Statics of bar under own weight.

[Chacabucogod]

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:

$w(x)=ρ*x*g[\itex] (where 0 is the end of the bar)<br /> <br /> now if you take a differential above x you will have that the downward force that it's experiencing is w(x), but above the dx is w(x+dx). Where did I go wrong here. Is there something I'm not taking into consideration?<br /> <br /> Thank you.$

 

[SteamKing]

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.