Question

A 50 N block rests on an adjustable inclined plane; the maximum angle just before it slips is 28 degrees with the horizontal. What is the coefficient of static friction between the block and the plane surfaces?

Answer 1

Well, call the angle alpha: The point when it slips is when the maximum frictional force and the component for gravity pulling the block down the slope are equal.
\[(1)mg \sin(\alpha) - F_{Fmax}=0\]
From the definition of static friction coefficient:
\[(2)\mu_{static} \times N = F_{Fmax}\]
N, the normal force, can be rewritten:
\[(3)N= mg \cos(\alpha)\]
substituting (3) into (2):
\[(4) \mu_{static} \times mg \cos (\alpha) = F_{Fmax}\]
And (4) into (1):
\[(5)mg \sin(\alpha)- \mu_{static}mgcos(\alpha)=0\]
Divide through by mg:
\[(6)\sin(\alpha)-\mu_{static} \cos(\alpha)=0\]
Rearrange:
\[(7)\mu_{static}=\tan(\alpha)\]
So...
\[\alpha \approx 0.531\]
An interesting result is that the friction coefficient is independent of the mass or surface area of the object, the 50N was unnecessary.

Answer 2

Sorry, that is supposed to be the friction coefficient above, not alpha. Mixing up my Greek alphabet