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Homework Help: How many bricks can be put without disordering the equilibrium

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  1. Mar 17, 2015 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    (L/2)/(L/5)=2.5

    3. 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.
     
  2. jcsd
  3. Mar 17, 2015 #2

    SammyS

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    That's not correct.

    Certainly, at least three bricks can be stacked this way.
     
  4. Mar 17, 2015 #3

    haruspex

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    What is the logic behind your (L/2)/(L/5) expression?
     
  5. Mar 18, 2015 #4
    And can you tell me what the real answer is? If I knew, I wouldn't ask it here...
     
  6. Mar 18, 2015 #5

    phinds

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    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.
     
  7. Mar 18, 2015 #6
    The logic behind is how many bricks can be put until reaching the middle of the brick, where the gravity force is applied
     
  8. Mar 18, 2015 #7

    phinds

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    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.
     
  9. Mar 18, 2015 #8
    I did it and I can't get to any conclusion. Look here http://imgur.com/sVHj4uB
     
  10. Mar 18, 2015 #9

    phinds

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    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.
     
  11. Mar 18, 2015 #10
    One says at least 3 you say four... can you give me a concrete and convincing way of solving this problem?
     
  12. Mar 18, 2015 #11

    phinds

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    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.
     
  13. Mar 18, 2015 #12
    I can't understand how a diagram can convince me if we can put 3 or 4
     
  14. Mar 18, 2015 #13

    phinds

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    What I'm asking is whether or not your diagram convinces you that 3 works. Forget 4. Forget the final answer. DOES 3 WORK ?
     
  15. Mar 18, 2015 #14
    Yes, I do. It has to do with the equation I said, right?
     
  16. Mar 18, 2015 #15

    CWatters

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    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.
     
  17. Mar 18, 2015 #16

    phinds

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    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.
     
  18. Mar 18, 2015 #17

    SammyS

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    sVHj4uB.png
    Hope this helps. It's now uploaded to PF.
     
  19. Mar 18, 2015 #18

    phinds

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    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.
     
  20. Mar 18, 2015 #19
    If I put four, it will be out of the mass center and the brick will fall
     
  21. Mar 18, 2015 #20

    phinds

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    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.
     
  22. Mar 18, 2015 #21
    Can you give me a hint? Cause I don't think the idea will come just like that
     
  23. Mar 18, 2015 #22

    phinds

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    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)?
     
  24. Mar 18, 2015 #23
    It forms a line that has an angle with the horizontal plane
     
  25. Mar 18, 2015 #24

    SammyS

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    Answer the rest of bhinds's questions, please
     
  26. Mar 18, 2015 #25

    phinds

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    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.
     
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