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Re: A free b.d. and a centripedal force question



Aha.. I see my mistake. The ghost force is only there if I take the
reference frame as M and observe m. Then taking M's reference frame
the net force acting on m is zero, kmg in +x direction and ma as the
ghost force in -x direction and they are equal in magnitude.

J. Denker has certainly a point in the argument F = ma <= kmg.
However, assume that the block m at the top of the other block M is
on the verge of slipping off. Then, when the frame of reference is
the table, which is stationary relative to the blocks, we can write
F-kmg=Ma for M, and kmg=ma for m.
Right?

Quoting "John S. Denker" <jsd@MONMOUTH.COM>:

Several people have written
F = ma
= kmg

The first part is indubitally correct.
Not so the second part.
Since this is static friction, and k is
the coefficient of static friction, we can
conclude that
F = ma
<= kmg

We cannot conclude any equalities involving k.