Card Stacking Problem, Redux
Date: 01/31/2017 at 15:19:47
From: Nikola
Subject: Inability to follow an explanation given by a mathmatician
I'm unable to follow what is happening here:
https://allmath.wordpress.com/solutionstochallengeproblems/ solutiontostackingdominoes/
Specifically, this passage:
"For any n, the sum of the torques of the first n blocks has to be 0.
Writing this out gives, without much difficulty, that if the ith block
is placed at x_i, then...."
How did the author get the next equation? How do we make that jump
"without much difficulty"? If x_(n  1) is the last displacement, then why
get all of that instead of just a sum of all the displacements?
I understand the derivation immediately prior, a normal integration
process; but I'm confused whether the author applies the same principle to
take the next step. (When I tried to do that, I did not get the right
answer.)
Maybe the author is making up a whole new thing. But if I cannot grasp the
next step conceptually, how can I grasp it mathematically?
I find the indexing notation confusing, too. Wouldn't using n itself make
more sense than i?
Date: 02/01/2017 at 21:39:38
From: Doctor Rick
Subject: Re: Inability to follow an explanation given by a mathmatician
Hi, Nikola,
Thanks for writing to Ask Dr. Math.
That explanation does look rather hard to follow. In fact, even if you get
past the part you asked about, it goes on to say, "From this one can
compute (this is not entirely straightforward, but not too difficult
either), ..."
The author punts, admitting that the next part isn't easy ... before
skipping over it anyway!
Maybe it will be more helpful if we find a better explanation of the
problem and solution.
Here is a page on a respected site that explains the problem nicely with
pictures:
Book Stacking Problem. From MathWorldA Wolfram Web Resource
http://mathworld.wolfram.com/BookStackingProblem.html
But it skips over most of the solution process with the words "it turns
out that ..."
The following document goes very deeply into the problem, discussing the
physics in detail ... before generalizing the problem to such a degree
that it gets VERY complex:
Overhang, by Mike Paterson and Uri Zwick
http://www.maa.org/sites/default/files/pdf/ upload_library/22/Robbins/Patterson1.pdf
In the archives of this very same askanexpert service, we do have one
related conversation, which falls in between these two levels of
explanation:
Card Stacking Problem
http://mathforum.org/library/drmath/view/69507.html
Even this, however, also glosses over the part you're asking about.
I am having a hard time locating anything that gives a clear, thorough,
simple explanation ...
... so I guess I'll have to write up something myself!
The question is: "How far can a stack of n blocks protrude over the edge
of a table without the stack falling over?"
Rather than discussing torques, I will work with this idea: The stack of
blocks (or books, or dominoes) will be stable if, for any number n of
blocks counting from the top, the center of mass of that set of blocks is
located over the block (or table) on which it rests. For maximum overhang,
the center of mass will be directly over the *edge* of the next block
down, or for the bottom block, the edge of the table.
Each block is uniform and symmetrical, so its center of mass is at the
center of the block.
Let's do as your source does, and take the width of the block to be 1
unit; and refer to the illustrations on the MathWorld page.
A single block can protrude beyond the table by a distance 1/2, as shown
in the top figure. That's because its center of mass is a distance 1/2
from the right edge of the block.
Now, we put another block under this one, with its edge where the table's
edge was. Where is the center of mass of the combination of two blocks?
The center of mass of an assemblage of objects is at the weighted average
of the centers of mass of the individual blocks. Let's work that out.
Measuring from the right edge of the top block, the center of mass of the
top block (which I'll call x[1]) is 1/2 unit from the right, as I said.
The center of mass of the next block down is 1/2 unit to the left of its
right edge, which in turn is x[1] = 1/2 unit to the left of the right edge
of the top block. Thus x[2] = 1/2 + 1/2 = 1 (as you can see in the second
figure).
Each block has the same mass, so their centers of mass are weighted
equally in the weighted average. Thus, the center of mass of the top two
blocks together is
x[3] = (1/2 + 1)/2 = 3/4
Note that this is called x[3] because it is where the edge of block number
3 will be placed.
Now, I'll generalize, going to x[n + 1].
Suppose we have x[n], the center of mass of the top n blocks. Block n + 1
has its right edge at x[n], and its center of mass is 1/2 unit to the left
of that; that is, at x[n] + 1/2. Now we take the weighted average of this
one block and the n blocks above it:
x[n + 1] = ((x[n] + 1/2) + n*x[n])/(n + 1)
= (x[n](1 + n) + 1/2)/(n + 1)
= x[n] + 1/(2(n + 1))
That's a recursive definition for x[n]. Working out successive values of
x[n], we find
x[1] = 1/2 (starting point of the recursion)
x[2] = 1/2 + 1/(2*2)
x[3] = 1/2 + 1/4 + 1/(2*3)
x[4] = 1/2 + 1/4 + 1/6 + 1/(2*4)
... and so on.
Notice that we can factor 1/2 out of all the terms, leaving:
x[n] = (1/2)(1 + 1/2 + 1/3 + 1/4 + ... + 1/n)
The sum in the last set of parentheses is known as the harmonic series.
You can look up more information about this series; the key point is that
it diverges. That is, if you choose a great enough value for n, you can
make the finite sum (nth partial sum of the infinite harmonic series) grow
as large as you wish. It takes a LOT of terms, even to reach 3, but it can
be done.
... *in principle*. It's worth noting that everything said here requires
ideal conditions. In practice, the blocks won't be perfect; a slight
rounding of the corners would be sufficient to make it impossible to reach
an overhang of 3.
Is this helpful?
 Doctor Rick, The Math Forum at NCTM
http://mathforum.org/dr.math/
