First of all I know there is a similar question to this one. However, it didn't help me resolve my problem.

Let's say there is a body on a angled plane and the system is in equilibrium (the body doesn't move). The forces on the body are friction, normal force and weight.

$\mu_s$ and $\mu_k$ are the same in my case (and in the question that I try to solve).

I chose that the axis $x$ is parallel to the plane.

Thus from First Law of Newton, $$\Sigma F_x = mg\sin\alpha - f = 0$$ $$\Sigma F_y = N - mg\cos\alpha = 0$$

From here I get that $$N = mg\cos\alpha$$ $$f = \mu N = mg\sin\alpha$$

I put what I found about $N$: $$\mu mg\cos\alpha = mg\sin\alpha$$

From here I can say that $$\mu = \frac{\sin\alpha}{\cos\alpha} =\tan\alpha$$

I found that $\mu$ depends in the angle of the plane. Nevertheless, I read that $\mu$, as friction co-efficent, will never change.

What are my mistakes or misunderstandings?

  • $\begingroup$ The value of the static frictional force can vary from zero up to a maximum value. The value of $\mu_s$ which is usually given is the maximum value that $\mu_s$ can have. $\endgroup$
    – Farcher
    Feb 13, 2016 at 14:12

1 Answer 1


There are two mistakes in your thinking. Considering static friction first, $\alpha$ is the minimum angle at which the body first starts to slide; if $\tan\alpha$ is less than $\mu_s$, the body won't slide at all. Considering kinetic friction, when $\tan\alpha=\mu_k$, the body can be moving at constant velocity; at larger angles, it will be accelerating.

  • $\begingroup$ But I'm not actually finding the fixed co-efficent (i.e. it should never change though). I know that $m = 2kg$, that in $\alpha = 8.6$ it doesn't move, and that in $\alpha = 22$ it begins to move. $\endgroup$ Feb 13, 2016 at 18:20
  • 1
    $\begingroup$ Your mistake is writing $f=\mu_sN$ for the case in which $\alpha = 8.6$ degrees. The static friction law is an inequality, not an equation. The correct version of the law is $f\leq \mu_sN$. The equal sign only applies when the body is at the verge of moving, in your case 22 degrees. $\endgroup$ Feb 13, 2016 at 19:33
  • $\begingroup$ Yes. But, in addition, I have read that $f$ will be always the value of $F$ force that is trying to move the object. $\endgroup$ Feb 13, 2016 at 20:30
  • $\begingroup$ Yes. That's what I'm saying. F (or f, if you prefer) is equal to the force that is trying to move the object, but F is less than or equal to the normal force times the coefficient of static friction. If F is less than the normal force times the coefficient of static friction, the body will not slide. If F is equal to the normal force times the coefficient of static friction, the body will begin to slide. $\endgroup$ Feb 13, 2016 at 20:55

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