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Imagine you have a cup of height $h$ which is filled with water up to $f\cdot h$, where $f\in(0,1)$. Let's say $f=\frac{2}{3}$ so it is easier to understand. Then suppose you have a straw of length $l>h$. What would be the minimum force needed for water to rise (also supposing there is an atmospheric pressure $p_0$)? I know that such a question involves pressures, so really what I'm confused about is how these pressures are added.

In the book I study from they say that $p_{\text{required}} = p_0 + D\cdot g \cdot (l-f\cdot h)$, where $D$ is the density of the water. What I don't understand is why the height of the water column is considered from the dry tip of the straw to the level of the water in the cup.

I know that in problems that involve calculating the pressure of the air that gets stuck in an upside-down cup when submerged it is simply the sum of the atmospheric pressure and that of the column of water between the level of water inside the cup and the surface of water outside. By the same logic I would expect a similar result with the straw, but that's obviously not the case. That being said, what is the idea behind adding pressures this way? Any help is much appreciated!

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The pressure is the normal force per unit area. Do a force balance on the liquid in the straw necessary for the liquid to be stationary: pressure at bottom - pressure at top keeps the height of liquid in the straw stationary. Pressure at bottom is $p_0 + Dgfh$ pressure at top is $p_{required}$. $p_0 + Dgfh - p_{required} = Dgl$.

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  • $\begingroup$ I kind of get what you are saying, but, physically, what is the $Dgl$? If I work, as you said, with forces such that water is stationary (considering that the axis is positive upwards) I get $F_{\text{required}} - G_{\text{water}} - F_{\text{of atmosphere}} = 0$. Dividing by the area of the straw: $p_r = p_0 + Dgl$. So, my guess would be that $Dgl$ is the gravitational "force" of the column of water, but I'm now missing $Dgfh$ which should be the gravitational force of the column? Which means I am ignoring a force from the start. If what I said is correct, what force am I not aware of? $\endgroup$
    – J__n
    Commented Apr 13, 2023 at 14:12
  • $\begingroup$ There are three forces acting on the stationary column of water of length l in the straw: Force up from pressure on bottom, force down from pressure on top, and force of gravity down from weight of water. The force of gravity is mg where m is mass of water. m = DAl where A is cross-sectional area of stream. Force balance is Fup = Fdown + DAlg, or using pressure Pup = Pdown + Dgl. Pup is po + Dgfh and Pdown is prequired. Note that Pup is the pressure at the bottom of the straw which is from both atmospheric pressure and the weight of water in length fh. $\endgroup$
    – John Darby
    Commented Apr 13, 2023 at 14:37
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$p_{required}$ is the pressure you are supposed to be sucking with. It has to be less than atmospheric to cause the water to rise a distance $l-fh$. so the correct answer is $p_{\text{required}} = p_0 - D\cdot g \cdot (l-f\cdot h)$.

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