How far can a stack of books lean over the edge of a table?

Last summer I helped at our school's second-hand book sale, and my job was to stack the unsold paperbacks at the end of each day. A younger student bet me that I could not build a pile in which the top book was completely clear of the table — that is, not above the table at all. I lost the bet on the day, because my pile fell over, but I was sure it was possible. When I tried again at home with one book per layer, I managed it with four books, and I wanted to know why four was enough, and how far the idea could go.

Aim. For a stack of \(n\) identical books, one book per layer, I will (i) find the largest possible overhang beyond the table edge and prove that no other arrangement of this kind does better, (ii) find how many books are needed for a given overhang, (iii) compare the theory with stacks I build and measure, and (iv) find the best order in which to stack books of different masses.

The overhang turns out to be connected to the harmonic series \(1 + \tfrac12 + \tfrac13 + \cdots\), which we met briefly in class as an example of a series that diverges even though its terms tend to zero. This exploration was my chance to find out what that surprising fact means for something I can hold in my hands.

I number the books from the top: book 1 is the top book and book \(n\) rests on the table. I use:

\(L\) = length of each book (cm); for my paperbacks \(L = 19.8\) cm;
\(e_k\) = horizontal position of the right-hand edge of book \(k\), measured from the table edge (positive beyond it), with \(e_{n+1} = 0\) for the table itself;
\(d_k = e_k - e_{k+1}\) = how far book \(k\) sticks out beyond the book (or table) under it;
\(D_n = e_1\) = the overhang of the whole stack.

Assumptions. (1) Each book is a rigid, uniform rectangular block, so its centre of mass is at its centre, a distance \(\tfrac L2\) behind its right edge. (2) All books have the same mass (until Section 7). (3) The books are pushed in one direction only, perpendicular to the table edge, so the problem is two-dimensional. (4) Friction is large enough that nothing slides; the only way the stack can fail is by tipping.

The balance condition. A set of blocks resting on a supporting edge does not tip as long as its combined centre of mass is not beyond that edge. So the stack does not tip outwards exactly when, for every \(k = 1, 2, \dots, n\), the centre of mass of the top \(k\) books is at or behind \(e_{k+1}\). (It could also tip backwards if that centre of mass were behind the left-hand end of the supporting book, but in a stack leaning outwards that never happens, so I only need the condition on the right.) Writing \(c_k\) for the horizontal position of that centre of mass,

\[ c_k = \frac{1}{k}\sum_{i=1}^{k}\left(e_i - \frac{L}{2}\right) \le e_{k+1}, \qquad k = 1, 2, \dots, n. \qquad (1) \]

Assumption (4) is the one I doubted most, because paperbacks are slightly soft and their spines are rounded. I come back to it when I test the model in Section 6.

My first idea was to build the stack from the bottom up, pushing each new book out as far as it would go. This fails: pushing the second book out as far as possible leaves no room to push the third one further out, because the pair would then tip. So I reversed the process. If I place each set of books so that its centre of mass is exactly on the edge of the book below — the critical position — every inequality in (1) becomes an equation.

Claim. In the critical stack, \(d_k = \dfrac{L}{2k}\) for every \(k\).

Proof by induction. Put the origin temporarily at \(e_{k}\), the right edge of book \(k\). If the top \(k-1\) books are critical, their combined centre of mass is on the edge of book \(k\), i.e. at position 0, with total mass \((k-1)m\). Book \(k\) has mass \(m\) and centre of mass at \(-\tfrac L2\). So the centre of mass of the top \(k\) books is at

\[ \frac{(k-1)m \times 0 + m \times \left(-\frac L2\right)}{km} = -\frac{L}{2k}. \]

For the top \(k\) books to be critical, this point must be exactly on the edge of book \(k+1\), so \(e_{k+1} = e_k - \frac{L}{2k}\), i.e. \(d_k = \frac{L}{2k}\). For \(k = 1\) the top book alone has its centre at \(-\tfrac L2\), so \(d_1 = \tfrac L2\), which is the base case. By induction the claim holds for all \(k\). \(\blacksquare\)

Adding the shifts gives the overhang of \(n\) books:

\[ D_n = \sum_{k=1}^{n} \frac{L}{2k} = \frac{L}{2}\left(1 + \frac12 + \frac13 + \cdots + \frac1n\right) = \frac{L}{2}H_n, \qquad (2) \]

where \(H_n\) is the \(n\)th harmonic number. For \(n = 4\), \(H_4 = \frac{25}{12}\), so \(D_4 = \frac{25}{24}L > L\): the top book is completely clear of the table, which is why four books won me the rematch. With three books, \(D_3 = \frac{11}{12}L\), just short.

The shifts get smaller as we go down, which matches what I saw: the top books stick out a long way and the bottom books hardly at all. It also explains why building from the bottom fails — the bottom book must stick out least, not most.

For \(n\) identical books with one book per layer, the largest possible overhang is \(\frac L2 H_n\). I proved by induction that the critical stack achieves it, and with a telescoping argument that no stable stack of this kind does better. Four books are enough for the top book to clear the table completely (\(H_4 = \frac{25}{12} > 2\)). Because the harmonic series diverges, any overhang is possible in theory, but the number of books grows like \(e^{2m-\gamma}\): three book lengths already needs 227 books. For books of different masses, the best order is lightest on top.

My experiment showed the most important limitation. Real stacks fall short of the ideal by a loss of about 0.3 cm per book, and because this loss grows linearly while the ideal overhang grows only logarithmically, a real stack of my books cannot do better than about 30.5 cm, reached with about 32 books. The model is therefore excellent for small stacks — which is what the bet was about — and misleading for large ones. My estimate of \(\delta\) comes from only ten stack sizes built by one person, and it probably depends on the books and on how patient the builder is, so the value 32 should be read as “about thirty” rather than as an exact number.

Other assumptions matter less at this scale: the books were close to uniform, and nothing slid on the worktop, although with glossy covers sliding could become the way the stack fails. The biggest limit of the whole exploration is the rule of one book per layer. The research I read shows that with several books per layer the overhang can grow like \(\sqrt[3]{n}\) rather than \(\ln n\) — so the answer to “how far can a stack of books lean?” depends as much on the rules of the game as on the mathematics.

What I will remember is that the same series can say “anything is possible” and, one small realistic assumption later, “about thirty books is the best you will do”. At the next book sale I intend to win the bet with four books and not attempt thirty.