Friction torque vs. contact area

[mstachowsky]

I promise this isn't homework, it's actually for my research Here goes:

When two bodies are in contact and *translate* with respect to each other, there is no dependence of friction on contact area, and Ff = mu*Fn.

Now, assume we have a block of wood on a table with mass m and we apply a torque to the block's center of mass, rather than a force. Assuming mass is constant and the coefficients of torsional friction are constant, does the friction torque depend on contact area? if so, why, and if not, why not?

 

[adjacent]

Now, assume we have a block of wood on a table with mass m and we apply a torque to the block's center of mass, rather than a force.

You need a force to produce a torque

 

[mstachowsky]

Yes...I may not have been clear. Here is the scenario:

I have two surfaces in contact, the block and the table. Let's now assume that I have pinned the block through its center so that it is free to rotate so long as I can overcome friction, but it cannot translate (this just makes life easier...). Now, apply a force to one end of the block using a lever of some kind. The lever is there so that we can make sure that the length of the moment arm is equal for the "large area" block and the "small area" block.

Will you need to apply a greater force to rotate the large area block compared to the small area block?

So far, I've read that the friction torque is mu*Fn*r, where r is the radius of contact (this only works for discs apparently, but it's good enough for now), so intuitively, a larger area means a larger radius and thus a larger friction torque. Thoughts?

 

[A.T.]

so intuitively, a larger area means a larger radius

Not necessarily. You can make two blocks with different area sizes but the same mean radius and mass. They will have the same frictional torque, so it doesn't depend on the area, but the mean radius.

 

[97313]

It is worth noting that the coulomb model of friction is not a true representation of how things behave. The coefficient of friction between two surfaces can actually change depending on the pressure involved, which can change the interactions between the surface of the materials. The pressure can vary even in the force is the same by adjusting the area, so yes, the contact area does matter, although the variations with pressure are more common in fabric/cloth, they may still have an effect on wood.

 

[DPHagar]

I promise this isn't homework, it's actually for my research Here goes:

When two bodies are in contact and *translate* with respect to each other, there is no dependence of friction on contact area, and Ff = mu*Fn.

Now, assume we have a block of wood on a table with mass m and we apply a torque to the block's center of mass, rather than a force. Assuming mass is constant and the coefficients of torsional friction are constant, does the friction torque depend on contact area? if so, why, and if not, why not?

 

[DPHagar]

You ask: Is torsional friction dependent upon the contact area?
Yes, sort of. The issue is not about the magnitude of area, but the distribution of the contact area about the axis of rotation. I'll try to explain the concept rather than show a bunch of math (I can provide that if you like).
PREFACE. Note that the reason typical friction does not depend on the area is because, since the contact area must be identically equal for the two bodies in contact, the Area drops out of the math. Further, the coefficient of friction, µ, is an empirical value determined experimentally to relate the normal force to the friction force; a block placed on a planar surface; the surface is raised from horizontal to a point where slippage is impending. The coefficient of friction is and turns out to be the tangent of the angle of the plane; tan(Θ) = rise/run; independent of the area.
First look at a solid, round disk. For an applied torque, the equivalent frictional force is a function of the distance from the axis of rotation.
Friction will cause an opposing torque to resist the torque you are imparting on the disk.
Think about converting that to an equivalent force. Torque=Force (F) x radial distance (r) Equation (1)
The distance, r, is a function of the geometry shape and position relative to the axis of rotation. After playing around with CAD and determining the value of r graphically, the light bulb went off and I realized that this radius, r is actually the Second Moment of Inertia, or Radius of Gyration (Polar)! Now everything started to make sense. Once I saw that, determining Polar Radius of Gyration for various shapes was simply a matter of looking up the formulae.
So, hopefully you see the answer I was trying to get to. The torsional friction is a function of the shape of the contact surface. You could have a plethora of "blocks" with the same area, but different shapes. How the area is distributed with respect to the axis of rotation determines the frictional resistance torque.
Another physical example that may help round disk with a hole in the middle. You can incrementally increase the inner and outer diameter such that the surface area remains constant. However, the Polar Moment of Inertia will continue to increase, as will the resulting frictional resistance torque.
Hope this helps! I was actually looking for similar answers myself; I just happened to get lucky and figure it out!