Force equation???????

Question

This question is a bit hard and I can't figure out how to get to the answer.

What minimum horizontal force F is needed to pull a wheel of radius R and mass M over a step of height h (R>h)?
a)Assuming the force is applied at the top edge of the wheel.
b) Assuming the force is applied at the wheel's center instead.

The ANSWER is
a) Mg [h/(2R-h)]^1/2
b) Mg [h/(2R-h)]^1/2 / (R-h)

I have tried to get to these answers but it has not worked. I don't know how to do it.

Answer 1

Consider the torque on the wheel around the point where the step contacts the wheel. (Draw a geometrical picture of the arrangement.) The torque from the horizontal force at the top of the wheel is F*(R+R-h) = F*(2R-h). [(R+R-h) is the vertical distance from the force to the contact point.] The torque from gravity acts downward at the center of the wheel and is M*g*√[R^2-(R-h)^2]. (You need the horizontal distance from the center of the wheel to the contact point. Draw a right triangle with the hypotenuse from the center of the wheel to the contact point. The bottom leg of that triangle is the moment arm for the gravitational torque.)

When these torques are equal, the wheel will turn

F*(2R-h) = M*g*√[R^2-(R-h)^2]

F = M*g*√[2Rh - h^2]/(2R - h)
F = M*g*√h*√[2R - h]/(2R - h)
F= M*g*√[h/(2R - h)]

For the second part, the torque from F is F*(R - h); the rest calculate the same way