Statics of bar under own weight.

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SUMMARY

The discussion focuses on the statics of a vertical rod under its own weight, specifically analyzing the weight distribution along the length of the rod. The weight at a point x is defined by the equation w(x) = ρ * x * g, where ρ is the density, g is the gravitational acceleration, and x is the distance from the free end. The key conclusion is that w(x) represents the partial weight of the rod from the free end to the point x, increasing from zero at the free end to the total weight at the ceiling attachment point.

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Chacabucogod
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Hi,


I was thinking about this and hit a "wall". If you have a vertical rod with constant area that is attached to the ceiling from one point its weight will be:

[itex]w(x)=ρ*x*g[\itex] (where 0 is the end of the bar)<br /> <br /> now if you take a differential above x you will have that the downward force that it's experiencing is w(x), but above the dx is w(x+dx). Where did I go wrong here. Is there something I'm not taking into consideration?<br /> <br /> Thank you.[/itex]
 
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w(x) represents the total weight of the rod only when x = L. Obviously, if x = 0 at the free end, w(0) = 0. As x runs from 0 to L, w(x), which represents the partial weight of the rod between the free end and the location x, according to your formula, must be increasing from zero to the total weight of the rod at the ceiling.
 

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