# What is friction?

Until now in physics, you've probably been ignoring friction to make things simpler. Now, it's time to include this very real force and see what happens.

## What are the forces of static and kinetic friction?

Parking your car on the steep hills of San Francisco is scary, and it would be impossible without the force of static friction.
The force of static friction $F_s$ is a force between two surfaces that prevents those surfaces from sliding or slipping across each other. This is the same force that allows you to accelerate forward when you run. Your planted foot can grip the ground and push backward, which causes the ground to push forward on your foot. We call this "grippy" type of friction, where the surfaces are prevented from slipping across each other, a static frictional force. If there were absolutely no friction between your feet and the ground, you would be unable to propel yourself forward by running, and would simply end up jogging in place (similar to trying to run on very slippery ice).
Now, if you park on a hill that is too steep, or if you are being pushed backward by a Sumo wrestler you're probably going to start sliding. Even though the two surfaces are sliding past each other, there can still be a frictional force between the surfaces, but this sliding friction we call a kinetic frictional force. This force of kinetic friction $F_k$ always opposes the sliding motion and tries to reduce the speed at which the surfaces slide across each other. For example, a person sliding into second base during a baseball game is using the force of kinetic friction to slow down. If there were no kinetic friction, the baseball player would just continue sliding (yes, this would make stealing bases in baseball difficult).
Most surfaces (like wood and plastics) seem pretty smooth to the naked eye. But if you were to zoom in, you would see that the surfaces are rough at a microscopic level. For instance, the surfaces of the box and floor as shown below are actually rough and jagged at a microscopic level.

Image Credit: Openstax College Physics
Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded view. In order for the object to move, it must rise to where the peaks can skip along the bottom surface. Thus a force is required just to set the object in motion. Some of the peaks will be broken off, also requiring a force to maintain motion. Much of the friction is actually due to attractive forces between molecules making up the two objects, so that even perfectly smooth surfaces are not friction-free. Such adhesive forces also depend on the substances the surfaces are made of, explaining, for example, why rubber-soled shoes slip less than those with leather soles. But to be completely honest, we still don't have a complete understanding of the microscopic causes of friction. More recent research in tribology (study of sliding surfaces) suggests that vibration in the surfaces, leading to energy loss through heat, is a main source of friction.
Concept Check: For each of the following cases of a car changing velocity described in the table below, choose whether it is more likely to be the force of static or kinetic friction causing the change in velocity.

Static frictional force
Kinetic frictional force
A car slows gently to a stop.
A car slams on the brakes and skids to a stop.
A car accelerates gently to a higher speed.
A car "floors it" and peels out of a stop light.
A car takes a turn gently.

If car tires are working the way they should (rolling without slipping) it's the grippy force of static friction that is acting on the tires. The force of static friction between the tires and the ground is what allows cars to speed up, slow down, and turn safely and gently. To speed up, the tires start spinning faster and grip the road using the force of static friction. The tires can then push backward on the street, and the third law dictates that the street has to push forward on the tires. It is the gripping force of static friction $\redD{F_s}$ that allows this to happen.
If the tires start spinning too quickly ("flooring it"), or stop spinning too quickly ("slam on the brakes"), there will be skidding (and smoke and skid marks) between tires and the ground. This skidding is an easy way to tell that the force acting is kinetic friction rather than static friction.
People often find it strange to think that a moving car tire can have static friction acting on it. Learners often erroneously think that static friction means the object can't be moving. But static friction just implies that two surfaces aren't slipping/sliding/skidding past each other, not that the objects have no motion. For the case of the car tire, the bottom of the tire that's in contact with the street is momentarily at rest with respect to the street since it isn't slipping. The rest of the tire rotates around the point of contact. The point in contact keeps changing as the tire continues to roll forward.

