# Lesson Video: Static Friction Physics • 9th Grade

In this video, we will learn how to analyze the forces acting on objects on frictional surfaces that are required to induce motion in the objects.

14:32

### Video Transcript

Static friction is our topic in this video. This is the sort of friction that exists between objects that are in contact with but not in motion relative to one another. Static friction is the force keeping all of these stacked objects from sliding down the incline or sliding past one another. As we’ll see, this is a force that has to do with the interfaces or the contact points between different objects.

So, static friction is a contact force. And interestingly, it only exists in response to other forces. Say, for example, we have this box at rest on a flat surface. The box we can see is in contact with the surface; let’s say it’s the top of a table. But even so, under these conditions, there’s no static friction force present. Rather, the only forces acting on the box are its weight force, its mass times the acceleration due to gravity, and the contact force or normal force acting on the box to oppose the weight force.

But then let’s imagine that we apply a steady and small force on the box, pushing it to the right. We can call this 𝐹 sub A to show that it’s an applied force, and it’s in response to this that a static frictional force will arise. That is, as we push on the box that would slide it across the tabletop, the microscopically rough interface between the box and the tabletop resists this. The large-scale effect of these microscopic interactions is a frictional force that acts to the left. And up to a point, this frictional force will exactly match and oppose the applied force 𝐹 sub A.

We can see this by looking at a graph of the frictional force, which arises as a response to the applied force. At the outset, before we’ve pushed on the box at all, we know the applied force is zero. And as we saw, so is the frictional force at that moment. But then, we do start to push on the box. And say that we evenly increase the applied force from zero up to one newtons. In response, the frictional force may also evenly match that applied force. And since all throughout this progression the frictional force magnitude equals our applied force magnitude, the box stays in place.

But let’s say we continue pushing harder on the box, smoothly increasing the force we apply up to two newtons. We may find it to still be the case that the frictional force matches this applied force. When this happens, we know that this frictional force refers to a static friction since the object under our applied force is not moving. We know from experience though that as we increase this applied force, eventually it will exceed the frictional force that the interface between our box and tabletop can exert. When this takes place, the box would start to move and we no longer have a static friction situation.

In this lesson, we’ll limit our discussion to those scenarios where the frictional force does match the applied force. Just know that if we kept increasing the force applied, eventually it would exceed the force of friction. In other words, we’ll require that 𝐹 sub f is equal to 𝐹 sub A. Now, we know that the applied force is somehow determined by an external condition. In the case of our box on the tabletop, it’s determined by how hard we push on the box. But what about the static friction force 𝐹 sub f? If we think about our own experience with friction, we can recall that heavier objects tend to require more force to get them moving across a surface. For example, if this box were one kilogram in mass, then it would be easier to start accelerating than if it were, say, 100 kilograms. This points us towards the idea that the static frictional force acting on an object is proportional to its weight.

And then, as we continue to think about static friction, we might also guess that it has something to do with the specific materials in contact with one another. For example, we know that if one material is metal, say the blade of an ice skate, and the other material is ice, then there’s very little friction between these two surfaces. On the other hand, say we’re imagining an asphalt road surface and a rubber tire on an automobile. We’re very glad for the fact that there is a lot of friction between these two surfaces. So, while the force of static friction does depend on an object’s weight as it rests on some surface, it also depends on what those particular surfaces in contact are.

And in fact, there’s a quantitative, that is, numbers-based, way of describing these interactions for various materials. It depends on what is called a coefficient of static friction, represented 𝜇 with an s subscript. The coefficient of static friction is a number often between zero and one, but not always. That gets higher for surfaces in contact that are less likely to slide past one another and lower for surfaces in contact that are more likely to slip past one another.

It’s possible to look up tables of coefficients of static friction for many different pairs of materials. For example, if we consider the coefficient of static friction between a wheel made of rubber and dry asphalt, we find that value to be approximately 0.9. Notice that a coefficient of static friction is unitless; it’s a pure number. On the other hand, if we look up 𝜇 sub s for when steel and ice are in contact, that value is actually close to zero. We bring all this up because it’s exactly this factor, the coefficient of static friction, that we’ll need to include in our equation for frictional force.

