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How much force $F$ (in terms of $G$) is needed to get through this sphere with the radius of $2r$, the weight of $G$, from the stair with the radius of $r$?
I had read the David Halliday's physics textbook, what I know are
$$\sum F_x = 0, \sum F_y = 0, \sum \tau= 0$$
Which tells us that the net force will be equal $0$ same as torque. However, I couldn't build the correct equation, let me show at least what I've written.
$$F - G + 2r + x = 0$$
I've gone truly wrong. Can you assist?
Regards!
Horizontal and vertical equilibrium are satisfied by the reactions in the corner.
The torque equilibrium with respect to the corner of the stair is
$$\sum \tau= Fr-Mgd=0\implies F=\frac{Mgd}{r}$$
The problem reduces to find the horzontal distance $d$ between the corner of the stair and the center of the sphere that is
$$d=\sqrt{(2r)^2-r^2}=r\sqrt 3$$