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STEP Mechanics from Siklos booklet

I have been working on a mechanics problem from the Siklos core booklet. The question is as follows:

A uniform solid sphere of radius aa and mass mm is drawn very slowly and without slipping from horizontal ground onto a step of height a/2a/2 by a horizontal force of magnitude FF which is always applied to the highest point of the sphere and is always perpendicular to the vertical plane which forms the face of the step. Find the maximum value of FF in the motion, and prove that the coefficient of friction between the sphere and the edge of the step must exceed 13\frac{1}{\sqrt 3}.

I am not sure how to show that the coefficient of friction must exceed the value stated. I got as far as showing that FF is equal to friction ff by taking moments, and even that R=mgR=mg. However, I do not know why this leads to the conclusion that μmg>Fmax\mu mg > F_{max}, where Fmax=mg3F_{max} = \frac{mg}{\sqrt 3} (hence result). Normally fmax=μRf_{max} = \mu R so I thought that in this case Fmax=μRF_{max} = \mu R also, but apparently this is not so. Can anybody explain this?
Reply 1
I haven't worked through the question, but in your notation, to prevent slipping as the question requires, we have to have that f<μRf < \mu R at all times, else the limiting value of friction is exceeded. I guess that gives the inequality.
Aside: In the diagram your direction for ff is incorrect. It should be in the opposite direction.
Reply 3
Original post by atsruser
I haven't worked through the question, but in your notation, to prevent slipping as the question requires, we have to have that f<μRf < \mu R at all times, else the limiting value of friction is exceeded. I guess that gives the inequality.


Thanks, that was the condition I needed.
Reply 4
Original post by ghostwalker
Aside: In the diagram your direction for ff is incorrect. It should be in the opposite direction.


You are right, of course. The diagram is straight out of the Siklos booklet. That definitely contributed to my confusion.

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