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!(adsbygoogle = window.adsbygoogle || []).push({});

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# Falling chimney internal torque

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