I've been trying to calculate the amount of pressure on a wall of a rectangular structure, that is under sea watter.

For example a dam. I can't come to see any example to be able to locate the answer to my problem.

Sea Water density= 1.025 g/m3
Sea Water specific gravity= 1025.8 Kg/m3
Sea Water specific weight= 10100 N/m3

//wall structure is:
base= 15000 m
height= 10 m
Area= 150,000 m

My questions are:

  • Does the volume of the vessel make a diference on the wall's pressure?
  • How much pressure do these messures exert?

I spent almost 8 hours researching information about hydrostatic pressure and viewing examples, but most of these explain about water on a dam's gate instead of the whole building itself

  • $\begingroup$ The pressure at a depth $h$ is $\rho gh$. This pressure acts against the wall and is irrespective of the total volume of water or anything else. $\endgroup$
    – lemon
    Mar 1, 2015 at 21:05
  • $\begingroup$ What is g on this formula? $\endgroup$ Mar 3, 2015 at 2:04
  • $\begingroup$ $g$ is the gravitational acceleration, $9.8$ m/s$^2$. And $\rho$ is the density of the water. $\endgroup$
    – lemon
    Mar 3, 2015 at 10:32

2 Answers 2


Since the pressure is dependent on depth, and your depth will vary if the structure (like a dam) is vertical, you will need to take the derivative of a vertical slice of the structure, and from there you will be able to multiply it by the horizontal (XY plane) area.

your dx will be the change in depth making up for the "height" of the area for that horizontal sliver of equal pressure. Then you need a width, and then the depth, plus density ($\rho$) and gravity.

$$\int \rho gh \times \text{area of slice you are finding}$$

and $$\text{area} = dx \text{ (the infinitesimally tiny height for which pressure will be equal)} \times \text{the width of the slice}$$

This is a calculus problem because the pressure varies as you go deeper.

Hope this helps.


The force on the dam is the integral of the pressure over the area. $$F_{dam} = \int_{A}{p(z)dA}=\int_{0}^{Z}p(z)dz\int_{0}^{L}{dx}$$ The pressure is a function of mass density, gravitational acceleration, and depth. $$p(z)=\rho g z$$ Then the force on the dam is $F_{dam}=\frac{\rho g L Z^2}{2}$


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