A classical mechanics problem by Md Zuhair

Note: This is not a proper solution

The limit of the force expression should be zero as h approaches zero, and infinity as h approaches R. (ii) is the only expression which meets these two requirements.

There are 2 contact points.Hence 2 normal forces.Let us name forces as $N_1$ that is normal reaction from ground.then $N_2$ the normal reaction from footstep will have $2$ components lets name them as $N_{2,y}$ and $N_{2,x}$ and of course the former is tangential to wheel.Then there is weight Mg acting downwards,and Force F acting horiontally We have 2 undesirables $N_{2,y}$ and $N_{2,x}$ ,so we will take torques about footstep to eliminate them.And in the right triangle forming we can see that moment arm of Mg satisfies

$x^2$ $=$ $R^2$ $-$ $(R-h)^2$ .We can get $x$ from here.Thus applying $2$ things

1) As wheel lifts up $N_1$ approaches $0$ .

2) Summing up torques about footstep $= 0$ .(No,we can;t write torque = I alpha because still wheel in on verge of lifting and has not acquired any angular acceleration.)

Solving For $F$ in terms of $M,g,R$ and $h$ we arrive at answer $(b)$ .