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Hanging over the edge

  1. Mar 1, 2005 #1
    first of all I would like to say that this is a question to which I do not have the answer...

    imagine that you are building a set of stairs for an ant out of a deck of n cards... what is the greatest horizontal distance from a table out over an abyss that can be covered over by the cards given that the structure must be freestanding, and have a height of n(given the height of one card = 1).

    for example... if you have one card, you can go half the length of the card over an edge by placing the card so that the edge of the table is exactly in the middle of the card.

    hope it makes sense...
  2. jcsd
  3. Mar 2, 2005 #2
    I believe 1
  4. Mar 2, 2005 #3
    If L is the length of one card, then you can go L*(2N-1)/(2N) over the edge.
  5. Mar 2, 2005 #4
    It really depends on the number of cards you have, can you use cards simply as weight? you oculd sandwich cards so that there is a stack of three with the middle one sticking out, and use weight on the back end to keep it sticking out?
  6. Mar 2, 2005 #5


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    Depends on how much glue you use :smile:
  7. Mar 3, 2005 #6
    no, no weights or glue can be used...

    and I thought that it would be 1 as well (well, i thought that the placement for the cards would be related to 1/2^n... but i tried it out... and can definately get more than 1 card length over the edge...

    any ideas?
    Last edited: Mar 3, 2005
  8. Mar 3, 2005 #7
    I dont see how !

    Remember you've said "the structure must have a height of n (given the height of one card = 1)".

    So, you can not have inclined cards. And besides, cards side by side at the same level are not allowed.

  9. Mar 3, 2005 #8


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    i remember from physics we had a similar problem, only it was about constructing bridges out of bricks. if thats what you mean, then as far as i know i *think* you can go as far out as you want, but its going to take alot of cards (or bricks). the method i think is to take the first card and have 1/2n (n is the total number of cards) of it sticking of the edge. then the next would be 1/(2n - 2) and so on until the last card is half way off the previous card. this keeps the center of mass over the original table or whatever it starts from.

    only problem i see with it is that if you construct it exactly (physics exactly), as soon as the ant starts walking it will collapse...

    *edit: i tried to draw it with dashes but it didnt save the formatting... :mad: :frown:
  10. Mar 3, 2005 #9
    so, more or less

    \sum_{n=1}^{[\frac{n}{2}]} \frac{1}{2n}

    (that is, if the weight of the ant = 0)...

    do you remember how you came up with that?

  11. Mar 3, 2005 #10
    You can use the diagonal of the cards to increase the reach of the cantilever significantly.

    The way to analyse this is to start with the top card, and work down. The top card can overhang by half its length (half the length of its diagonal, if you're greedy) then the top but one card can overhang by a quarter, the next one by a sixth and so on.

    With an infinite number of cards, there is no limit to the overhang.
    Last edited: Mar 3, 2005
  12. Mar 3, 2005 #11
    Put some {code} tags around your text to preserve formatting {/code} Use square brackets in place of the curly ones, and you get something like this:
    Code (Text):
  13. Mar 3, 2005 #12


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    thanks ceptimus.

    to answer mssmith12's question, its really all about keeping the center of mass just on the edge of the table. try it for one book, and its just half on and half off. add another *below* that one and it should be about 1/4 off the table. etc etc. i suppose there is a way to formally prove it by going through the whole center of mass thing, but it doesnt seem that hard. you shouuld actually try it with books (like encycolpedias) and its kind of cool how the last book is completely over the edge of the table. trying it diagonally would be alot harder...
  14. Mar 15, 2007 #13
    If you continue the sequence, you can reach infinity. It's a harmonic sequence.

    The first card can get half-way over the card under it, the second card can get 1/3 of the way off the card under that (the total overhang now being 1/2+1/3); by performing some simple center of mass equations, you can figure out that the pattern goes like this: [tex]\sum_{n=1}^\infty [\frac{1}{n}][/tex]

    It's a harmonic series and equals infinity when its limit is reached.

    Edit: I suck with latex... It's supposed to be the sum of 1/n as n goes from 1 to infinity.

    Edit2: Sorry for bringing up a 2-year old topic... I forgot I was searching through the ancient brain teasers when I saw this. :-X
    Last edited: Mar 16, 2007
  15. Apr 24, 2007 #14


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    i have a bunch of old baseball cards I just tried this with.

    I was able to get about 1 foot over the table using about 300 cards. about 2-/3 of them were stacked on the tables edge holding down the rest that were overhanging. so if you are allowed to use the cards as weight, im sure you could go on forever given ideal conditions.

    Without using cards as weights well ill leave that up to you math wizzards
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