Toppling Towers

A block tower consists of a solid rectangular prism whose height is $2$ and whose base is a square of side length $1$. A second prism, made of the same material, and with a base that’s $L$ by $1$ and a height of $1$, is attached to the top half of the first block, resulting in an overhang as shown below.

When $L$ exceeds some value, the block tower tips over. What is this critical length $L$?

While the problem was stated in three-dimensions, the depth dimension is inconsequential since we will go out on a limb and assume that the solid material that comprises the blocks is uniformly dense. Let's define the coordinate system as the right corner of the 1 x 2 block is at the origin, so that the left corner of the 1x2 block is at $(-1,0)$ and far left corner of the horizontal block is at $(-L-1, 2).$ The block tower tips over if and only if the center of gravity of the tower falls outside of the support the 1x2 block. That is, if $(\tilde{x}(L), \tilde{y}(L))$ is the center of gravity of the full tower, then the tower will fall over if and only if $\tilde{x}(L) \lt -1.$

So we've taken a three dimensional problem and reduced it down to a one dimensional problem. All we have left to do is to compute $\tilde{x}(L),$ so .... let's do that. Let's conveniently define the density so that the weight of the 1x2 block is $2$ units and the other block is $L$ units. The $x$-coordinate of the 1x2 block is at $-\frac{1}{2},$ while the $x$-coordinate of the second block is at $-\frac{L+2}{2}.$ Therefore we have $$\tilde{x}(L) = \frac{\left(-\frac{1}{2}\right) 2 + \left(-\frac{L+2}{2} \right) L}{L + 2} = -\frac{L^2 + 2L + 2}{2(L+2)}.$$ Therefore, we need to solve when $$-\frac{L^2 + 2L + 2}{2(L+2)} \lt -1,$$ which is equivalent to $L^2 - 2 \gt 0,$ so the critical length above which the tower will topple over is $L^* = \sqrt{2}.$