Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.

Video: Static Friction

Ed Burdette

In this video we learn what static friction is and what causes it, and also about the coefficient of static friction between pairs of materials.

11:33

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 will 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 tires 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 tires and dry concrete we’ll call 𝜇 sub 𝑠. We want to solve for the maximum acceleration magnitude possible so that the truck’s tires 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 tires, we know the tire rotates and it interacts with the dry concrete surface. We know that the tire pushes on the concrete. And the concrete responds through friction, pushing the tire ahead.

This frictional force is a static frictional force, because even though the tire is rotating, at any given instant, the interface between the tire and the concrete doesn’t involve slipping.

With this frictional force being the only horizontal force acting on the tire, 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 tires 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.