Simple Coefficient of Friction Calculation / Principle Behind Friction Test Machine 1

[Joccob]

Hi,

I worked in medical catheter industry and recently had some problems with simple CoF calculation. Because we normally work with catheters and tubes, instead of using standard ASTM D1894 testing method, we usually use so called "Pinch Test" to determine the CoF of the catheter surface. I simplified it into a diagram and attached below. Or you can search "Harland Friction Testing Machine" to get an idea of how it works.

It looks like a very simple problem, you clamp the catheter with force "N" and measure the pulling force. Because you need two pads to clamp it, there are then two friction forces. so the final calculation, as shown in the diagram, should be CoF=F-pull/2*N. But then one of my colleagues told me that there should be no denominator of 2 and CoF simply equals to F-pull/N. I also reached out to some friction testing machine supplier, and they also said that it should be F-pull/N (although they didn't explain in details on what is N). Now I'm totally confused cause I couldn't find out wwhy my calculation is wrong.

I'd really appreciate it if some one experienced with friction calculation to help me out.

 

[3DDave]

This is no different than for something like a single rotor disk brake. There are two pads, and each supplies a normal force to resist sliding. The force applied is the same on both sides so the sliding resistance is doubled.

In aircraft brakes, and clutch packs, layers of pads and disks are used to multiply the effect of the brakes.

If the factor of 2 is neglected then one could see the oddity that adding more pads and rotors would increase the coefficient towards infinity when the individual pads are seeing no change.

The trick is this - is it measuring the CoF of the contact between the tubing and the pad or is this a measurement meant to compare all similar setups and the dropping of the "/2" removes a potential source of error?

I think it is the latter because all such testing will be performed in a similar way for this product type and single-sided application of friction won't happen.

The only place it becomes a problem is if one is designing a system based on single friction pad data that is usually reported for such as "steel on brass" and "nylon on steel" and wondering why this machine reports much different results than expected.

Fun experiment - off topic - take two paperback books and interleave the pages, left, right, left, right, for about 100 page pairs. Then try to pull them apart by grabbing the spines. It used to be a fun thing to try with phone books, but those are tough to come by anymore. It shows the mechanical advantage of interleaving, even when the normal force is zero.

 

[MintJulep]

Your FBD and calculations look correct.

Early in my career I faces a big problem because my predecessor neglected that 2 in the calculations for a brake disc.

Although Harland calls it a "Friction Testing Machine", it seems the result it reports is pull force, not coefficient of friction.

https://harlandmedical.com/products-and-services/testing-equipment/fts7000

 

[desertfox]

Hi

Imagine a block on a flat plate resting on a surface, assuming that the block has a downward force of Fn then the CoF would be Fp/Fn where Fp is the required pull force.
Imagine now that you double Fn so 2Fn would be the downward force, so the CoF for the materials doesn’t change and on that basis the force to slide one material over the other would be 2Fp
and so the CoF for the latter case would be 2Fp/2Fn.
So simplifying the equation the numerical number 2 cancels out.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[MintJulep]

Imagine a block on a flat plate resting on a surface,

Different case because there is only one friction plane.

 

[desertfox]

Not really if you have two planes exerting Fn on each plane then twice the pulling force to overcome friction would be required when compared with two planes which only apply Fn/2 on each plane, unless the CoF of friction between the two materials changes.

Basically I am saying the same as 3DDave in that if you double the normal force ie N then you automatically double the force to overcome friction provided that the CoF remains constant.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[IRstuff]

The trick is this - is it measuring the CoF of the contact between the tubing and the pad or is this a measurement meant to compare all similar setups and the dropping of the "/2" removes a potential source of error?

That is the crux of the question; are you measuring a system CoF, or a fundamental property of the surface. For the purposes of comparing different tubing, machts nichts which way you go, so long as you are consistent in the measurements and presentation of the comparisons. For characterizing materials, which gets to the fundamentals of the surface, then division by 2 is a must; otherwise, you get a goofy answer.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

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http://www.eng-tips.com/forumlist.cfm

 

[desertfox]

Hi Joccob

There is nothing wrong with your workings out, if you look at the third line of the equations you have written on the right hand side of your post you state Fpull = 2μFn, which aligns with what others and I have said, if you double the normal or applied force you double the pull force to overcome friction.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[MintJulep]

What matters is the total value of the applied force and not the number of surfaces that the total applied force is divided across.

Sorry, but that's simply wrong.

 

[desertfox]

Okay please expand on that

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[MintJulep]

Sure.

Two friction planes hold double the pull force for the same normal force.

You could do the same experiment at home. As you see, it doesn't need much equipment.

[URL unfurl="true"]https://res.cloudinary.com/engineering-com/video/upload/v1724590158/tips/Friction_yox8by.mp4[/url]

 

[desertfox]

Well that’s interesting however I am not sure that the static laws of friction apply to paper sheet because after the first test with one friction plane any static charge would affect the next experiment.
For the experiment to happen as it did on the video then the coefficient of friction must have changed, I’ve pretty much stated in the posts that the coefficient of friction is assumed to stay the same.
I am basing statement on the first law of friction:-

First law of friction: The amount of friction is proportional to the normal force exerted between the surfaces. Second law of friction: Friction does not depend on the area of contact between the object and the surface.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Joccob]

Thanks everyone for the help. I'm glad my calculations are correct. It makes perfect sense that they dropped the "/2" to simplify the equation since it doesn't impact the comparison of different tubes, which is the case 99% of the time. That might also be the reason why no one brought this up in the past. In fact, this might make CoF differences more noticeable (maybe that's a marketing strategy?). Anyway, I brought this up because I have to write formal test reports, and I don't feel comfortable including incorrect numbers, even if they don't affect the overall results.

 

[desertfox]

Hi joccob

I have amended my post 25h August regarding the number of surfaces, my statement was incorrect.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[btrueblood]

desertfox, kudos to owning the error and fixing it. Star for Mint.