I will take a stab, though it needs a solid state physics background, in which I am weak.
a) in the case of crystals, as the two surfaces touch the crystal symmetry will take over with maybe some dislocation plane. i.e. the van der Waals forces will make the two one and you will not be able to easily separate them.
b) if amorphous,
1) inorganic example glass on glass, the attraction is not strong enough to create one body. The two will be separable.
2) organic surfaces with strong static fields example two cling film layers ,would adhere as one body, depending on the material .
etc.
So it will depend on the specific materials under study.
In mechanics no. Torque is not a fundamental quantity. it's only job is to describe where in space a force is acting through (the line of action). Torque just describes a force at a distance. Given a force $\boldsymbol{F}$ and a torque $\boldsymbol{\tau}$ you can tell that the force acts along a line in space with direction defined by $\boldsymbol{F}$, but location defined by $\boldsymbol{\tau}$ as follows $$ \boldsymbol{r} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau} }{ \| \boldsymbol{F} \|^2 } $$
In fact, you can slide the force vector anywhere along its line and it won't change the problem, so the $\boldsymbol{r}$ calculated above happens to be the point on the line closest to the origin.
It might be easier to discuss angular momentum first, since torque is the time derivative of angular momentum, just as force is the time derivative of linear momentum.
For a single particle with linear momentum $\boldsymbol{p} = m\boldsymbol{v}$ located at some instant at a point $\boldsymbol{r}$ the angular momentum is $$ \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$$
So where is the momentum line in space? The momentum line is called the axis of percussion. It is located at
$$ \boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ \| \boldsymbol{p} \|^2 } = \frac{\boldsymbol{p} \times ( \boldsymbol{r} \times \boldsymbol{p})}{\| \boldsymbol{p} \|^2} = \frac{ \boldsymbol{r} (\boldsymbol{p} \cdot \boldsymbol{p}) - \boldsymbol{p} ( \boldsymbol{p} \cdot \boldsymbol{r}) }{\| \boldsymbol{p} \|^2} = \boldsymbol{r} \frac{ \| \boldsymbol{p} \|^2}{\| \boldsymbol{p} \|^2} = \boldsymbol{r} \; \checkmark $$
provided that the point $\boldsymbol{r}$ is perpendicular to the momentum $\boldsymbol{p}$. Let me elaborate. Imagine the direction of the line being $\boldsymbol{\hat{e}} = \boldsymbol{p} / \| \boldsymbol{p} \|$, and consider a point $\boldsymbol{r} + t \boldsymbol{\hat{e}}$ for some arbitrary scalar $t$. The angular momentum will be $\boldsymbol{L} = ( \boldsymbol{r} + t \boldsymbol{\hat{e}}) \times \boldsymbol{p} = \boldsymbol{r} \times \boldsymbol{p} $. So where along the line (the value of $t$) doesn't matter. Finally, if $\boldsymbol{r}$ is not perpendicular to $\boldsymbol{p}$ you can always find a value of $t$ that makes the point perpendicular. Set $t = -(\boldsymbol{r} \cdot \boldsymbol{p}) / \| \boldsymbol{p} \|$ and the point will be perpendicular.
Such a point can always be found, and it is the point on the line closest to the origin.
The conservation law for angular momentum (coupled with the conservation law for linear momentum) just states that not only the magnitude and direction of momentum is conserved but also the line in space where moment acts through is also conserved. So not only which direction is momentum point, by where is space it exists.
To visualize this, consider a case where you want to remove the momentum of a freely rotating body that is moving in space. You have a hammer, and you need to find out the following in order to completely stop the body. a) how much momentum to hit it with (the magnitude), b) in which direction to swing (direction) and c) where to hit it (location).
In summary, the common quantities in mechanics are interpreted as follows
$$ \begin{array}{r|l|l}
\text{concept} & \text{value} & \text{moment}\\
\hline \text{rotation axis} & \text{rot. velocity}, \boldsymbol{\omega} & \text{velocity}, \boldsymbol{v} = \boldsymbol{r}\times \boldsymbol{\omega} \\
\text{line of action} & \text{force}, \boldsymbol{F} & \text{torque}, \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} \\
\text{axis of percussion} & \text{momentum}, \boldsymbol{p} & \text{ang. momentum}, \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}
\end{array} $$
The stuff under the value column are fundamental quantities that give us the magnitude of something (as well as the direction). The stuff under the moment column are secondary quantities that depend on where they are measured and give use the relative location of the fundamental quantities. Hence the terms torque = moment of force, velocity = moment of rotation and angular momentum = moment of momentum. All that means is that these quantities are $\boldsymbol{r} \times \text{(something fundamental)}$ and they describe the moment arm to this something.
The location of the line in space is always the same formula
$$ \text{(location)} = \frac{ \text{(value)} \times \text{(moment)}}{ \text{(magnitude)}^2} $$
where $\text{(magnitude)}$ is always the magnitude of the $\text{(value)}$ vector.
In statics, for example, we learn to balance forces and moments, which should be interpreted as balancing the force magnitude and the line of action of the force.
Best Answer
When two bodies come in close contact with each other, the irregularities on the surface touch each other and exert enormous pressure as the area of contact between surfaces is very small. The atoms of the objects come very close and start experiencing electrostatic force and in a way get cold welded to each other. When the objects try to move relative to each other, work has to be done to break these electrostatic cold welds. This is the origin of frictional force. It's worth mentioning that if we bring two perfectly smooth bodies in contact, they will exhibit very high coefficients of friction when moving relative to each other.
Tension in ropes is another example of electrostatic force.