How far can a stack of books lean?
Tonight's idea
Use an ideal model: identical, rigid, uniform books of length 1. Ignore deformation and tiny disturbances. A single book is supported while its middle lies above the table; with its middle exactly above the edge, it is at the tipping limit. Every exact balance below means this limiting position.
In a stack the same rule applies at every level: the books above any book are supported while their combined balance point is no farther out than that book's end. Push every level to the tipping limit, working from the top down.
Let the outer end of book \(k\) be \(x = 0\), with positive \(x\) pointing out over the air. Balance the top \(k - 1\) books at that end, so their combined centre is at \(x = 0\). Book \(k\)'s own centre is half a length back, at \(x = -\tfrac12\). Since every book has the same weight, the centre of all \(k\) books is
\[x_{\mathrm{centre}} = \frac{(k - 1) \times 0 + 1 \times (-\tfrac12)}{k} = -\frac{1}{2k}.\]
Line up the edge of whatever is below with this combined centre. The outer end of book \(k\) is then \(\tfrac{1}{2k}\) farther out.
\(n\) books reach an overhang of \(\tfrac12 + \tfrac14 + \tfrac16 + \cdots + \tfrac{1}{2n} = \tfrac12\left(1 + \tfrac12 + \tfrac13 + \cdots + \tfrac1n\right)\).
The bracket is the harmonic series, and you have seen that it passes every number. So there is no limit to the overhang.
Watch one: four books
From the top down, the books stick out by \(\tfrac12, \tfrac14, \tfrac16\) and \(\tfrac18\). The bottom one moves least, because it carries the most.
\[\tfrac12 + \tfrac14 + \tfrac16 + \tfrac18 = \tfrac{12 + 6 + 4 + 3}{24} = \tfrac{25}{24}.\]
That is more than 1. The whole of the top book is past the edge of the table, with nothing but air below it. Three books give \(\tfrac{11}{12}\), not quite enough.
Your turn
- Find the overhang this method gives for two books, and for five books.
- Stack three books this way. Measured from the table edge, their outer ends are at \(\tfrac16\), \(\tfrac{5}{12}\) and \(\tfrac{11}{12}\). Find where the middle of each book is (negative means over the table). Show that the average of the three middles is 0. What does that tell you?
- The overhang reaches 2 book lengths when \(1 + \tfrac12 + \cdots + \tfrac1n\) passes 4. The sum up to \(n = 24\) is 3.7760. Keep adding terms. How many books are needed?
- Grouping the terms shows that \(1 + \tfrac12 + \cdots + \tfrac{1}{2^k} \ge 1 + \tfrac{k}{2}\). Use this to find a number of books that is certain to give an overhang of 10 book lengths. What does doubling the number of books add to the overhang, at least?
Stretch: a very tall stack
For large \(n\), the sum \(1 + \tfrac12 + \cdots + \tfrac1n\) is close to \(\ln n + 0.577\), where \(\ln\) is the logarithm to base \(e\). Estimate how many books give an overhang of 5 book lengths. If each book is 2 cm thick, how tall is the stack?
Research-level surprise. This harmonic staircase is best possible under the one-on-one rule, where the books go one on top of another, a single book at each level. If books may sit side by side as counterweights, the best overhang grows on the order of \(n^{1/3}\), far faster than \(\ln n\). The balance rule is the same; the geometry is much wilder.
Answers
- Two books: \(\tfrac12 + \tfrac14 = \tfrac34\). Five books: \(\tfrac{25}{24} + \tfrac1{10} = \tfrac{137}{120} \approx 1.14\).
- Each middle is \(\tfrac12\) back from the outer end: \(\tfrac16 - \tfrac12 = -\tfrac13\), \(\tfrac5{12} - \tfrac12 = -\tfrac1{12}\) and \(\tfrac{11}{12} - \tfrac12 = \tfrac5{12}\). Their sum is \(\tfrac{-4 - 1 + 5}{12} = 0\), so the average is 0. The balance point of the whole stack is exactly above the table edge: it is on the point of tipping, and no further out is possible with this design.
- \(3.7760 + \tfrac1{25} = 3.8160\), then \(3.8544,\ 3.8915,\ 3.9272,\ 3.9617,\ 3.9950\) at \(n = 30\), and \(4.0272\) at \(n = 31\). So 31 books.
- An overhang of 10 needs the sum to reach 20. \(1 + \tfrac k2 \ge 20\) when \(k = 38\), so \(2^{38} \approx 2.7 \times 10^{11}\) books are certainly enough. (Far more than needed: the estimate in the stretch puts the real number at roughly 270 million.) Going from \(2^k\) to \(2^{k+1}\) books adds at least \(\tfrac12\) to the sum, so at least \(\tfrac14\) of a book length to the overhang. The overhang grows without limit, but very slowly.
Stretch. The sum must reach 10, so \(\ln n \approx 10 - 0.577 = 9.423\) and \(n \approx e^{9.423} \approx 12\,400\). (The exact answer is 12,367.) At 2 cm each the stack is about 250 m tall, to move the top book 5 book lengths sideways.