Can You Topple the Tower?

A block tower is a solid rectangular prism of height $2$ on a square base of side $1$. A second prism of the same material, with a base $L$ by $1$ and height $1$, is fixed to the top half of the first block, sticking out to one side as an overhang. When $L$ grows past some value, the tower topples. What is that critical length $L$?

The Fiddler, Zach Wissner-Gross, December 12, 2025(original post)

Solution

Take the ground edge of the base as $x=0$ and the far edge as $x=1$; the overhang juts out to $x>1$. The tower tips when the combined centre of mass passes the supporting edge $x=1$. Density is uniform, so mass is volume: the lower block has mass $2$ with its centre at $x=\tfrac12$, and the overhang has mass $L$ with its centre at $x=1+\tfrac L2$. Height does not enter, since toppling about a horizontal edge depends only on the horizontal position of the centre of mass. Setting that position to the pivot, $$\frac{2\cdot\tfrac12 + L\bigl(1+\tfrac L2\bigr)}{2 + L} = 1 \;\Longrightarrow\; 1 + L + \tfrac{L^2}{2} = 2 + L \;\Longrightarrow\; L^2 = 2 .$$ So the tower topples once $$L > \boxed{\sqrt2}.$$

The computation

Solve the toppling condition itself: set the combined centre of mass (mass $2$ at $x=\tfrac12$, mass $L$ at $x=1+\tfrac L2$) equal to the pivot $x=1$ and solve for $L$.

from sympy import symbols, solve
L = symbols('L', positive=True)
print(solve((2*0.5 + L*(1 + L/2)) - 1*(2 + L), L))   # [sqrt(2)]

Extra Credit

Now the tower is an annular sector: the region between two arcs of angle $\theta$ on circles of radius $1$ and $2$. For small $\theta$ it rests happily on one of its two flat radial faces; past some angle it can no longer balance on a flat face. What is that critical $\theta$?

Solution

Rest the sector on its radial face along the $x$-axis from $(1,0)$ to $(2,0)$, the body curving up through angles $0$ to $\theta$. Using polar coordinates with area element $r\,dr\,d\varphi$, the horizontal centre of mass is $$\bar x = \frac{1}{A}\int_0^\theta\!\!\int_1^2 (r\cos\varphi)\,r\,dr\,d\varphi = \frac{14\sin\theta}{9\theta}.$$ As $\theta\to0$ this is $\tfrac{14}9\approx1.56$, safely inside the face $[1,2]$; balance is lost when $\bar x$ reaches the inner edge $x=1$, that is when $\sin\theta=\tfrac{9}{14}\theta$, whose positive root is $$\theta = \boxed{1.555\ldots\text{ rad}}\;\approx 89.1^\circ.$$

The computation

Integrate the sector’s geometry directly: for each $\theta$ compute the horizontal centroid $\bar x(\theta)$ on a polar grid, then root-find the angle at which it slips back to the inner edge $x=1$.

import numpy as np
from scipy.optimize import brentq
def centroid_x(theta):
    r = np.linspace(1, 2, 500); phi = np.linspace(0, theta, 500)
    R, PHI = np.meshgrid(r, phi); w = R                # area element r dr dphi
    return (R*np.cos(PHI)*w).sum()/w.sum()
theta = brentq(lambda t: centroid_x(t) - 1, 0.5, 3.0)
print(round(theta, 3), round(np.degrees(theta), 1))    # 1.555  89.1