Rack of blocks held horizontally under pressure

[silverrahul]

You are getting sidetracked. Instead of badgering me to agree with you about intuition, just buckle down and write the equations. If you trust intuition more then you can use it to check your math, but you cannot use it to avoid the math altogether.

I am not badgering you to agree to anything. You are free to disagree with everything and anything. I am asking you to share what your intuition tells you would happen. Me asking for your intuition in no way implies that i am putting intuition on a higher pedestal. I am also fully agreeing that what the math tells us will happen can be counter to intuition. The math and the intuition are 2 completely different issues. And of course, the ultimate judge of which is correct is experiment, i do not disagree with that

 

[sophiecentaur]

I do not have to define anything

Well, someone has to. We should be aiming at the very simplest model which would rely just on friction to start with and no deformation.

I am asking you to share what your intuition tells you would happen.

Frankly, I don't see where intuition should come into this at all. Intuition is strictly personal and always relies on past learning, which is different from person to person. The whole idea of Science is to try to eliminate intuition so we can all agree where an agreed set of input parameters produce an agreed output result.

 

[silverrahul]

I agree with you,

If the OP confirms all our assumptions, we can dedicate them to look for a solution and draw some conclusions.

Okay, i might have asked a question that seemed simple to me at first glance but looking at all the different answers, i guess it is much more complicated. I am grateful to everyone who tried to answer. I am sorry that i had not watched this space recently, and so could not answer some questions that i think were directed towards me.

After reading the responses i realize , i could have been more precise, but in my defense i had not thought this would be as complicated.
So, i would just like to explain what was my intention with this question and the motivation behind it. So,
i just wanted to know what would happen in a real life case. I was playing with a kid and he was holding a few blocks in his hand by pressing them between his palms and asked me to put another toy on the blocks . When i did it a few times, i felt that the middle blocks could support less weight than the blocks at the 2 ends ( which honestly was my intuition as well ). Obviously, this was far from a controlled experiment. We don't know if the force applied through palms was uniform and constant in each case. The result may be different based on different coefficients of friction , different weights etc.
So, i am just wondering can we model what would happen in real life through the torque and 2nd law equations. I was just considering everything in terms of "spherical cows ". And obviously, real life considerations might change things.
But , will they change them drastically , as to give completely opposite results i.e. central blocks weakest vs central blocks strongest. If they give similar results, then i want to take the idealised case. I do not care if the real case might support a few more Newtons of force, due to some other considerations of deformation etc.
But , if the real life case , gives completely opposite results compared to the idealised case, then of course i do not want to consider the idealised case.
For example, while studying projectile motion, we make so many approximations, but it is okay, because the real life results are approximately the same. But, if real life considerations , completely change the results, then we should consider the real life factors.
So, to summarise my long spiel, My simplified assumption was incompressible , perfect cubes, no deformation, bending etc. Constant and equal horizontal force. and any other idealised assumption required.
I am assuming the real life results would be either of these 2 . Is there any other alternative ?
1) central block is strongest and it becomes weaker the farther we go towards the ends
2) central block is weakest and it becomes stronger the farther we go towards the ends

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.
And also if you want to say it will be (1) for a certain range of coefficient or a certain range of weights etc. and it will be (2) otherwise, i would be interested to see how you reached that conclusion as well.

 

[silverrahul]

Frankly, I don't see where intuition should come into this at all. Intuition is strictly personal and always relies on past learning, which is different from person to person. The whole idea of Science is to try to eliminate intuition so we can all agree where an agreed set of input parameters produce an agreed output result.

Okay, when asking about intuition, i do not mean we should reject the maths JUST BECAUSE it disagrees with intuition. I am interested in intuition , because whenever the maths disagrees with intuition , it is very interesting to sometimes consider why our intuition is different. Most of the times, when the math disagrees with our intuition it is a great learning experience to figure out, through the maths, why our intuition failed us. So, i do not think we need to shy away from admitting what our intuition tells us. I freely admit, that my intuition here tells me that the central block would be the weakest and i am also well aware, that intuitions can be false. That is why i want to see what the maths tells us would happen in real life. That is all i meant when i was asking about the intuition. I am in no way doubting the maths and physics JUST BECAUSE it may go against our daily life intuition.You talked about intuition being different from person to person . That was kind of what i was curious about , to be frank, when i asked about intuition. I just wanted to know, if this was just my intuition or others shared this intuition as well ( irrespective of what the truth is which obviously is not beholden to our intuition and can be opposite of it ). If others shared this same intuition, and this intuition indeed turned out to be wrong, i think looking at the maths would tell us why our intuition failed, which i think is a good learning. So, if anyone feels that i am doubting the maths or physics involved because it runs counter to intuition, i want to say that was not what i meant when asking about intuition.

 

[silverrahul]

Also, i agree with the following very much

We should be aiming at the very simplest model which would rely just on friction to start with and no deformation.

We should aim for the simplest model . If the conclusions from the simplest model are drastically inconsistent with the reality, then we should include as many real life complications as are reasonable that gives a conclusion consistent with reality,

 

[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 ?

 

[Dale]

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.

 

[jbriggs444]

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]

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 $x=nL$

 

[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 ?

 

[A.T.]

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.

 

[A.T.]

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).

 

[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

 

[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.

 

[A.T.]

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 ?

 

[Dale]

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.

 

[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.