Pressure Difference in a Drinking Straw

  • Thread starter Thread starter Zanathyne
  • Start date Start date
AI Thread Summary
The discussion revolves around the physics of pressure differences in a drinking straw, specifically how fluid statics apply when a straw is submerged in liquid and then removed with a finger over the top. The initial confusion stems from the assumption that the forces acting on the fluid are in equilibrium, leading to the incorrect conclusion that the pressures at the top and bottom of the straw must be equal. Clarification is provided that there are three forces at play: the atmospheric pressure on the bottom, the pressure at the top, and the weight of the fluid column. The correct understanding reveals that the pressure difference is indeed due to the gravitational force acting on the fluid, leading to the conclusion that the pressures differ by the term Dgh. Ultimately, the resolution highlights the importance of considering all forces involved in the equilibrium of the fluid system.
Zanathyne
Messages
2
Reaction score
0
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?
 
Physics news on Phys.org
Zanathyne said:
...
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
...

at this stage an error kreeps in ...

I would prefer if you would rather call the "bouyant force" the upwards force say Fu, since bouyancy has another meaning in physics. This force is caused by the atmospheric pressure on the bottom of the water column in the straw. It can be found by multiplying your eqaution

Pt = Po - Dgh

with A giving

APo = APt + ADgh - which can be written as APo = APt + mg via the density giving

Fu = APt + mg or Fu = Ft + mg

the first term gives Fu (or your Fbo). So you have missed a term in your reasoning - the force due to the air pressure on top of the liquid in the straw (I chose Ft for the force caused by this air pressure).
 
Last edited:
andrevdh said:
at this stage an error kreeps in ...

I would prefer if you would rather call the "bouyant force" the upwards force say Fu, since bouyancy has another meaning in physics. This force is caused by the atmospheric pressure on the bottom of the water column in the straw. It can be found by multiplying your eqaution

Pt = Po - Dgh

with A giving

APo = APt + ADgh - which can be written as APo = APt + mg via the density giving

Fu = APt + mg or Fu = Ft + mg

the first term gives Fu (or your Fbo). So you have missed a term in your reasoning - the force due to the air pressure on top of the liquid in the straw (I chose Ft for the force caused by this air pressure).


Thanks for responding to my thread. I see where I went wrong! This says there are three forces not two as I first believed, those being the forces brought on by the pressure at the bottom and the pressure at the top and the one that I can't believe I missed (although I stated it) the force caused by the mass of the water! ha ha :smile: so these three forces are in equilibrium, from your equation:
Fu = Ft + mg
Fu - Ft - mg =0
perfect sense. Thanks.
 
It's a pleasure.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top