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# 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, start subscript, s, end subscript 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, start subscript, k, end subscript 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).
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 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.

## What is the formula for the kinetic frictional force $F_k$F, start subscript, k, end subscript?

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, start subscript, n, end subscript).
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, start subscript, k, end subscript. The parameter mu, start subscript, k, end subscript 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, start subscript, k, end subscript.
We can put these ideas into a mathematical form with the following equation.
F, start subscript, k, end subscript, equals, mu, start subscript, k, end subscript, F, start subscript, n, end subscript
Note that we can rewrite this equation as mu, start subscript, k, end subscript, equals, start fraction, F, start subscript, k, end subscript, divided by, F, start subscript, n, end subscript, end fraction, which shows that the coefficient of kinetic friction mu, start subscript, k, end subscript is a dimensionless quantity.

## What is the formula for the static frictional force $F_s$F, start subscript, s, end subscript?

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, start subscript, n, end subscript), 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, start subscript, s, end subscript) 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.
F, start subscript, s, start text, space, m, a, x, end text, end subscript, equals, mu, start subscript, s, end subscript, F, start subscript, n, end subscript
Be careful, the quantity F, start subscript, s, start text, space, m, a, x, end text, end subscript 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, start subscript, s, start text, space, m, a, x, end text, end subscript, equals, 50, start text, space, N, end text. If you were to try and budge the washing machine with 30, start text, space, N, end text, the static frictional force will only be 30, start text, space, N, end text. If you increase the force you exert to 40, start text, space, N, end text, the static frictional force will also increase to 40, start text, space, N, end text. 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, start text, space, k, g, end text refrigerator sits on the floor. The coefficient of static friction between the refrigerator and the floor is 0, point, 60, and the coefficient of kinetic friction between the refrigerator and the floor is 0, point, 40. The person pushing on the refrigerator tries to budge the fridge with the following forces.
i. start color #1fab54, F, start subscript, start text, p, u, s, h, end text, end subscript, end color #1fab54, equals, 400, start text, space, N, end text
ii. start color #1fab54, F, start subscript, start text, p, u, s, h, end text, end subscript, end color #1fab54, equals, 600, start text, space, N, end text
iii. start color #1fab54, F, start subscript, start text, p, u, s, h, end text, end subscript, end color #1fab54, equals, 800, start text, space, N, end text
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, start subscript, s, start text, space, m, a, x, end text, end subscript, equals, mu, start subscript, s, end subscript, F, start subscript, n, end subscript, start text, left parenthesis, s, t, a, r, t, space, w, i, t, h, space, t, h, e, space, f, o, r, m, u, l, a, space, f, o, r, space, t, h, e, space, m, a, x, i, m, u, m, space, s, t, a, t, i, c, space, f, r, i, c, t, i, o, n, a, l, space, f, o, r, c, e, right parenthesis, end text
F, start subscript, s, start text, space, m, a, x, end text, end subscript, equals, left parenthesis, mu, start subscript, s, end subscript, right parenthesis, left parenthesis, m, g, right parenthesis, start text, left parenthesis, t, h, e, space, n, o, r, m, a, l, space, f, o, r, c, e, space, w, i, l, l, space, b, e, space, e, q, u, a, l, space, t, o, space, t, h, e, space, f, o, r, c, e, space, o, f, space, g, r, a, v, i, t, y, space, i, n, space, t, h, i, s, space, c, a, s, e, right parenthesis, end text
F, start subscript, s, start text, space, m, a, x, end text, end subscript, equals, left parenthesis, 0, point, 60, right parenthesis, left parenthesis, 110, start text, space, k, g, end text, right parenthesis, left parenthesis, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, start text, left parenthesis, p, l, u, g, space, i, n, space, t, h, e, space, c, o, e, f, f, i, c, i, e, n, t, space, o, f, space, s, t, a, t, i, c, space, f, r, i, c, t, i, o, n, comma, space, m, a, s, s, comma, space, a, n, d, space, v, a, l, u, e, space, o, f, space, g, right parenthesis, end text
F, start subscript, s, start text, space, m, a, x, end text, end subscript, equals, 647, start text, space, N, end text, start text, left parenthesis, c, a, l, c, u, l, a, t, e, right parenthesis, end text
Now that we know the maximum amount of static frictional force is 647, start text, space, N, end text, 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 start color #1fab54, F, start subscript, start text, p, u, s, h, end text, end subscript, end color #1fab54, equals, 400, start text, space, N, end text there will be a matching static frictional force of F, start subscript, s, end subscript, equals, 400, start text, space, N, end text preventing the refrigerator from budging. There will be no kinetic friction since the refrigerator will not slide.
ii. If the person pushes with start color #1fab54, F, start subscript, start text, p, u, s, h, end text, end subscript, end color #1fab54, equals, 600, start text, space, N, end text there will be a matching static frictional force of F, start subscript, s, end subscript, equals, 600, start text, space, N, end text preventing the refrigerator from budging. There will be no kinetic friction since the refrigerator does not slide.
For case iii, the force start color #1fab54, F, start subscript, start text, p, u, s, h, end text, end subscript, end color #1fab54, equals, 800, start text, space, N, end text 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, start subscript, k, end subscript, equals, mu, start subscript, k, end subscript, F, start subscript, n, end subscript, start text, left parenthesis, u, s, e, space, t, h, e, space, f, o, r, m, u, l, a, space, f, o, r, space, k, i, n, e, t, i, c, space, f, r, i, c, t, i, o, n, right parenthesis, end text
