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With an object sitting at the top of an inclined plane, the static coefficient friction can be calculated with the simplified equation of: μ = tan(θ). In the same way, with an object moving down an inclined plane at constant velocity, the kinetic coefficient of friction can be calculated with the simplified equation μ = tan(θ). How does that make sense if the kinetic coefficient of friction is supposed to be less than the static coefficient of friction? For example, if an object stays still at 15°, but moves at a constant velocity at 20°, the kinetic coefficient of friction will be greater than the static coefficient of friction. How does that make sense?

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  • $\begingroup$ The static friction coefficient leads to the maximum friction that will be seen before the object slips, and that amount of friction only applies at that one point (e.g., at 15 degrees). For lower angles, the force of static friction is smaller than what is given by the formula. $\endgroup$ – David White Mar 1 at 20:36
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For example, if an object stays still at 15°, but moves at a constant velocity at 20°, the kinetic coefficient of friction will be greater than the static coefficient of friction. How does that make sense?

If the object is still at 15$^0$ all you know is the static friction force up the incline equals the force down the incline. The static friction force continues to match the downward force until the maximum possible static friction force is reached. The only way to find that out is to keep increasing the angle until motion just begins.

You could theoretically determine both static and kinetic coefficients of friction by conducting a thought experiment along the following lines:

Slowly increase the incline angle until motion just begins. The tangent of that angle is roughly the coefficient of static friction. Once motion begins the friction force abruptly transitions from static to kinetic. For a diagram of this transition, see the "Friction Plot" in the following link: http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin

Since the coefficient of kinetic friction is generally less than static friction, if there is no change in the angle, the friction force acting up the incline will now be less than the gravitational force acting down the incline for a net downward force and acceleration of the object.

Now imagine you are able to slowly lower the angle while monitoring the velocity of the object until the velocity becomes constant. At that lower angle the upward kinetic friction force exactly matches the force down the incline. The tangent of the lower angle would theoretically equal the coefficient of kinetic friction.

Hope this helps.

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  • $\begingroup$ Yeah this is a good answer, so I won't make another, but I also want to emphasize that the OP is asking about two different physical situations, which is why the direct comparison doesn't work. If $\mu_s=\tan\theta$ is the max angle before slipping, then $\mu_s>\mu_k$. If $\mu_k=\tan\theta$ for constant velocity, then still $\mu_s>\mu_k$ it's a different object or ramp. $\endgroup$ – levitopher Mar 1 at 18:45
  • $\begingroup$ @levitopher What I am saying is if the coefficient of kinetic friction is less than static friction, then it is impossible for the object to remain still at 15 $^0$ yet move at constant velocity at 20$^0$. The latter angle has to be smaller (the point of my thought experiment). Oh well, I guess I'm interpreting the OP 's statement differently than you are. $\endgroup$ – Bob D Mar 1 at 19:22

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