A uniform solid cylindrical roller of mass \('m'\) is being pulled on a horizontal surface with force \(F\) parallel to the surface and applied at its centre. If the acceleration of the cylinder is \('a'\) and it is rolling without slipping then the value of \('F'\) is 

\(ma = F - f     \)      ...(i)

And, torque \(\tau  = I\alpha \)

\(\begin{array}{l}
\frac{{m{R^2}}}{2}\alpha  = fR\,\\
\,\frac{{m{R^2}}}{2} - \frac{a}{R} = fR\,\,\left[ {\alpha  = \frac{a}{R}} \right]\\
\frac{{ma}}{2} = f\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)
\end{array}\)

Put this value in equation \((i),\)

\(ma = F - \frac{{ma}}{2}\,\,or\,\,F = \frac{{3ma}}{2}\)