0
$\begingroup$

My teacher is very strict about copying his work so I will have to explain it.

We have a wall with two fluids of different constant densities and depths. The wall is 'L' long. The wall has a section that goes sideways a certain depth in and then down all the way to the bottom of the tank. We have already solved for the pressures as a function of z on both sides. Previous exercises had us find the resultant force on different sections of the wall. Now we are asked to find the torque on the lower portion of wall about point P.

I see that 2 things will be changing with z - the pressure force exerted from both fluids and the moment arm. I am not too great with integration - If I could get pointers on where to start with my integration that would be appreciated. refer to the picture.

Improve this question
$\endgroup$
2
  • $\begingroup$ 'I am not too great with integration' You can get through this without actually computing any integrals. Translate the co-ordinate axes so the origin is halfway up the wall, then with (quite a lot of) algebraic rearrangement, the torque can be written as a weighted sum of: one integral that goes to zero by symmetry (so you don't have to compute it in detail); one integral that is the definition of the surface area of the wall (and you know how to compute the area of a rectangle without doing any integration);... $\endgroup$
    – Daniel Hatton
    Commented Nov 20, 2021 at 11:22
  • $\begingroup$ ... and one integral that is the definition of the second moment of area of the wall about its centre line (and you can look up the second moment of area of a rectangle about its centre line in a table without doing any integration). $\endgroup$
    – Daniel Hatton
    Commented Nov 20, 2021 at 11:25

2 Answers 2

Reset to default
0
$\begingroup$

I believe I need to set my limits of integration so that point P is zero and then my torque will be P(z)*z --> (integral from 0 to -z+depth of P) P(z)zdz I can do that for both sides and add the torques

Improve this answer
$\endgroup$
0
$\begingroup$

Assume point "P" is rigid, calculate the fluid pressures on both sides of the wall, from the respective free water surface down to point "P", then sum the moments about it. I assume you know how to calculate the resultant forces, and the locations of their application (1/3 height of the respective pressure diagram above the base).

enter image description here

Improve this answer
$\endgroup$
3
  • $\begingroup$ Just idle curiosity, why are you ignoring the contribution to the torque from the horizontal pb ? $\endgroup$
    – Greg Locock
    Commented Dec 15, 2022 at 22:30
  • $\begingroup$ Only the pressures above point "P" causes torque at that point. $\endgroup$
    – r13
    Commented Dec 15, 2022 at 22:59
  • $\begingroup$ Oh, got it, yes, i hadn't realised the thing was not able to rotate about P. $\endgroup$
    – Greg Locock
    Commented Dec 15, 2022 at 23:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.