# Rack of blocks held horizontally under pressure

silverrahul
One more point, one of the answers seems to be getting confused and making it unnecessarily more complicated by considering the coefficient of friction between block and hand . There is no need to have an additional complication of a human hand. I think this confusion came about by my use of the word "held"
There is a rack of blocks held horizontally by applying equal and opposite pressure at both ends.
I did not mean held by hand. I just meant held by two walls or pistons etc. which are pushing against the two ends and have the exact same surface as the blocks, so the same coefficient of friction applies between the surface of the end block and this wall/piston

silverrahul
In particular, the support force from the left hand piston and from the right hand piston need not be divided 50/50.
How is this possible ? Shouldnt it be always 50/50 ? Given that the structure is in a stable condition , everything is at rest. So, no matter , additional complications of bending, deformation, torque etc. shouldn't we be able to conclude that the two pistons exert idential forces ? Otherwise, the system would not stay in rest along the horizontal direction.
What am i missing here ? Where am i going wrong ?

Mentor
I am asking you to share what your intuition tells you would happen.
My intuition is that the system is probably statically indeterminate and will need math to analyze in any more detail.

How is this possible ? Shouldnt it be always 50/50 ? Given that the structure is in a stable condition , everything is at rest. So, no matter , additional complications of bending, deformation, torque etc. shouldn't we be able to conclude that the two pistons exert idential forces ? Otherwise, the system would not stay in rest along the horizontal direction.
What am i missing here ? Where am i going wrong ?
The support forces he is talking about are vertical, not horizontal. The vertical forces on the left and right need not be the same.

Homework Helper
How is this possible ? Shouldnt it be always 50/50 ? Given that the structure is in a stable condition , everything is at rest. So, no matter , additional complications of bending, deformation, torque etc. shouldn't we be able to conclude that the two pistons exert idential forces ? Otherwise, the system would not stay in rest along the horizontal direction.
What am i missing here ? Where am i going wrong ?
If the two ends can apply torque, you can have a shear force on the bridge. It is tempting to make the right and left hand interfaces into point supports, perhaps at the midpoint of those faces.

Dale
Mentor
If the two ends can apply torque, you can have a shear force on the bridge. It is tempting to make the right and left hand interfaces into point supports, perhaps at the midpoint of those faces.
Oh, that is even more general than I had thought.

Richard R Richard
I redevelop the ideal case,

• ## n ## blocks
• All of the same mass ## m ##
• All cubic of side ## L ##
• All of friction coefficient ## \mu_s ##
• The force of normally holding ## N ## in directions opposite to compression.
• ## F_R ## friction force on the right side
• ## F_L ## friction force on the left side
• ## P ## weight of an object whose CM is placed at a distance ## x ## from the left side taken as the origin of the reference system.

By Newton's second law

$$F_L+F_R=nmg+P$$

The sum of moments at the origin.

$$\displaystyle \sum M_L = 0 = mgL \dfrac {n ^ 2} {2} -nL \, F_R + P \, x$$

Later

$$F_R = \dfrac {mgL \dfrac {n ^ 2} {2} + P \, x} {nL}$$

Let's find the maximum by deriving ...

$$\dfrac {\partial F_R} {\partial x} = \dfrac {P} {nL} \neq 0 \quad \to \quad \nexists$$

But let's force the situation

$$\left \{\begin {array} {ccc} x & = 0 \quad \to \quad F_R & = \dfrac {mgn} {2} \quad \to \quad F_L & = \dfrac {mgn} {2} + P \\ x & = nL \quad \to \quad F_R & = \dfrac {mgn} {2} + P \quad \to \quad F_L & = \dfrac {mgn} {2} \\ x & = \dfrac {nL} {2} \quad \to \quad F_R & = \dfrac {mgn + P} {2} \quad \to \quad F_L & = \dfrac {mgn + P} {2} \end {array} \right.$$

the maximum values of ## F_R ## and ## F_L ## are obtained at the ends when the load is placed at the same (*edited Thanks @A.T. ) end.

If the system fails, it will do so when the load is placed at one end and the external force is

$$N <\dfrac { \dfrac {mgn} {2} + P} {\mu_s}$$

or when

$$P>\mu_sN-\dfrac {mgn} {2}$$

where ##x=0## or ##x=nL##

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Arjan82 and silverrahul
silverrahul
the maximum values of FR and FL are obtained at the ends when the load is placed at the opposite end.

If the system fails, it will do so when the load is placed at one end
Like i said before, If the result in case of real life is same as result in case of simplified assumptions, then i feel comfortable with making those assumptions. If the real life case is the complete opposite, then i feel we need to ditch as many assumptions as are required and reasonable so as to make it consistent with real life results.

So, since you have taken the simplified ideal case, you found out that the center block will be the strongest, right ? So, are you claiming that this will be what happens in real life too ? i.e. center block will be the strongest ? If the whole thing were on top of lava, and you had to stand on one block, you would choose to stand on the center block ?

the maximum values of ## F_R ## and ## F_L ## are obtained at the ends when the load is placed at the opposite end.
Why on the opposite end? Your math says that if the load is placed at the left end (##x = 0##) then ## F_L ## is maximal. Which makes perfect sense.

So, are you claiming that this will be what happens in real life too ? i.e. center block will be the strongest ?
If the blocks were indestructible and perfectly rigid, then the center block could carry most extra weight. But real life can be arbitrarily more complicated than that.

Real life blocks can deform, as shown below. Here the center block becomes a wedge, that is being squeezed out downwards by the normal forces acting on its sides (adding to the downwards force on it), while the outer blocks are squeezed out upwards (against gravity).

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silverrahul
silverrahul
But real life can be arbitrarily more complicated than that.
I agree that real life can be more complicated. My question is : Is the real life complicated enough to change the conclusion from what Richard concluded to its opposite, i.e In real life, will center block be weakest , rather than the answer reached by Richard that center block will be strongest ?

If so, then i think we have to consider the deformation etc. and how it affects the free body diagram . If not, then i think we can live with taking the idealised case

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Richard R Richard
I agree that real life can be more complicated. My question is , Is the real life complicated enough to change the conclusion to its opposite.
Yes, but there are many dependencies to say it this way. In some cases, yes, in others, no.

My question is , Is the real life complicated enough to change the conclusion to its opposite. i.e center block will be weakest , rather than the answer reached by Richard that center block will be strongest ?
For some parameter combination, possibly.

silverrahul
In some cases, yes, in others, no.

For some parameter combination, possibly.

Is it possible to get some understanding of what kind of cases, would reverse the conclusion and make center blocks the weakest ? I mean is it an unlikely extreme scenario where this would happen ? Or is it rather more common to have center block be the weakest ?

Mentor
Is the real life complicated enough to change the conclusion to its opposite. i.e center block will be weakest , rather than the answer reached by Richard that center block will be strongest ?
The issue has nothing to do with changing the intuitive conclusion. If the simple model has no solutions or one solution then you can use the simplified model. If there are multiple solutions then you must use a more complicated model. Only actually doing the math can tell you that.

silverrahul
The issue has nothing to do with changing the intuitive conclusion
Okay, i think you misunderstood what i meant. When i said "Is the real life complicated enough to change the conclusion to its opposite " , i did not mean the intuitive conclusion. I meant the conclusion reached by Richard in his answer where he takes the idealised case. I thought it was obvious from the second part of the sentence "i.e. center block will be weakest , rather than the answer reached by Richard that center block will be strongest ?" I will edit the sentence to make it more explicit what i meant.

Homework Helper
Gold Member
2022 Award
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.
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
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
Gold Member
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.