A cylinder of $\textrm{500 grams}$ in mass has a very thin flexible non elastic tin wire of negligible weight winded around it as shown in the figure from below. By how much a force must be applied to pull the wire so that the cylinder spins and keeps in place?. Assume that the coefficient of friction is $0.3$ and the acceleration due gravity is $9.8\,\frac{m}{s^2}$

The alternatives given in my book are as follows:

$\begin{array}{ll}

1.&\textrm{3 N}\\

2.&\textrm{2.5N}\\

3.&\textrm{1.5N}\\

4.&\textrm{0.15N}\\

5.&\textrm{4.5N }\\

\end{array}$

I'm not sure exactly if I'm understanding this problem correctly. What I've attempted to do here was to assume that the condition which must be met is given by:

$\sum ^n_{i=1}\tau_{i}=0$

Therefore:

$-F\cos 37^{\circ}\cdot R-F\sin 37^{\circ}\cdot R + f_R\cdot R = 0$

Hence:

$f_R=F\cos 37^{\circ}+F\sin 37^{\circ}=F\frac{4}{5}+F\frac{3}{5}$

$f_R=\frac{7}{5}F$

But:

$f_R=\mu N$

$N=mg-F\sin 37^{\circ}$

$f_R=\frac{3}{10}(mg-F\sin 37^{\circ})=\frac{3}{10}\left(0.5\times 9.8-\frac{3F}{5}\right)$

Solving this thing yield:

$\frac{3}{10}\left(0.5\times 9.8-\frac{3F}{5}\right)=\frac{7}{5}F$

Hence:

$F=0.93\,N$

But it doesn't check with any of the alternatives given. What did I missunderstood?. Can someone help me here please?.