The static frictional force acting on some object is equal to that object’s weight multiplied by the coefficient of static friction relevant for the material the object is made of and the material that it rests on. And often, we’ll see a negative sign precede this expression. And this is to express the fact that a frictional force in direction always opposes an applied force. This then is the mathematical relationship we can use to calculate the force of friction on an object that is not in motion, that is, is static. And notice that because the magnitude of a static frictional force is always equal to the applied force tending to put the object in motion, we can say that the applied force is equal to the magnitude of the right-hand side of this expression, in other words, 𝑚 times 𝑔 times 𝜇 sub s.

Before we go on to try an example exercise, there’s one subtlety to this equation that we should mention. Recall on our opening screen that instead of having a flat surface for our objects to rest on, we had a surface at an incline. If we put our box on this slope, then we know that the weight force 𝑚𝑔 once again acts straight downward. But now, the normal or reaction force does not directly oppose the weight force. This distinction actually has an effect on the static frictional force between the box and the inclined plane. That’s because the force of friction, rather than depending directly on the weight force, actually depends on the normal or reaction force. Most generally then, in our equation for static frictional force, we would replace 𝑚 times 𝑔 with 𝐹 sub N.

This way, whether our object is on a flat surface or on a sloped one, we’re still able to accurately calculate the static frictional force acting on it. In cases where our object is on a flat surface, 𝐹 sub N is equal to 𝑚 times 𝑔. But in other scenarios, we’ll need to use the normal force rather than the weight force in this expression. Knowing all this about static friction, let’s now look at an example.

The figure shows two blocks, block A and block B. The blocks have equal masses. Block A is on a rough surface, and block B is on block A. A force 𝐹 acts on block B, parallel to the surface. Initially, block B does not move, and 𝐹 is steadily increased. When 𝐹 has increased enough to make block B start to move, both blocks move in the direction of 𝐹 at the same speed. Which of the following is a correct statement about the relation of 𝜇 sub one, the coefficient of static friction between block A and the surface, to 𝜇 sub two, the coefficient of static friction between the blocks? (A) 𝜇 sub one is equal to 𝜇 sub two. (B) 𝜇 sub one is less than 𝜇 sub two. (C) 𝜇 sub one is greater than 𝜇 sub two.

Alright, in this example, we have these two blocks: block B rests on block A and then block A is resting on this rough surface. While the two blocks are stationary, an applied force 𝐹 is exerted on block B, parallel to the interface between the blocks. At first, nothing happens. But then, as 𝐹 is increased, eventually block B does start to move. At that exact instant though, block A is also set in motion. And both blocks move to the left at the same speed we’re told. So, this is interesting. Even though the force 𝐹 is being applied to block B, what ends up happening is both blocks move as a unit to the left.

Knowing this, we want to make a comparison between the coefficient of static friction that exists between the two blocks, that’s called 𝜇 sub two, and also the coefficient of static friction that exists between block A and the rough surface; that is called 𝜇 sub one. Basically, we want to make a comparison between these two coefficients of static friction. As we think about what’s happening physically in this situation, we see that even though we’re applying a force just to block B, apparently, the frictional force between block B and block A is strong enough that the whole entire mass of the two blocks together begins to move before block B starts to slide across the top of block A.

We could say then that this static frictional force here that acts between block A and the surface is weaker than the static frictional force that acts between the two blocks. We can write out equations for these two forces of static friction. We can recall that the magnitude of the static frictional force acting on some object that’s at rest on a flat surface is equal to that object’s mass times the acceleration due to gravity times the coefficient of static friction between the object and the surface it rests on. If we call the frictional force in our scenario that exists between block A and the surface 𝐹 sub one, then we know that’s equal to the mass of block A plus the mass of block B, because block B rests on top of block A, multiplied by 𝑔 times 𝜇 sub one, the coefficient of static friction between block A and the surface. Compared to this, if we call the static frictional force between the two blocks 𝐹 sub two, then that’s equal simply to the mass of block B times 𝑔 times 𝜇 sub two.

