Pressure due to a Liquid on container Wall

AI Thread Summary
The discussion revolves around calculating average pressure on container walls, specifically for cuboidal and triangular shapes. For a cuboidal container, the average pressure is derived as Pavg = pgh/2, while for a triangular wall, it is Pavg = pgh/3 due to the different locations of the geometric center. The participants clarify that the average pressure is calculated as the total force divided by the area of the surface. There is confusion regarding the total force calculations and the correct application of geometric centers for different shapes. Ultimately, the correct average pressures for the respective shapes are confirmed, emphasizing the importance of understanding geometric centers in pressure calculations.
SpectraPhy09
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Homework Statement
What is the average pressure on the triangular wall by the liquid ?
given :
density of liquid p
gravitational acc is g
For Dimensions of container please check the figure(i have attached please Check)
.( Assume Patm = 0)
Relevant Equations
Pressure at a point, x distance below the free surface of the liquid = pgx
I have just started this topic so i Don't have much clearity about it .
In our school it was tought that
Pavg = (Pi + Pf)/2 ...(i)
If we have a cuboidal shape container then it should be pgh/2 ryt ?

Screenshot_2021-11-02-10-14-54-296.jpeg


But also Pavg = Ftotal on the wall /A ...(ii)
= (pgh²l/2)/lh = pgh/2

It it was a triangular wall then Pavg comes different

Using (i) Pavg = pgh/2
(ii) Pavg = pgh/3

Which is correct ? 😓

Using
 

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  • Screenshot_2021-11-02-10-21-00-120.jpeg
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SpectraPhy09 said:
Ftotal on the wall /A ...(ii)
= (pgh²l/2)/lh
How do you get that for the total force?
And are you saying lh is the area?
 
@haruspex
Total force on a container wall
= Pressure at geometrical centre × Area of the Surface Or Pavg × Area
For Cuboidal Container
Geometric centre is at h/2 from surface
P = pgh/2
Area = l × h
F = pgh²l/2
For Triangular Surface shaped Container
Geometric centre is at centroid i.e h/3 from surface
P = pgh/3
A = h×l/2
 
SpectraPhy09 said:
Geometric centre is at h/2 from surface
Not so.
 
The average pressure on a flat surface is the total force on the surface divided by total area of the surface. What is the total area of the surface that you are interested in? Do you know how to determine the total pressure force on this surface?
 
Chestermiller said:
The average pressure on a flat surface is the total force on the surface divided by total area of the surface. What is the total area of the surface that you are interested in? Do you know how to determine the total pressure force on this surface?
Yes got it now
I was finding Pavg on the triangular wall

What I could do is this -
IMG_20211102_231132.jpg


There's a correction in the image
Pavg = (pglh^2/6)/(lh/2) = pgh/3
now is it correct ?

haruspex said:
Not so.
But why? For a rectangular surface isn't the Geomecal center at h/2?
 
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