# I Stacking bricks-Physics puzzle

1. Aug 13, 2016

### Andreas C

Part 1:
You have an unlimited supply of indestructible bricks that are 10cm long and 4cm high. After some thougt, you realize that building a huge wall to keep immigrants out is a dumb idea, you decide to do something more creative with them. Since sticking things together, reinforcing structures and using concrete are all techniques foreign to you, you decide to put one brick on the ground, put another one on top of if, then another one on top of that, etc. How far to the right can you get the edge of the last brick from the center of the first one without making the whole structure colapse?

Part 2
You now decide to make a sort of an arc. You start stacking bricks leaning towards the right, and then you stop and start stacking them towards the left, until the center of the last brick is over the center of the first brick. If you draw a line from the bottom brick to the top brick in this structure, what can the maximum distance of a brick from that line be?

2. Aug 15, 2016

### haruspex

If this is homework, please repost to the homework->precalc forum.

3. Aug 15, 2016

### Andreas C

That is not homework, that is a puzzle intended to be solved by the members of this forum! If you know the answer, feel free to post it, but with a spoiler, so that others can solve it themselves!

4. Aug 15, 2016

### haruspex

Oh, ok. The first part is very well known, so I won't post a solution to that.
For part 2, I offer an answer to a different question.
The number of bricks in the lower section, heading to the right, is roughly the square of the number of bricks in the upper section.

Last edited: Aug 15, 2016
5. Aug 16, 2016

### Andreas C

That counts as an answer I guess. Can anyone prove this conclusion?

6. Aug 16, 2016

### Staff: Mentor

Single-sided stack: brick n from the top can be 1/(2n) brick lengths ahead of brick n+1 to stay in balance, that is the optimal strategy. The harmonic series diverges, therefore arbitrary overhead length can be reached, but the number of bricks needed is about n = e2x up to a numerical prefactor, where x is the overhang length in multiples of the brick length. Note that doubling x means squaring n.

Two-sided: The top stack is like the single-sided stack, at the transition region we can go back by nearly one brick length in one step, afterwards we are stuck in the same pattern again. Neglecting this one brick length, we get the same situation as for the single-sided stack, where the bottom part doubles x and therefore squares the number of total blocks. For large n we can approximate n(n-1) as n2.

7. Aug 16, 2016

### Andreas C

Good answer, but did you mean "nearly half a brick length" instead of "nearly one brick length"? You can't really go more than half a brick length, unless of course you mean precisely AT the transitional step...

8. Aug 16, 2016

### Staff: Mentor

Yes, I mean the transition step: the first in the opposite direction, as seen from the top. We can go backwards by 1 - 1/(2n) where n is the number of bricks that were stacked in one direction.