In our problem statement, we were told that as these blocks move, they move as a unit. We saw that this implies that the frictional force between the blocks must be greater than that between block A and the surface. In other words, 𝐹 sub two must be greater than 𝐹 sub one. We can then replace 𝐹 sub two in this inequality with this expression and 𝐹 sub one with this one. And once we’ve done that, let’s recall an important bit of information given to us earlier on. We weren’t told what the masses of blocks B and A are, but we were told that they’re equal. Off to the side then, we can write that 𝑚 sub B is equal to 𝑚 sub A, and we can call them both just 𝑚. If we make that substitution in our inequality, then we have this expression where 𝑚 and 𝑔 are common to both sides.

If we then divide both sides by 𝑚 times 𝑔, then this cancels out those terms on either side of the equality while keeping its direction the same. And we find that 𝜇 sub two is greater than two times 𝜇 sub one. This means that even if we double 𝜇 sub one, it’s still less than 𝜇 sub two. Looking at our three answer options, we see that this corresponds to option (B) 𝜇 sub one is less than 𝜇 sub two.

Let’s look now at another example exercise.

A brick is initially at rest on top of a rough horizontal surface, as shown in the diagram. The brick is pushed by a force 𝐹 parallel to the surface. The weight of the brick equals 30 newtons. The coefficient of static friction of the brick with the surface is 0.75. What must the magnitude of 𝐹 exceed to start the brick moving across the surface?

Okay, so we have this brick initially at rest on a rough horizontal surface. And we see that this brick is being pushed by a force 𝐹 parallel to the surface. For small enough values of 𝐹, the brick actually won’t move. And that’s due to frictional effects between the brick and the surface, leading to a frictional force on the brick. This frictional force is in opposition to the applied force 𝐹. But the frictional force has some maximum value it can achieve. If the applied force 𝐹 exceeds that maximum value, then the brick will start to move.

To figure out the greatest possible static frictional force that acts on the brick, we can recall that the magnitude of the frictional force acting on some object on a flat surface is equal to the object’s weight multiplied by the coefficient of static friction between the object and the surface it rests on. We can say then that the maximum static frictional force acting on the brick is equal to its weight, here we have the magnitude of its weight, multiplied by the coefficient of static friction between the brick and the rough surface.

Just as a side note, we say that this is the maximum static frictional force because the force actually could be less than this. When the force 𝐹 that we apply to the brick is very small, yet increasing, the static frictional force on the brick keeping it in place increases with 𝐹. But as we’ve said, this happens only up to the point where that applied force is either less than or equal to the weight of this brick multiplied by the coefficient of static friction.

When 𝐹 exceeds this value, it’s at that exact moment that the brick will start to move. And that’s why we’re solving for 𝐹 sub f max. Substituting in our given values, this is equal to 30 newtons times 0.75. And if we keep just two significant figures from our answer, that ends up as 23 newtons. Once our applied force 𝐹 exceeds this value, the static frictional force will no longer be able to keep up, so to speak, and the brick will start to move.

Let’s summarize now what we’ve learned about static friction. In this lesson, we saw that static friction is a contact force. That is, it’s a force between two objects that are in physical contact. It keeps a given object from moving, and it does so by acting in response to other forces. The force of static friction acting on an object depends on two factors: its weight, or the normal force acting on it if the object is on an incline, and the coefficient of static friction. We saw that this second factor, represented with 𝜇 sub s, depends on the materials that the object and the surface it rests on are made of.

All this comes together into an expression for the static frictional force acting on some object. It’s equal to the normal or reaction force acting on the object, which is equal to 𝑚 times 𝑔 if the object is on a flat surface, multiplied by the relevant coefficient of static friction. And we say this force is negative because it opposes the direction of an applied force tending to make the object in question move. This is a summary of static friction.