Video Transcript
In this video, we’re going to learn
about kinetic friction. We’ll learn what it is, how it
compares to static friction, and how to work with it practically.
To get started, imagine you are an
athlete in training for the upcoming Winter Olympic Games. In your chosen Olympic sport of
curling, the goal is to slide a smooth stone onto the center of a bullseye mark. To slide the stone so that when it
comes to a stop it’s exactly at the center of the bullseye requires a steady hand, a
practiced eye, and an intuitive understanding of kinetic friction.
Kinetic friction is a force
measured in units of newtons that resists object motion when an object slides along
a surface. Kinetic friction occurs when a car
comes skidding to a stop with its wheels locked. When you slide a can of your
friend’s favorite type of soda across a tabletop to them. Or when you’re simply doing the
moonwalk dance, walking backwards and sliding your feet across the floor. All these are examples of objects
moving across a surface while sliding.
To understand kinetic or moving
friction better, let’s consider an example that begins with a box at rest on a flat
rough surface. Imagine that you stand by the box
and begin to press on it with your foot, pushing it to the right. Let’s further imagine that we’re
able to plot the frictional force between the box and the floor and the applied
force of your foot on the box on a pair of axes.
When we first start out, before you
apply any force with your foot, there is no frictional force between the box and the
floor. But as you gradually increase the
pressure of your push, the applied force increases and so does the friction force of
the box. It’s important to realize that, at
this point, the box still hasn’t started to move. You’re pressing harder and harder
on it, but the box remains stationary as static friction keeps it in place.
But as you press harder and harder,
something interesting happens. We’ve all had the experience of
starting an object moving from rest. Right after it starts moving, it
jolts a bit. And that’s because we overcome the
force of static friction and then settle into a steady state of what we can call
kinetic friction. As the box moves along under our
increasing applied force, the kinetic friction force that resists its movement stays
roughly constant.
The magnitude of the kinetic
frictional force, the one that resists the movement of the box while it’s sliding,
is equal to the coefficient of kinetic friction — we call it 𝜇 sub 𝑘 — multiplied
by the normal force exerted on the box. If you like to know more about this
coefficient of friction 𝜇, take a look at the static friction video where we
explain it more in depth.
For our purposes, it’s enough to
say about the coefficient of kinetic friction, 𝜇 sub 𝑘, that, for one thing, it
varies between materials. 𝜇 sub 𝑘, for example, between
steel and ice will not be the same as between wood and glass. We can also say that 𝜇 sub 𝑘 is
typically greater than or equal to zero and less than or equal to one. And also, and this is strange, 𝜇
sub 𝑘 largely does not depend on object’s speed or the force being applied to the
moving object. So this means that if our box was
moving at half a meter per second, its coefficient of kinetic friction will be the
same as if we’re moving at a million meters per second.
And as far as depending on the
applied force, if we look back to our graph, we can see that the flat line of the
frictional force, when it becomes kinetic friction, goes on and on even as our
applied force increases. So when it comes to solving for the
force of kinetic friction on our object, we need to know two things: the coefficient
of kinetic friction, called 𝜇 sub 𝑘, and the normal force acting on our
object. Knowing this, let’s get some
practice with kinetic friction through a few examples.
A box of mass 12.4 kilograms slides
from rest downward along a 1.58-meter-length ramp that is inclined at 23.3 degrees
below the horizontal. The coefficient of kinetic friction
between the box and the ramp is 0.0346. What is the magnitude of the box’s
acceleration? What is the speed of the box at the
bottom of the ramp?
We can call the box’s acceleration
𝑎 and the box’s speed at the bottom of the ramp 𝑣 sub 𝑓. Let’s start on our solution by
drawing a diagram of the moving box. In this example, we have a box of
mass 𝑚 — 𝑚 is 12.4 kilograms — on a ramp of length 𝑙, where 𝑙 is 1.58
meters. The ramp is inclined at an angle
we’ve called 𝜃, where 𝜃 is 23.3 degrees above the horizontal. Under the influence of gravity, we
expect the box to slide down the ramp. and we want to solve for its acceleration,
𝑎, when it does so. We also wanna find out how fast the
box is moving when it gets to the bottom of the ramp.
