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?

  • $\begingroup$ limiting friction? remember $f\leq\mu R$... the friction won't always be maximal as you increase the angle - at some stage it will be /on the point/ of slipping $\endgroup$
    – danimal
    Mar 2, 2017 at 16:58
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    $\begingroup$ Which different answers do you get for different $u$? Setting $\alpha=90^\circ$ causes the whole negative term to disappear (zero) and $u$ disappears and has no influence. $\endgroup$
    – Steeven
    Mar 2, 2017 at 17:00
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    $\begingroup$ Your expression simplifies to, $g(sin \alpha-u cos\alpha)=a$. This expression is maximised when the value inside the parenthesis is maximum. The function inside the bracket is maximum when $u cos \alpha=0$, which happens at $\alpha=90˚$. It is very easy to see it this way intuitively rather than differentiate and try to find maximum points $\endgroup$
    – Sumant
    Mar 2, 2017 at 17:04

2 Answers 2


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.

  • $\begingroup$ But then why does using the Maxima Calculator from Wolfram alpha give me different values for different $u$'s $\endgroup$
    – xasthor
    Mar 3, 2017 at 1:15
  • $\begingroup$ (wolframalpha.com/widgets/…) $\endgroup$
    – xasthor
    Mar 3, 2017 at 1:23
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    $\begingroup$ @xasthor I just played around with it a bit and see where the problem is. You need to use the absolute value $-|\mu \cos \alpha|$ in the calculator because you know that friction will always oppose the motion (making it negative). Really you have to constrict your problem between 0° and 90­° because outside of that range you aren't really representing the situation anymore. That equation only holds while the block is on the slope, which only happens between 0° and 90°. $\endgroup$
    – JMac
    Mar 3, 2017 at 1:28

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.


$$ (mg\sin {a})' - (umg \cos {a} )' = (ma)' $$ $$mg\cos {a} + umg\sin {a} = 0$$ $$mg\cos {a} = -umg\sin {a}$$ $$\tan {a} = \frac {-1}{u} $$ $$ a = \arctan{\frac {-1}{u}} $$

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.

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    $\begingroup$ To find a maximum or minimum using differentiation, the extremum must be localizable. This function doesn't have that because beyond 90 degrees the function becomes meaningless since the $\mu m g \cos \alpha$ term no longer exists. The acceleration increases monotonically. $\endgroup$
    – Bill N
    Mar 2, 2017 at 20:39
  • $\begingroup$ What about the limiting friction part ? Is there anything wrong with it as far as you can tell ? $\endgroup$
    – Ege Keyvan
    Mar 2, 2017 at 20:54
  • $\begingroup$ I don't know what is meant by limiting friction. $\endgroup$
    – Bill N
    Mar 3, 2017 at 1:16

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