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I am particularly confused about finding pressure forces (specifically vertical force) on a curve surface. Although I understand the general method is to use sum of forces in the x/y direction (ie by considering the liquid weight above and so on) and find resultant force that way. I have been practising a lot of problems lately and this is one of them:

enter image description here enter image description here

So here are all my attempts:

For the pressure diagram I did, this because pressure vary linearly.

enter image description here

For the horizontal force: $F_H=\:\frac{1}{2} \times \rho \times g \times h^2 \times b$ Or in this case, $h = D = 4$ and $b = B = 5$

Vertical force (my confusion):

Is the sum of forces in the y-direction equals the weight of the liquid? Or does this also include the unknown $F$? In other words, is $F_y = F + W$ or just $F_y = W$?

Also what is the equation $x = \frac{y^2}{A}$ for? I am guessing you can use that to find the self weight? ie $W = mg = \rho V g = \gamma A w$, in this case $A = A_2$

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for $D$ = 4, $x = \frac{D^2}{4}=4$

$A_1 + A_2 = x \times D$

$A_1$ can be found by integration, $A_1=\:\int _0^D\:\frac{y^2}{4}dy\:\:=\:\frac{1}{12}D^3,\:\:D\:=\:4,\:A_1\:=\:\frac{1}{12}\left(4\right)^3$

So is$F_y = W$ Only?

d) For the last part, I assume you need to make moment at the hinge? But then how can I calculate this distance from the self weight of the fluid to the hinge?

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    $\begingroup$ Hmmm.I having the same issue too with figuring out the vertical forces, but in my problem the curve of the dam looks a bit different. I definitely want to see a solution to this question. $\endgroup$
    – Sarah V.P
    Mar 26 at 15:01
  • $\begingroup$ @SolarMike Hi solar Mike, yes I have been practising a lot. I upvoted and accepted the answer from r13 for my recent problem. I hope you don't mind, I just want to understand the concepts and I put a lot of effort into those questions. $\endgroup$
    – CountDOOKU
    Mar 26 at 15:04
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    $\begingroup$ I suggest reviewing this article. Pay attention to case 2 of the subject labeled "Total Hydrostatic Force on Curved Surfaces". mathalino.com/reviewer/fluid-mechanics-and-hydraulics/… $\endgroup$
    – r13
    Mar 26 at 22:51
  • $\begingroup$ @r13 Thanks for the article really helps for finding the vertical force, But now I don't know how to find the centroid of the irregular shape, I think this is my last part as now I can take moment about hinge and solve. Many thanks again $\endgroup$
    – CountDOOKU
    Mar 27 at 4:34
  • $\begingroup$ @r13 Actually I think I solved it. I found the centroid of the area above the boundary. But I am still unsure what the pressure diagram looks like. $\endgroup$
    – CountDOOKU
    Mar 27 at 6:44

2 Answers 2

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Hope the sketches below help the understanding of this problem. Note, the concept below holds true for any shape of the gate (straight/curved concave up/down). The difference lies in finding the volume and weight/force center of the respective shapes.

enter image description here

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$$x=y^2/4m \quad y=1/2\sqrt{x} :\ for\ y=4,\quad x=4 $$

The vertical force is the sum of infinitesimal small forces that are maximum at x=0, h=D and decrease by the following

$$F_v=\rho(d-1/2\int \sqrt{x}dx)$$. Assuming $$rho=1000kg/m^3 \quad F_v=1000(4-1/2\frac{2}{3}x^{3/2})=1000/3*/4^{3/2})=1000(4-2.66)=1333.3kg$$

This force works at a height of the Sum of the

$infinitesimal$ $\int\rho*ydx$ forces multiplied by their centroids, $y/2$ divided by the total force $F_v$.

$ \bar{D}= 1/2 *1/2\frac{ \int_{0}^{D} x \sqrt{x} dx}{F_v} =1/4*2/5 \frac{x^{5/2}}{2.666}=0.1*32/1.333=1.203m$

Now we calculate the Horizontal force on the gate and its centroid.

Then we set the sum of the moments equal to zero $M_{F_v}+ M_{F_h}+M_F=0$

Check my arithmetics. I did not have time to double-check.

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  • $\begingroup$ Hi Kamran thanks for the answer, does the pressure diagram that I have drawn make sense? Thanks. $\endgroup$
    – CountDOOKU
    Mar 27 at 4:07
  • $\begingroup$ @CountDOOKU, No! You are confused by the curve of the gate. The water pressure is independent of the shape of its container. It just is always P= rhoh. In this case, too you drop a vertical line and graph the triangular pressure distribution. if you like to lay it on the gate you should substitute the x in Y=1/2sqrt(x) to x+rhoD so y= 1/2 sqrt(x+rho*D). $\endgroup$
    – kamran
    Mar 27 at 4:24
  • $\begingroup$ So I don't draw my pressure diagram on the gate? But my pressure diagram includes both x and y components no? Not just the x component so why is it always $\rho g h$? I though I think my diagram is wrong. Many thanks again sir. $\endgroup$
    – CountDOOKU
    Mar 27 at 4:45
  • $\begingroup$ Check your math $\endgroup$
    – Phil Sweet
    Mar 27 at 14:17

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