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
![](http://bdml.stanford.edu/twiki/pub/Brunelleschi/FirstMomentOfaTriangle/_MathModePlugin_1c702dd49509b2424e6bfca113a6df4d.png)
. Then
![](http://bdml.stanford.edu/twiki/pub/Brunelleschi/FirstMomentOfaTriangle/_MathModePlugin_274e76804c129834d29525b56e78eadd.png)
for the area, which is also equal in magnitude to the applied
load,
P. The first moment is calculated as
![](http://bdml.stanford.edu/twiki/pub/Brunelleschi/FirstMomentOfaTriangle/_MathModePlugin_a34c2f979af4d95e6137b3afb11162f6.png)
where
![](http://bdml.stanford.edu/twiki/pub/Brunelleschi/FirstMomentOfaTriangle/_MathModePlugin_4153ca28f28c4bba60e31cc77fff746b.png)
. So we have
![](http://bdml.stanford.edu/twiki/pub/Brunelleschi/FirstMomentOfaTriangle/_MathModePlugin_02a59b21fbe100fc077eb718f5ba7cc0.png)
where
A = P, the applied load.
So it's at 2/3 of the width,
W.
--
MarkCutkosky - 11 Jan 2005
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