The problem is as follows:
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?.
The module of the torque is $F·R·\sin\alpha$ being $\alpha$ the angle between the force and the arm and this is not what you wrote that it rather seems the decomposition of some force. So, the condition of rotation equilibrium is:
$$f_R·R-F·R=0$$
But this doesn't matter because one of your implicit assumptions is wrong. You assumed that it is a problem of statics, but it isn't. If the (center of mass of the) cylinder has to be in place, it has to have an angular acceleration around its center of mass: the above equations doesn't hold.
By Hypothesis, the center of mass is not moving, thus, the total force along an horizontal axis is zero:
$$F\cos 37º-f_R=0$$
Then
$$f_R=\frac{4F}{5}$$
that, with the equation you got,
$$f_R=\frac{3}{10}\left(0.5\times 9.8-\frac{3F}{5}\right)$$
Gives $F=1.5$, approx.
We can get the angular acceleration from here:
$$I\alpha=F·R-f_R·R$$
Being $\alpha$ the angular acceleration and $I$ the moment of inertia of the cylinder around its symmetry axis.