Friction on an inclined plane

[paul_harris77]

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

 

[Dale]

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

The correct equation is $$F \leq \mu_s R$$. The static friction force is whatever force is required for the object to not move up to a maximum given by $$\mu_s R$$

 

[astrokreng]

It is correct that as the angle θ increases, both the parallel force and the reaction force will increase. However, it is important to remember that the static friction force is a maximum value and it will only increase up to the point where it is equal to the parallel force. Beyond that point, the object will start to slide down the incline.

The equation for static friction force, F = μR, assumes that the object is on the verge of slipping. This means that the coefficient of static friction, μ, is a maximum value and it will only increase to the point where it is equal to the parallel force. So in your equation, as θ increases, the value of R will also increase until it reaches the maximum value allowed by the coefficient of static friction, and then it will remain constant.

Therefore, there is still equilibrium because the parallel force and the maximum static friction force are equal. The reason why R and the friction force do not increase at the same rate as θ increases is because of the assumption that the object is on the verge of slipping. Once it reaches that point, the friction force will remain constant, even as the angle θ continues to increase.

I hope this explanation helps to clarify the concept of friction on an inclined plane. It is important to remember that friction is a complex force and it depends on various factors such as the angle of the incline, the coefficient of friction, and the weight of the object. Keep in mind that the equations and assumptions used in this problem are simplified and may not accurately represent real-world scenarios.