Answer
$\sqrt{15}Mg$
Work Step by Step
We can find the required minimum horizontal force as follows:
$sin\theta=\frac{\frac{1}{4}R}{R}$
$sin\theta=\frac{1}{4}$
$\theta=sin^{-1}(\frac{1}{4})$
$\theta=14.48^{\circ}$
Now $MgRcos\theta-F_{min}(\frac{1}{4}R)=0$
This simplifies to:
$F_{min}=4Mgcos\theta$
We plug in the known values to obtain:
$F_{min}=4Mgcos(14.48^{\circ})$
$F_{min}=\sqrt{15}Mg$
