What is the longest straw you could possibly drink from?

[jumbogala]

Homework Statement

What is the longest vertical straw you could possibly drink from?

Homework Equations

The Attempt at a Solution

The solution says to use p = ρgd, but I don't understand why or how you can use it.

I have a suspicion p = 101 325 Pa (atmospheric pressure)
and ρ = 1000 kg/m^3, the density of water

And it actually gives the right answer, which is 10.3 m. I just don't understand why this works or even why you use that formula. Thanks!

 

[collinsmark]

Rather than give you the answer, this might be a rewarding exercise for you to work out.

What you're trying to do (when deriving the equation), is set the pressure at the very base of the straw (caused by the column of water on top of it) equal to the atmospheric pressure.

This concept, by the way, is how standard barometers work. So it is useful.

Intermediate questions:
(a) What is the mass of the column of water in terms of density ρ, length of column d, and cross sectional area A?
(b) What is the weight of that column of water in terms of g, ρ, d and A?
(c) What pressure is caused at the base of the straw by the weight of the column (the base of the straw has a cross section area of A, by the way)?

Have phun!

[Edit: Allow me to make a correction for clarity. I said "...pressure at the very base of the straw," but what I meant to say, or should have said, was "...pressure at the very base of the water column, where this "base" is defined as the same height as the water surface in the water container from which the straw is drawing the water.]

 

[vela]

You need to think about how drinking through a straw works. Where does the force that propels the water up the straw come from?

 

[jumbogala]

Part of the reason I was confused was that I thought the pressure at the top of the water would be the atmospheric pressure. So if you have a glass of water, the pressure at the bottom of the glass should be more than atmospheric pressure... shouldn't it?

I suppose the pressure at the top of the straw should be 0 though.

Anyway, mass of the water is ρDA, and its weight is ρDAg.

P = F/A, so pressure at base of straw is P = ρDAg / A = ρDg! Cool :)

I'm still confused why the pressure at the bottom of the glass is atmospheric pressure though.

 

[Borek]

Actually 10.3m is not true - mouth muscles are not strong enough to suck that strong. So while this is kind of a theoretical limit of how far up water can be suck at the atmospheric pressure, it has nothing to do with the reality.

It is not the pressure at the bottom of the glass that counts, rather that at the water surface (outside the straw).

 

[collinsmark]

Part of the reason I was confused was that I thought the pressure at the top of the water would be the atmospheric pressure. So if you have a glass of water, the pressure at the bottom of the glass should be more than atmospheric pressure... shouldn't it?

I suppose the pressure at the top of the straw should be 0 though.

Anyway, mass of the water is ?DA, and its weight is ?DAg.

P = F/A, so pressure at base of straw is P = ?DAg / A = ?Dg! Cool :)

I'm still confused why the pressure at the bottom of the glass is atmospheric pressure though.

Hello Jumbogala,

As Borek points out below, it is a theoretical limit. It assumes the "suction" device is so strong that it can create a complete vacuum at the top of the water (even continually sucking out any residual vapor pressure, before it has a chance to reach equilibrium). Once there is a complete vacuum at the top portion of the straw, there is no way to draw the water column any higher without increasing the pressure at the bottom of the column (such as increasing the atmospheric pressure, for example). So yes, to find how high the column of water can possibly rise, one must assume zero pressure at the top.

Actually 10.3m is not true - mouth muscles are not strong enough to suck that strong. So while this is kind of a theoretical limit of how far up water can be suck at the atmospheric pressure, it has nothing to do with the reality.

Hello Borek,

Yes, it is a theoretical limit. With that I agree. But that doesn't, and shouldn't, keep physics students from discussing theoretical possibilities in the physics, restricting the discussion to the applicable physics, even if it might violate practical physiology. Examples such as Einstein's full size locomotive trains moving near the speed of light come to mind -- nothing to do with reality but still very useful as an analogy. Still, Einstein's analogies do have direct real-world applications for fast moving electrons and GPS/GLONASS satellites. Similarly, this drinking straw exercise has real-world applications for modeling barometers.

It is not the pressure at the bottom of the glass that counts, rather that at the water surface (outside the straw).

I agree completely. The bottom of the water column in question is at the same height as the surface of the water in the glass. This is the vertical point were the water in the column equals atmospheric pressure -- so we define this point as the bottom of the column. And this is true even if part of the straw sinks further down to say, the bottom of the non-empty glass.