Rack of blocks held horizontally under pressure

[haruspex]

The safe way to approach any idealised problem, such as inelastic strings, is to start with at least an approximation of the general case then take the limit.
For simplicity, I'll just analyse the case where the blocks are massless, and treat them as a continuum. So we just have a point loaded beam, except that its resistance to shear comes only from friction.

Let the weight W be placed at distance S, at least half way along a beam of length L and depth H. Say the horizontal force F through the beam was negligible before W was applied, but the ends of the beam are kept fixed.
The weight creates shear forces and torques in the beam. The torques are balanced by the effective line of action of the horizontal force changing along the beam. This comes about because each vertical slice through the beam of thickness dx (a block, in the original problem) gets rotated a fraction by the torque from the shear forces.
If we assume the blocks are sufficiently rigid and sufficiently tall that they cannot rotate completely out of position then this line of action will be at the base of the beam at each end and at the top of the beam at S.
This leads to $FH=W\frac{(L-S)S}L$.
If nothing slips then $\mu F$ must exceed the maximum shear force, i.e. $W\frac SL$ (since I chose S>=L-S).
W disappears, leaving $\mu\geq \frac{H}{L-S}$.
This is minimised at S=L/2.

Note that the compression force depends upon the placement of the weight, so taking that force to be a given could lead to the wrong answer.

Reintroducing a mass for the blocks leads to a piecewise parabolic line of action for the compression force.

 

[Richard R Richard]

The factors that can influence the change of the conclusion, when going from an ideal model to a real one, I summarize them in the following

The friction force between blocks is not a constant, it depends on the value of the static coefficient of friction between two contiguous blocks, and it does not have to be a constant, the materials, although they may have similar characteristics, when the number of blocks increases. it is more likely that one surface does not have the same coefficients as the others

$$\mu_{si} = \mu \pm \Delta \mu s$$

Similarly, internal resistance to deformation, related to Young's modulus, can also vary from block to block.

$$E_i = E \pm \Delta E$$

The consequence is not a perfect parabolic deformation. So the internal stresses vary from block to block, and can then collapse into some weakness.

The wedge shape of the intermediate blocks are prone to slip due to projections of the compression force N in the direction of the cut between blocks.

The shape of the weight $P$, the deformation on the blocks, depends on their shape, a large flat base will not be the same as a small one, that is, the density of P influences the deformation produced.

The regular stacking, when we place a block next to the other, they will not necessarily be aligned vertex to vertex, so there will be an alignment error that reduces the area where the normal tension of the blocks is transmitted.

$$A_i = A \pm \Delta A \quad \to \quad \sigma_i = \dfrac {N} {A_i} \quad \to \quad \sigma_i = \sigma \pm \Delta \sigma$$

The combination of these factors between the faces of the same pair of blocks, located in any position can create a weakness in the structure, and resist less there than at the ends.

Sure there are mor factors that i don't see in this moment

 

[256bits]

One more,
End blocks can 'dig in' ( at a corner due to the moment ) to the end surfaces due to deformation of the surface, reducing their tendency to slip along the surface.