Question

THE uniform solids wheel to mass m and th e radius r is halted at a step of hight h .what is the minimum horizantal force F is applied at the centre of the wheel ,necessary to rise the wheel over the step?

Answer 1

HERE IS THE DIAGRAM OF ABOVE QUESTION

Answer 2

consider the net torque \(\tau\) about O at t = 0. Set it to zero:

\(\tau = F \sin \phi ~ r - W d\)

\(\implies F \sin \phi ~ r = W d\)

\(d = r \cos \phi\)

\(F \sin \phi ~ r = W r \cos \phi\)

\(F = W \dfrac{d}{r-h}\)

\(F = W \dfrac{\sqrt {r^2 - (r - h)^2 } }{r-h}\)

\(F = W \dfrac{\sqrt { h (2r - h) } }{r-h}\)

You can then reason that the thing accelerates up the step because of the way the torque changes: \(\sin \phi\) increases whereas \(d\) decreases.

IOW the trick is to get it moving in the first place and F is the min horizontal Force needed