The load is balanced by a pressure. If the load is off-center the pressure must be non-uniform. Assuming a linear distribution, it would be a triangle.

So where is the center of mass of a triangle? Equivalently, where is the line of action for a triangular pressure distribution?

Let triangle be of width *W* and height . Then for the area, which is also equal in magnitude to the applied
load, *P*. The first moment is calculated as where . So we have where *A = P*, the applied load.

So it's at 2/3 of the width, *W*.

-- MarkCutkosky - 11 Jan 2005

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