Force of friction keeping velocity constant | Physics | Khan Academy

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

Watch the next lesson: https://www.khanacademy.org/science/physics/forces-newtons-laws/inclined-planes-friction/v/intuition-on-static-and-kinetic-friction-comparisons?utm_source=YT&utm_medium=Desc&utm_campaign=physics

Missed the previous lesson? https://www.khanacademy.org/science/physics/forces-newtons-laws/inclined-planes-friction/v/correction-of-force-of-friction-keeping-the-block-stationary?utm_source=YT&utm_medium=Desc&utm_campaign=physics

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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.