How many bricks can be put without disordering the equilibrium

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Homework Help Overview

The problem involves stacking homogeneous bricks on a horizontal surface, with each subsequent brick shifted horizontally by L/5 from the one below. The main question is how many bricks can be stacked without disrupting equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the center of gravity of the bricks and how it affects stability. There are differing opinions on the maximum number of bricks that can be stacked, with some asserting at least three, while others suggest four might be possible. Questions arise about the logic behind the calculations and assumptions made regarding the center of mass.

Discussion Status

The discussion is ongoing, with participants sharing diagrams and attempting to clarify their reasoning. Some guidance has been offered regarding the analysis of mass distribution and the importance of understanding the progression of the center of mass as more bricks are added.

Contextual Notes

Participants are encouraged to share their diagrams for better insight into their reasoning. There is a focus on understanding the underlying principles rather than simply arriving at a numerical answer.

MaiteB
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Homework Statement


A homogenous brick with length L, is on a horizontal surface. On this brick are put one after another other bricks, which are the same with the first. Every brick is moved along L/5 units horizontally from the predecessor one. How many bricks can be put without disordering the equilibrium?

Homework Equations


(L/2)/(L/5)=2.5

The Attempt at a Solution


Since the bricks are homogeneous then the center of gravity is in the middle. So in the point L/2. So (L/2)/(L/5)=2.5, so there are 2 bricks.
 
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MaiteB said:

Homework Statement


A homogenous brick with length L, is on a horizontal surface. On this brick are put one after another other bricks, which are the same with the first. Every brick is moved along L/5 units horizontally from the predecessor one. How many bricks can be put without disordering the equilibrium?

Homework Equations


(L/2)/(L/5)=2.5

The Attempt at a Solution


Since the bricks are homogeneous then the center of gravity is in the middle. So in the point L/2. So (L/2)/(L/5)=2.5, so there are 2 bricks.
That's not correct.

Certainly, at least three bricks can be stacked this way.
 
What is the logic behind your (L/2)/(L/5) expression?
 
SammyS said:
That's not correct.

Certainly, at least three bricks can be stacked this way.
And can you tell me what the real answer is? If I knew, I wouldn't ask it here...
 
MaiteB said:
And can you tell me what the real answer is? If I knew, I wouldn't ask it here...
we're actually a lot more interested in helping people figure out how to get their own answers than we are in just spoon-feeding answers. How about you answer haruspex's question.
 
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haruspex said:
What is the logic behind your (L/2)/(L/5) expression?
The logic behind is how many bricks can be put until reaching the middle of the brick, where the gravity force is applied
 
MaiteB said:
The logic behind is how many bricks can be put until reaching the middle of the brick, where the gravity force is applied
I'm not following what you mean by that. It's possible that this somewhat muddled statement is just a reflection of your confusion. Probably you'd get a better handle on the problem by drawing a diagram with force vectors. Certainly a diagram would immediately show you the truth of Sammy's statement.
 
phinds said:
I'm not following what you mean by that. It's possible that this somewhat muddled statement is just a reflection of your confusion. Probably you'd get a better handle on the problem by drawing a diagram with force vectors. Certainly a diagram would immediately show you the truth of Sammy's statement.
I did it and I can't get to any conclusion. Look here http://imgur.com/sVHj4uB
 
MaiteB said:
I did it and I can't get to any conclusion. Look here http://imgur.com/sVHj4uB
Are you saying that your diagram does not convince you of the truth of Sammy's statement?

Taking it a step further, it's fairly obvious, if you continue the diagram, that you can do at least 4 bricks, but that may not be the end of it. I haven't worked it out in detail.
 
  • #10
phinds said:
Are you saying that your diagram does not convince you of the truth of Sammy's statement?

Taking it a step further, it's fairly obvious, if you continue the diagram, that you can do at least 4 bricks, but that may not be the end of it. I haven't worked it out in detail.
One says at least 3 you say four... can you give me a concrete and convincing way of solving this problem?
 
  • #11
MaiteB said:
One says at least 3 you say four... can you give me a concrete and convincing way of solving this problem?
So, you ARE saying that your diagram does not convince you that at least 3 bricks can be stacked? I've asked this a couple of times now and you have not answered.

Try a diagram w/ 4 bricks and see if it's isn't fairly obvious that 4 will work.
 
  • #12
phinds said:
So, you ARE saying that your diagram does not convince you that at least 3 bricks can be stacked? I've asked this a couple of times now and you have not answered.

Try a diagram w/ 4 bricks and see if it's isn't fairly obvious that 4 will work.
I can't understand how a diagram can convince me if we can put 3 or 4
 
  • #13
MaiteB said:
I can't understand how a diagram can convince me if we can put 3 or 4
What I'm asking is whether or not your diagram convinces you that 3 works. Forget 4. Forget the final answer. DOES 3 WORK ?
 
  • #14
phinds said:
What I'm asking is whether or not your diagram convinces you that 3 works. Forget 4. Forget the final answer. DOES 3 WORK ?
Yes, I do. It has to do with the equation I said, right?
 
  • #15
MaiteB - Forum rules discourage us from just giving you the answer.

It would help a lot if you posted your diagram so we can see what if any mistakes you are making. For example it's tempting to think that it's only the centre of mass of the top brick that matters. That would be a mistake.
 
