drag/friction on a cylindrical surface

[pinketon101]

First some backstory....having graduated quite a few years ago and going to work directly out of school (SolidWorks VAR) I feel as though I might be brain dead for having to ask this but I'm absolutly stuck now that I'm back in the "real world" and even my reference materials at home don't seem to be of much help.

The issue at hand is as follows: I've been asked to find the "drag" generated between two surfaces of a torsional suspension system and can't quite figure out where to start....even my initial free body diagrams look those from my freshman statics class!

The system at hand consists of a "roller" (1.41 inch diameter 2" in length, hardened steel etc...) that for all intensive purposes only is capable of rotation, no translation etc... Really the roller has a bolt through the center and is located between two ridgid walls. On top of said roller is the "arm" of the torsional spring system...the arm acts as the level that acts upon a torsion bar. The face of the arm is tangent to the outer cylindrical face of the roller at all times....but as the arm moves up and down the system is such that is also moves forward and backward on the roller. If my description of the system is a bit fuzzy, imagine laying a beer can on its side and laying your ruler on top of it...that's pretty much the system.
My problem continues....now since the bottom face of the torsion arm is only 0.875" wide the actualy contact patch between the roller and the arm is very small. To make matters worse....the purpose of this exercise is to attempt to figure out the difference in resistance, drag, friction (whatever you would like to call it....) generated between the two faces, but also to compare the resistant forces when the bottom face of the torsion arm is changed so that its profile is rounded with a 0.375" radius.......essentially find the difference when you roll a flat piece of stock on the roller, or a piece of round rod on the roller......

I hope someone out there can follow my rambling....I need to get started on this ASAP so a nudge in the correct direction would be fantastic!

 

[rb1957]

liked the beer can analogy ...

it sounds like the roller is fixed in possition and the arm is moving up and down, driven by something ...

if this is right, what's causing the arm to move ? and what's resisting it

 

[hydtools]

It sounds like the friction between the arm and roller would be rolling friction. Friction between the roller and the through bolt would be sliding friction. The friction force is only dependant on the force normal to the contact surfaces and not on the size of the contacting surfaces.

Friction force = friction coefficient * normal force.

The friction force acts tangent to the contacting surfaces.
The sliding friction force will be higher than the rolling friction force unless there is lubricant present on the bolt. If there is roller contact with the 'walls', then there is that friction, too.

Ted

 

[MechEng2005]

Following up on Hydtools post... It is correct that the friction force does not depend on the area and therefore the friction should be the same regardless of the contact area. Putting a radius on the arm will cause more wear though and a larger stress concentration (if that's a concern).

If we assume all you want to know is the friction force between the arm and roller set up a free body diagram with the roller as a fixed point. Find the reactions at the roller, specifically the force normal to the arm. This gives you the normal force needed for

FrictionForce = Coeff * NormalForce

As far as finding the coefficient of friction, you can probably find some reference for steel on steel. Though I would imagine it depends partly on the speeds involved.

-- MechEng2005
I LOVE the free body!

 

[sreid]

Hydtools has the right approach. Assuming the arm rolls on the cylinder, the losses will be the sliding between the bolt (axle) and the cylinder.