Does smaller friction area cause less 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

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.