The maximum force of static friction between two surfaces is roughly proportional to the magnitude of the normal force $N$ pressing the two surfaces together.

In many problems involving rolling cylinders, the rolling object may not be under the influence of the maximum possible frictional force; but they always assume the coefficient of static friction is maximum while deriving the equations or solving problems. Why does this work ? (Shouldn't the magnitude of the frictional force be equal to the force with which the object is being pulled ? ($mg\sin\theta$ in below example))

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    $\begingroup$ Neither of the two options is correct. Friction force is not the maximum static force and is not equal $mgsin\theta$ (unless $a=0$). Friction force is initially unknown, as well as the acceleration. Assuming rolling without slipping, you can set up two equations to find both variables. $\endgroup$
    – npojo
    Oct 9, 2018 at 9:43
  • $\begingroup$ I have made more in depth edits to my answer. $\endgroup$ Oct 9, 2018 at 13:24

2 Answers 2


The friction force does not have to be at its maximum, nor does it have to be equal to the component of the weight down the ramp (it actually can't be equal to this. See below.).

Let's set up the problem. You already have the free body diagram in your question, so I will not make a new one. Of course the forces balance out perpendicular to the ramp, which you already have: $$N=mg\cos\theta$$

Parallel to the ramp we have, using Newton's second law: $$\sum F_x=mg\sin\theta-F_f=ma$$

Also, using the Newton's second law for rotation we have $$\sum \tau=RF_f=I\alpha=I\frac aR$$

Where we have assumed rolling without slipping, so $\alpha=\frac aR$. From here you will find that you don't actually need a value for the friction force to find the acceleration. I will leave this up to you to solve from here.$^*$

Shouldn't the magnitude of the frictional force be equal to the force with which the object is being pulled ?

This would mean that $mg\sin\theta=F_f$, so then $a=0$. But if $a=0$, then by our torque equation we must have $F_f=0$, which is a contradiction. Therefore, it is impossible for our static friction force to be equal to the component of the weight down the ramp. The rolling ball will always accelerate down the ramp, and so $F_f<mg\sin\theta$.

...they always assume the coefficient of static friction is maximum while deriving the equations or solving problems

This is not the correct way to do the problem. The way to do the problem is above, but let's explore this a bit more:

You can actually solve for $F_f$ here too by eliminating $a$ from the above equations. The result ends up being, if we want no slipping $$F_f=\frac{mgI\sin\theta}{I+mR^2}\leq\mu mg\cos\theta$$

Therefore: $$\tan\theta\leq\frac{I+mR^2}{I}\mu$$

If we have the special case where $I=\gamma mR^2$, then we have $$\tan\theta\leq\frac{\gamma+1}{\gamma}\mu$$

We can compare this to a block on an incline, where the angle of no sipping is determined by $\tan\theta\leq\mu$, which is when $\gamma\gg1$ (i.e. an object that is very hard to roll, kind of like a block on an incline (the analogy is not perfect, so don't go too far into it)).

So we see here that the static friction force will not always be equal to the maximum static friction force. If we have our angle at exactly $\tan^{-1}\left(\frac{\gamma+1}{\gamma}\mu\right)$, then we can say the static friction force is at a maximum. Any angle past this will then produce rolling with slipping.

$^*$Note: This is the standard procedure for solving problems like these.

  • $\begingroup$ Hey ty for the detailed explanation :) Sorry I've still got a stupid question : since the object is not slipping, the velocity at the contact point (between object and plane) is $0$. Doesn't this mean the static friction is also $0$ ? $\endgroup$
    – AgentS
    Oct 9, 2018 at 11:40
  • $\begingroup$ The $0$ friction case, I'm referring to the case when the plane is horizontal... I don't yet know much about the inclined plane case $\endgroup$
    – AgentS
    Oct 9, 2018 at 11:43
  • $\begingroup$ Here is the short video if it helps understand my question better youtube.com/watch?v=hxa6jAYA980 $\endgroup$
    – AgentS
    Oct 9, 2018 at 11:44
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    $\begingroup$ @rsadhvika on a flat surface with no other forces acting on the ball there isn't anything trying to slide the surfaces past each other. On the incline gravity is trying to slide the surfaces past each other, so you have non-zero friction. $\endgroup$ Oct 9, 2018 at 12:00
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    $\begingroup$ If you push on the ball on the flat surface you will have non-zero friction. You are right. $\endgroup$ Oct 9, 2018 at 12:04

static friction is the friction that acts on static bodies i.e bodies which are not moving even though force is acting on it, static friction increases with the force until the force is enough to set the body to motion. so when the body is already in motion we just take the static friction to be maximum. at the same time if the body is at rest even though "x" amount of force is acting on it then the static friction will be x itself.

  • $\begingroup$ there is another friction that's called kinetic friction which acts on moving bodies which will be constant depending on the surface $\endgroup$ Oct 9, 2018 at 10:51
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    $\begingroup$ This is not correct $\endgroup$ Oct 9, 2018 at 11:24
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    $\begingroup$ Kinetic friction is not necessarily equal to the maximum static friction (it's usually less). If you're ignoring the relevant concept of rolling without sliding, at least avoid saying incorrect things. $\endgroup$
    – user191954
    Oct 9, 2018 at 11:35
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    $\begingroup$ @rsadhvika Don't look too far into this answer. The ball will be moving (rolling) without slipping even if we have not reached the maximum static friction force. $\endgroup$ Oct 9, 2018 at 13:32
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    $\begingroup$ @rsadhvika Rolling is a special case of static friction. The object can move while still under static friction, because the point of contact has no sliding or slipping. There is no relative velocity between the wheel contact and the road (ideally), so the friction is static, not kinetic. $\endgroup$
    – JMac
    Oct 9, 2018 at 14:44

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