# Finding the coefficient of static friction on slope

## My Problem

1.0kg block is sitting on a horizontal incline plane. The plane is tilted into the block begins to slide. You notice that the plane is inclined at an angle of 32 degrees above the horizontal when the block begins to slide. Calculate the coefficient of static friction that acts between the block and the plane.

## My Solution

I tried solving this by doing FgSinθ=μCosθ and manipulating it into Sinθ/Cosθ=μ which is Tanθ=μ. This comes out to Tan(32)=0.62

## My Question for You

But my question is that it's already moving at 32 degrees, so am I finding the coefficient of KINETIC friction the way I did it. If so then how do I find the coefficient of STATIC friction given the information I have.

You have found the critical angle $\theta_c$ at which the block begins to slide. That gives you the coefficient of static friction $\mu_s = \tan\theta_c$. Kinetic friction comes into consideration when the block is actually moving.
Before the block can move, the force $mg\sin\theta$ acting down the incline must be at least equal to the maximum possible value of the static friction force, which is $\mu_s mg\cos\theta$, acting up the plane. When the block just begins to move at angle $\theta_c$ these two forces are equal :
$mg\sin\theta_c = \mu_s mg\cos\theta_c$
$\mu_s = \tan\theta_c$.
When it moves the sliding block might - and usually does - accelerate down the slope because kinetic friction $\mu_k$ is often less than static friction $\mu_s = \tan\theta_c$. The rate of acceleration $a$ is given by
$a = g\sin\theta_c - \mu_k g\cos\theta_c = (\tan\theta_c - \mu_k)g\cos\theta_c = (\mu_s - \mu_k)g\cos\theta_c.$
If the block does not accelerate down the slope ($a=0$) but moves at constant velocity then $\mu_k = \mu_s = \tan\theta_c$. If the block does accelerate ($a\ne 0$) then you can rearrange this relation to find $\mu_k$.