Static coefficient of friction decreasing with normal force?

In summary: That would also reduce the force that you're getting from the rubber.I'm not sure how you would measure the force if the rubber is deforming.
  • #1
Kinetic95
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In my experiments, on a flat plane, when I incrementally increase the normal force acting on a rubber block, I am measuring a decrease in the static coefficient of friction. I am also measuring the friction force, which increases as normal force increases BUT this is not a proportional increase, hence the drop in coefficient of friction value. For instance

Mass: 5kg, 10kg, 15kg, 20kg,
COF: 1.8, 1.6, 1.4, 1.3

Why is this? Shouldn't the static coefficient of friction increase with normal force? Or is the roughness of the surface sample I am using playing a role?
 
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  • #2
Kinetic95 said:
Shouldn't the static coefficient of friction increase with normal force?
Why do you expect this?

How do you measure the force? How do you vary the mass? Do you have a sketch of the setup?

Did you plot mass against force? That should give a hint what could have gone wrong.
 
  • #3
mfb said:
Why do you expect this?

How do you measure the force? How do you vary the mass? Do you have a sketch of the setup?

Did you plot mass against force? That should give a hint what could have gone wrong.

I expect an increase of the actual area of contact (at the surface asperity level) as I increase the normal force, therefore more asperity junctions would need to be broken leading to a higher COF.

I am using a pneumatic ram to apply a horizontal force to a block of metal with a rubber underneath. A force transducer measures the ram force, which will be equal to the friction force I believe. I obtain these readings on my laptop. The normal force is changed by adding or subtracting weights from the metal block./

No I don't think there is any problem with mass vs force, I think the values are correct there.

http://imgur.com/a/QmZ9b
 
  • #4
Kinetic95 said:
The normal force is changed by adding or subtracting weights from the metal block./
What about the weight of the metal block itself?

Your values are perfectly linear - just with an offset. If there are 4 kg of unaccounted mass (or a force measurement offset by 40 N), then the coefficient of friction is constant.
 
  • #5
mfb said:
What about the weight of the metal block itself?

Your values are perfectly linear - just with an offset. If there are 4 kg of unaccounted mass (or a force measurement offset by 40 N), then the coefficient of friction is constant.

Yes I've included the weight of the metal block itself. I'm not sure what you mean by unaccounted mass?

Sorry those values were just example values to show the the COF was decreasing. The type of values I am actually getting are:

Normal force: 69.5N, 118.56N, 167.6N, 216.7N
Static COF: 1.503, 1.375, 0.968, 0.769

I'm just wondering physically why this is. If friction partly comes from asperity junctions, won't an increasing normal force raise the number of these junctions (areas of contact), increasing the friction value/force? This would at least, as you say, keep the COF constant. Or is hysteresis in the rubber contact having an effect? I'm quite confused as I just don't see why the COF would be falling; I think there is another variable influencing the interaction, but I'm unsure what.
 
  • #6
Kinetic95 said:
I'm not sure what you mean by unaccounted mass?
If all your mass values would actually be 4 kg higher, you would get a constant force to mass ratio.
Kinetic95 said:
Normal force: 69.5N, 118.56N, 167.6N, 216.7N
Static COF: 1.503, 1.375, 0.968, 0.769
That looks quite odd, and more values in between would help to see what is going on.
Kinetic95 said:
If friction partly comes from asperity junctions, won't an increasing normal force raise the number of these junctions (areas of contact), increasing the friction value/force?
As long as that is proportional to the force, it doesn't increase the coefficient of friction.

Rubber can deform, that could change the shape of the contact area, and therefore the angle between that and your pulling direction.
 
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