0
$\begingroup$

I had a question about calculating the static coefficient of friction of an object on an inclined plane (classic physics question). We can simplify those calculations to $\mu = \tan \theta$. Since $\theta$ can be $0$ (i.e. the object is on a flat surface), does this mean that the minimum static coefficient is always $0$?

Also, when calculating the static friction coefficient, is it best to take the average? Like take the maximum static coefficient and divide it by 2 to obtain the average. Or should I just take the maximum static coefficient?

Thanks a lot for the help.

$\endgroup$

3 Answers 3

0
$\begingroup$

The other answers are fine. This one is just to give you another perspective that may help in combination with the other answers.

First of all, the static friction force parallel to a surface always matches the opposing force parallel to the surface up until the maximum possible static friction force is reached, at which point relative motion between the object and the surface is impending.

So if you think of the object resting on a horizontal surface, there is no force parallel to that surface for the static friction force to oppose. So the static friction force is zero. Now if you start to increase the angle of the surface relative to the horizontal, the component of the gravitational force acting on the object down and parallel to the incline increases to $mgsin\theta$. The static friction force acting up the incline increases an equal amount preventing a net force downward and a downward acceleration of the object, as long as the maximum possible static friction force is not exceeded, which equals $\mu N$ where $N$ is the normal force to the incline.

We can simplify those calculations to $\mu = \tan \theta$

The following inequality means that in order for impending motion not to occur at an incline angle of $\theta$, the coefficient of static friction has to be

$$\mu \ge tan\theta$$

Or

$$tan^{-1}\theta \lt \mu$$

So, for example, if the coefficient of static friction $\mu =0.5$ then impending motion of the object will not occur as long as $\theta \lt 26.66^0$

Also, when calculating the static friction coefficient, is it best to take the average? Like take the maximum static coefficient and divide it by 2 to obtain the average. Or should I just take the maximum static coefficient?

There is no average static friction force. As indicated above, the static friction force matches the opposing force on the object until the maximum possible static friction force is reached. That maximum force is based on a single value of the coefficient of static friction.

Hope this helps.

$\endgroup$
1
  • $\begingroup$ oh my goodness. Thanks a lot for the amazing answer. $\endgroup$
    – ahmed33033
    Mar 1, 2021 at 7:36
0
$\begingroup$

I had a question about calculating the static coefficient of friction of an object on an inclined plane (classic physics question). We can simplify those calculations to $μ=\tanθ$.

Your statement is a little incomplete. I assume you want the object to stay immobile? In that case it is required that:

$$\mu\geq \tan\theta$$

But for $\theta=0$ (the horizontal case), the gravity vector $m\vec{g}$ doesn't have a horizontal component, so no friction is needed because there can be no acceleration since as $F_{net}=0$. So no minimum value of $\mu$ is needed.

As regards your second question, what do you understand by "maximum static coefficient"?

$\endgroup$
4
  • $\begingroup$ What I was getting at is when calculating the static friction coefficient, you'll end up with something that looks like this: mgsinθ - mgcosθ*μ = 0, which can be simplified to μ=sinθ/cosθ and eventually μ=tanθ . And yes, I see my error with assuming you could have θ = 0. As for my second question, what I meant was the largest value possible for the static friction coefficient for a given material. As per my knowledge, the static coefficient of friction can be a range of numbers depending on the angle of the incline. So should I assume the highest value, or should I take the average? $\endgroup$
    – ahmed33033
    Feb 27, 2021 at 22:04
  • $\begingroup$ Well, re. this 'maximum FC', it appears you know more than me! $\endgroup$
    – Gert
    Feb 27, 2021 at 22:07
  • $\begingroup$ my apologies if I came off the wrong way. I'm just learning about this, so a lot of the things I know about it may be false, or something that I came to misunderstand. I really appreciated your help! $\endgroup$
    – ahmed33033
    Feb 27, 2021 at 22:19
  • $\begingroup$ No offence was take at all. Thanks. $\endgroup$
    – Gert
    Feb 27, 2021 at 22:20
0
$\begingroup$

For a body at rest, impending motion is countered by the force of static friction. For impending motion the maximum force of friction is determined by $\mu_s$, the coefficient of static friction. For your problem, if $tan\theta$ exceeds $\mu_s$ static friction cannot keep the body from moving. For any smaller angle, the force of friction varies with the angle as necessary to keep the body from moving; at zero degrees the force of friction is zero.

$\mu_s$ determines the maximum force of static friction, but the force of static friction can vary from zero to the maximum as necessary to keep the body from moving.

Once the body starts to move, the force of friction is determined by $\mu_k$ the coefficient of kinetic friction, which is a fixed value, regardless of the angle of inclination of the plane.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.