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Imagine I have a vertical straw, and that it is partially filled with water. If I were to decrease the radius of the straw at the bottom end, eventually surface tension would allow the water to remain in the straw (i.e. the force of gravity would not be enough to overcome the surface tension and for water to flow out of the straw).

What I'm curious to understand is for a certain (small) radius at the bottom, how much force/pressure would be required at the top of the straw to overcome surface tension and force the water out of the bottom of the straw at a certain (volume flow) rate.

Any help would be greatly appreciated!

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At the critical condition(i.e when the surface tension is not enough), it is easy to see that the radius of curvature of the meniscus becomes equal to the radius of curvature of the straw, or equivalently, angle of contact becomes zero degrees.Thus from free body diagram of straw we can easily see that$$P=P_o-\frac{2T}{R}+\rho gh$$, where T is surface tension, R is radius of straw, h is length of straw, other variables have usual meaning.( NOTE that I have taken a different case, for your or any other case, free body diagrams may be different, thus make suitable changes accordingly.)enter image description here

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  • $\begingroup$ You forgot to draw surface tension at the bottom in free body diagram. $\endgroup$
    – lucas
    Commented Apr 21, 2016 at 4:30
  • $\begingroup$ Sorry, but i mentioned that I have taken a different case( to simplify the question, and provide a basic model). $\endgroup$
    – oops
    Commented Apr 21, 2016 at 4:32

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