# Falling chimney internal torque

1. Aug 6, 2010

### dEdt

Suppose a (uniform) chimney began falling. Where would the chimney break? I approached the problem by considering the torque acting on a point along the chimney, at a distance x from the point of rotation. I got the internal torque + external torque = (mx^2)(3gcosA/2L), where A is the angle between the chimney and the ground, m is the mass of the point, and L is the length of the chimney. The external torque is just mgcosA; using this, I get that the internal torque is maximized at x = L/3. This, at least, was the given solution. But thinking some more, I began doubting my solution. Shouldn't I be trying to maximize the internal force rather than the internal torque? But doing that yields silly answers, so that seems wrong too. Help please!

2. Aug 6, 2010

### zhermes

Figuring out exactly how a chimney would break would be an extremely complicated problem! To solve it--as you clearly know--you have to make simplifying assumptions.
For instance, you have to guess why it would break--i.e. from shear forces? compression? expansion? torque?

Conceptually, torque seems like a good bet--i.e. the chimney would snap, in which case your analysis is great.

3. Aug 6, 2010

### dEdt

I can't see why torque would make it snap though. So what if the torque is maximized there? Worse still, the second derivative is actually positive at x = L/3, so I think it's a minimum in actuality.