rotational-dynamicsstaticstorque
Consider this problem, which ask for the value of the two normal forces (friction is negligible).
Easy, right? You calculate the moments around G, the center of mass, and equate them to 0 (the object slides without rolling), and get the left and right normal forces to be 359N and 664N, respectively. But why do you have to calculate the moment around the center of mass?
As it turns out, if I calculate the moment around any other points and equate it to 0, the result will turn out wrong.
First lets give the box some dimensions: $w$ for width and $h$ for height.
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 $F\frac{h}2 - N\frac{w}2 = 0$.
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 $F_\text{friction}>\mu_\text{static}N$.
In these cases it always helps to draw a diagram:
The green vectors represent the force of gravity $w=mg$ (dashed) and its components along the inclined plane and perpendicular to it. The red forces are the normal force of the plane on the ball $n$, the force of friction $F$, and their vector sum (dashed).
Now the sphere rotates about the contact point - that is the point that doesn't move. In that frame of reference, noting that the red vectors all pass through the center of rotation we compute the torque as the force of gravity $w$ times the perpendicular distance to the pivot point $d= r\sin\theta$, i.e. $$\Gamma = w\cdot r \sin\theta$$ and we consider the moment of inertia of the ball about this pivot to be $$I = \frac25 mr^2 + mr^2=\frac75 mr^2$$ (by the parallel axes theorem).
As you pointed out, by considering the motion about the contact point, the value of $F$ doesn't seem to come into play. But remember that the center of mass of the sphere must accelerate as though all forces are acting on it; after canceling out the normal forces, that leaves us with $mg\sin\theta$ down the slope, and $F$ going the other way. The difference between these two forces gives rise to the acceleration of the sphere's c.o.m. so we can compute $F$ from
$$mg \sin\theta - F = m a$$
To compute $a$, we first need the angular acceleration $\dot\omega$which is found from
$$\dot \omega = \frac{\Gamma}{I} = \frac{mgr\sin\theta}{\frac75 m r^2} = \frac{5g\sin\theta}{7r}$$
The linear acceleration $a$ is of course the angular acceleration multiplied by the radius of the sphere, so
$$a = \frac57 g\sin\theta$$
From which it follows that
$$F = \frac{2}{7} m g \sin \theta$$
And if we know that, we can now compute the angular acceleration of the sphere about its center. The torque seen in the frame of reference of the sphere is
$$\Gamma' = Fr = \frac{2}{7} m g r \sin\theta$$
Now we use the moment of inertia of the sphere about its center in order to compute the angular acceleration, and find
$$\dot \omega = \frac{\Gamma'}{\frac25 mr^2} \\ = \frac{\frac{2}{7} m g r \sin\theta}{\frac{2}{5} m r^2}\\ =\frac{5 g \sin\theta}{7 r}$$
which is the same result as before.
So there is no contradiction. The forces of friction and gravity work together to cause the rotation - the difference in apparent torque comes about from the fact that you are working in different (and non-inertial) frames of reference, but if you do the calculation carefully you get the same answer.
Best Answer
Because the system is in equilibrium only in the CM frame. If you calculate the moment about any other point, you need to consider pseudo force as well, viz., 133.4N pointing to the left passing through the CM. Then you will get the correct answer again.
If you do not want to work in the CM frame, then you should notice that the angular momentum is
$${\bf L}={\bf r} _{CM} \times M{\bf v}_{CM}+{\bf L}_{CM}$$
Since the object is not rotating,
$${\bf L}_{CM}={\bf 0}$$
However, it is accelerating and hence ${\bf r}_{CM} \times M{\bf v}_{CM}$ is not a constant in general (unless ${\bf r}_{CM} // {\bf v}_{CM}$), and therefore, you cannot equate the moment to $0$.