prove that a tangent angle is equal to coeficient kinetic friction when the block slides down the incline?

Question

with a constant speed. (By using Symbols). And if an angle is the max. angle of incline just before the block moves, what is the coeficient kinetic friction in terms of angles?

Answer 1

Fg = Wsinθ
Ff = μWcosθ
If the block is not accelerating,
Ff = Fg
μWcosθ = Wsinθ
μ = tanθ

The second part of your question deals with static friction, not kinetic, and the angle will be larger just before the block moves than it must be after the block starts moving. The coefficient of kinetic friction cannot be given solely in terms of angle in this case.

Answer 2

i think of of you're caught considering you're lacking a cost for the mass. you are able to desire to understand if the gravitational stress exceeds or would not exceed the frictional stress at that attitude. Fg=mgcos(theta) Ffriction = ok*N*sin(theta) the two equations require a mass the area ok is the coefficient of friction and N is the classic stress which you will are available the time of by means of technique of finding on the gravitational stress edit. lol, oops plenty cancel :)

Answer 3

Let θ be the angle of incline when the block just begins to slide down.

Since the block just begins to slide down the friction is maximum and corresponds to maximum static friction.



The force pulling down is the component of mg and is mg sin θ.
The component ‘mg cos θ’ is opposite and equal to the reaction of the plane.

Frictional force F is opposite and equal to mg sin θ and

It is also equal to μ times normal reaction‘mg cos θ’
F = mg sin θ = μ mg cos θ

sin θ = μ cos θ


μ = tan θ

The coefficient of kinetic friction will be less than this static friction and hence the angle will be slightly less than this angle.

Once the body begins to slide down we can reduce this angle slightly so that uniform motion is maintained.