This is what I did:
[tex]F_x\leq\mu_s N=\mu_s\left(mg+F\cos{\theta}\right)[/tex]
...and then I would simplify. But unfortunately, I was expecting the "F"s to cancel out, meaning the force didn't matter at that critical angle. They didn't obviously. Did I set this up correctly? If not, where did I go wrong?
Thanks, I appreciate it.
Alex

Is the normal component of F really in the same direction as mg? You need to check your coordinate system and make sure you are consistent with + and - directions.

It seems to be. The force F is broken down as follows:
[tex]F_x=\left<-F\sin{\theta},0\right>[/tex]
[tex]F_y=\left<0,-F\cos{\theta}\right>[/tex]
Now the normal force for Fy is -Fy:
[tex]N_{F_y}=\left<0,F\cos{\theta}\right>[/tex]
Adding this to the normal from gravity clearly gives:
[tex]N=mg+F\cos{\theta}[/tex]
I don't see where I'm going wrong :grumpy:
Thanks again,
Alex

Now I can solve for θ and come up with an inequality for that, but it still depends on F! I am confused as to why I still have an F here because it seems obvious that at that critical angle (or anything less), the applied force wouldn't make a difference; the mop would stay still. Is what I posted above correct, or have I gone wrong? I really appreciate it

By applying a vertical force only (θ=0) or if the force is less than or equal to the max. static friction. If it is less than or equal to the max. static friction, then the forces will cancel out and there won't be any x-force. ?

Sorry for being difficult here. I see what you're saying. So if the angle is not zero, there will always be a force that is greater than the static friction. Because of this, the only angle that fits this problem is θ=0. Is this what you're saying?

First of all, there is a sign mistake on the left side (like Cyclovenom pointed out). Gather all terms with F and theta on the left side and factor. Take a look at the left side and consider how to make the inequality valid (Hint: one possibility is that the left size is zero. How could that happen?)