F, start subscript, k, end subscript, equals, left parenthesis, 0, point, 40, right parenthesis, left parenthesis, 110, start text, space, k, g, end text, right parenthesis, left parenthesis, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, start text, left parenthesis, p, l, u, g, space, i, n, space, t, h, e, space, c, o, e, f, f, i, c, i, e, n, t, space, o, f, space, k, i, n, e, t, i, c, space, f, r, i, c, t, i, o, n, space, a, n, d, space, n, o, r, m, a, l, space, f, o, r, c, e, right parenthesis, end text
F, start subscript, k, end subscript, equals, 431, start text, space, N, end text, start text, left parenthesis, c, a, l, c, u, l, a, t, e, space, t, h, e, space, k, i, n, e, t, i, c, space, f, r, i, c, t, i, o, n, a, l, space, f, o, r, c, e, right parenthesis, end text
iii. So if the person pushes with start color #1fab54, F, start subscript, start text, p, u, s, h, end text, end subscript, end color #1fab54, equals, 800, start text, space, N, end text there will be a kinetic frictional force of F, start subscript, k, end subscript, equals, 431, start text, space, N, end text 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, point, 3, start text, space, k, g, end text box of frozen chocolate chip waffles is pulled at constant velocity across a table by a rope. The rope is at an angle theta, equals, 60, start superscript, o, end superscript and under a tension of 4, start text, space, N, end text.
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, start subscript, k, end subscript, equals, mu, start subscript, k, end subscript, F, start subscript, n, end subscript 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, start subscript, x, end subscript, equals, start fraction, \Sigma, F, start subscript, x, end subscript, divided by, m, end fraction, start text, left parenthesis, s, t, a, r, t, space, w, i, t, h, space, N, e, w, t, o, n, apostrophe, s, space, s, e, c, o, n, d, space, l, a, w, space, i, n, space, t, h, e, space, h, o, r, i, z, o, n, t, a, l, space, d, i, r, e, c, t, i, o, n, right parenthesis, end text
0, equals, start fraction, start color #1fab54, T, start subscript, x, end subscript, end color #1fab54, minus, start color #7854ab, F, start subscript, k, end subscript, end color #7854ab, divided by, 1, point, 3, start text, space, k, g, end text, end fraction, start text, left parenthesis, p, l, u, g, space, i, n, space, h, o, r, i, z, o, n, t, a, l, space, f, o, r, c, e, s, comma, space, a, c, c, e, l, e, r, a, t, i, o, n, comma, space, a, n, d, space, m, a, s, s, right parenthesis, end text
0, equals, start fraction, T, start text, c, o, s, end text, 60, start superscript, o, end superscript, minus, mu, start subscript, k, end subscript, F, start subscript, n, end subscript, divided by, 1, point, 3, start text, space, k, g, end text, end fraction, start text, left parenthesis, p, l, u, g, space, i, n, space, h, o, r, i, z, o, n, t, a, l, space, c, o, m, p, o, n, e, n, t, space, o, f, space, t, e, n, s, i, o, n, space, a, n, d, space, f, o, r, m, u, l, a, space, f, o, r, space, k, i, n, e, t, i, c, space, f, r, i, c, t, i, o, n, right parenthesis, end text
0, equals, T, start text, c, o, s, end text, 60, start superscript, o, end superscript, minus, mu, start subscript, k, end subscript, F, start subscript, n, end subscript, start text, left parenthesis, m, u, l, t, i, p, l, y, space, b, o, t, h, space, s, i, d, e, s, space, b, y, space, m, a, s, s, right parenthesis, end text
mu, start subscript, k, end subscript, equals, start fraction, T, start text, c, o, s, end text, 60, start superscript, o, end superscript, divided by, F, start subscript, n, end subscript, end fraction, start text, left parenthesis, a, l, g, e, b, r, a, i, c, a, l, l, y, space, s, o, l, v, e, space, f, o, r, space, t, h, e, space, c, o, e, f, f, i, c, i, e, n, t, space, o, f, space, k, i, n, e, t, i, c, space, f, r, i, c, t, i, o, n, right parenthesis, end text
At this point you might think we should plug in the normal force as m, g, but since the rope is also pulling upward on the box, the normal force will be less than m, g. 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, start subscript, y, end subscript, equals, T, start text, s, i, n, end text, 60, start superscript, o, end superscript. So the normal force in this case will be F, start subscript, n, end subscript, equals, m, g, minus, T, start text, s, i, n, end text, 60.
Now we can plug this expression for the normal force F, start subscript, n, end subscript into our formula for the coefficient of kinetic friction we found above.
mu, start subscript, k, end subscript, equals, start fraction, T, start text, c, o, s, end text, 60, start superscript, o, end superscript, divided by, F, start subscript, n, end subscript, end fraction, start text, left parenthesis, u, s, e, space, f, o, r, m, u, l, a, space, w, e, space, f, o, u, n, d, space, a, b, o, v, e, space, f, o, r, space, c, o, e, f, f, i, c, i, e, n, t, space, o, f, space, k, i, n, e, t, i, c, space, f, r, i, c, t, i, o, n, right parenthesis, end text
mu, start subscript, k, end subscript, equals, start fraction, T, start text, c, o, s, end text, 60, start superscript, o, end superscript, divided by, m, g, minus, T, start text, s, i, n, end text, 60, start superscript, o, end superscript, end fraction, start text, left parenthesis, p, l, u, g, space, i, n, space, e, x, p, r, e, s, s, i, o, n, space, f, o, u, n, d, space, f, o, r, space, n, o, r, m, a, l, space, f, o, r, c, e, right parenthesis, end text
mu, start subscript, k, end subscript, equals, start fraction, left parenthesis, 4, start text, space, N, end text, right parenthesis, start text, c, o, s, end text, 60, start superscript, o, end superscript, divided by, left parenthesis, 1, point, 3, start text, space, k, g, end text, right parenthesis, left parenthesis, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, minus, left parenthesis, 4, start text, space, N, end text, right parenthesis, start text, s, i, n, end text, 60, start superscript, o, end superscript, end fraction, start text, left parenthesis, p, l, u, g, space, i, n, space, v, a, l, u, e, s, space, f, o, r, space, t, h, e, space, t, e, n, s, i, o, n, space, a, n, d, space, m, a, s, s, right parenthesis, end text
mu, start subscript, k, end subscript, equals, 0, point, 216, start text, left parenthesis, c, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, right parenthesis, end text

