Solved Does The Coefficient Of Kinetic Friction Depend On Weight Of The Block?

It’s perhaps the second week of your introductory physics course. Your instructor starts talking about friction and writes the following two formulas on the board. Then there is probably some sort of lecture like this: Friction is a contact force when two surfaces interact. The second equation is the kinetic frictional force that is used <…>


Is a block on an inclined plane the most boring physics problem ever? Maybe not.Is a block on an inclined plane the most boring physics problem ever? Maybe not. Rhett Allain

It's perhaps the second week of your introductory physics course. Your instructor starts talking about friction and writes the following two formulas on the board.

You are watching: Does the coefficient of kinetic friction depend on weight


Then there is probably some sort of lecture like this:

Friction is a contact force when two surfaces interact. The second equation is the kinetic frictional force that is used when two surfaces are sliding against each other. The frictional force in this case depends on the two types of materials interacting (described by the coefficient μk) and how hard these two surfaces are pushed together (the normal force). The static friction case is similar for when the two surfaces are stationary relative to each other. In static friction, the frictional force is whatever value it needs to be to prevent sliding up to some maximum value.

Technically, this is called Amontons' First and Second Law of Friction. See, it's not just Newton that has laws. Notice that both of these friction formulas ONLY depend on the coefficient of friction and the normal force. It does not depend the area of contact, it doesn't depend on the sliding speed.

Next, there will probably be some type of friction laboratory experiment. In this lab, students will measure coefficients of friction and show that the frictional force doesn't depend on surface area in contact. Also, the coefficient of friction doesn't depend on the mass of the object. Pretty standard stuff here.

Friction Is Just a Model

How about another experiment? In this experiment, I am going to put an object on an moveable plane. I can then increase the angle of inclination until this block just starts to slide. At the moment it starts to slide, I can calculate both the normal force (pushing the plane against the object) and the friction force (the maximum static friction force).


Just at the instant this thing starts to slide, all of these forces still have to add up to the zero vector (object is in equilibrium). That means that the component of the gravitational force perpendicular to the plane must be equal to the magnitude of the normal force and the component parallel to the plane must be equal to the frictional force.


With just the mass and the sliding angle, I can get both the frictional force and the normal force. How can I calculate the coefficient of friction? What if I made a plot of friction vs. normal force for the same surface but with different masses? If the normal force and the frictional force are really proportional (like in the model above) then this data should be linear with the slope of the line being the coefficient of friction.

It's simple, right? Ok. Let's do this. In order to keep everything the same except for the mass, I am going to put masses into one of these small boxes.


This box has a teflon bottom with an open top so you can put masses inside (oh, it's from PASCO). There is also a variable angle inclined plane. This one in particular has a large angle measurement on the side and here you can see the friction box with a large amount of mass both inside and on top of it.


Actually, there is also a similar plane that is made of metal instead of wood. I tried this experiment both with a felt-bottomed box on wood and a teflon box on metal. For each mass, I slowly lifted the incline until the box slipped and then recorded the angle. I repeated the experiment for the same mass 5 or 6 times so that I could get an average angle and a standard deviation in the angle measurement.

Here is a plot of friction force vs. normal force for both surfaces. The error bars are calculated (using the crank three times method) from the standard deviation in angle measurements.

What's going on here? Let's look at the data for the teflon (the blue data). I fit a linear function to the first 4 data points and you can see it is very linear. The slope of this line gives a coefficient of static friction with a value of 0.235. However, as I add more and more mass to the friction box, the normal force keeps increasing but the friction force doesn't increase as much. The same thing happens for friction box with felt on the bottom.

This shows that the “standard” friction model is just that – a model. Models were meant to be broken.

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A More Detailed Look at Friction

Really, what is friction? You could say that when two surfaces come near each other (call them surface A and surface B), the atoms in surface B get close enough to interact with surface A. The more atoms that are interacting in the two surfaces, the greater the total frictional force. How do you get more atoms to interact from the two surfaces? Well, if you push the surfaces together you can get more atoms from A to be close enough to the atoms from B to interact. Yes, I am simplifying this a bit. However, the point is that contact area does indeed matter.

I am talking about contact area, not surface area. Suppose you put a rubber ball on a glass plate. As you push down on the rubber ball, it will deform such that more of the ball will come in “contact” with the glass. Here is a diagram of this.


Greater contact area means greater frictional force. If the contact area is proportional to the normal force, then this looks just like Amontons' Law with the frictional force proportional to the normal force. Of course this model “breaks” when the contact area can no longer increase. As I add more and more mass onto the friction box, there is less and less available contact area to expand into. In a sense, the contact area becomes saturated. I suppose that if I kept piling on the weight, the friction force would eventually level out and stop increasing.

It's Just a Model

This really isn't a big deal. The Amontons' Law isn't a law at all (ok – it depends on your definition of Law). It's just a model. A model is not THE TRUTH, it's just something that works some of the time. Let me give an example.

Gravitational Model. Near the surface of the Earth, we can calculate the gravitational force on an object using the following model.


The g vector is the local gravitational field. On Earth, it points “down” and has a magnitude around 9.8 N/kg. We often call this gravitational force the weight and it's a very useful model.

Even though this model is useful, we still know it's wrong. The above gravitational model says that it doesn't matter how high above the surface of the Earth you are, the weight is the same. Of course that's not true, but it's approximately true when close to the surface.

Here is a better gravitational model.


This says that the gravitational force decreases as the two interacting objects get further away from each other. If you put in the mass of the Earth and the radius of the Earth you get a weight that looks just like the mg version. So, at some point the two versions of gravity agree.

The same is true for friction. The introductory physics version of friction works for some stuff and a more complicated version of friction works for other cases. Of course you could still use the complicated version of friction for simple cases – but why make your life difficult?


Rhett Allain is an associate professor of physics at Southeastern Louisiana University. He enjoys teaching and talking about physics. Sometimes he takes things apart and can't put them back together.

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