## What is the formula for the kinetic frictional force $F_k$?

If you press your hands into each other hard and rub them together, the force of kinetic friction will be larger than if you were only pressing your hands together lightly. That's because the amount of kinetic frictional force between two surfaces is larger the harder the surfaces are pressed into each other (i.e. larger normal force $F_n$).
Also, changing the types of surfaces sliding across each other will change the amount of kinetic frictional force. The "roughness" of two surfaces sliding across each other is characterized by a quantity called the coefficient of kinetic friction $\mu_k$. The parameter $\mu_k$ depends only on the two surfaces in contact and will be a different value for different surfaces (e.g. wood and ice, iron and concrete, etc.). Two surfaces that do not slide easily across each other will have a larger coefficient of kinetic friction $\mu_k$.
We can put these ideas into a mathematical form with the following equation.
$\Large F_k=\mu_kF_n$
Note that we can rewrite this equation as $\mu_k=\dfrac{F_k}{F_n}$, which shows that the coefficient of kinetic friction $\mu_k$ is a dimensionless quantity.
Since $\mu_k=\dfrac{F_k}{F_n}$, we know the units of the coefficient of kinetic friction $\mu_k$ are the same as the fraction $\dfrac{F_k}{F_n}$. But both $F_k$ and $F_n$ have units of newtons, so the ratio of $\dfrac{F_k}{F_n}$ ends up with no units (the units cancel).
This shows that the coefficient of kinetic friction $\mu_k$ is a number with no units. We call these types of quantities dimensionless quantities since they have no units (i.e. no physical dimensions).

## What is the formula for the static frictional force $F_s$?

The static frictional force is a little different from the kinetic frictional force. For one, the static frictional force will change its value based on how much force is being applied to the unbudging object. Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do. It increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion it is easier to keep it in motion than it was to get it started, indicating that the kinetic frictional force is less than the maximum static frictional force.
If you add mass to the crate, say by placing a box on top of it (increasing the amount of normal force $F_n$), you need to push even harder to get it started and also to keep it moving. Furthermore, if you oiled the concrete (reducing the coefficient of static friction $\mu_s$) you would find it to be easier to get the crate started (as you might expect).
We can put these ideas in a mathematical form by writing the following formula that lets us find the maximum possible static frictional force between two surfaces.
$\Large F_{s\text{ max}}=\mu_sF_n$
Yes. Since this equation only gives the maximum value for the force of static friction $F_{s\text{ max}}$, we can instead choose to write a formula for the actual value of the force of static friction $F_s$ by writing,
$F_s \leq \mu_s F_n$
This equation says that the actual magnitude of the force of static friction $F_s$ can be as large as the quantity $\mu_s F_n$.
Be careful, the quantity $F_{s\text{ max}}$ only gives you the maximum possible static frictional force, not the actual static frictional force for a given scenario. For instance, suppose that between a washing machine and a tile floor the maximum possible force of static friction was found to be $F_{s\text{ max}}=50\text{ N}$. If you were to try and budge the washing machine with $30\text{ N}$, the static frictional force will only be $30\text{ N}$. If you increase the force you exert to $40\text{ N}$, the static frictional force will also increase to $40\text{ N}$. This continues until the force you apply is greater than the maximum static frictional force, at which point the washing machine budges and starts sliding. Once the washing machine starts sliding, there is no longer static frictional force but only kinetic frictional force.

## What do solved examples involving the force of friction look like?

### Example 1: Push the fridge

An initially stationary $110 \text{ kg}$ refrigerator sits on the floor. The coefficient of static friction between the refrigerator and the floor is $0.60$, and the coefficient of kinetic friction between the refrigerator and the floor is $0.40$. The person pushing on the refrigerator tries to budge the fridge with the following forces.
i. $\greenD {F_\text{push}}=400 \text{ N}$
ii. $\greenD {F_\text{push}}=600 \text{ N}$
iii. $\greenD {F_\text{push}}=800 \text{ N}$
For each individual case listed above, determine the magnitude of the frictional force that will exist between the bottom of the refrigerator and the floor.

