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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. 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
Copyright &© by the contributing authors. All material on this collaboration platform is the property of the contributing authors. Ideas, requests, problems regarding TWiki? Send feedback