Rack of blocks held horizontally under pressure

  • #1
silverrahul
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Let us assume that there are 5 blocks , each of weight W. The blocks are held horizontally by applying a high enough force, ( say N ) to the leftmost and rightmost blocks.
Let us say the blocks are numbered 1 to 5 from left to right.
So, if i try to draw the free body diagrams, then block 3 will have upward friction force acting on it through each of its surfaces and this force on each surface should be equal to W/2.
By action -reaction principle , friction force on block 4 at the 3-4 surface would be a downward friction force of W/2 and hence friction on block 4 at the 4-5 surface would be an upward force of 3W/2.
Similarly, friction on block5 at 4-5 surface would be a downward force of 3W/2 and hence friction on rightmost surface of block5 would be an upward force of 5W/2.

So, by this logic we can see that the friction force between the blocks at the end is more than the friction between blocks at the middle. So , we know that the surface at the rightmost end will reach limiting friction before the surfaces near the middle. Also, if an additional weight M is put on block 3 , then by this chain of reasoning, the friction force at the rightmost block will increase by M/2.
However , if the additional weight is put on any of the other blocks to the right of block 3, then the friction force at the rightmost block will increase by M.

So, this leads to the somewhat counterintuitive conclusion that :
1. Block 3 i.e. the center most block is the block which can hold the highest amount of additional weight before the whole structure collapses.
2. Apart from Block 3, for all other blocks, they would be able to support the exact same amount of additional weight before the whole structure collapses.

Are these conclusions correct ?
 

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  • #2
A.T.
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However , if the additional weight is put on any of the other blocks to the right of block 3, then the friction force at the rightmost block will increase by M.
What about moment balance?
 
  • #3
PeroK
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So, this leads to the somewhat counterintuitive conclusion that :
1. Block 3 i.e. the center most block is the block which can hold the highest amount of additional weight before the whole structure collapses.
2. Apart from Block 3, for all other blocks, they would be able to support the exact same amount of additional weight before the whole structure collapses.

Are these conclusions correct ?
Friction between the blocks is not fundamentally different from the internal forces holding the blocks together. The contact points are effectively simply points of weakness in the structure.

It should be clear that weighting such a structure in the middle is harder to sustain than weighting it at either end, so your calculations must be wrong.
 
  • #4
silverrahul
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Friction between the blocks is not fundamentally different from the internal forces holding the blocks together. The contact points are effectively simply points of weakness in the structure.

It should be clear that weighting such a structure in the middle is harder to sustain than weighting it at either end, so your calculations must be wrong.
" It should be clear that weighting such a structure in the middle is harder to sustain than weighting it at either end "
This is exactly what my intuition also tells me, which is why i am confused.
Which is why i am asking where my calculations are wrong
 
  • #5
PeroK
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" It should be clear that weighting such a structure in the middle is harder to sustain than weighting it at either end "
This is exactly what my intuition also tells me, which is why i am confused.
Which is why i am asking where my calculations are wrong
It's the concept of torque that you are missing.
 
  • #6
silverrahul
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What about moment balance?
It's the concept of torque that you are missing.
I tried the whole thing again with moments around the left most end.
I got even more paradoxical results.
If i try to find the friction f between block 3 and 4 , then using moments approach, if length of each block is d and i try to consider the block 1, 2 and 3 as one body
then, using moments around the left end.
W x .5d + w x 1.5d + w x 2.5d = f x 3d
i.e f = 1.5 W

But if friction between 3 and 4 is 1.5w, then friction between 3 and 2 will also be 1.5 w.
Then net upwards force on block 3 becomes 3w , which is absurd because net upward force on block 3 must be w.
 
  • #7
A.T.
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I tried the whole thing again with moments around the left most end. I got even more paradoxical results.
You have to balance both: forces and moments.
 
  • #8
silverrahul
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You have to balance both: forces and moments.
Yes, i am not able to understand how to do this. balancing forces gives one result. Balancing moments gives another result. I do not understand how to do both
 
  • #9
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Yes, i am not able to understand how to do this. balancing forces gives one result. Balancing moments gives another result. I do not understand how to do both
You have to set up a system of simultaneous equations to solve. The system may have no solutions, one solution, or many.
 
  • #10
silverrahul
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How can it have many solutions or no solutions ? Given a real situation, there will be definite values for each of the forces , right ?
 
  • #11
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How can it have many solutions or no solutions ? Given a real situation, there will be definite values for each of the forces , right ?
No solutions means the system will collapse. Many solutions means that you have to consider the deformation of the material to determine the forces.

