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5:36 AM
@PinkAura I'm not sure what calculation you're referring to. Can you give me a link describing the calculation?
 
6:24 AM
@JohnRennie sir i don't have the link but i can send you the texts , i searched it on chatGPT
im able to derive this for simple projected area, but it gets complicated for shapes such as semi circular arc,.. Also, is centroid same as COM?
 
I can't remember if the centroid is always the same as the COM or not.
I'm busy at the moment. I'll ping you as soon as I'm free.
 
ok sir
 
 
2 hours later…
8:22 AM
@PinkAura Hi
 
@JohnRennie hi sir
 
The way the calculation works is you divide up the liquid above the base into infinitesimal columns with area dA (area measured at the base).
Then the force is dF = ρgh dA where h is the height of the infinitesimal column.
 
yes
 
Then you integrate to get the total force.
and the column area dA is the projection of the surface at the top of the column onto the base, which is where the projection come in.
 
but i dont find the reason it is nothing but the pressure at the centroid
 
8:28 AM
@PinkAura I must admit that isn't obvious to me either.
It must just work out that way.
 
hmm
 
Suppose you can define an average height h₀ for the surface of the liquid, and we won't worry yet exactly what we mean by this average.
 
yes
 
Then we can write h = x + h₀ where x is positive if h > h₀ and negative if h < h₀
 
yes
 
8:30 AM
Then our integral becomes:
F = ∫ ρgh dA
 
yes
 
F = ∫ ρg(h₀ + x) dA
F = ∫ ρgh₀ dA + ∫ ρgx dA
 
i see
 
Now in the second term x can be negative and positive because some parts of the surface are below h₀ and some parts will be above h₀.
 
yes
sir can we take an example?
 
8:34 AM
Suppose we choose h₀ so that the positive and negative x terms cancel to make the second integral zero, our integral becomes just:
F = ∫ ρgh₀ dA
And since ρ, g and h₀ are all constants this is just:
F = ρgh₀A
And if h₀ is the height of the centroid then this matches what the ChatGPT post said.
Yes?
 
yes
 
So the definition makes sense if the centroid happens to be the point where:
∫ ρgx dA = 0
Physically this is saying suppose we start with the surface horizontal and draw an axis through the centroid,
then we rotate the surface about this axis.
The does the volume below the horizontal equal the volume above the horizontal?
I don't know if that's a property of the centroid or not.
 
sir but the fact is centroid coincides with com...
but com is believe is the point about which mass moments of all points is zero
is centroid's definition same?
 
@PinkAura I don't know. It shouldn't be hard to Google.
 
ok sir
 
8:42 AM
For simple objects it looks like the centroid and COM are the same.
 
ohh yes sir
i wonder the definition of centroid is same as com : en.wikipedia.org/wiki/Centroid
basically centroid is related to the shape of the lamina but com is actually related to mass distribution : we defined com for uniform density and then we defined centroid and we fixed the definition for centroid for a particular case of com, i believe
 
The articles I found say the centroid and the COM are same if the area density is constant.
So that makes sense. With a constant area density the shape is directly related to the mass distribution.
I'm not sure how you would derive this from first principles, but it seems plausible.
 

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