Slipping between friction wheels 2

[AntoSilv]

Dear community,

I am planing to use friction wheels on my current design but before starting the selection of the elements, I was reading a little bit about them. I found online the following information:

I would like to ask if you agree with the statement of the last bullet point. In my opinion, P>F is the required condition to initiate the movement of the wheels, nevertheless, the author claims that this is the condition for slipping.
I can imagine that there is a flaw in my reasoning but I cannot identify it.

Thank you for your time!

 

[Compositepro]

You will get slip. There are microscopic deformations that occur when the surfaces contact. These accumulate throughout the rotation as slip. Compare a toothed belt between two spockets to a flat belt between two pulleys. Belts stretch under load. With a toothed belt this causes a phase shift between the two spockets, but they always stay in synch, hence they are also referred to as synchronous belts. With a flat belt, tension changes result in a speed change between the two pulleys, which is the same as slip.

 

[3DDave]

This is misreading the statement. Which is - if the friction force is less than the required transmitted tangential force then it will slip. Assume the output is locked. Then no matter how large P is there will be no transmitted motion. If P<F then the input will be stalled. If P>F the input will turn while the output remains locked.

 

[Compositepro]

True, it is not clear what point the snippet is trying to make. Upon rereading it seems to say that the rolls will slip if you exceed the static frictional force between the rolls. Well, duh!. My point was that there will always be slip between the rolls. It may be small, but it will never be a synchronous drive.

 

[mfgenggear]

what the function of using friction drive rolls, why not at least flat belt

 

[AntoSilv]

Thank you for the feedback everyone.

Can I assume then, in the case that the output is not locked, that an infinitesimal small torque will already move the two wheels? (ignoring the inertia of the elements).
I believe my misunderstanding was to think that I required a minimum torque to overcome the friction, nevertheless, the friction is the required effect that allows the system to work.

 

[Compositepro]

Yes, all you need is sufficient coefficient of friction and contact force for this to work.

 

[hydtools]

The frictional force F is the result of radial force Rn, the force pushing one wheel against the other. F is not conditional on tangent force P, only on the radial force Rn.
As long as the driving torque causes a tangential force P < friction force F there will be no slip. When the driving torque causes a tangential force P>F the wheels will slip.

Ted