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I am needing to be educated about one aspect of all this. One way to
determine the max value of µ(s) is to increase the slope until the mass
starts to slide. The algebra tell us that µ(s) = tanø. All is independent of
mass. So I am confused by the statements that one of the blocks will slide
before the other, based on mass. I have not been able to find that in all
the discussions. Maybe I missed it.That's really only true in that one situation. The trick here is that the ribbon (the surface) is accelerating. Just like the tablecloth, if you pull hard enough, static friction won't be able to pull the objects on top hard enough to keep up, and they'll slip. This is dependent upon the object's mass in this case, because the acceleration is dependent upon the mass in a more complex way than simple proportionality, so the mass doesn't cancel out.
The way that I did this was to look first at the case with no slippage and determine a. This situation works as long as the friction force required is less than u m g cos (theta) for EACH block. This means that the static friction coefficient must be greater than or equal to (2M/(m+M))* tan (theta).
This condition fails first for the small block, because of its smaller normal force. Note that the static friction forces on the two blocks are always equal, because of the 'light' ribbon condition.
If we examine the case where the small block slips and the large one doesn't, we can also find the acceleration.
That model's valid only if the acceleration returned by the solution is positive. That requires that the kinetic friction coefficient must be less than (M/m)tan(theta).
This range overlaps with the static friction coefficient range from the first case, and since mu_s > mu_k, this must cover all possibilities.