Into this scenario, we set up an
𝑥- and a 𝑦-coordinate axis so that positive motion in the 𝑥-direction is down the
incline and positive motion in the 𝑦-direction is perpendicular to the plane. Next, let’s draw a free body
diagram of the mass 𝑚, that is, the forces that act on it. We know there’s a gravitational
force acting on 𝑚 straight down of magnitude 𝑚 times 𝑔, where 𝑔, the
acceleration due to gravity, is 9.8 meters per second squared. There’s also a normal force acting
on the mass to keep it from moving into the plane. And lastly, as the box slides down
the plane, there’s a kinetic frictional force that resists its motion.
Looking at this free body diagram,
we can divide up the weight force into its 𝑦- and 𝑥-components so that we can
write out force balance equations in the 𝑦- and 𝑥-directions. And we know that, with these force
components at a right angle to one another, a triangle is formed, where the upper
angle is the angle 𝜃.
Recalling Newton’s second law of
motion, that the net force on an object equals its acceleration times its mass, we
can consider the forces in the 𝑥-direction in our scenario. Those are the kinetic friction
force and the 𝑥-component of the weight force. Since we’ve decided that motion
down the ramp is motion in the positive direction, our force balance equation for
the 𝑥-forces is 𝑚𝑔 sin 𝜃 minus the kinetic friction force is equal to the mass
of the object times its acceleration.
The kinetic friction force, 𝑓 sub
𝑘, is equal to the coefficient of kinetic friction, 𝜇 sub 𝑘, multiplied by the
normal force acting on an object. This means we can replace 𝑓 sub 𝑘
with 𝜇 sub 𝑘 𝑓 sub 𝑛 in our 𝑥-force balance equation. And we remember that we’re given 𝜇
sub 𝑘 in the problem statement.
Looking at this force balance
equation, we see that if we divide both sides by the mass 𝑚, we see we have an
expression for the acceleration 𝑎 we want to solve for. The one challenge is, we don’t yet
know the normal force, 𝑓 sub 𝑛, that acts on our mass. To find that out, let’s look at the
forces in the 𝑦-direction on our mass.
Applying Newton’s second law to
forces in this direction, we write that the normal force minus 𝑚𝑔 cos 𝜃, the
𝑦-component of the weight force, is equal to the mass of the box times its
acceleration in the 𝑦-direction. But because the box doesn’t leave
the surface of the plane, this acceleration is zero. And our equation simplifies to the
normal force is equal to 𝑚𝑔 times the cos of 𝜃. This is perfect because now we can
take this term for 𝑓 sub 𝑛 and substitute it into our equation in the
𝑥-direction.
With this expanded expression for
the acceleration 𝑎, we see that the box’s mass 𝑚 appears in all the terms in the
numerator and the denominator. So it cancels out. Factoring out the acceleration due
to gravity, 𝑎 is equal to 𝑔 times the quantity sin 𝜃 minus 𝜇 sub 𝑘 cos 𝜃. Substituting in for 𝑔, 𝜃, and 𝜇
sub 𝑘, when we calculate 𝑎, to three significant figures, we find it’s 3.56 meters
per second squared. That’s the acceleration of the box
as it descends the ramp.
Next, we wanna solve for the speed
the box has when it reaches the bottom of the 1.58-meter-long ramp. To figure this out, we know that
the acceleration the box undergoes during its journey is constant. This means that the kinematic
equations of motion apply to the motion of this box. As we look over these four
equations of motion, we seek one that lets us solve for final velocity, 𝑣 sub 𝑓,
in terms of information we know.
