dianad
Mechanical
- Dec 27, 2007
- 66
Hi,
In hydrostatics everyone learned that P=F/A
p=Pressure
F=force
A=area
From this relation we can determine for a cilinder(for example):
P=(m x g)/A = (mv x V x g)/A = (mv x A x h x g)/A = mv x h x g
m=mass (kg)
mv=specific heigth (kg/m3)
V=volume
h=height
g=gravity acceleration
So, from this late relation: P=mv.h.g , wich corresponds to STEVIN theorem.
My problem is that with Stein theorem, says that Pressure only depends on the depth and always is presented the image that i've attached.
But if we use for both images the relation P=F/A, instead of the Stevin relation, to determine the pressure at the bottom, we have the following:
m=10kg (both)
A1=2 m2
A2=4 m2
P1=10kg * 10m/s2 / 2m2 = 50 Pa
P2=10kg * 10m/s2 / 4mw = 20 Pa
If we used the relation of Stein, we would have the same pressure for both.
What is your comment about this?
Thanks
In hydrostatics everyone learned that P=F/A
p=Pressure
F=force
A=area
From this relation we can determine for a cilinder(for example):
P=(m x g)/A = (mv x V x g)/A = (mv x A x h x g)/A = mv x h x g
m=mass (kg)
mv=specific heigth (kg/m3)
V=volume
h=height
g=gravity acceleration
So, from this late relation: P=mv.h.g , wich corresponds to STEVIN theorem.
My problem is that with Stein theorem, says that Pressure only depends on the depth and always is presented the image that i've attached.
But if we use for both images the relation P=F/A, instead of the Stevin relation, to determine the pressure at the bottom, we have the following:
m=10kg (both)
A1=2 m2
A2=4 m2
P1=10kg * 10m/s2 / 2m2 = 50 Pa
P2=10kg * 10m/s2 / 4mw = 20 Pa
If we used the relation of Stein, we would have the same pressure for both.
What is your comment about this?
Thanks