# Pressure on the sides of a container?

Sorry if this is an incredibly basic question for these categories. Basically, I don't understand why I'm not getting these types of problem. I'm sure it's something really simple I'm missing.

So let's say there's an open swimming pool with width $w$, length $l$, depth $d$, and density $\rho$ (equal to water's density).

So basically, I'm quite sure the formula I should be using for this is $P = \rho gd$. This turns into $\frac{F}{A}=\rho gd$

How would I go about finding the force the of the water onto the sides ($w$ and $l$)? In a problem like this, what would the area $A$ represent? If I wanted to find the pressure on a $w$ side, would I use $w\times h$? I try this but it doesn't work. I get $F = \rho gwh^2$. But this answer is double the actual answer. It seems like I'd need to integrate (that's what I tried first) but it didn't work either.

There has to be some really basic concept I didn't understand.

• Define the terms first. What are h, w and l? – UKH May 12 '16 at 8:31
• d = 2.5m, w = 14m, l = 24m. Is that what you mean by define? – hhh May 12 '16 at 8:33
• No, what these terms indicate – UKH May 12 '16 at 8:35
• if you integrate you should get a $\frac{1}{2}$ factor – Jolie May 12 '16 at 8:42
• w h and l are multiplied to create different possible areas. d (or h) is the depth of the water from the surface. – hhh May 12 '16 at 8:42

The pressure at a particular depth, $h$, is $h\rho g$.

Force on a rectangular strip of infinitesimal height $\text dh$ and width $w$ is $h\rho g\text dhw$ which is when integrated for the entire surface becomes $\frac{1}{2}\rho gh^2w$.

Similarly for other sides.

• I understand this for the most part, but why is the h from pgh excluded in the integration? Is that h supposed to be constant or something? – hhh May 12 '16 at 9:25
• It is not excluded..integration of $h\text dh$ is $h^2/2$. $\rho, g, w$ are constant – Jolie May 12 '16 at 12:13

At a certain depth a liquid exerts the same pressure on all sides. So if you want to calculate the force exerted by water on sides w and l then you have to put in the value of depth in the formula. Whatever the answer will come will be the pressure exerted by water on the sides w and l for that particular depth.

As for the ratio F/A is concerned, it states the amount of force F applied on the area A.

For eg. if the pressure is 5 pascals (SI unit of force) then it means that 5 Newtons of force is being applied on 1 m^2 of area.

Summary : the pressure on sides w and l vary with the depth so at different depths the pressure will be different. And you have to calculate them separately.

Hope it helps !

Pressure is indeed F/A however force varies upon the depth of the water therefore to find the force you need to

"divide each area into small bands of areas, each having its own pressure (rho)gh, then add up those little forces you have got to gain the total force on the wall"

Which this is actually the concept of integration

Therefore you will get (integrate)(rho)gh x wdh from 0 to h

So from this you would get 0.5rho g w h^2

If you aren't familiar with integration, look up distributed loadings, it gives the same answers with less time :D

• "(integrate)(rho)gh x wdh from 0 to h" Where are you getting "wdh" from? Wouldn't it just be wh? Also, I looked at my old notes and it kind of makes sense to me. Basically, the notes said "dF = PdA", and I can easily turn dA into "wdh". However, why is P (aka "pgh") excluded from this integration if it includes h? – hhh May 12 '16 at 9:19
• Wdh is simply the small band of area i told you, then why isn't it hdw you might asked, that is because the pressure varies with h, therefore our procedures are dividing h into small h, each having their unique value of pressure, in fact P (pgh) is also not excluded but infact is integrated using h as a variable (as we are integrating dh), feel free to ask if you are still unclear :D – Derpson May 12 '16 at 9:31

You can indeed write $F=\rho gdA$, but the problem is that the force isn't constant throughout the entire area. You need to find an area with constant force, which would be a horizontal strip, because the depth is the same. The force on small strip (at the $w$ side) at depth $x$ with height $\Delta x$ is $\rho g x A=\rho g xw\Delta x$. The force isn't actually constant, but the smaller $\Delta x$, the more accurate it becomes. By letting $\Delta x$ get so small it becomes $dx$ and integrating over $x$ you get $$\int^{h}_{0}{\rho g wx\ dx}=\Big[\tfrac{1}{2}\rho gwx^2\Big]^{h}_{0}=\tfrac{1}{2}\rho g wh^2$$ Because $$f(x)=x^a \rightarrow F(x)=\tfrac{1}{a+1}{x^{a+1}+C}$$