The first step is to write the system of equations
 
  • #12
silverrahul
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Okay, i am considering the case of high enough coefficients of friction and high enough horizontal reaction force and low enough weights that system does not collapse. Think of you pushing the blocks with your 2 palms on the 2 ends
Regarding, the deformation, i was assuming no deformation.
Can we not calculate assuming no deformation ? We should still get consistent results, right ?
 
  • #13
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Can we not calculate assuming no deformation ? We should still get consistent results, right ?
You won’t know until you set up the system of equations. It is not at all obvious to me one way or the other.
 
  • #14
silverrahul
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You won’t know until you set up the system of equations. It is not at all obvious to me one way or the other.
What about the intuitive conclusion that the central block would be able to support less weight than the end blocks.
" It should be clear that weighting such a structure in the middle is harder to sustain than weighting it at either end " as said by the user PeroK


Do you agree with this intuitive conclusion, or would you say this intuitive conclusion may not be true ? Can i use this as a "postulate" to check if my answers are correct ?
 
  • #15
PeroK
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What about the intuitive conclusion that the central block would be able to support less weight than the end blocks.
" It should be clear that weighting such a structure in the middle is harder to sustain than weighting it at either end " as said by the user PeroK


Do you agree with this intuitive conclusion, or would you say this intuitive conclusion may not be true ? Can i use this as a "postulate" to check if my answers are correct ?
It's not a postulate, but there is a greater torque on a pivot the further a mass is from the pivot. It's the whole structure that may become unstable if you load the middle. To balance everything, you need to consider forces with the ground where the structure is anchored.

In any case, this is perhaps not the simplest statics problem to start with. Especially if you are considering weighting the structure at different points with variable masses.
 
  • #16
silverrahul
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It's not a postulate, but there is a greater torque on a pivot the further a mass is from the pivot. It's the whole structure that may become unstable if you load the middle. To balance everything, you need to consider forces with the ground where the structure is anchored.

In any case, this is perhaps not the simplest statics problem to start with. Especially if you are considering weighting the structure at different points with variable masses.
Why do i need to consider the ground forces ? Surely, the blocks and the forces on the blocks can be considered as one system.
Although ,i can see why this system itself can be more complex than i initially thought. Especially , it seems i cannot reach a solution consistent with real life results, unless i include deformation etc.
Considering each block as rigid seems to not allow for any solution consistent with itself or with the real world
 
  • #17
PeroK
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Why do i need to consider the ground forces ? Surely, the blocks and the forces on the blocks can be considered as one system.
Your force and torques won't balance without the external forces and torques.
 
  • #18
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Do you agree with this intuitive conclusion, or would you say this intuitive conclusion may not be true ? Can i use this as a "postulate" to check if my answers are correct ?
Intuitive conclusions may not be correct. It is much better to set up the equations correctly and use them to check your intuition. Going the other way can cause problems. If you have a conflict between your intuition and your equations and both seem correct, then the only way to resolve it is through experiment.
 
  • #19
silverrahul
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Intuitive conclusions may not be correct. It is much better to set up the equations correctly and use them to check your intuition. Going the other way can cause problems. If you have a conflict between your intuition and your equations and both seem correct, then the only way to resolve it is through experiment.
I agree that the ultimate solution to judge between intuition and equations is experiment.
Either my intuition is wrong or equations are wrong. But since, my intuition is built up through daily life experience and i am not so confident in how correct my equations are i was trying to ask about the equations here. And surely, some intuitions can be trusted more than others . I mean , this is not like time dilation, length contraction etc, which happens at speeds which are not the domain of our intuition. This example, is a much more daily life example, hence i was willing to give more credibility to intuition here.
 
  • #20
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hence i was willing to give more credibility to intuition here.
Sure, but you asked my opinion and I give the math more credibility than the intuition. Plus, I can check your math, but not your intuition (except through experiment)
 
  • #21
silverrahul
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Yes, i would always prefer your opinion on the math first of all. But , i would also like to hear your intuition as well. With the full acceptance that intuition can be wrong. I do not disagree with that
 
  • #22
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Yes, i would always prefer your opinion on the math first of all. But , i would also like to hear your intuition as well. With the full acceptance that intuition can be wrong. I do not disagree with that
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.
 
