Physics Friday 98: A Classic Problem

Consider a set of identical books (or other rectangular objects), stacked flat one atop another. By offsetting the books, one can have a “leaning” tower that angles to one side.

Is it possible to make such a tower where no part of the top book is over any part of the base book, so that the “lean” is larger than the length of a book? What is the maximum possible horizontal offset between the top and bottom books?

Consider the top book; with (roughly) uniform density, its center of gravity is the geometric center. To keep its center of gravity over the book supporting it, its maximum offset is half the book length; in units of book lengths, we have a maximum offset of h=1/2 for a stack of n=2 books.

Now, that set of two books has a center of gravity that, via symmetry, has horizontal position in the middle of the overlap, and so the maximum offset between the lower book of this pair and a third book supporting the pair is 1/4 of a book length, so we have total offset of the top book of .

If we number the books from the top down, and let be the displacement of the ith book relative to the one below, so we have and . Now, consider the top k books, and their displacements relative to the k+1 book. The center of mass of the top k books must be offset from the k+1 book by 1/2. The displacement of book k relative to book k+1 is , that of book k-1 relative to book k+1 is ; book i is deplaced relative to book k+1 by . Then the center of mass of the top k books (relative to book k+1) is the average of these displacements. Reordering the sum of these, we see appears in the sums for the displacements of books 1 through i, and so the sum of these is
. Thus, we see
and so
and subtracting this from the above, we see
And for a stack of n books, we have total offset
where is the nth harmonic number.

To answer our first question, then, we have h>1 when Hn-1>2. The harmonic numbers pass 2 with H4=25/12, so only five books are needed for this tower. As to the second question, the harmonic series diverges; so long as we stack enough books, any finite offset can be produced. For large n, we have , where γ is the Euler-Mascheroni constant. Thus, for larger displacements, we have
and thus
and the number of books needed to achieve a given displacement grows (roughly) exponentially with that displacement.


Tags: , , , ,

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: