How do the coefficients of static friction and kinetic friction change on a flat surface vs an inclined plane? When the angle of incline increases, what happens to the coefficients?

Answer 1

Hi Steve!

You touch on a very important point here-- what is the coefficient of friction? We know from our equation for friction that F_f=µN, where N is the normal force (i.e. the force of the surface pushing into our object). The coefficient of friction is always constant, as it is a property of the surface-- we can think of it as a measure of the surface's roughness. It tells us how large the friction force will be in relation to our normal force. This makes intuitive sense-- imagine lightly brushing your face across a carpet: you're not pushing very hard into the carpet, so the carpet isn't pushing very hard back on you (i.e. small normal force), and you aren't affected very much. Now, imagine a vindictive sibling grinding your face into the carpet: as they are pushing quite hard, the carpet exerts a very large normal force on you, and the friction causes rug burns! From this, we can see that it is not the coefficient of friction that is changing (i.e. how scratchy/rough the type of carpet is), but the normal force (how strongly the carpet is pushing into your face).

So what does this have to do with the angle of incline of an inclined plane? When you change the angle of incline, the relation of the pull of gravity on our object to the angle of the plane changes. We can understand this through vector decomposition: if the force of gravity is always pointing straight down, and the incline is angled at θ above the horizontal, then if you do a bit of clever geometry, you can see that the angle that the vector of the force of gravity with a line drawn perpendicular to the surface would also be θ. This means that the perpendicular component of the force of gravity (ie how much the gravity is pushing the object into the inclined plane, as opposed to down it) is equal to mgcos(θ). Since the object is not falling through the inclined plane, nor is it launching off of it, we can tell that there is no acceleration perpendicular to the plane-- this means that the normal force must exactly equal the perpendicular component of gravity. Since the force of friction is given by F_f=µN, we can therefore see that F_f=µmgcos(θ). It is not the coefficient of friction that is changing with the angle of incline, but the normal force! This, and the fact that the pull of gravity parallel to the plane (i.e. down the plane) is equal to mgsin(θ), are the reason why the hold of friction begins to slip as the angle of incline gets steeper.

I hope this helps, and please reach out to me if you would like to schedule a lesson for more in depth help!

Best,

Jade