# Does smaller friction area cause less friction?

• posted on November 21, 2005, 11:12 am
From:
http://www.boltscience.com/pages/faq.htm

Is this relationship linear? If the friction area is halved so is the friction?
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• posted on November 21, 2005, 1:26 pm

The coefficient of friction does not change, but the friction force does.
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• posted on November 22, 2005, 3:42 am

Actually, neither the coefficient of friction nor the friction force does change. See my next message to Michael Pardee.
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• posted on November 21, 2005, 11:05 pm

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
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• posted on November 22, 2005, 2:44 am

'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.
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• posted on November 22, 2005, 4:11 am

It is the translation of that into angular resistance (torque) that varies proportionally with the radius (diameter).
Mike
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• posted on November 22, 2005, 3:43 am

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.
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• posted on November 22, 2005, 6:28 am
karl wrote:

snip
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.