# Is it possible to find the of coefficient of kinetic friction given only the angle?

## The Attempt at a Solution

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No. Since $F_{friction} = μF_n$, in order to find the coefficient of friction for a surface, you'd need the normal force and the actual force that the friction exerts. I'm not exactly sure what you mean by the angle. Are you talking about some mass being on a slope and knowing the angle of that slope, or are you talking about something exerting force down on something at an angle? Be more specific about that.

a metal disk is placed on a metal sheet and the sheet is tilted on an angle until right before the disk slides. The angle is then measured and using the tan of the angle the coefficient of static friction can be found. So now the experiment is repeated until the disk begins to slide down the metal sheet. The variables measured at this point is the hypoteneus (length of sheet) the angle it began to slide at, and we have the static friction coefficient from prior. How do I now find the coefficient of kinetic friction?

I can see how you get the coefficient of static friction. You can just set the force along the ramp which is mgsinθ equal to the force of static friction. Then plug in mgcosθ for the normal force and get the coefficient of static friction. The problem with finding the coefficient of kinetic friction is that you can no longer set the force going down the ramp and the friction equal to each other. Instead, you have to set the sum of those two equal to mass * acceleration. So I think after the disk starts sliding down the hill, you need to know the acceleration. As before, mass doesn't matter because it cancels on both sides of the equation. But you need the acceleration of the disk. Once you find that, then you can set the sum of the forces I mentioned earlier equal to the acceleration (since the mass with the acceleration already cancelled out). and find the coefficient of kinetic friction the same way you found the coefficient of static friction. Tell me if I'm missing something here, but I think you need the acceleration.