  • #23
Richard R Richard
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Hello, if you hold between your hands ## n ## blocks without deformation only sustained by friction, the second Newton's law must be fulfilled for each block

For block 1 on the left

$$ Fr_{L} -m_1g-F_{12} = 0 $$

For block 2

$$ F_{12} -m_2g-F_{23} = 0 $$

For each intermediate block

$$ F_{(i-1) i} -m_ig-F_{i (i + 1)} = 0 $$

For block no right

$$ F_{(n-1) n} -m_ng + Fr_{R} = 0 $$



If we add the ##n## equations assuming that all masses ##m## are equal

$$ Fr_{L} -nmg + Fr_{R} = 0 $$

If we take moment on the left contact surface and assume that each block has a length ##L##

$$ \displaystyle \sum M = 0 = \sum \limits_{i = 1} ^ n mgL \dfrac {2i-1} {2} - nLFr_{R} $$

$$ 0 = mgL \dfrac {n ^ 2} {2} - nLFr_{R} $$

$$ Fr_{R} = \dfrac {mgL n ^ 2} {2nL} = \dfrac {nmg} {2} $$

In the same way we get that $$ Fr_{L} = \dfrac {nmg} {2} $$

But then we know that $$ | Fr_{L} | = | Fr_{R} | \leq \mu_sN $$

Where ##N## is the normal force of tightening the hands

You can see that none of the internal contact forces is greater than those of contact with the hands, that is, if the hand –block friction coefficient is the same as that of the block –block, then conclude that the system will fail if you reduce N below

$$ N <\dfrac {nmg} {2 \mu_s} $$

It will be released from either the left or the right hand, at one end.

Well so far the idealization of the calculation.

But what happens in reality? The static friction coefficient between hand - block is greater than block-block, so it will fail between the first and the second block, or between the last and the second to last.

But also if the blocks do not have the same roughness, the one in which

$$ F {(i-1) i}> \mu_{si} N $$

That is, when the static friction cannot support the weight of the row either to the left or to the right

When you assume the deformable blocks, then the blocks try to follow a catenary curve, the ones in the center will be the ones that do not receive all the normal force applied at the ends, and they will be the ones prone to slide first.
 
  • #24
jbriggs444
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When you assume the deformable blocks, then the blocks try to follow a catenary curve
You seem to be shifting here from a system in which the components are supported by shear forces (friction) with torque and compression ignored to a system in which the components are supported by tension alone and in which torque is naturally zeroed.
 
  • #25
sophiecentaur
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You seem to be shifting here from a system in which the components are supported by shear forces (friction) with torque and compression ignored to a system in which the components are supported by tension alone and in which torque is naturally zeroed.
This question is too open ended. You might have a chance if the blocks became plates, in which case you could probably ignore torque - or perhaps concentric cylinders, with the outer one fixed.
But the original model of blocks squeezed together by hand implies that the experimenter would be able to restrain himself from pushing the bottom harder than the top and that there was no distortion.

In real life (and for the past thousands of years) we use a thing called an ARCH, which can genuinely use just normal forces. Any friction is just there to help with stability.
 
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  • #26
Richard R Richard
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You seem to be shifting here from a system in which the components are supported by shear forces (friction) with torque and compression ignored to a system in which the components are supported by tension alone and in which torque is naturally zeroed.
Please tell me where you disagree with my point.
Where will it collapse from the center? where do the outer blocks fall first? Or I don't understand that.

compres.png

Schematic graphic not full scale

I have not come up with equations for the deformation block model, as they are much more difficult, and I don't think they will be a topic for this thread.
So that the blocks do not fall, they must have an original length greater than the ideal blocks, since when applying pressure each one must be compressed following Hooke's law, the final individual length will depend on the length of the arc.
It is known that the upper part of the faces exposed to friction shear are more compressed than the lower part, any small slip will change the coefficient of friction from stratum to dynamic, which if it is not large enough for the load, will cause collapse the stacked structure.
 
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  • #27
jbriggs444
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I have not come up with equations for the deformation block model, as they are much more difficult, and I don't think they will be a topic for this thread.
You need to at least define the problem properly.

Start with five identical rectangular slabs in a horizontal row from left to right. They have height h, left-to-right width w and front-to-back depth d. Each has mass m. Gravity is given by g. We look at the row of slabs from the side.

Obviously we can ignore depth d. The bridge of slabs just has to have enough depth that it doesn't buckle in the horizontal direction.

At the left and right ends, the slabs are are pressed between pistons that exert a fixed force ##f_x## in the horizontal direction and maintain a fixed position in the vertical direction. The piston faces are vertical and are not free to rotate. The face of each piston is flat and mates with a flat surface of the end slabs.

All the faces are held together with pressure and friction alone.

The slabs do not deform at all. However, the pistons can move. For instance, if the center slab descends so that the row of slabs arches downward (as shown in your second drawing), the pistons will be forced apart slightly.

Simplify the situation to pretend that each slab has exactly three forces between it and each of its neighbors.

1. Horizontal pressure on its top corner.
2. Horizontal pressure on its bottom corner.
3. Sum of vertical friction on the face.

If five slabs is problematic, make your life easier. Start with two slabs and see if you can solve that.

