The coefficient of friction at a specific angle.

[texan14]

Homework Statement

A block of mass 'm' is placed on an inclined plane which has an adjustable angle with the horizontal. The angle is increased slowly from zero until the block just starts to slide. This angle is called the critical angle, θ_s.

Express the coefficient of friction in terms of θ_s.

Homework Equations

F = m*a
friction = μ*(normal force, N)
μ = coefficient of friction

The Attempt at a Solution

F_x = m*a_x
m*a_x = m*g*sinθ_s - μN
m*a_y = m*g*cosθ_s - N

After I set both equations equal to the normal force, N, I got:

μ = (m*g*sin_s-m*a_x) / (m*g*cosθ_s-m*a_y)

If I am allowed to set the acceleration equal to zero, I would get μ = tanθ_s, which is the correct answer. But since the block "just starts to slide" at θ_s, I don't think the acceleration would equal zero... Help please?

 

[TaxOnFear]

It is assumed that acceleration is equal to 0 as the block just starts to slide. Hence the forces in play are equal, otherwise the friction force would be less than ma and the equation would become an inequality.