# Calculating pressure forces (vertical) on a curve surface

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:

So here are all my attempts:

For the pressure diagram I did, this because pressure vary linearly. 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 $$A = A_2$$ 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?

• 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. Mar 26 at 15:01
• @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. Mar 26 at 15:04
• 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/…
– r13
Mar 26 at 22:51
• @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 Mar 27 at 4:34
• @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. Mar 27 at 6:44

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

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.

• Hi Kamran thanks for the answer, does the pressure diagram that I have drawn make sense? Thanks. Mar 27 at 4:07
• @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). Mar 27 at 4:24
• 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. Mar 27 at 4:45
• Check your math Mar 27 at 14:17