Why Is There a Limit to Drinking Water Through a Straw?

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

My understanding is that while sucking , the pressure inside the straw decreases .Atmospheric pressure on the water surface pushes the water down such that it rises in the straw, into the mouth .

I don't understand how there could be a limit on the depth of water which could be drunk .

How should I proceed ?

 

[haruspex]

how there could be a limit on the depth of water which could be drunk .

The question could be clearer.
It means the maximum height difference between the drinker's mouth and the surface of the water in the glass (outside the straw).
You will need a figure for atmospheric pressure.

 

[Jahnavi]

You will need a figure for atmospheric pressure.

760 mm of Hg .

It means the maximum height difference between the drinker's mouth and the surface of the water in the glass (outside the straw).

OK . How should I move ahead with this interpretation ?

 

[haruspex]

760 mm of Hg .
OK . How should I move ahead with this interpretation ?

What is the pressure inside the straw at the height of the water outside the straw?
How does that relate to the pressure in the mouth?

 

[Jahnavi]

What is the pressure inside the straw at the height of the water outside the straw?

Atmospheric pressure 760mm of Hg .

How does that relate to the pressure in the mouth

This is what I don't understand .

 

[haruspex]

Atmospheric pressure 760mm of Hg .This is what I don't understand .

How does pressure inside a static fluid vary with height?

 

[Jahnavi]

How does pressure inside a static fluid vary with height?

OK . Let me try .

Considering the straw is completely vertical and pressure at the top of the water ( inside the mouth ) is P_M . Pressure in the liquid at the level of the water outside the straw is P _A . Water is standing up to a height 'h' .

Using hydrostatics , P_M +hρg = P_A , we can find h .

Is that correct ?

 

[haruspex]

OK . Let me try .

Considering the straw is completely vertical and pressure at the top of the water ( inside the mouth ) is P_M . Pressure in the liquid at the level of the water outside the straw is P _A . Water is standing up to a height 'h' .

Using hydrostatics , P_M +hρg = P_A , we can find h .

Is that correct ?

Yes.

 

[Jahnavi]

Thanks .

But at some places the solutions involve buoyancy . Please see couple of the accepted solutions I found on the web . I wonder how buoyancy play any role in this .

Is it that we are interpreting the question differently ?

 

[Orodruin]

I think the most instructive way of solving this problem is to simply look at the definition of the unit mmHg, i.e., the pressure that can support a column of mercury of 1 mm. If you were trying to drink mercury (not recommended) with a differential pressure of 10 mmHg, you could therefore lift the mercury at most 10 mm. You can compute the height of the water column by computing how high the water column would have to be to have the same mass as this hypothetical mercury column. To do this you can compare the densities. In essence, it is a question of unit conversion from mmHg to mmH₂O.

 

[haruspex]

Thanks .

But at some places the solutions involve buoyancy . Please see couple of the accepted solutions I found on the web . I wonder how buoyancy play any role in this .

Is it that we are interpreting the question differently ?

Those two "solutions" are clearly derived one from the other, and they are gobbledegook. They get the right equation, but it has nothing to do with the explanations offered.

 

[Jahnavi]

Thanks !

gobbledegook

I had to actually look up what it meant

I was literally taken aback reading those explanations wondering how air was displacing water .