Integrating to find moments?

  • #1
I am trying to get more familiar with using calculus in unfamiliar situations, although I am stuck when thinking about moments. I am considering a wall that is depressed 0.7m into the ground and sticks out above ground by 2.0m (and has a width of w metres) and I am assuming that wind speed varies linearly with height about ground so that the wind speed at the ground is zero and at 2m is a max speed of V_0, so that V=V_0 h/2.

Now assuming the wind is stopped by the wall, the pressure of the wind at a given height is [itex]P=\frac{Force_{on small area,dA}}{dA}=\frac{mass per second_{hitting area,dA} v_{wind speed}}{dA}=\frac{\rho_{air}vdA v}{dA}=\rho v^2 = \rho v_0^2 h^2/4[/itex]

Now to find the total force of the wind, I would integrate the pressure with respect to the area, and since the width of the wall w is constant, this turns out to be
[itex]\int_{0}^{2}Pwdh =\rho w v_0^2/4 \int_{0}^{2}h^2dh[/itex]

But then for the moment about the lowest edge of the wall wedged into the ground, it would be [itex]\int_{0.7}^{2.7}Fdh[/itex] but the force is an integral of h because it is pressure x w x dh, so how do I deal with integrating with respect to h twice? Could someone please explain the principles behind this as well because I am really trying to improve my skills in applying calculus...

Note: I realise that this sounds like a homework question, but it is not. It is just something I was thinking about and trying to fiddle with myself. If it would be better suited to the homework thread though, I would be happy to move it, it's just that my main question is about applied calculus methods rather than just finding an answer to a question :)

Answers and Replies

  • #2
I also think I may have made a mistake because in my calculation for the pressure at a given height, I assumed that the height h=0 occurs at ground level, and I used this resulT of pressure in my caulculation for the moment, however in my calculation for the moment I assumed that the point 0.7m below ground was zero, so then I think it pressure equation no longer applies. Is there a way I can go about this?