Quote by tricky_tick
can you please explain how you arrived at that conclusion?

You have four forces. Suppose the xacis is along the ramp and the force F for pulling the object is aligned along this xaxis. Yaxis perpendicular to xaxis.
gravity
[tex]mg\sin(\theta)*e_{x}  mg\cos(\theta)*e_{y}[/tex]
normal force
[tex] N * e_{y} [/tex]
friction
[tex]{\mu}N*e_{x} [/tex]
pullforce
[tex]F*e_{x} [/tex]
The sum of ycomponets yield :
[tex] N = mg\cos(\theta) [/tex]
The sum of xcomponets yield :
[tex] {\mu}N mg\sin(\theta) + F = 0 [/tex]
So we have that [tex] F = {\mu}mg\cos(\theta) + mg\sin(\theta) [/tex]
or [tex] F = mg(\mu\cos(\theta) + \sin(\theta)) [/tex]
let's take the derivative with respect to the angle theta and set this equal to 0. Thus we get :
[tex] 0 = mg({\mu}\sin(\theta) + \cos(\theta)) [/tex]
or
[tex] 0 ={\mu}\sin(\theta) + \cos(\theta) [/tex]
or
[tex] \mu = \frac {\cos(\theta)}{\sin(\theta)} = \cot(\theta)[/tex]
Thus [tex]\tan(\theta) = \frac {1}{\mu}[/tex]
marlon.