Teetering Book Tower

What a nice problem with a lovely counterintuitive answer! I hope my solution will convince you that the tower can indeed be extended as far as we please.

The answer seems so impossible because we imagine building the tower from the bottom up (naturally!). At some stage our tower of n books is tottering on the brink of collapse - so if the tower is already way out beyond the edge of the table how could we possibly add an $(n+1)^{th}$ book without disaster? But imagine adding a new book to the bottom of the tower, and things are not quite so crazy. For instance suppose our tottering tower of n books is built to overhang the top of the $(n+1)^{th}$ book which is sitting flush with the edge of the table. Now it does not seem so impossible that the new book could be pushed out a little from the edge of the table, and so the tower could be extended.

My solution is based on building up the tower by introducing new books one by one at the bottom, finding the centre of gravity of the new tower and then sliding the whole wobbly thing right a bit so that the centre of gravity is exactly over the edge of the table. In this way we can find the maximum extension beyond the edge of the table with n books. Suppose the books have length 1 and weight w, and start by sliding the first book out as far as it will go without falling, in other words shift it to the right by $\frac{1}{2}$ .

Now introduce a new book underneath this book, the edge of the new book being flush with the table. The centre of gravity of this two book tower is to the left of the edge - but by how much? Well suppose the centre of gravity is a distance $x$ from the edge. Then the turning moment (about the edge of the table) of the two block tower of weight 2w is 2wx. This turning moment is made up of two parts - the moment of the new book (which is $\frac{w}{2}$ ) and the moment of the first (top) book which is zero, because at the moment its centre of gravity is above the table's edge! Thus we have

$2wx=\frac{w}{2} \implies x=\frac{1}{4}$

so we can get a two book tower tottering on the edge by sliding everything to the right by $\frac{1}{4}$ .

Now introduce a third book underneath this two book tower, the edge of the new book being flush with the table. The centre of gravity of this three book tower is to the left of the edge - and again we ask, 'by how much?' Well suppose the tower's centre of gravity is a distance $x$ from the edge. Then the turning moment (about the edge of the table) of the three block tower of weight 3w is 3wx. This turning moment is made up of two parts - the moment of the new book (which again is $\frac{w}{2}$ ) and the moment of the first two books which is zero, because the centre of gravity of the two book tower is, at this point in the construction, above the table's edge! Thus we have

$3wx=\frac{w}{2} \implies x=\frac{1}{6}$

so we can get a three book tower tottering on the edge by sliding everything to the right by $\frac{1}{6}$ .

By repeating the argument you can see that for n books the total distance slid to the right is

$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots + \frac{1}{2n}=\frac{1}{2}\left(1+ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \dots +\frac{1}{n}\right)$

The series in brackets is the 'harmonic series' which does not converge - in other words it can be made as large as you like by taking enough terms.

Since the extension of the tower can be approximated by $\frac{1}{2} \ln( n)$ , and the logarithm grows so slowly, you will need whole libraries of books if you want to go more than a couple of metres from the edge of the table!

The center of mass of the books on the table cannot lie beyond the edge of the table or the books will topple. As with @Peter Macgregor below, this means the top book cannot overhang the book beneath it by more than $\frac{1}{2}$ .

Let's measure the center of mass of all our books from the right edge of the top book. The book underneath the top book can be offset by $\frac{1}{2}$ to the left and the top book won't topple.

Now suppose we have n-1 books at center of mass c. Another book can be added beneath these books offset by $\frac{1}{2}$ from the combined center of mass of n-1 books above so they don't topple. The center of mass of the n books is

$\frac{n-1}{n}$ * c + $\frac{1}{n}$ * (c+ $\frac{1}{2}$ ) = c + $\frac{1}{2n}$

Writing out this sequence as @Peter Macgregor, the center of mass is the sum of $\frac{1}{2n}$ , which diverges as noted.