want to make a
quick clarification to the last video, and then
think about what's friction up to when the block
is actually moving. So in the last
video, we started off with the block being stationary. We knew that the
parallel component of the force of
gravity on that block was 49 newtons downwards,
down the slope. And when the block
was stationary, we said there must be
an offsetting force. And we said that's
the force of friction, and it must be 49
newtons upwards. And so they completely
net out in that direction. Now, what we said is we're going
to keep applying a little bit more force until we
can budge this block to start accelerating downwards. And I said I kept applying
a little bit more force, a little bit more force,
until I get to 1 newton, and then the block
started to budge. So at that point, when
it started to budge, I'm applying this 1 newton
over here, right over here. There was already
49 newtons of force, or the component of
gravity, in this direction. So combined, we're
providing 50 newtons to just start budging
it, to just overcome the force of friction. The one thing I
want to clarify here is this whole time
the force of friction was not constant at 49 newtons. When I wasn't messing
with this block, and the parallel component of
the force was 49 newtons, then the force of friction
was 49 newtons. When I started to press
on it a little bit, apply a little bit
of force, maybe I applied a tenth of a
newton on top of that, then the force of friction
was 49 and 1/10 newton, because it was still
providing enough force so that this block was not moving. Then maybe I applied
half a newton. And so the total force
in the downward direction would have been 49
and 1/2 newtons. But if it still was not moving,
then the force of friction was still completely
overcoming it. So the force of
friction, at that point, must have been 49
and 1/2 newtons, all the way up to the combined
force in the downward direction being 49.999999 newtons. And then the force of friction
was still 49.99999 newtons, all the way until I hit 50
newtons and then the block started to budge, which tells us
that the force of friction now, all of a sudden, or at least
the force of static friction all of a sudden now
couldn't keep up and it started to accelerate downwards. So in that static scenario,
the force of friction changed as I applied
more or less force in this downward direction. Now with that out of
the way, let's take a different scenario. Let me just redraw that same
block, just since all of this is getting messy. So we have the same block, and
as we said in the last video, we're now assuming that
this is wood on wood. So this is the wedge. This is the block
right over here. We know that the component
of gravity that is parallel to the plane right
there is 49 newtons. We know that this is 49 newtons. We know the component
of gravity that is perpendicular to the
plane-- we figured out this two videos ago-- is 49
square roots of 3 newtons. We know that this block
is not accelerating in this normal
direction, so there must be some force counteracting
gravity in that direction. And that's the normal force
of the wedge on the block. So that is going
in that direction at 49 square roots of 3 newtons. And now instead of assuming
that this block is stationary, let's assume that it's moving
with a constant velocity. So now we're dealing
with-- let me do that in a different color. So now we're dealing
with a scenario where the block has
a constant velocity. And for the sake of
this video, we'll assume that that constant
velocity is downward. And so the constant velocity,
v, is equal to-- I don't know. Let's say it is 5 meters
per second down the wedge, or down the ramp. Or I guess we could say in
the direction that is parallel to the surface of the ramp. So it's in this direction
right over here. So that's the constant velocity. So what are all
the forces at play? And be very careful here. There might be a temptation
that says, OK, there's a net force here. We're moving. So maybe that's the net force
that's causing the move. But remember, this
is super important. This is Newton's first law. If you have a net force, if
you have an unbalanced force, it will cause it to accelerate. And we are not
accelerating here. We have a constant velocity. We are not accelerating here. So if you're not accelerating
in that direction, then that means that the
force in that direction must be balanced. So there must be
some force acting in the exactly
opposite direction that keeps this thing from
accelerating downwards. And so it must be
exactly 49 newtons in the opposite direction. And as you can imagine, this
is the force of friction. This right over here is
the force of friction. And the difference between
this video and the last video is last time
friction was static. Even at 49 newtons,
the box was stationary. You had to keep nudging
until you get to 50 newtons, and then it started moving. Here we're just jumping
into this picture where we just see a
box that's moving down the slope at 5
meters per second. So we don't know
how much force it took to overcome
static friction. But we do know that there is
some force of friction that is keeping this box from
accelerating, that's keeping it at a constant velocity,
that is completely negating the parallel component
of the force of gravity, parallel to the
surface of this plane. So given this, let's
calculate another coefficient of friction. But this is going to
be the coefficient of kinetic friction, because now
we are moving down the block. And I'll do a video
on why sometimes a coefficient of
static friction can be different than
the coefficient of kinetic friction. So the coefficient of kinetic
friction-- we'll write it. So this is the Greek letter
mu, and we put this k here for kinetic, or we can kind
of say moving friction. It's going to be equal
to the force of friction, or I should say the magnitude
of the force of friction over the normal force. I should say the magnitude
of the normal force. And you can derive
this experimentally. One, if you just observe
this whole thing going on and you knew the
mass of the block, so you knew this
component of gravity that's going in this direction. If you knew this
angle was 30 degrees from the last situation,
you could figure out this coefficient of
kinetic friction. And what's cool
about this is this is in general going to be true
for any two materials that are like this. So maybe this is a
certain type of wood on a certain type of wood, or
a certain type of sandpaper on a certain type of sandpaper--
whatever you're talking about. And then you can use
that to make predictions if the incline was different,
or if the mass was different, or even if you were
on a different planet, or if someone was pressing
down on this block. That would change
the normal force. So given this right
here, let's figure out-- for the sake of doing it-- the
coefficient of kinetic friction here. The force of friction
here, completely offsetting the parallel force of gravity
parallel to the surface, is 49 newtons. And the normal force
here, the force of contact between these
two things, this block and this wedge, is 49
square roots of 3 newtons. So we get 1 over the
square root of 3. And let me get
the calculator out to get an actual number here. So we have 1 divided by
the square root of 3, which gives us 0.5--
I'll just round. 0.58. It is equal to 0.58. And there's no units here,
because the units cancel out. It's a unit-less measurement. Now the interesting thing here
is that the way I've set up this problem, the coefficient
of kinetic friction is lower, if we assume
the same materials, than the coefficient
of static friction was. And for some materials, they
might not be that different. But for other materials,
the kinetic friction can be lower than
static friction. You never see a situation
where the coefficient of static friction-- at
least that I know of-- is lower than kinetic friction. But you do see situations
where the coefficient of kinetic friction is
lower than the coefficient of static friction. Once something is
moving, for some reason-- and we'll theorize why
that might be-- friction is a little less potent than
when something is stationary. So we can say this generally,
that the coefficient of kinetic friction is less
than or equal to the coefficient of static friction. It's a little bit easier, or
friction provides a little less than or equal to the force
when something's moving than when something
is stationary. So I'll think about
that a little bit deeper in the next video.