Write force balance equations and torque balance equations for each block.

You will find that if the system is stable at all (if the squeezing force is inadequate, it may not be), that it is statically indeterminate. In particular, the support force from the left hand piston and from the right hand piston need not be divided 50/50. So do yourself a favor: assume that the support forces from the end pistons are divided 50/50.

First step: Determine the pattern of forces that must exist if the structure is to be stable. Allow the coefficient of friction to be large enough that no slipping occurs.

Second step: Determine the minimum compressive force from the pistons needed for stability.

Third step: Determine the minimum coefficient of friction needed for stability under that much compressive force.
 
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  • #28
Richard R Richard
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Hi,
You need to at least define the problem properly.

I do not have to define anything about the statement of the problem, in any case the author of the thread will do it.
I made the outline for you to give your opinion on what things you disagree with, but I understand you want to change the meaning of the problem.

Start with five identical rectangular slabs in a horizontal row from left to right. They have height h, left-to-right width w and front-to-back depth d. Each has mass m. Gravity is given by g. We look at the row of slabs from the side.

Obviously we can ignore depth d. The bridge of slabs just has to have enough depth that it doesn't buckle in the horizontal direction.

At the left and right ends, the slabs are are pressed between pistons that exert a fixed force ##f_x## in the horizontal direction and maintain a fixed position in the vertical direction. The piston faces are vertical and are not free to rotate. The face of each piston is flat and mates with a flat surface of the end slabs.

All the faces are held together with pressure and friction alone.

If this way you define a new problem, I am also waiting for the solution, since you have not answered my question yet.
I believe that the idea of whoever opened the thread is to justify that the block will resist more weight, the exteriors or the interiors, I have already given my opinion, I hope yours, then we will see calculations.

The slabs do not deform at all. However, the pistons can move. For instance, if the center slab descends so that the row of slabs arches downward (as shown in your second drawing), the pistons will be forced apart slightly.

Simplify the situation to pretend that each slab has exactly three forces between it and each of its neighbors.

1. Horizontal pressure on its top corner.
2. Horizontal pressure on its bottom corner.
3. Sum of vertical friction on the face.

Are you looking for a solution with or without deformation?
The failure or collapse of the system, without deformation, is due to exceeding the maximum friction stress, but with deformation the break occurs due to the action of bending caused by gravity, precisely when N is insufficient, but I read it and understand that you are interpreting another thing, as if that system is going to fail to compress it, say by buckling.

If five slabs is problematic, make your life easier. Start with two slabs and see if you can solve that.

If you have already read me, you will have seen that my proposal solves the problem for ##n## non-deformable blocks, because then you think that it would be problematic for me to raise those equations, since I have read you in other threads I am convinced that they are not problematic for you either,
I only said that they are more difficult, and that perhaps they are unnecessary for an informative thread.

Write force balance equations and torque balance equations for each block.

You will find that if the system is stable at all (if the squeezing force is inadequate, it may not be), that it is statically indeterminate. In particular, the support force from the left hand piston and from the right hand piston need not be divided 50/50. So do yourself a favor: assume that the support forces from the end pistons are divided 50/50.

First step: Determine the pattern of forces that must exist if the structure is to be stable. Allow the coefficient of friction to be large enough that no slipping occurs.

Second step: Determine the minimum compressive force from the pistons needed for stability.

Third step: Determine the minimum coefficient of friction needed for stability under that much compressive force.

The initial conditions of the problem were defined by whoever opened the thread,
• The pattern of forces is constant, the masses are equal, he just wants to know where to put more weight and not collapse.
• The compression force is also data, it does not have to be determined.
• The materials are given and the friction coefficient is what it is since it is data of the problem ...
Are we reading the same problem ?.

Regards, ##R^3##
 
  • #29
jbriggs444
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The initial conditions of the problem were defined by whoever opened the thread,
• The pattern of forces is constant, the masses are equal, he just wants to know where to put more weight and not collapse.
If you review post #1, torque was not mentioned. Yet, as others have guessed, torque considerations lurked behind the contradiction sensed by OP. Post #1 is not explicit that the supports on either end are constant position vertically and constant force horizontally. But that is the natural reading.

The hope was that with a clear problem statement in hand, analysis could move forward. A problem well asked is half solved.
 
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  • #30
Richard R Richard
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I agree with you,

If the OP confirms all our assumptions, we can dedicate them to look for a solution and draw some conclusions.
 
  • #31
silverrahul
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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
 
  • #32
sophiecentaur
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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.
 
  • #33
silverrahul
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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.
 
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  • #34
silverrahul
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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.
 
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  • #35
silverrahul
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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,
 

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