Friction on an inclined plane I have an inclined plane with angle $\alpha$ with a block on it and a friction coefficient of $\mu$. I want to find $\alpha$ for which its acceleration, $a$, is maximum. Intuitively, 90$^o$ makes sense (regardless of the value of $\mu$) and anything above it would have the same acceleration as 90$^o$ since the block would fall off.
Mathematically, $$mg\sin\alpha - \mu mg\cos\alpha = ma,$$ and trying to find the maxima(using Wolfram alpha) is giving me different answers for different values of $\mu$. What am I missing here?
 A: As Sumant pointed out in a comment, this equation does have its maximum when the angle is 90 degrees.
At 90 degrees the friction term becomes 0 regardless of $\mu $.  The sin term also becomes 1, giving you $mg =ma$.  This is the same as saying it is in free fall, which is what would happen when the slope is perpendicular to the ground.
Edit: By using Wolfram Alpha it modeled something outside of the situation.  $$mg\sin \alpha - \mu mg \cos \alpha = ma$$
only applies when $$0 ° \ge \alpha \ge 90 °$$ outside of that it is no longer on a slope.  The equation in wolfram alpha was considering friction acting with the motion, which makes no sense in the real situation.
A: I believe you should find the derivative for the $a $. When you differentiate a differentiable function, one local minima point of the derivative should point the worst or best value for the first equation. In other words the $a $ value which makes the derivative of the function $0$ is your answer i believe.
Derivation:
$$ (mg\sin {a})'  - (umg \cos {a} )' = (ma)' $$
$$mg\cos {a} + umg\sin {a} = 0$$
$$mg\cos {a} = -umg\sin {a}$$
$$\tan {a} = \frac {-1}{u} $$
$$ a = \arctan{\frac {-1}{u}} $$
Another way to compute would be finding the limiting friction and equalize it to the $mg $.
$$ k A \vec {V}^2 = ma $$
Angles solving this equation must be consistent with derivation way.
Hope this helps.
