This is a great follow on to Cantilever bridges, if they've seen that experiment this expands on one of the initial things many people try.

Your challenge is given N identical blocks place them in a stack (one on top of each other) and try and maximise the overhang over a table edge.

If they've seen Cantilever they may try the cantilever solution, however the twist here is you're only allowed one brick per layer (this means the tower must be N bricks tall).

We'll now talk through some physics that we'll need to figure out how well we can do.

Centre of Mass - We can view gravity as acting through one point of the block, we call this the CoM, if the block is uniform it'll be in the centre, sort of where you'd expect.

Try placing the block on the edge of the table and moving it's CoM around, where can it balance? You'll find that you can get out until the CoM is right on the edge but no further. This means for N=1 we've solved the problem! The answer is 1/2 (times the length of the block).

Moments - A Moment is a turning force, this is what we get when the block is on the table. For a moment, we need a pivot. With one block our only pivot is the point on the edge of the table, friction holds the block enough so that what we experience here is just a turning force (initially you'll notice a slip later once it's started falling, see the falling toast experiment for more on this though). The forces are the gravitation acting through the CoM and a reaction force from the table supporting the brick.

Levers - levers increase the force with distance. With moments we need to consider a levered force by timising the force by distance from the pivot. If the forces clockwise and counterclockwise (ccw) balance then there's no tip. If we have an unbalanced force twisting us into the table then it won't tip but we could extend further!

So going for N=2 case now. We can keep extending like before but we need to decide which order is going to give us the best overhang. As we need to consider the CoM it would be better if there was the least total overhang, this means that the overhang between blocks on one layer and the one above should increase as we get higher up.

Obviously don't tell them this let them try it. If they place something on top of the N=1 solution it'll pivot (unless it's not more overhanging). They'll actually need to place something underneath it.

The maximum turns out to be 1/2 + 1/4 = 3/4.

Ask them how far out they think they can reach? Can they get arbitarily far if they can make N large?

It turns out you can, the optimum is 1/2*\sum_{i=1}^{N} 1/i. This is half the harmonic series so it does in fact diverge, which you can show as 2^n terms in a row are all greater than 1/2^n so sum to 1. However it doesn't diverge quickly! To span a gap of 1,2,3,4,5,6,... you need N=4, 31, 227, 1674, 12367, 91380, ...

Asymptotically this is roughly log(N)

You can allow multiple blocks per level if you like, this copies cantilever bridges and the solution is the same. Counterbalancing gives the optimum and it scales like N^{1/3}.