
#1
Mar207, 12:21 AM

P: 1,370

Let's say there is a Ushaped tube, both ends of the tube open, filled with some fluid. The fluid is still because the atmospheric pressure pushing on both openings of the tube are the same.
Suppose the column of fluid, when streched into a straighttube, measures x. Given any crosssection of the fluid in the tube, the pressure on it measures 0. Let's look at this in detail: Consider a crosssection of fluid at some distance h below the surface of the fluid. The pressure on one side of the fluid is [tex]P_0 + \rho h g[/tex] where [itex]P_0[/itex] is the atmospheric pressure and [itex]\rho[/itex] is the density of the fluid. The pressure on the other side is [tex]P_0 + \rho (x  h) g[/tex] Since the pressure on both sides are equal, equating the two equations above yields h = x  h. This, of course, is only possible if the crosssection is in the middle of the tube (h = x/2). Hmm...did I miss something here? 



#2
Mar207, 03:58 AM

P: 70

You are wrong when you assert that the upward pressure is due to:
[tex]P_0 + \rho (x  h) g[/tex] This implies that the remainder of the fluid is all above the cross section in question. This is true at the mid point only so your calculations work there. If the cross section is not at the mid point then some of the fluid is below the cross section. Fluid below the cross section does not contribute to the pressure. The only fluid above the cross section is the bit in the other arm of the u tube which is the same depth as the fluid in the first arm. Hence the fluid is in equilibrium at all points. 



#3
Mar207, 09:22 AM

P: 1,370





#4
Mar207, 09:40 AM

Sci Advisor
HW Helper
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P: 25,161

Pressure in a Ushaped Tube
The pressure at a point in a fluid is a scalar. It doesn't point in any direction. It's the same in all directions. Is that enough of a first principle?




#5
Mar207, 11:15 AM

P: 70

The formula p= density *g*depth works out pressure at depth. Only fluid above the cross section counts. The pressure increase is due to the weight of the fluid above the area over which pressure is created. Consider a column of fluid of crossectional area A density rho and height H.
Its mass is volume*density= A*rho*H Its weight is mass*gravity = A*rho*g*H pressure = weight/area = rho*g*H Fluid below the cross section creates a reduction in pressure if its in the same side as the cross section. This balances out with pressure from the fluid below the cross setion in the other arm. Therefore no net contribution from fluid below the cross section. (Pressure at a point is a scalar but the force exerted by pressure on a surface is a vector normal to that surface) 



#6
Mar207, 12:06 PM

P: 1,370

It seems that the concept of pressure is the cause of my confusion, so for now let's forget about pressure. Consider a column of fluid in the tube. What are the forces acting on this column of fluid? The sum of the forces pushing against the top of the column are the weight of the fluid above the column and the force due to atmospheric pressure right? What about the forces pushing on the bottom side of the column? What are they?
Oh, and let's not forget about the weight of the column itself. 



#7
Mar207, 04:55 PM

P: 70

If the column rests on something ie its in a sealed container, then a reaction force from the base holds it up. If we have an open bottom to the tube then the fluid can be held up if the top of the tube is sealed. The sealed region above the column could contain a vacuum so there is no downwards force on the liquid other than its own weight. This can be held up by the force created by atmospheric pressure pushing up on the bottom of the column. Atmospheric pressure can support a 10m (30 foot) column of water. This is how most trees get water to their top branches. The height of the supported column varies from day to day because air pressure does too. Some measurements of pressure are linked to the height of fluid supported by air pressure eg mmHg (millimeters of Mercury). If there were some air in the region above the column of fluid then this would exert a downward force. If this force is larger enough it could push the column down. But as the column moves down the air expands and its pressure drops until an equilibrium is reached (assuming we have a long tube). At this point the weight of the trapped air plus the weight of the fluid column is equal to the force created by atmospheric pressure upwards on the base of the fluid column. 



#8
Mar307, 10:50 AM

P: 1,370

[tex]P_0 A + \rho g (x  x_2)  (P_0 A + \rho x_2 g) = \rho g (x  2x_2)[/tex] Since the net force is 0, then [itex]x_2 = x / 2[/itex]. This is exactly the same kind of result I derived in my first post. Hmm...I guess it is pointless to analyse the forces on a column of fluid that does not have one of its ends at exactly [itex]x / 2[/itex]. Why is this? 



#9
Mar307, 12:25 PM

PF Gold
P: 4,081





#10
Mar407, 01:41 PM

P: 1,370

I'm still at a loss though as to why it is not possible to calculate the forces on any small column of fluid within the whole column of fluid. 



#11
Mar407, 05:35 PM

PF Gold
P: 4,081

I don't understand your question. If the fluid is in equilibrium all forces cancel out. 



#12
Mar507, 09:31 AM

P: 1,370





#13
Mar507, 10:14 AM

PF Gold
P: 4,081

I don't think I need to. If nothing is moving  there are no forces.




#14
Mar507, 04:36 PM

P: 1,370





#15
Mar607, 12:00 AM

PF Gold
P: 4,081

The weight of the liquid on the left cancels out the weight on the right  so no movement.




#16
Mar607, 09:33 AM

P: 1,370





#17
Mar607, 03:36 PM

PF Gold
P: 4,081

Because the levels are the same left and right  and  there is no movement.




#18
Mar607, 03:58 PM

P: 1,370

Also, what if the levels weren't the same? How would that affect the analysis of the forces on the column of fluid? 


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