## Want to join the conversation?

• what happens to co-efficient of friction ,when a body weight is doubled? •   the coefficient of friction does not depend upon weight.... as far as I know it is a function only of the materials?
• So, coefficient of kinetic / static friction cannot be a negative number? • Friction helps us to walk but in water why we cannot walk • We can walk on Land because the number of irregularities present on the surface are very plentiful, and differ from area to area, whereas in water, there are no irregularities and it is actually often used only as a lubricant, which is why the wet floor is extremely difficult to walk on. So, technically the only reason we can walk on land and not on water is actually because the number of irregularities is present in solids and is absent in liquids
• I don't understand why when force applied is less than the maximum value of static friction then its value is the same as the value of frictional force. Can someone explain?, thanks. • A good question

Imagine you are pushing against a table trying to slide it across the floor.

The maximum value of static friciton is 100N. If you reach this force or exceed it, the table will slide.

OK??

So you start pushing with a small force; say, 15N. The table does not slide.

Why? because the force of friction is opposing your force.

how much force is the table pushing back on you with?

This force you feel pushing back on you is the force of friciton.

Now increase your force to 60N. The table still does not slide.

How much force will you feel from the table? again, this the force of friciton. now increased.

as you increase your force, the force opposing you (friciton) will also increase according to Newtons third law. Until you exceed the force that friction can push back with.... when you reach or exceed 100N then the table will slide.

OK??
• In Example 1 : iii) before the fridge starts moving the person must exert a force of 647 N to overcome the static friction, So, the force just after begining of motion would be 800-647=153 N !
Am I right! • It would seem so but no. The coefficient of friction changes from static to kinetic once the fridge starts moving, and as long as it stays moving it won't change no matter how much you accelerate/decelerate (as long as you don't go back to cero velocity). As you can see in the formula of kinetic friction there is no mention of the force you're aplying, it's only proportional to the kinetic friction coeficient and to the normal force. If you calculate that you get that the force of the kinetic friction is (110kg*9.8m*s^-2)*(0.4)=431.2N. You can then calculate the real magnitude of the force your'e applying using that value: 800N-431.2N=368.8N. At least I believe this is correct :d
Be carefull though since the normal force is not allways the same as the weight, in this case it is since there are no other vertical forces.
• how can cars accelerate if its static friction that acts on them. cause the net force would always equal 0 ? plz explain thanks. • can someone please explain how the formulas for static and kinetic frictional forces were derived? is there a way to understand the formula w/o just memorizing it? thanks :) • They're not derived, they're based on observation and on the definition of the coefficient of friction.

The harder surfaces press against each other, the bigger the frictional force is. That's all the formula says. The coefficient says that the amount of the force depends on the specific combination of materials that are being pressed together.   