How does ground interact with a box rotating around its corner? I have some questions about how forces $F$, friction force $F_{friction}$, and normal $N$ interact in order to the box below to turn (clockwise) about the red corner, without sliding. The force $F$ is acting horizontally on the center of gravity (CG) of the box:

My attempt to understand this scenario:
There must be a non-zero clockwise torque in order to the box to turn. And I want to analyze the rotation about the CG, not about the red corner.
There are two forces which can affect the torque around the CG: $F_{friction}$ and normal force $N$. 
Let the net torque about the CG be:
$$\tau_{net} = \tau_{friction} + \tau_{N}$$
My first question: Should I expect the box start to rotate (clockwise) only when $\tau_{net}$ is non-zero and clockwise?
My further analysis:
When friction $F_{friction}$ acts on the red corner, it contributes to the net linear acceleration of CG, but, when, summed to $F$, it yields zero net linear acceleration. That means that the CG would remain static.
When torque $\tau_{net}$ is clockwise, the red corner acquires an angular acceleration about the center of gravity. Disregarding the ground for a moment, the position and orientation of the box on a next instant would be the green cube below, on the left. The green arrow denotes the infinitesimal displacement of the red corner - a bit downwards, a bit leftwards):

Now comes the part that puzzles me: what happens next? Obviously, the ground will not allow the box to have such displacement - it will prevent the box to "penetrate". But how?
My first thought about it is that, once the red corner starts to accelerate to penetrate the ground, the ground will increase the normal force $N$, thus increasing the (counter-clockwise) torque of force $N$, keeping the red corner on the ground. The problem with this approach is that the box would never start to rotate, as the torque of $N$ would be always able to counter-torque the torque of $F_{friction}$!
So, my second question: how does the ground acts to prevent the red corner to penetrate it, and at the same time allow the box to rotate?
Maybe I am analyzing using the wrong frames of reference. I'm stuck on this point.
My third question: as the square turns clockwise without sliding, does the friction force $F_{friction}$, in this case, always have the same magnitude as force $F$? Well, at first sight, I think it does, but I'm somewhat uncomfortable with it: I'm wonder if the net torque $\tau_{net}$ is also able to slide the red corner on the ground.
My fourth question: as we've seen above, the net linear acceleration of the CG is zero. I would conclude from that that the CG is static - but it is moving anyways. It describes an arc of a circle as the box turns around the red corner. How does it make sense? Should I look it from a different frame of reference?
Any help on this will be very appreciated.
Thanks!
 A: First lets give the box some dimensions: $w$ for width and $h$ for height.
A level, stationary box
The scenario you described above is almost correct if $F\,h\leq m\,g\,w$.
The net torque on the box would be $F\frac{h}2 - N\frac{w}2$, which if $N=m\,g$ would result in a net counter clockwise (negative in my chosen reference frame) torque. This would mean the corner not in red would be pushed into the floor, supporting some of the weight of the box, reducing $N$ until $F\frac{h}2 - N\frac{w}2 = 0$.
An accelerating box
If there is a net torque on the box (i.e. $F\frac{h}2 - N\frac{w}2 \geq 0$) then the box will be accelerating around the red corner. Note that to accelerate around the red corner would accelerate the center of mass. Since it's accelerating we can no longer claim that the net forces are zero. In particular now $F\gt F_\text{friction}$ and $N\gt m\,g$.
So Your first and second questions are answered by the fact that yes $N$ will increase as the box starts rotating, but that increase will allow it to overpower the weight of the box allowing the center of mass to accelerate upwards. Once the center of mass has moved upwards, there is now room for the corner to rotate without penetrating the ground.
As for your third question. No $F\neq F_\text{friction}$ once the box starts accelerating. $F_\text{friction}$ will reduce once the box starts rotating.
For your forth question, the net linear acceleration of the CG is not zero. You are correct that the CG describes an arc of a circle.
If you would like to calculate these values I would proceed as follows:
The moment of inertia of a box about its corner is
$$I=m\frac{w^2+h^2}3 \, .$$
The angular equivalent of $F=ma$ is
$$\tau=I\,\alpha=I\,\dot\omega=I\,\ddot\theta \, .$$
Now before we looked at the net torque about the center of gravity. While it's possible to solve this problem using that origin, choosing the red dot as our center allows us to skip a few steps:
\begin{align}
\tau &= F\frac{h\,\cos(\theta)+w\,\sin(\theta)}2-m\,g\frac{w\,\cos(\theta)-h\,\sin(\theta)}2 \\
\ddot\theta &= F\frac{h\,\cos(\theta)+w\,\sin(\theta)}{2I}-m\,g\frac{w\,\cos(\theta)-h\,\sin(\theta)}{2I} \, .
\end{align}
Unfortunately, this is equivalent to the large oscillation pendulum problem by a rotation of coordinate system. As such there is no analytic solution, but a numerical solution could get you $\theta(t)$.
Of course this solution would only be valid up to the point where friction would give way $F_\text{friction}>\mu_\text{static}N$.
