Introductory textbooks explain that friction does not depend on surface area. They usually illustrate this with an image like figure 10.41 on the left shown below:

They then also explain that a block of mass $m$ will be able to attain a maximum angle $\theta_{max}$ without slipping down hill. At this maximum angle, the force of static friction will be as follows $F^f_{s,max} =\mu_s F^n$ where $\mu_s$ is the coefficient of static friction. But now imagine the following scenario depicted in the rightmost image above where we have a block of mass $m$ at rest placed on an incline at its maximal angle before slipping occurs

If we now take the exact same situation but replace the block of mass $m$ with a wheel or sphere of mass $m$ as shown in the image above, intuitively, we know that the wheel will not be at rest unlike the block. It will roll down the incline. But the force of gravity for the block and the wheel are same. So I must infer that the frictional force on a wheel is less than the frictional force on an otherwise equal block ($\theta_{block}=\theta_{wheel}$, $m_{block}=m_{wheel}$ etc..).

I have two explanations for this but I'm not sure of their validity. The first is to realize that for the wheel/sphere, the center of mass is not directly above its 'point of contact with the incline' and hence is unstable and will "topple over" due to gravity. This "toppling occurs" immediately and hence the static frictional force possibly does not have enough time (?) to grow and become equal and opposite to the x component of the force of gravity. The second explanation is to suppose that the coefficient of friction does in fact depend on surface area. So do either of these explanations validly explain why the static frictional force is lower for the sphere than it is for the block or is there some other reason?

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    $\begingroup$ @The first explanation is correct $\endgroup$
    – paul230_x
    Commented Aug 17, 2021 at 10:20
  • $\begingroup$ @KP99 I though as much however it can't be a complete explanation since it doesn't explain why the static friction force for the wheel/sphere is lower than for the block $\endgroup$ Commented Aug 17, 2021 at 10:52

1 Answer 1


Sure, the wheel will not remain at rest. But it also will not slide. And that was the condition you set up for the block. So there is no difference in terms of friction between the two.

There is only a difference in terms of toppling tendency, which you also touched at in your next paragraph. This is due to the mismatched centre-of-mass.

Imagine placing the brick on one end instead. If its centre of mass is located beyond the edge of the contact area, then it will topple over. This does not mean that static friction can't hold on to it but rather that the brick is "lifting itself off" of the surface by hinging over the contact point so that static friction doesn't apply anymore. After toppling over, the momentum it gains might keep it toppling (depending on the surface it lands on). If the brick then happens to continue toppling all the way down, then at no point during this tumbling did it slide. It just "let go" of the previous spot to which it was stuck (it was stuck along the parallel direction only) due to static friction.

A wheel does the same thing just with an infinitely small contact area. It basically "topples" constantly but never slides - this is what we call rolling. (This is why the invention of the wheel was quite a paradigm shift, theoretically at least, in engineering: motion without kinetic friction means motion without energy loss ideally. That's quite something.)

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    $\begingroup$ Thanks for the great answer. Okay just one issue. The block and the wheel are both subject to an identical force of gravity $F^G$. This means that although the block is stationary and the wheel rolls, we must have that $F^G_{x, b}=F^G_{x,w}$ and $F^G_{y, b}=F^G_{y,w}$ at all times. We also know that both objects do not fall downwards through the ramp so both objects experience the same normal force. But the wheel must be subject to a net force in the x-direction (conversely $F_{net,b,x}=0$) to agree with observation meaning $F^f_{static,w} <F^f_{x,static,b}$. I still dont ... $\endgroup$ Commented Aug 17, 2021 at 10:35
  • $\begingroup$ understand how $F^f_{static,w} <F^f_{x,static,b}$ comes about. Both objects have the same normal force and the same x-component of gravitational force "urging" them downward. So the coefficients of friction must be different for the two cases? (else both object would move down hill or both would stand still) . The only way I can account for the rolling of the wheel on a force diagram is if the wheel experiences a lower frictional force which allows it to have a net force. $\endgroup$ Commented Aug 17, 2021 at 10:39
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    $\begingroup$ @SalahTheGoat Good questions. Remember that static friction is a variable force. It is only as large as it has to be. In the case of the block, it grows to hold the entire block still. In the case of the wheel, you might be right that static friction will be slightler lower than in the block case since it is not holding still the entire object but only the contact point. $\endgroup$
    – Steeven
    Commented Aug 17, 2021 at 10:52
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    $\begingroup$ Okay I think I understand now. The static friction experienced by a wheel of mass $m$ placed on an incline must be less than the static friction of block of mass $m$ placed on the same incline. This discrepancy is what allows the wheel to accelerate in the x direction but keeps the block stationary. The reason the static friction is less in the case of the wheel is because it only counters the smaller mass $m_{small}$ of the contact region due to the no slip condition. The reduced static friction is not able to counter the mass of the whole block though which is why the ball rolls. $\endgroup$ Commented Aug 17, 2021 at 11:24

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