To start we'll solve for the maximum possible amount of static frictional force.
$F_{s\text{ max}}=\mu_sF_n \quad \text{(start with the formula for the maximum static frictional force)}$
$F_{s\text{ max}}=(\mu_s)(mg) \quad \text{(the normal force will be equal to the force of gravity in this case)}$
$F_{s\text{ max}}=(0.60)(110\text { kg})(9.8 \dfrac{\text{m}}{\text{s}^2}) \quad \text{(plug in the coefficient of static friction, mass, and value of g)}$
$F_{s\text{ max}}=647 \text{ N} \quad \text{(calculate)}$
Now that we know the maximum amount of static frictional force is $647\text{ N}$, we know that any force the person exerts below this amount will get matched by the force of static friction. In other words,
i. If the person pushes with $\greenD {F_\text{push}}=400 \text{ N}$ there will be a matching static frictional force of $F_s=400\text{ N}$ preventing the refrigerator from budging. There will be no kinetic friction since the refrigerator will not slide.
ii. If the person pushes with $\greenD {F_\text{push}}=600 \text{ N}$ there will be a matching static frictional force of $F_s=600\text{ N}$ preventing the refrigerator from budging. There will be no kinetic friction since the refrigerator does not slide.
For case iii, the force $\greenD {F_\text{push}}=800 \text{ N}$ is above the maximum force of static friction, so the fridge will start to slide. Now that the fridge is sliding there will be a kinetic frictional force exerted on it. We can find the force of kinetic friction as follows.
$F_k=\mu_kF_n \quad \text{(use the formula for kinetic friction)}$
$F_k=(0.40)(110\text{ kg})(9.8\dfrac{\text{m}}{\text{s}^2}) \quad \text{(plug in the coefficient of kinetic friction and normal force)}$
$F_k=431 \text{ N} \quad \text{(calculate the kinetic frictional force)}$
iii. So if the person pushes with $\greenD {F_\text{push}}=800 \text{ N}$ there will be a kinetic frictional force of $F_k=431\text{ N}$ exerted on the fridge. There will be no static frictional force since the fridge is sliding.

### Example 2: Box pulled across a rough table

A $1.3 \text{ kg}$ box of frozen chocolate chip waffles is pulled at constant velocity across a table by a rope. The rope is at an angle $\theta=60^o$ and under a tension of $4 \text{ N}$.
What is the coefficient of kinetic friction between the table and the box?
Since we don't know the coefficient of kinetic friction we can't use the formula $F_k=\mu_kF_n$ to directly solve for the frictional force. However, since we know the acceleration in the horizontal direction (it's zero since the box moves at constant velocity) we should start with Newton's second law.
Whenever we use Newton's second law we should draw a force diagram.
$a_x=\dfrac{\Sigma F_x}{m} \quad \text{(start with Newton's second law in the horizontal direction)}$
$0=\dfrac{\greenD{T_x}-\purpleD{F_k}}{1.3\text{ kg}} \quad \text{(plug in horizontal forces, acceleration, and mass)}$
$0=\dfrac{T\text{cos}60^o-\mu_kF_n}{1.3\text{ kg}} \quad \text{(plug in horizontal component of tension and formula for kinetic friction)}$
We can find the horizontal component of tension by using trigonometry and the definition of sine and cosine.
$\text{cos}60^o=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{{T_x}}{ T}$
$T_x={T}\text{cos}60^o$
$0=T\text{cos}60^o-\mu_kF_n \quad \text{(multiply both sides by mass)}$
$\mu_k=\dfrac{T\text{cos}60^o}{F_n} \quad \text{(algebraically solve for the coefficient of kinetic friction)}$
At this point you might think we should plug in the normal force as $mg$, but since the rope is also pulling upward on the box, the normal force will be less than $mg$. The normal force will be reduced by the amount we pull up on the box. In this case, the vertical component of the tension is $T_y=T\text{sin}60^o$. So the normal force in this case will be $F_n=mg-T\text{sin}60$.
We can use Newton's second law in the vertical direction to find the normal force between the box and the table.
$a_y=\dfrac{\Sigma F_y}{m} \quad \text{(use Newton's second law in the vertical direction)}$
$0=\dfrac{F_n+T\text{sin}60^o-mg}{1.3\text{ kg}} \quad \text{(plug in vertical forces, acceleration and mass)}$
$F_n+T\text{sin}60^o-mg=0 \quad \text{(multiply both sides by mass)}$
$F_n=mg-T\text{sin}60^o \quad \text{(solve for normal force)}$
Now we can plug this expression for the normal force $F_n$ into our formula for the coefficient of kinetic friction we found above.
$\mu_k=\dfrac{{T}\text{cos}60^o}{F_n} \quad \text{(use formula we found above for coefficient of kinetic friction)}$
$\mu_k=\dfrac{{T}\text{cos}60^o}{mg-T\text{sin}60^o} \quad \text{(plug in expression found for normal force)}$
$\mu_k=\dfrac{{(4\text{ N})}\text{cos}60^o}{(1.3\text{ kg})(9.8\dfrac{\text{m}}{\text{s}^2})-(4\text{ N})\text{sin}60^o} \quad \text{(plug in values for the tension and mass)}$
$\mu_k=0.216 \quad \text{(calculate and celebrate)}$