
#1
Jul1409, 08:30 AM

P: 9

Hi all,
I was asked today by a friend a simple question but couldn't answer so I'm asking you. Let's assume we have a vertical cylinder full of water, what is the pressure on the sides of the cylinder? If it was the pressure downwards it simply L*g*rho but to the sides I'm not sure... 10x a lot, Adam. 



#2
Jul1409, 09:36 AM

PF Gold
P: 37

Hi.
The preassure in the cylinder on depth [tex]z[/tex] is [tex]p_0+\rho g z[/tex], where [tex]p_0[/tex] is an atmosphere. The force exerting on a small ring with height [tex]dz[/tex] on depth [tex]z[/tex] inside the cylinder is: [tex](p_0+\rho g z)2 \pi r dz[/tex] For the whole cylinder this force is: [tex]\int_0^L (p_0+\rho g z)2 \pi r dz=2 \pi r L (p_0 + \frac{1}{2}\rho g L)[/tex] For the bottom of the cylinder the force is: [tex](p_0+\rho g L)\pi r^2[/tex] 



#3
Jul1509, 03:57 AM

P: 9

The force you have calculated is the force downwards.
I'm not sure it is the same answer for the sides of the cylinder. How is it reasonable that the pressure downwards is the same as the pressure to the sides. Adam. 



#4
Jul1509, 05:31 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Simple question about water pressure in a cylinder 



#5
Jul1509, 06:04 AM

P: 9

Ah, now I get it.
It's funny, you can learn QFT and GR but you realize you sometimes have shortage in fundamental physics... 10q very very much. 


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