Question: How far horizontally can you lean a stack of blocks?

Answer: As far as you want. (Sort of.)

Here's the rigorous version of the derivation.

Let's number the blocks 1,2,3, ... , i-1, i, i+1, ..., N, where 1 is on the bottom and N is on the top. The first task, as in nearly any statics problem, is to set up a co-ordinate system in which to place all the forces.

Each block feels exactly three forces: I) an upward normal force F_i from the block below it, II) a downward normal force F_i+1 by the block above it, and III) a gravitational force W downward. The normal forces come in equal and opposite pairs, of course, as each block pushes on and is pushed by its neighbors. I neglect friction forces, which shouldn't change the final result much.

The block and force positions are shown below. I let the length of each block to be 2 (so that the weight W is applied at a distance of 1 from either edge), and define distances x_i = the distance block i hangs over block (i-1) and y_i = the point where the upward normal Fi is applied to block i, as measured from the right edge of block i. Note that these variables must always satisy 0 < x_i < y_i < 2. Any other block length L can be handled by scaling all distances by L/2.

Now we're ready to write the equations of static equilibrium for the i'th block. We require that (I) the forces add to zero, and (II) the torques about any point be zero. I set the origin for the torque equation at the right bottom corner of each block, since all distances are measured from there. Writing all these equations as exact equalities will give us an extreme stack which is teetering on the edge of collapse; in practice one would need to keep all the x_i and y_i slightly smaller than the ones we will find.

Our equations are then

(I) F_i = W + F_i+1 (i.e. upward = downward forces) ,

(II) y_i * F_i = 1 W + (y_i+1- x_i+1) * F_i+1 (clockwise = counter-clockwise torques).

Using these definitions, the distance ofthe total overhang of N blocks, each with length 2, is

Max Overhang = x₁ + x₂ + x₃ + ... + x_N = Sum( x_i ) .

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To solve the equations, just start at the top N'th block and work our way down. With no block above the, its the equations reduce to

(I : top block) F_N = W ,

(II : top block) y_N * F_N = W ,

which gives y_N = 1. Since x_N < y_N , the maximum overhang of the first block is 1, or half the size of the block. This makes perfect sense.

The forces F_i for the other blocks now follow directoy from equation (I): each normal force is W more than the previous one, to support the weight of all the blocks above it. Therefore,

(I*) F_i = ( N + 1 - i ) * W ,

which gives F_N = W and F₁= N * W, as expected.

Plugging this into (II) and dividing out the weight of the block W gives

(II*) y_i * ( N + 1 - i ) = 1 + ( y_i+1 - x_i+1 ) * ( N - i ) .

Sliding out each x_i as far as possible will just set x_i = y_i , since any further would require that the normal force be exerted beyond the edge of the lower block. (In practice we will need x_i slightly smaller than y_i . This is discussed a bit further down.) Thus

(II**) y_i ( N + 1 - i ) = 1 ,

which implies that

(III) x_i = y_i = 1 / ( N + 1 - i ) .

Our series is then xN = 1, x[N-1] = 1 / 2, x[N-2] = 1 / 3, and so on until the last x₁ = 1 / N.

The total overhang is

Max Overhang of N length 2 blocks = Sum( x_i ) = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/N

Thus the answer to the original question is depends on what the limit of this series is for large N. But this is the famous harmonic series, which has no limit! The series turns out to be an approximation to log(N), which tends towards infinity as N does. However, it does get there very slowly, which means that we will need a very tall stack to have a substantial overhang.

Therefore, if N is large enough, we can make the stack lean over as much as we wish. Even if we slide all the blocks in a bit, say by 10% of the overhang, and set x_i = 0.9 ( 1 / (N+1-i) ) , so that we aren't "teetering on the edge", we can *still* grow the stack out as large as we please. Slick, eh?

The top block, block N, with length=2 and mass=1, hangs over the one below it by 1.

The top two blocks should be positioned so that their center of mass is at the right edge of block three. So where is that center of mass? Halfway between the centers of blocks N and N-1. Block N's center is at the right edge of block N-1; N-1 has it's center a distance 1 away. So the center of mass of the pair is at distance 1/2 from the right edge of the bottom one.

The top three blocks (N, N-1, N-2) then hang over the fourth (N-3) such that their center of mass is at the edge of the fourth block from the top. Where is that center of mass? The weight of the top two blocks is at the right edge of the third. The third's center is a distance 1 away. So the center of all three is 1/3 from the bottom edge.

This pattern continues all the way down. The top J blocks have their center of mass at the right edge of the next (J+1), whose center is distance 1 away. So the center of those J+1 blocks is shifted by 1/(1+J).

The total overhang distance is thus again 1 + 1/2 + 1/3 + ... .

Let's consider a somewhat realistic case. Say we're use an astronomy textbook off my shelf, which has length 30 cm and the thickness 3 cm. And suppose we pick a specific distance that we wish to extend the stack off the edge of a table, say 90 cm - nearly a meter - or three times the length of the book itself. That should be far enough to be interesting. How many books will we need, and how high will the stack be?

Well, since we chose units in our original formulation in which the length of a block was 2, three times the length of a block would be 6 units. And let's also give ourselves that 10% margin of error, so that we set the distances to 90% of their "tip over" point. Thus we need to know for what value of N we have 0.9 * (1 + 1/2 + 1/3 + ... + 1/N) = 6.

A bit of playing with a calculator shows that with 441 books, the sum is 1 + 1/2 + ... + 1/441 = 6.667. (Note that if we hadn't wanted that extra margin of error, we'd only need 227 books to get to 6.) The stack will then be ( 3 cm ) * 441 = 13.23 meters = 43.4 feet high, which is certainly something that you could do with a lot of books and a tall ceiling. That's not so bad.

I tried this with a few dozen books off my bookshelf. I was able to get the top book past the edge of the table; however, I ran into a "tilting" problem : the books are not stiff enough, and compress on the outside edge. This lets the the higher books tilt out of the horizontal plane, which starts to bring friction into an important role, and generally messes things up. I expect that metal blocks or bricks would work much better.

To lean the stack over much further, though, you need lots of blocks. Even a stack as high as a mountain, say, 16,000 ft or about 5,000 m, which would have 167,000 book-sized blocks, would only lean over by 6.3 block lengths (since 1 + 1/2 + 1/3 + ... + 1/1.67e5 = 12.6), or less than two meters. And that's without any margin of error or cheating at all.

Soon after a moutain-sized stack (logarhythmically speaking), the gravitational field of the earth starts dropping off, and begins changing direction as you go out "horizontally" - whatever that means out in space somewhere. The equations become much messier.

Nevertheless, in theory the answer is infinite, assuming you have as many incrompessible blocks as you need, as well as a nice, tall (light-years and beyond), uniform gravitational field, and of course a nice steady base to stack them on.

In practice, you can with some effort go out a few times the length of the blocks. Three times the length of the block requires 250 - 450 blocks. Six times requires over 100,000 blocks, which is starting to sound difficult. But who knows? I've seen the government fund things that sounded stranger...