Pressure due to a Liquid on container Wall

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Homework Help Overview

The discussion revolves around the calculation of average pressure exerted by a liquid on the walls of containers with different shapes, specifically cuboidal and triangular containers. Participants are exploring the relationships between pressure, force, and area in the context of fluid mechanics.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulas for average pressure in cuboidal and triangular containers, questioning the validity of different expressions for average pressure. There are attempts to derive total force and average pressure based on geometric centers and area calculations.

Discussion Status

The discussion is active, with participants questioning assumptions about geometric centers and the application of pressure formulas. Some have provided insights into calculating total force and average pressure, while others are seeking clarification on these concepts. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are navigating through potential misunderstandings regarding the geometric center of different shapes and how it affects pressure calculations. There is mention of specific formulas and their applicability to different container shapes, indicating a need for clarity on these points.

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
 

Attachments

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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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