Coefficient of friction of a sandwiched body

[vamsi]

Homework Statement

How to determine the coefficient of friction of a sandwiched body with experimental methods

Relevant Equations

F=μbottom∗Nbottom+μtop∗Ntop

Consider a test setup as shown in Image below, where I have a blue object sandwiched between a material inside the container (for eg. sand) and a surface. The container is fixed and the material inside is assumed to not move.

Using a force guage, I can measure the peak amount of force F needed to overcome the static friction at both the top and bottom surfaces. For known weights of the blue object, material inside container and a known μ_bottom between blue object and surface (also determined by measuring seperately the force needed to pull only the blue object on the surface), can I determine the unknown μ_top
at the top surface between blue object and material inside the container using the following formula?

F = μ_bottom ∗ N_bottom + μ_top ∗ N_top

and solving for unknown μ_top where we have known values of

F = Force measured by force gauge in N
μ_bottom = static COF at the bottom contact area (determined by the same test setup without the container)
N_bottom = Normal force pointing up at bottom surface (sum of weights of blue object and material)
N_top= Normal force pointing up at top surface (weight of material)

 

[Chestermiller]

Are you allowed to vary the weight of the blue object?

 

[vamsi]

Are you allowed to vary the weight of the blue object?

Unfortunately not. However, I have thought of adding more normal force by placing an object on top of the material in container. Something like this.

 

[Chestermiller]

Are you allowed to lower the coefficient of friction between the blue object and the base by putting butter in between??

 

[vamsi]

Are you allowed to lower the coefficient of friction between the blue object and the base by putting butter in between??

I have plans to use a PTFE sheet as the base surface. I found out that the COF between steel and PTFE material is around 0.04

 

[Chestermiller]

Are you free to change the elevation at which the force F is applied?

 

[vamsi]

Are you free to change the elevation at which the force F is applied?

I cannot imagine how I could do that. By elevation do you mean if the force is applied at the top/centre/bottom of the side surface? Can you please elaborate.

 

[Chestermiller]

I cannot imagine how I could do that. By elevation do you mean if the force is applied at the top/centre/bottom of the side surface? Can you please elaborate.

If it’s not a rigid body and the location at which you apply the load determines the share stress distribution at the top and bottom surfaces. Are you considering the blue body rigid body?

 

[vamsi]

If it’s not a rigid body and the location at which you apply the load determines the share stress distribution at the top and bottom surfaces. Are you considering the blue body rigid body?

Yes it is a rigid body made of steel. It does not change shape upon stress.

 

[Chestermiller]

Yes it is a rigid body made of steel. It does not change shape upon stress.

If it’s a rigid body, then both surfaces have to release at the same time. In order for that to happen, the normal force on the top surface. The coefficient of friction at the top surface must match friction on the bottom surface times the normal force on the bottom surface, this would be very unusual and unexpected

 

[vamsi]

I also do not think that the surface release at top and bottom side happen at the same time. I have thought of a simplification of the problem by spreading the material inside the container on a surface, and making the blue object slide on the material while measuring the peak force F. Do you think it might be better?

 

[Chestermiller]

Unfortunately not. However, I have thought of adding more normal force by placing an object on top of the material in container. Something like this.

View attachment 352364

This is getting more complicated.

 

[vamsi]

This is getting more complicated.

Okay. Can you think of some simple test setup which can be used to measure the coefficient of friction between a non-rigid material (for eg. heap of sand/salt) and a rigid steel body? I would be most greatful for that!

 

[haruspex]

If it’s a rigid body, then both surfaces have to release at the same time. In order for that to happen, the normal force on the top surface. The coefficient of friction at the top surface must match friction on the bottom surface times the normal force on the bottom surface, this would be very unusual and unexpected

I don't see a problem. The blue object can only move macroscopically when both static frictions are overcome, implying that the force F is distributed across the upper and lower surfaces accordingly. This would be achieved by a combination of elastic deformation of the blue object and the adjacent surfaces, and a little creep.

 

[haruspex]

F = μ_bottom ∗ N_bottom + μ_top ∗ N_top

and solving for unknown μ_top where we have known values of

F = Force measured by force gauge in N
μ_bottom = static COF at the bottom contact area (determined by the same test setup without the container)
N_bottom = Normal force pointing up at bottom surface (sum of weights of blue object and material)
N_top= Normal force pointing up at top surface (weight of material)
[/QUOTE]
Looks good to me. What do you think is doubtful?

 

[vamsi]

F = μ_bottom ∗ N_bottom + μ_top ∗ N_top

and solving for unknown μ_top where we have known values of

F = Force measured by force gauge in N
μ_bottom = static COF at the bottom contact area (determined by the same test setup without the container)
N_bottom = Normal force pointing up at bottom surface (sum of weights of blue object and material)
N_top= Normal force pointing up at top surface (weight of material)

Looks good to me. What do you think is doubtful?
[/QUOTE]

I had a doubt in the formula. But just checked today using a test setup and a force guage. I was able to calculate the μ_top with the same arrangement as the picture as 0,526. I have tried to slide the blue object on top of the yellow material surface, and I got a μ_bottom value of 0,616.