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Physics
Classical Physics
Mechanics
Friction of a Wheel on the Axle?
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[QUOTE="jbriggs444, post: 6031208, member: 422467"] The relevant notion is that of work. Work = force times distance. Or torque times rotation angle. We typically model kinetic friction as depending only on the normal force and not on the slip rate. So a bronze on bronze bearing supporting the weight of a wagon should have the same linear frictional force as a bronze-shod tire on a bronze rail supporting the same wagon. If the wheels were locked and the wagon was skidding along the rails, that frictional force is what the horses would need to overcome. The wagon tires skid a long distance against the frictional force, dissipating a large amount of work. With the wheels released, the moving surfaces are now at the hub. The frictional force is the same, but the hub moves a small distance against the frictional force, dissipating a small amount of work. It is a direct proportion. A mechanical advantage of 8 to 1 translates to a reduction in work dissipated of 8 to 1. [/QUOTE]
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Classical Physics
Mechanics
Friction of a Wheel on the Axle?
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