u r old...
anyway:
work done by frictional forces (kinetic energy of the car converted into heat):
Work = F_friction x L distance traveled by the pad.
F_friction = �µ x F
�µ -- coefficient of friction,
F is the force applied to the pad.
Area of the pad is not in the picture, as you can see. Now, it's more complicated than this of course, because the pressure on the pad material plays its role too.
pressure x Area = F
Extremely high pressure (as in small Area) would cause glazing, overheating and other bad effects.
The above formulae helps to explain why area of the pad is immaterial, but it's not useful in practice. After all, you can transfer the same amount of kinetic energy into heat by increasing distance traveled. That's not what you want
That is why MFE quoted the equation for torque. Let's look at the other side of the pad first:
The other side of the pad is more interesting. Pascal's law states that the pressure in the liquid is equal in all directions, so the force applied to the pad is
F = PS x AP.
Two pads: PS x AP x 2
Friction: PS x AP x 2 x �µ
Tprque F_friction x R (effective radius which can be calculated as a distance from the axis of rotation to the center of mass of the pad's working surface)
TW = PS x AP x 2 x �µ x RE
amarv12:
we posted almost at the same time. your argument about pressure in the system applies to the other side of the pad i.e. caliper size. see above.