The second kinematic equation
written matches our conditions well. Rewriting it in terms of our
variables, the final speed of the box squared is equal to the box’s initial speed,
which is zero because the box starts from rest, plus two times the box’s
acceleration times the length of the ramp, 𝑙. If we take the square root of both
sides of this equation, we find that 𝑣 sub 𝑓 is equal to the square root of two
times 𝑎 times 𝑙.
We know 𝑎 from part one and 𝑙 is
given information. So we’re ready to plug those in to
our expression. When we do, when enter these values
on our calculator, we find that 𝑣 sub 𝑓 is 3.36 meters per second. That’s the speed of the box when it
reaches the bottom of the ramp.
Let’s try a second example
involving kinetic friction.
A car is moving along a highway
when the driver has to make an emergency stop. The wheels of the car lock and
cease rolling. The car skids and leaves
straight-line skid trails between the point where the breaking began and the point
where the car comes to rest. The skid trails are 27.9 meters
long. The coefficient of kinetic friction
between the car’s tires and the road is 0.512. At what speed was the car moving
when its wheels locked?
Given the distance the car skidded
with its wheels locked and the coefficient of kinetic friction between the car’s
tires and the road, we want to solve for the speed of the car when the car’s braking
started. If we draw an overhead view of the
car after it’s come to a stop, it’s created skid marks, which are 𝑙, 27.9 meters
long. We want to solve for the speed, 𝑣,
of the car just as it started to create those marks, that is, just as its wheels
locked.
If we set up the convention that
motion in the direction of the car’s travel is positive motion, we see that the only
force acting on the car along that axis as it skids to a stop is the force of
kinetic friction. And by Newton’s second law of
motion, that the net force on an object is equal to its mass times its acceleration,
we can write that negative the force of kinetic friction is equal to the car’s mass
times its acceleration while it stops.
When we recall further that the
force of kinetic friction equals the coefficient of kinetic friction multiplied by
normal force, we can rewrite our expression for 𝑓 sub 𝑘. And we’re also able to observe that
𝑓 sub 𝑛, the normal force, is equal to the car’s mass multiplied by the
acceleration due to gravity, 𝑔. The car’s mass cancels out from
both sides of the equation. And we now have an expression for
the car’s acceleration while it breaks.
Looking at this expression, we see
that it doesn’t vary in time. 𝜇 sub 𝑘 is a constant, and so is
𝑔, 9.8 meters per second squared. This constant acceleration tells us
that the kinematic equations effectively describe the motion of the car. In particular, we can say that its
final speed squared equals its initial speed squared plus two times its acceleration
times the distance it travels.
Applying this expression to our
scenario, we see that the car’s final speed squared, which we know is zero since the
car comes to a rest, is equal to its initial speed squared plus two times 𝑎 times
𝑙, the distance it skids. Rearranging to solve for 𝑣
squared, 𝑣 squared is equal to negative two times 𝑎 times 𝑙. And we’re able to substitute in our
expression for 𝑎 in terms of 𝜇 sub 𝑘 and 𝑔. When we do, the minus signs cancel
out. And when we take the square root of
both sides of this equation, we see we’ve been given the value for 𝜇 sub 𝑘 and
𝑙. And 𝑔 is a constant we know. So we’re ready to plug in and solve
for 𝑣. Entering these values on our
calculator, we find that 𝑣 is 16.7 meters per second. That’s how fast the car was going
just as it started to brake.
Let’s summarize what we’ve learned
so far about kinetic friction. We’ve seen that kinetic friction is
a force that resists object motion when an object slides along a surface. As an equation, we can write the
force of kinetic friction as the product between the coefficient of kinetic
friction, 𝜇 sub 𝑘, and the normal force acting on the object in motion. We’ve seen that this coefficient of
kinetic friction varies depending on the types of materials involved.
And finally, recalling our plot of
force of friction versus applied force for an object which starts out still and then
gains motion, we see that kinetic friction force magnitude is typically less than
the maximum static friction magnitude on an object. Kinetic friction describes that
force of friction when an object is in motion.
