Minimum force required 

Below is a cylindrical drum of mass $m$. The coefficient of static friction of all contact surfaces is $u$. We need to find the minimum force required so that the drum is just about to spin.
I calculated the torques with respect to point $A$. Here $Fr$ and $f_kr$ rotate the drum counterclockwise and $mgr$ rotates it clockwise. Just at the moment of spinning, both the counterclockwise and clockwise torques must be equal,so $(F+f_k)r=mg$ or $F+f_k=mg$ or $F+umg=mg$ or $F=mg-umg$. Is my solution correct? If not i am hoping for a correct solution.
 A: Not going to answer your homework for you. But I will give several strong hints. This will show some of the method I used to solve it.
You can get a good idea about how correct your answer might be by considering the extreme possible values of $u$. If it is zero (no friction) then the drum should turn with zero force, and you have it turning with mg. Also, more friction should mean it requires more force, and you have it requiring less. And, is there a value of friction such that no possible force on the drum will make it rotate?
Which way does the force $F$ want to rotate the drum? Suppose friction were near to zero. Then which way would the drum rotate?
Friction will resist the motion of an object. So which way will the frictional forces be pointing if the drum is just moving?
When the drum is at rest, all forces on it must balance. This includes both friction forces and linear forces. So what is the contact force at the bottom? What is the contact force with the wall? And then what is the friction force at each location? And then, what are the equations that these must satisfy to be in balance?
And finally, what is the condition such that the drum is just at the point of moving?