  • #16
CWatters said:
MaiteB - Forum rules discourage us from just giving you the answer.

It would help a lot if you posted your diagram so we can see what if any mistakes you are making. For example it's tempting to think that it's only the centre of mass of the top brick that matters. That would be a mistake.
He posted a link to a diagram w/ 3 bricks, which is why I kept asking him if that diagram convinced him that 3 would work.
 
  • #17
MaiteB said:
I did it and I can't get to any conclusion. Look here http://imgur.com/sVHj4uB
sVHj4uB.png

Hope this helps. It's now uploaded to PF.
 
  • #18
MaiteB said:
Yes, I do. It has to do with the equation I said, right?
I still don't know where you got the formula, but for the moment, forget the formula. Think about WHY are you convinced that 3 will work. Can you extend your logic to 4? This kind of progression often leads to an understanding of just what is going on and what a reasonable math process might be to solve the problem.

EDIT: Actually, I shouldn't say that I don't know where you got the formula, what I should say is that I haven't even looked at the formula. I've been trying to get you to think about it logically so you can figure out if your formula or some other is the general solution. I don't care about formulas, I care about understanding how you GET formulas.
 
  • #19
phinds said:
I still don't know where you got the formula, but for the moment, forget the formula. Think about WHY are you convinced that 3 will work. Can you extend your logic to 4? This kind of progression often leads to an understanding of just what is going on and what a reasonable math process might be to solve the problem.
If I put four, it will be out of the mass center and the brick will fall
 
  • #20
MaiteB said:
If I put four, it will be out of the mass center and the brick will fall
OK, there's your problem. You are not analyzing the mass distribution properly. Try again.

And by the way, to reiterate what cwaters said, we are not trying to give you a hard time or annoy you or make you jump through hoops, we are trying to help you understand how to solve such problems.
 
  • #21
phinds said:
OK, there's your problem. You are not analyzing the mass distribution properly. Try again.

And by the way, to reiterate what cwaters said, we are not trying to give you a hard time or annoy you or make you jump through hoops, we are trying to help you understand how to solve such problems.
Can you give me a hint? Cause I don't think the idea will come just like that
 
  • #22
Here's a hint. Draw a dot at the center of mass of each brick and connect the dots. I say again, it's a matter of looking at the progression. Think about the dot-line for 2 bricks, 3 bricks, 4 bricks. Where is the total center of mass for each set (of 2, then 3, then 4)?
 
  • #23
phinds said:
Here's a hint. Draw a dot at the center of mass of each brick and connect the dots. I say again, it's a matter of looking at the progression. Think about the dot-line for 2 bricks, 3 bricks, 4 bricks. Where is the total center of mass for each set (of 2, then 3, then 4)?
It forms a line that has an angle with the horizontal plane
 
  • #24
MaiteB said:
It forms a line that has an angle with the horizontal plane
Answer the rest of bhinds's questions, please
 
  • #25
MaiteB said:
It forms a line that has an angle with the horizontal plane
Attempting to help you with this problem is beginning to feel like pulling teeth. MY teeth. As Sammy said, please answer the rest of my questions. You continue to not answer questions as I ask them. I am not just throwing out random questions for the hell of it. If you don't know the answer or don't understand the question, then say so. Continuing to just ignore my questions is really dragging this out.
 
  • #26
phinds said:
Attempting to help you with this problem is beginning to feel like pulling teeth. MY teeth. As Sammy said, please answer the rest of my questions. You continue to not answer questions as I ask them. I am not just throwing out random questions for the hell of it. If you don't know the answer or don't understand the question, then say so. Continuing to just ignore my questions is really dragging this out.
I answered your last question. Are you asking me to do the same picture I showed before just joining the center of mass points? Yes I did it... And what? it is a line that has an angle with the plane.
And about you attempt, I really appreciate it, but try to understand me too. I am a person that does not have english as her native language and some terms are difficult for me
 
  • #27
phinds said:
Here's a hint. Draw a dot at the center of mass of each brick and connect the dots. I say again, it's a matter of looking at the progression. Think about the dot-line for 2 bricks, 3 bricks, 4 bricks. Where is the total center of mass for each set (of 2, then 3, then 4)?
MaiteB said:
It forms a line that has an angle with the horizontal plane
phinds said:
... As Sammy said, please answer the rest of my questions. ...
MaiteB said:
I answered your last question.
This is the question you didn't address:
"Think about the dot-line for 2 bricks, 3 bricks, 4 bricks. Where is the total center of mass for each set (of 2, then 3, then 4)?. Where is the total center of mass for each set (of 2, then 3, then 4)?"​
 
  • #28
SammyS said:
This is the question you didn't address:
"Think about the dot-line for 2 bricks, 3 bricks, 4 bricks. Where is the total center of mass for each set (of 2, then 3, then 4)?. Where is the total center of mass for each set (of 2, then 3, then 4)?"​
I don't know
 
  • #29
MaiteB said:
I don't know

Hint:

It's along that dot-line, somwwhere.

Try to figure out where that is.
 
  • #30
SammyS said:
Hint:

It's along that dot-line, somwwhere.

Try to figure out where that is.
In the middle of it?
 

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