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

[SSHS KPHOX <kphox@MAIL.CCSD.K12.CO.US>]

Forum for Physics Educators <PHYS-L@lists.nau.edu> writes:
> When I consider the forces
> acting on m (Free body diagram) in x-axis, i can identify the
> friction force acting in +x direction which equals normal force
> (equal to mg) times the coefficient of friction k, and the fictious
> force (ghost force) in -x direction which is equal to ma. How would
> you write newtons second law for m: kmg-ma=ma ?

When I do this, I see friction in positive x, mg down, normal force up.
The latter are equal and opposite and thus are not needed in determining
the net force. The remaining (net)force, and application of Newton's
Second law,  is friction = kmg (in your notation) = ma.

I do not see the need for the ghost force as it is observed in the frame
of the accelerating box, not the laboratory frame.

Ken Fox