Friction on an inclined plane

  • #1
I am having problems with a simple problem with a mass (m) stationary on an inclined plane of angle θ.

I have obtained the equations for the reaction/normal force and the force acting on the mass parallel to the plane:

Reaction force / normal force (R) = mgcosθ
Parallel force = mgsinθ

Also, I know that the static friction force is given by Friction force = (mu static) x R.

My question is this:

For the mass to be stationary, obviously the parallel force must equal the static friction force in magnitude. But if the angle, θ of the inclined slope is increased, logically both forces must also increase in magnitude by the same amount to keep equilibrium.

Since mu static is a constant, this means that as the parallel force increases with θ, R must increase in the (mu static x R) equation to provide a frictional force of equal magnitude to the parallel force. But R, and hence the frictional force, do not increase with θ as the cosine in the reaction force equation decreases with an increase in θ between 0 and 90 degrees, also decreasing the frictional force.

So how can there be equilibrium (which there obviously is!)?

As far as I can see, since the two forces involve sine or cosine, they cannot both increase at the same time between 0 and 90 degrees.

Any help would be greatly appreciated.

Many thanks

Paul
 
Last edited:

Answers and Replies

  • #2
Dale
Mentor
Insights Author
2020 Award
31,811
8,656
Also, I know that the static friction force is given by Friction force = (mu static) x R.
The correct equation is [tex]F \leq \mu_s R[/tex]. The static friction force is whatever force is required for the object to not move up to a maximum given by [tex]\mu_s R[/tex]
 

Related Threads on Friction on an inclined plane

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
2K
Replies
8
Views
2K
Replies
3
Views
2K
Replies
1
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
885
Top