- #1
Zanathyne
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Hi, firstly allow me to say that this is not a homework question but the question fits mostly in the introductory physics section. I’m sorry if this is truly the wrong section for my question. Please direct me to right section and I will delete this thread from this section immediately. The question is just something I’ve been pondering and I would like for someone to help me figure out where I’m going wrong.
It’s a fluid static’s question I guess. I’ve been thinking about the old dipping a straw into a liquid and then holding your finger over the top, and then you extract the straw and the fluid remains inside. Of course the pressure at the top is lesser than the pressure at the bottom where the pressure at the bottom is atmospheric pressure. Using some equations I get:
Where Pb = Pressure at the bottom,
Pt = Pressure at the top,
Po= atmospheric pressure,
D = density of the fluid,
g = gravitational acceleration,
h = length of the colunm of fluid in the straw
Pb = Pt + Dgh
Pb =Po
Po= Pt +Dgh
So
Pt = Po - Dgh
The pressure at the top is lesser than the pressure at the bottom by Dgh, which makes sense.
But this is where I become confused and the math doesn’t add up.
First I’m thinking that since the water is at rest that means it is at equilibrium so the net force is zero, gravitational force = the buoyant force (I assume the force acting upward is the buoyant force)
Where A= cross sectional area of the straw (equal everywhere; meaning the straw is just a straight tube)
Ft =force acting downwards on the top of the straw
Fb = force acting upwards on the bottom of the straw
Fg = gravitational force
Fbo = buoyant force
First:
Ft = Fg
Fb = Fbo
Since they are in equilibrium:
Fbo = Fg
So Pb = Fbo/A
And Pt = Fg/A
Since Fg = Fbo
Then this must mean Pb = Pt
Which is incorrect!
To put it in plain words, the pressures differ by Dgh and since the areas are the same that means that the forces must differ (where Fb – Ft = ADgh = mg where m is the mass of the fluid causing a downward acceleration) but how can they differ if the net force = 0 which means the forces are equal. Can someone please help me out with this?
It’s a fluid static’s question I guess. I’ve been thinking about the old dipping a straw into a liquid and then holding your finger over the top, and then you extract the straw and the fluid remains inside. Of course the pressure at the top is lesser than the pressure at the bottom where the pressure at the bottom is atmospheric pressure. Using some equations I get:
Where Pb = Pressure at the bottom,
Pt = Pressure at the top,
Po= atmospheric pressure,
D = density of the fluid,
g = gravitational acceleration,
h = length of the colunm of fluid in the straw
Pb = Pt + Dgh
Pb =Po
Po= Pt +Dgh
So
Pt = Po - Dgh
The pressure at the top is lesser than the pressure at the bottom by Dgh, which makes sense.
But this is where I become confused and the math doesn’t add up.
First I’m thinking that since the water is at rest that means it is at equilibrium so the net force is zero, gravitational force = the buoyant force (I assume the force acting upward is the buoyant force)
Where A= cross sectional area of the straw (equal everywhere; meaning the straw is just a straight tube)
Ft =force acting downwards on the top of the straw
Fb = force acting upwards on the bottom of the straw
Fg = gravitational force
Fbo = buoyant force
First:
Ft = Fg
Fb = Fbo
Since they are in equilibrium:
Fbo = Fg
So Pb = Fbo/A
And Pt = Fg/A
Since Fg = Fbo
Then this must mean Pb = Pt
Which is incorrect!
To put it in plain words, the pressures differ by Dgh and since the areas are the same that means that the forces must differ (where Fb – Ft = ADgh = mg where m is the mass of the fluid causing a downward acceleration) but how can they differ if the net force = 0 which means the forces are equal. Can someone please help me out with this?