Video Transcript
In this video, we’re going to learn
about static friction, what it is, what causes it, and how to calculate it
practically.
To start out, imagine that, as an
expert in factory workplace efficiency, you have been hired by ABC tool company to
improve the efficiency of their factory assembly line. Specifically, the company wants to
know if it’s possible to compress its assembly line to a smaller size and occupy a
smaller footprint. As you survey the assembly line,
you noticed that there are many segments of the line that connect different vertical
elevations. To consider how it might be
possible to shorten these segments and the overall assembly line length, we’ll want
to know something about static friction.
Static friction is a force measured
in newtons that keeps an object at rest relative to a surface. Say that there is a gigantic heavy
box at rest on a flat surface. If you walked up to this box and
tried to move it by exerting a force pushing to the right, we know from practical
experience that the box would remain still. But why? After all, Newton’s second law says
that if we apply a net force to an object, then that object accelerates.
The reason this box doesn’t start
moving when we push on it is that there’s another force at work. This force originates in the
interaction between the bottom of the box and the surface that it’s resting on. And this force directly opposes the
pushing force that we apply. The name of this force is static
friction. And we abbreviate it 𝐹 sub 𝑠.
Here’s something interesting to
consider. If the surface that our box is
resting on is wood, what if we were to change out that surface to something
smoother? What if our box was on ice
instead? In that case, if we were able to
apply the same amount of pushing force, our box may well start to move. This tells us something about
static friction.
Intuitively, we recognize that
static friction between objects depends on the materials the objects are made
of. The interaction of one material
type with another has been captured by a variable we represent by the Greek symbol
𝜇 with a subscript 𝑠 when we’re talking of static friction. This symbol is the coefficient of
static friction.
This table shows us how 𝜇 sub 𝑠
comes about from material pairings. In the first column, we have one
type of material, and in the second the second type. The coefficient of static friction
between these particular material pairs is given in the far right column. The normal range that the
coefficient of static friction can take on is anywhere from zero to one. It’s possible for 𝜇 sub 𝑠 to
exceed one, but it’s very rare.
Let’s take another look at our box
sliding along ice to get an understanding of static friction at the microscopic
level. We know that, at the interaction
between the box and the ice, there’s friction that occurs. If we were to zoom very, very far
in into this interaction point, we would see that, rather than our two materials
being smooth, like they may feel to our hand, they’re actually quite rough. It’s these rough microscopic teeth
interacting with one another that give rise to static friction. And the more jagged the teeth are,
the less likely the objects are to slide past one another.
We’ve seen in our table an example
of a very high coefficient of static friction. Concrete and rubber are very
unlikely to slide past one another. But what about at the other end of
the spectrum? You may have heard of teflon as a
cooking material. Teflon on teflon has a coefficient
of static friction of only 0.04. And the synovial joints in our
bodies have even lower friction.
We’ve talked about the coefficient
of static friction, 𝜇 sub 𝑠, but what about the force of friction, 𝐹 sub 𝑠, we
mentioned earlier? How do we actually calculate the
force of static friction? We can guess that this force has
something to do with 𝜇 sub 𝑠. And that’s right. But there’s another element
involved.
Consider again our box pushing
example. And imagine that our box was much
smaller now than it was originally. We know intuitively that this box
would be easier to push along the surface than the larger box. The resistive frictional force is
less. And that’s because the weight force
on this box is less than that on a larger one. It turns out that the static
friction force isn’t directly dependent on the weight force. But it is directly dependent on the
normal force that the surface responds to the object with.
If we take this normal force, 𝐹
sub 𝑁, and multiply it by the coefficient of static friction, then it’s this
product that’s equal to the static friction force. Keeping this relationship in mind,
let’s try a few examples involving static friction.
The coefficient of static friction
between the rubber eraser of the pencil and the tabletop is 𝜇 sub 𝑠 equals
0.80. If the force 𝐹 is applied along
the axis of the pencil, as shown below, what is the minimum angle at which the
pencil can stand without slipping? Ignore the weight of the
pencil.
Knowing the coefficient of static
friction between the pencil and the tabletop, we wanna solve for the smallest angle,
𝜃, at which the pencil can still be standing upright. To figure out this angle, we can
consider the forces acting on the pencil at the place where it meets the
tabletop. We’re to ignore the weight force of
the pencil.
At this location, there are three
forces in action. There’s the force 𝐹 applied along
the length of the pencil, the normal force pushing the pencil up from the tabletop,
and the static friction force keeping the pencil from sliding. We want to solve for 𝜃 such that
these three forces are in equilibrium.
