
#1
Jun1905, 02:39 PM

P: 39

A piston having a crosssectional area of 0.07 m^2 is located in a cylinder containing water as shown in the figure. An open utube manometer is connected to the cylinder as shown. For h1=60 mm and h=100 mm, what is the value of the applied force, P, acting on the piston? The weight of the piston is negligible.
I'm confused because there is more than one "u"shape in the figure, so, it's unclear to me how to proceed. Help would be appreciated! legend for figure : p = piston first weird looking h = h1 second weird looking h = h piston is present on top of the water. Force P downwards on piston. 



#2
Jun1905, 03:59 PM

Sci Advisor
HW Helper
P: 3,033

The U on the right is the manometer. I've never seen a manometer filled with water on one side, but that's the way your diagram looks. Is there any air between the water and the mercury? It's not clear to me why the 60mm is given, but the height of the water relative to the top of the mercury is important




#3
Jun1905, 07:19 PM

P: 39

Here's a scan of the piston I'm talking about.




#4
Jun1905, 07:32 PM

Mentor
P: 40,907

Fluid Mechanics  Pressure  UTube Manometer
This fact may help: Since the fluid is connected (and in hydrostatic equilibrium) the pressure at any height within the same fluid must be the same throughout.




#5
Jun1905, 07:53 PM

P: 39

Ok, so far this is what I've got. Let me know if I'm on the right track.
If you take the pressure at the piston as P1 then, P1 + [tex]\gamma[/tex]H2O*h1  [tex]\gamma[/tex]Hg*h = 0 Solve for p1 and then P(Force)= p1*Area 



#6
Jun1905, 08:08 PM

Mentor
P: 40,907

Don't forget that the tube is open to the air. (But you are on the right track.)




#7
Jun1905, 08:17 PM

P: 39

Hmmm. I did take that into account when I did [tex]\gamma[/tex](Hg)*h




#8
Jun1905, 08:26 PM

Mentor
P: 40,907





#9
Jun1905, 08:29 PM

P: 39

well, isn't the pressure at the open end supposed to be equal to zero?




#11
Jun1905, 08:53 PM

P: 39

hmmm then what would it be? because my book says "at the open end the pressure is zero."




#12
Jun1905, 09:40 PM

Sci Advisor
HW Helper
P: 3,033

Don't forget to calculate the force. If all you have done is caclulate the added pressure, you are not finished. 



#13
Jun1905, 09:47 PM

P: 39

so the force P = p1*A, correct?
I stated that above in my post...is this incomplete? And in any case, if Pair was not neglected, how would you write it into the equation? 



#14
Jun1905, 09:56 PM

Sci Advisor
HW Helper
P: 3,033

Sorry. I had missed your force calculation. That is OK.
If you had to include air pressure in the force you would need to look up the normal pressure of air (1 atmosphere) and multiply by the area of the piston to find the force the air exerts on the piston. 


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