Static coefficient of friction is 1.6 between rubber and steel. Let's say you put a steel ball on a rubber incline at 45 degrees. Friction is more than gravity so it can't accelerate transitionally down the ramp. But there's clearly torque by friction * radius around the center of mass, so it has an angular acceleration. Also, it has to be nonslip motion because slipping only occurs with low coefficient of friction. However, this is a contradiction. How do you resolve this?


Think about a rectangular object on an inclined plane. If there is enough friction it won't slip down. But it will topple when its centre of mass is moved past the pivot point. Your ball is effectively a many-sided polygon that is toppling over its edges. Think about where the centre of mass of a circle is compared to its point of contact with the plane. You will find that there is a turning effect around that point of contact.

Note also that the coefficient only helps us find the maximum possible value of friction. Often friction is less than that value. Consider the situation with an object resting on a horizontal surface. There could be friction up to the limiting value, but if I'm not providing any sideways force, the friction will be zero.

  • $\begingroup$ So are you saying that the value of friction is not the same as the coefficient of friction * the mass * gravity * cosine of the angle, but rather less? $\endgroup$ – John Targaryen Jan 17 '17 at 9:05
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    $\begingroup$ Static friction is always less than or equal to $\mu_s \times$ the normal force. The static friction force is whatever is needed to make the ball roll without slipping, i.e. the angular velocity $\omega$ of the ball is equal to $vr$. If the friction coefficient is very small, it might be impossible to have enough static friction force to do that. In that situation, you have dynamic friction which is always equal to $\mu_d \times$ the normal force, and the ball will partly slide and partly roll. (The static and dynamic friction coefficients are always such that $\mu_d \le \mu_s$.) $\endgroup$ – alephzero Jan 17 '17 at 9:32
  • $\begingroup$ Right so when a circle is rolling down an incline with no kinetic friction and only static friction (nonslip), when does mgsin theta - mu_static mg cos theta = ma not hold? Clearly it doesn't hold in the posted example, but it usually is correct so where is the line drawn? $\endgroup$ – John Targaryen Jan 17 '17 at 19:10

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