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No.ChrisBrandsborg said:∑F = Ft + Fc - Fs
Centripetal force is not an applied force. It is that resultant of the applied forces which leads to the centriptal acceleration:
##\vec F_c=\Sigma\vec F=\vec F_t+\vec F_s##
I note that there has been a change of notation in the thread. Originally you defined Ft as the force applied by the bully. Now it is being used for the tangential force applied to the girl, and the bully is applying force Fbully. That's ok, just as long as we all understand that.
The static frictional force only equals μs multiplied by the normal force (mg here) when just about to slip. Of course, here you are interested in the case where it is about to slip so the equation is valid; just pointing out that it is not in general true.ChrisBrandsborg said:ΣF = mrα + mω2r - μsmg
Also, you may be confused that gneill seems to be taking quite a different approach from mine. It's just a matter of the order of steps. At some point, we have to connect Fbully with Ft, and connect Ft with α. I was starting with the first of those, whereas gneill has started on the second. Both have to be done.