九月初九是什么节日

Question

Assume a cylinder / ball of moment of inertia $I = \beta m r^2$ rolls without slipping on an incline with an angle $\theta$. A classic problem is to show that the coefficient of static friction must then satisfy $\mu_s \ge \left(1 + \frac{1}{\beta}\right)^{-1}\tan\theta$. For example, for a uniform ball $\beta = \frac{2}{5}$ and so $\mu_s \ge \frac{2}{7}\tan\theta$.

The usual derivation proceeds as follows:

1. Considering the force of friction acting along an incline $F_f$, and the component of gravity acting along an incline $mg\sin\theta$, the linear acceleration along the incline is $a_{CM} = g\sin\theta -\frac{F_f}{m}$. The angular acceleration $\alpha = \frac{F_f r}{\beta mr^2}$. For rolling without slipping to occur, $a_{CM} = \alpha r$, which translates to $$ F_f = \left(1 +\frac{1}{\beta}\right)^{-1}mg\sin\theta $$

We see a difference compared to the behaviour of a friction force acting on a static point mass / block: if a force $F_{ext}$ is applied to a static block, static friction is equal in magnitude and opposite to $F_{ext}$ until $F_{ext} = mg\mu_s\cos\theta$, at which point the block starts to move and the friction becomes kinetic. For a ball rolling down an incline without slipping, friction is clearly not equal to and opposite to the applied force $mg\sin\theta$, but smaller by a factor of $1 + \frac{1}{\beta}$.

Coming back to the ball on the incline, most derivations use the relation $F_f \le mg\mu_s \cos\theta$, which together with $F_f = \left(1 +\frac{1}{\beta}\right)^{-1}mg\sin\theta$ yields $\mu_s \ge \left(1 + \frac{1}{\beta}\right)^{-1}\tan\theta$.

Naively, I would expect all 3 of the following basic laws of frictions to work:

1. Static friction is equal to and opposite in magnitude to the applied force ($F_f = mg\sin\theta$) as long as...
2. $mg\sin\theta \le F_{f, max}$ (friction is static until applied force exceeds the maximum possible value of static friction)
3. $F_f \le F_{f, max} = mg\mu_s\cos\theta$ (maximum magnitude of static friction)

However, the usual solution assumes that only the law 3. holds in this case, and 1. and 2. do not: (1) doesn't hold as $F_f = \left(1 + \frac{1}{\beta}\right)^{-1}mg\sin\theta < mg\sin\theta$. (2) does not necessarily hold as if $mg\sin\theta \le mg\mu_s\cos\theta$, then $\mu_s \ge \tan\theta$, without the factor of $1 + \frac{1}{\beta}$.

How to reconcile the fact that 1. and 2. do not hold while 3. does hold with the standard situation of a point mass, in which case all relations 1., 2., and 3. hold?

I realize this is most likely to the differences between handling rotational vs. translational motion, but I would prefer as rigorous / mathematical answer as possible.

This question is not a duplicate of similar questions (e.g., Acceleration of a ball rolling down incline without slipping), as they do not discuss the apparent contradiction with the situation of a block on an inclined plane.

Answer 1

Your "basic laws" 1 and 2 are just not true. Your confusion is natural as this topic is often taught poorly. If you were taught that the friction force is equal and opposite to the applied force, forget that -- it's false.

The magnitude of the static friction force at the interface between two objects that are not sliding against each other is the smaller of the following:

1. $\mu N$, where $N$ is the normal force at the interface and $\mu$ is the coefficient of static friction between the objects; and
2. The force required to prevent the two objects from sliding against each other.

The law does not prescribe a universal numerical relationship between the friction force and other applied forces. As you have just seen, the force required to prevent sliding depends on the details of the objects.

As a side note, you might be wondering if "the force required to prevent the two objects from sliding" is always possible to calculate, or even well-defined. Of course it is not. Like other empirical force laws (e.g. Hooke's law for springs), the law of static friction is only an approximate model of reality, and the model does not work for all situations. The situations where it fails will not generally arise as problems in elementary physics classes.