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 FsF_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 FkF_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).

What is the formula for the kinetic frictional force FkF_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 FnF_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 μk\mu_k. The parameter μk\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 μk\mu_k.
We can put these ideas into a mathematical form with the following equation.
Fk=μkFn\Large F_k=\mu_kF_n
Note that we can rewrite this equation as μk=FkFn\mu_k=\dfrac{F_k}{F_n}, which shows that the coefficient of kinetic friction μk\mu_k is a dimensionless quantity.

What is the formula for the static frictional force FsF_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 FnF_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 μs\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.
Fs max=μsFn\Large F_{s\text{ max}}=\mu_sF_n
Be careful, the quantity Fs maxF_{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 Fs max=50 NF_{s\text{ max}}=50\text{ N}. If you were to try and budge the washing machine with 30 N30\text{ N}, the static frictional force will only be 30 N30\text{ N}. If you increase the force you exert to 40 N40\text{ N}, the static frictional force will also increase to 40 N40\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 kg110 \text{ kg} refrigerator sits on the floor. The coefficient of static friction between the refrigerator and the floor is 0.600.60, and the coefficient of kinetic friction between the refrigerator and the floor is 0.400.40. The person pushing on the refrigerator tries to budge the fridge with the following forces.
i. Fpush=400 N\greenD {F_\text{push}}=400 \text{ N}
ii. Fpush=600 N\greenD {F_\text{push}}=600 \text{ N}
iii. Fpush=800 N\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.
Fs max=μsFn(start with the formula for the maximum static frictional force)F_{s\text{ max}}=\mu_sF_n \quad \text{(start with the formula for the maximum static frictional force)}
Fs max=(μs)(mg)(the normal force will be equal to the force of gravity in this case)F_{s\text{ max}}=(\mu_s)(mg) \quad \text{(the normal force will be equal to the force of gravity in this case)}
Fs max=(0.60)(110 kg)(9.8ms2)(plug in the coefficient of static friction, mass, and value of g)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)}
Fs max=647 N(calculate)F_{s\text{ max}}=647 \text{ N} \quad \text{(calculate)}
Now that we know the maximum amount of static frictional force is 647 N647\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 Fpush=400 N\greenD {F_\text{push}}=400 \text{ N} there will be a matching static frictional force of Fs=400 NF_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 Fpush=600 N\greenD {F_\text{push}}=600 \text{ N} there will be a matching static frictional force of Fs=600 NF_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 Fpush=800 N\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.
Fk=μkFn(use the formula for kinetic friction)F_k=\mu_kF_n \quad \text{(use the formula for kinetic friction)}
Fk=(0.40)(110 kg)(9.8ms2)(plug in the coefficient of kinetic friction and normal force)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)}
Fk=431 N(calculate the kinetic frictional force)F_k=431 \text{ N} \quad \text{(calculate the kinetic frictional force)}
iii. So if the person pushes with Fpush=800 N\greenD {F_\text{push}}=800 \text{ N} there will be a kinetic frictional force of Fk=431 NF_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 kg1.3 \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 θ=60o\theta=60^o and under a tension of 4 N4 \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 Fk=μkFnF_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.
ax=ΣFxm(start with Newton’s second law in the horizontal direction)a_x=\dfrac{\Sigma F_x}{m} \quad \text{(start with Newton's second law in the horizontal direction)}
0=TxFk1.3 kg(plug in horizontal forces, acceleration, and mass)0=\dfrac{\greenD{T_x}-\purpleD{F_k}}{1.3\text{ kg}} \quad \text{(plug in horizontal forces, acceleration, and mass)}
0=Tcos60oμkFn1.3 kg(plug in horizontal component of tension and formula for kinetic friction)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)}
0=Tcos60oμkFn(multiply both sides by mass)0=T\text{cos}60^o-\mu_kF_n \quad \text{(multiply both sides by mass)}
μk=Tcos60oFn(algebraically solve for the coefficient of kinetic friction)\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 mgmg, but since the rope is also pulling upward on the box, the normal force will be less than mgmg. 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 Ty=Tsin60oT_y=T\text{sin}60^o. So the normal force in this case will be Fn=mgTsin60F_n=mg-T\text{sin}60.
Now we can plug this expression for the normal force FnF_n into our formula for the coefficient of kinetic friction we found above.
μk=Tcos60oFn(use formula we found above for coefficient of kinetic friction)\mu_k=\dfrac{{T}\text{cos}60^o}{F_n} \quad \text{(use formula we found above for coefficient of kinetic friction)}
μk=Tcos60omgTsin60o(plug in expression found for normal force)\mu_k=\dfrac{{T}\text{cos}60^o}{mg-T\text{sin}60^o} \quad \text{(plug in expression found for normal force)}
μk=(4 N)cos60o(1.3 kg)(9.8ms2)(4 N)sin60o(plug in values for the tension and mass)\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)}
μk=0.216(calculate and celebrate)\mu_k=0.216 \quad \text{(calculate and celebrate)}