The problem with attempting to do the analysis with the forward point of contact on the box when it is sliding is that the box is accelerating. This makes a non-inertial frame and there's more moving parts. Besides the force of gravity on the center of mass, there will be fictitious forces.

First, lets assume friction is zero. If so, we can calculate the acceleration of the box.
$$F_{x'} = mg\sin \theta$$
$$a_{x'} = g\sin \theta$$
In an accelerating frame, we get fictitious forces.
$$ F_{fict} = -ma_{frame}$$
$$ F_{fict} = -mg\sin \theta$$
The sign indicates the force is opposite the direction of the acceleration. The case where we expect the box to tip most likely is where it is tall and not wide. Let's assume the width is minimal (a rod sliding on its end). If so, the torque from gravity will be the COM at half the height, and leaning forward $\sin \theta$
$$\tau_g = \frac12 hmg \sin \theta$$
Meanwhile the fictitious force is located at half the height and points opposite the acceleration vector, so it has a lever arm of exactly half the height.
$$\tau_{fict} = F_{fict} d$$
$$\tau_{fict} = -\frac12 hmg\sin \theta$$

$$ \tau_{net} = \tau_{g} + \tau_{fict}$$
$$ \tau_{net} = \frac12 hmg \sin \theta - \frac12hmg \sin \theta$$
$$ \tau_{net} = 0$$

If you add friction, you will reduce the acceleration of the box, reducing the fictitious force in this frame. When that happens, the net torque would be sufficient to tip a rod, but would depend on the dimensions if a box would tip over.

I didn't do the math, but $\tau_{fict}$ should be pretty simple to calculate, and then $\tau_{g}$ becomes $\frac12h\sin \theta - \frac12w\cos \theta$ (I think)...

Sorry, I wanted to reply to this comment, and I couldn't do it in the comment space.

If we instead take the static plane to be the frame of reference (and
a minimal width box), wouldn't we get the same results anyways?

The problem is that in the frame of the plane, the axis you want to consider in the question (the front corner of the box) is accelerating. I don't think that's okay to do. Let's imagine a pendulum being accelerated via a force on the axis. And we let it reach steady state so the angle isn't changing.

The solid pendulum has a force accelerating it to the right, and it has gravity pulling it down. The force is applied directly at the axis and contributes no torque. Gravity pulls the weight down, so is applying a torque of $(mg \times L\sin \theta)$. Yet the pendulum does not rotate. Perhaps someone else has more information about this situation. My intuition is that this is simply the wrong technique for an accelerating axis, but I'm not sure what the correct analysis would be.

## Best Answer

It makes the problem easier to visualize if you collapse to 2 dimensions and imagine a circle in the same situation with homogeneous distribution of mass effectively centered at the center of the circle. Constrained initially by contact with the table's edge, and after your nearly forceless nudge the center will begin to accelerate through an angle under its own weight. Then when the rotation reaches 90 degrees the circle becomes a free falling body with fixed angular velocity that was imparted when it lost contact with the table - at 90 degrees.