If we break the applied force 𝐹 up
into its vertical and horizontal components, we see these components help to form a
right triangle with that applied force. Because our forces are in
equilibrium, we can write that 𝐹 times the cos of 𝜃 is equal to the frictional
force 𝐹 sub 𝑠. Recalling that, in general, the
friction force 𝐹 sub 𝑓 is equal to the product of the coefficient of friction and
the normal force, we can replace 𝐹 sub 𝑠 with 𝜇 sub 𝑠 times 𝐹 sub 𝑁.
We now have an equation that
balances out the forces in the horizontal direction. Let’s look now in the vertical
direction. The vertical forces on the pencil
are the applied force 𝐹 times the sin of 𝜃 and the normal force. And in magnitude, they equal one
another. This means we can replace 𝐹 sub 𝑁
in our first expression with 𝐹 times the sin of 𝜃.
And when we do that, we see that
the applied force 𝐹 appears on both sides and, therefore, cancels out. If we then divide both sides of our
equation by the cos of 𝜃, the left-hand side becomes one. And looking at the right-hand side,
we can recall the trigonometric identity that the sine of an angle divided by the
cosine of the same angle equals the tangent of that angle. Our equation simplifies to one
equals 𝜇 sub 𝑠 times the tan of 𝜃, or the tan of 𝜃 equals one over 𝜇 sub
𝑠.
If we then take the arc tangent of
both sides, we find an expression for the angle we want to solve for, 𝜃. We’re ready to plug in for 𝜇 sub
𝑠 with our given value of 0.80. When we solve for 𝜃, to two
significant figures, our result is 51 degrees. That’s the smallest angle between
the pencil and the tabletop, with the pencil staying upright.
Now let’s look at an example
involving static friction for a rolling object.
If half the weight of a small
utility truck with a mass of 1.00 times 10 to the third kilograms is supported by
its two drive wheels, what is the maximum magnitude of acceleration that the truck
can achieve on dry concrete, if the coefficient of static friction of the truck’s
tyres on dry concrete is 1.0?
We can call the mass of this small
truck 𝑚. And the coefficient of static
friction between the truck’s tyres and dry concrete we’ll call 𝜇 sub 𝑠. We want to solve for the maximum
acceleration magnitude possible so that the truck’s tyres don’t slip on the dry
concrete. We’ll label this acceleration 𝑎
and start off by drawing a diagram of the situation.
In this example, we have a small
utility truck with front wheel drive. That is, its engine turns the two
front wheels in the drivetrain. Looking at an expanded view of one
of these tyres, we know the tyre rotates and it interacts with the dry concrete
surface. We know that the tyre pushes on the
concrete. And the concrete responds through
friction, pushing the tyre ahead. This frictional force is a static
frictional force, because even though the tyre is rotating, at any given instant,
the interface between the tyre and the concrete doesn’t involve slipping.
With this frictional force being
the only horizontal force acting on the tyre, recalling Newton’s second law of
motion that the net force acting on an object equals its mass times its
acceleration, we can apply this by writing 𝐹 sub 𝑠, the frictional force, is equal
to the mass of the truck multiplied by its acceleration. Remember, we also know that the
force of static friction is equal to the coefficient of static friction times the
normal force.
And in this case, the normal force
of the dry concrete acting on the two front wheels is equal to the mass of half of
the truck multiplied by 𝑔, where we assume that 𝑔 is equal to exactly 9.8 meters
per second squared. When we replace 𝐹 sub 𝑁 in our
equation with the half the mass of the truck times 𝑔, when we consider the equation
that results, we see that the mass of the truck 𝑚 cancels out from both sides. So, the acceleration of the truck
𝑎 is equal to 𝜇 sub 𝑠 times 𝑔 over two. Plugging in for our given value of
𝜇 sub 𝑠 and our given value for 𝑔, we calculate 𝑎 to be 4.9 meters per second
squared. That’s the maximum possible
acceleration of the truck without its tyres slipping on the concrete.
Let’s summarize what we’ve learned
so far about static friction. We’ve seen that static friction is
a force that keeps an object at rest relative to a surface. Written as an equation, the force
of static friction is equal to the coefficient of static friction between the object
and the surface multiplied by the normal force of the surface on the object.
We’ve seen that 𝜇 sub 𝑠, the
coefficient of static friction, varies depending on the materials involved in the
interaction. And we’ve also seen that static
friction is caused by microscopic surface irregularities between the surface and the
object on that surface. When an object is resting on a
surface, under an applied force, static friction is the force responsible for it
remaining at rest.
