# Frictional force doesn't depend on surface area, but why does this application work?

I know friction doesn't depend on surface area and the professor has been demonstrating the same in all the previous lectures. But in this lecture he shows an application where the friction helps in balancing a large weight (T2) with a much smaller weight (T1).

He further says increasing the angle the rope is in contact with the cylinder increases the frictional force and helps in balancing a much larger weight $$T_2$$. Doesn't this contradict the fact that friction doesn't depend on surface area?

Hope I gave all the details so seeing the video is not required but here it is. • It is a fixed cylinder in the lecture, adding more loops of rope around the cylinder lessens the amount of force needed to keep the rope from slipping on the cylinder and lowering the weight Nov 30, 2019 at 5:21
• @AdrianHoward before slipping, each unit segment of the rope in contact with the cylinder produces the same amount of frictional force. Since the rope isn't slipping, this force on each segment is less than the max possible. Adding up the forces from these unit segments gives the overall frictional force. Increasing the rope contact length thus increases the overall frictional force. I think I finally get this. Thank you so much:) Nov 30, 2019 at 6:21
• Nov 30, 2019 at 7:06
• Please don't post videos. Write a summary so we can tell what the argument is without watching a video.
– user4552
Dec 1, 2019 at 2:41
• @BenCrowell I'd amend that to "Please don't just post videos". Otherwise it could give people the impression that we have a blanket ban on links to videos (which is not the case, of course). Dec 1, 2019 at 4:32

Frictional force does not directly depend on surface area, but it does depend on the normal reaction force.

Consider two cubic bodies made of same material. The bigger body will have more weight, and higher friction will act on it. This is not a direct consequence of the fact that the bigger body has higher surface area. Generally bodies with higher surface area have more weight, so the amount of frictional force acting on them is higher.

• $F=\mu N$ so friction increases with normal force. This is clear. In the application, the professor says increasing the number of turns of the rope increases the friction. The normal force is not increasing here.. then how can the friction increase? Nov 30, 2019 at 4:44
• @pooja The normal pressure may be constant, but the normal force is not. This is very clear for small wrapping angles if you treat similarly to finding equilibrium conditions when pushing in the middle of a rope. Nov 30, 2019 at 5:30
• Ahh I get it. When balanced, inegrating $\mu N$ along the rope that is in contact with the cylinder gives the weight of the body. Thank you all so much you're awesome:) Nov 30, 2019 at 5:42

Winding the rope around the cylinder, you are not only adding a contact surface. You are also adding a "normal" force (a force pressing the rope to the cylinder).

p.s. frictional force non-dependence of the surface area is a gross simplification, but it does not generally relate to the rope-cylinder problem.

No, friction force still doesn't depend on contact area.

Here, true that friction does not depend on the surface area, but any part of the rope in contact with the surface is not just a single body, each part has its own term of friction force, which adds up to give a huge amount of friction force.

• Oh so this application works because cylinder rotates about an axis? What if we were to make the cylinder fixed, immovable? Nov 30, 2019 at 4:40
• @pooja yes because of friction the cylinder rotates about the axis Nov 30, 2019 at 4:46
• It is a fixed cylinder in the lecture, adding more loops of rope around the cylinder lessens the amount of force needed to keep the rope from slipping on the cylinder and lowering the weight. Nov 30, 2019 at 5:15
• I am sorry for the incorrect explanation given earlier Nov 30, 2019 at 5:43
• @SKDash it's all good. In real sailor application, it seems they use a rolling cylinder and your explanation is still valid. Thank you so much:) Nov 30, 2019 at 5:45

The "capstan formula" written on the board in your picture demonstrates clearly that the result does not depend on the contact area.

The result is independent of the radius of the cylinder. It only depends on the change of angle of the rope around the cylinder.

For a fixed change of angle, the contact area is proportional to the radius, but the ratio of the tensions in the rope does not change.

No frictional force does depend on the surface area of contact (which itself is dependent on the normal reaction). Here is a para from Halliday Resnic Krane:

On the atomic scale even the most finely polished surface is far from flat. Figure 5-13, for example, shows an actual profile, highly magnified, of a steel surface that would be considered to be highly polished.One can readily believe that when two bodies are placed in contact, the actual microscopic area of contact is much less than the true area of the surface; in a particular case these areas can easily be in the ratio of $$1:10^4$$.
The actual (microscopic) area of contact is proportional to the normal force, because the contact points deform plastically under the great stresses that develop at these points. Many contact points actually become “cold-welded” together. This phenomenon, surface adhesion, occurs because at the contact points the molecules on opposite sides of the surface are so close together that they exert strong intermolecular forces on each other.