# Friction and forces in rollers

1. Feb 5, 2012

### bootsnbraces

HI all wonder if anyone can help im banging my head on this one! If i have 2 discs of a given material with smooth edges (imagine 2 gears side to side but no teeth) How do i calculate the required force to create enough friction to transmit a given amount of torque between the 2? i have the coefficent of friction for the 2 materials and the torque required but im struggling with the area of contact between the 2 edges (a straight line contact i think?) and the force required to stop slip. Ive got a feeling this has something to do with the normal force?

Last edited: Feb 5, 2012
2. Feb 5, 2012

### McLaren Rulez

Your force of friction would be $f=\mu N$ where $N$ is the normal force. The product fR where R is the radius of the second disc will be equal to the torque being transmitted.

So you need R here. And just one thing, I wouldn't call the edges smooth because that usually means frictionless in most contexts. But I know what you mean here, its just gears with no teeth.

3. Feb 6, 2012

### Lsos

And notice you don't need to know the area of contact.

4. Feb 6, 2012

### ZealScience

What do you meant by

Where and in which direction are you applying the force?

If coefficient of friction μ is given, I think you can compute the normal force required.

5. Feb 6, 2012

### bootsnbraces

zeal science- the force is pushing the 2 rollers together side on like over meshing a pair of gears (only no teeth on the rollers)

So f=μN where N is the normal force. The product fR where R is the radius of the second disc will be equal to the torque being transmitted.

So for example radius 0.5m and torque of 10kg/m the force required would be 20N? and with a coefficient of friction of 1 that would equate to a normal force of 20N?

Sorry if thats all wrong im trying to work in these modern units to make it mroe readable and it doesnt help!

6. Feb 7, 2012

### bootsnbraces

Hi all, well some time studying the equations and a bit of experimentings put me on the right tracks! from the fact that no area of contact is included in the equations im gathering that the length of the line of contact between the cylinders has no effect on the required force?

I.e wether the discs were cd's or 6ft long cylinders they will require the same force to stop them from slipping at a given torque?
In which case im guessing the length of the cylinders would only need be long enough to keep the surface stress caused by the force low enough and the cylinder thick enough not to buckle under the force?

Thanks
harry