Short Answer:
For the ideal scenario, the path the cuboid takes is independent of friction. The primary way your desk experiments differ from the ideal scenario is your assumption that the applied force is constant (namely in direction).
Long Answer:
Let's lay out some assumptions for the ideal scenario:
- The cuboid is uniform in mass.
- The cuboid has a uniform coefficient of friction (no 'sticky' spots).
- The force applied to the cuboid is truly constant.
There is a common assumption about frictional forces as well:
- Frictional forces are always opposite in direction to motion (i.e. $ \hat f_k = - \hat v $ )
Claim 1:
Under assumptions 1 and 3 above, the cuboids path will be a combination of rotation about the center and translation of the center.
With a single force $\vec F$ applied to the cuboid, Newton's 2nd law will non-zero acceleration of the center of mass and non-zero rotational acceleration about the center of mass (provided $\vec F$ is offset from the CM).
$$
\begin{array}{|l|cc|}
\hline
& \text{Translation} & \text{Rotation} \\
\hline
\text{No Friction} & \vec F = m \vec a & \vec R \times \vec F = I
\vec \alpha \\
\hline
\end{array}
$$
Claim 2:
Under assumption 2 and 4 above, the path the cuboid takes is independent of friction.
$$
\begin{array}{|l|cc|}
\hline
& \text{Translation} & \text{Rotation} \\
\hline
\text{Friction} & \vec F + \vec f_k = m \vec a & \vec R \times \vec F + \vec \tau_k= I \vec \alpha \\
\hline
\end{array}
$$
This can be understood intuitively from assumption 4: if friction is always a drag force (i.e. directed behind you) it never has a component to the left or right. Consequently, it only speeds you up or slows you down; it doesn't change the path you take. A mathematical proof is given below.
If friction can't change the path the cuboid takes, it still does the same amount of rotation and translation regardless of the coefficient of friction.
Problems with real world examples
Assumption 3 is the weakest assumption when applied to experiments. Take for example pushing a book's corner with a finger. In that case, the book likely only experiences a contact (a.k.a Normal) force perpendicular to the face:
To truly apply a constant force, you would need some sort of linkage. For instance, a push-rod that connects to a pin in the cuboid. Otherwise you are missing out on part of the applied force parallel to the face.
Without that linkage setup, the force will start performing translation of the cuboid, but then predominantly rotate as you get further into the path.
Proof frictional forces don't change the path an object takes:
Begin with Newton's law, but break the forces up into components parallel to the velocity and perpendicular to the velocity:
$$ \Sigma \vec F_{||v} + \Sigma \vec F_{\perp v} = m \frac{d \vec v}{dt} $$
Let velocity be written as $\vec v = v \hat v$ (i.e. magnitude and direction). Then:
$$ \frac{d \vec v}{dt} = \frac{dv}{dt} \hat v + v \frac{d\hat v}{dt}$$
Applying the expansion of $\frac{d\vec v}{dt}$ and taking the dot product with $\hat v$:
$$\Sigma \vec F_{||v} \cdot \hat v = m \frac{dv}{dt}\hat v \cdot \hat v + mv\frac{d \hat v}{dt} \cdot \hat v$$
The dot product $\vec F_{\perp v} \cdot \hat v = 0$ by definition, while $\frac{d \hat v}{dt} \cdot \hat v = 0$ is true because $\hat v \cdot \hat v = 1$ by definition. Thus:
$$\Sigma F_{||v} = m \frac{dv}{dt}$$
Shows that Forces parallel to v can only change the magnitude of the velocity, not it's direction.
