Does smaller friction area cause less friction?

From:

http://www.boltscience.com/pages/faq.htm

Is this relationship linear?
If the friction area is halved so is the friction?

 

Typically 50% of the torque is used to overcome friction

The coefficient of friction does not change, but the friction force does.

 

If I understand what they are saying, the smaller radius results in less
rotational friction, which would be a linear relationship with respect to
parasitic torque. That is, if the original torque was 50% on the threads and
50% on the bolt head, cutting the radius in half would cause only 25% of the
torque to be used to overcome bolt head friction and would put the remaining
75% of the torque on the threads.

Note that this is not the area but the radius that is changed. In theory,
friction is independent of area. What they seem to be describing is
leverage.

Mike

 

'Frictional Force' is calculated by multiplying the coefficient of friction
times the normal force. Normal force is the perpendicular force in a system.
It can be quite difficult to determine the correct coefficient of friction
to utilize when there is plating involved, two different types of materials,
etc.
This same problem is why some of these highway patrolmen's estimates of a
car's speed that was involved in an accident can be total b.s. Many factors
such as road film, tire condition, loose gravel and brake conditions should
be taken into account, technically, and there's simply no way that this
could be calculated accurately at the scene of an accident.

Ron M.

 

Actually, neither the coefficient of friction nor the friction force
does change. See my next message to Michael Pardee.

 

Got it. It is the integral of the travel. That is, friction at the bolt
head (washer) is proportional to,

friction = D(l)^2/2 - D(s)^2/2

D(s) = small diameter
D(l) = large diameter

Note that D(l) is the smaller of bolt head diameter or large diameter
of the washer.

 

As karl clarifies, the linear friction is indeed constant, whatever it is.
It is the translation of that into angular resistance (torque) that varies
proportionally with the radius (diameter).

Mike

 

This is wrong, it relates to the area. But relevant is the length of
the travel. Friction then is proportional to,

friction = D(l) - D(s)

I hope I got it right this time.