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(III) A common effect of surface tension is the ability of a liquid to rise up a narrow tube due to what is called capillary action. Show that for a narrow tube of radius $r$ placed in a liquid of density $\rho$ and surface tension $\gamma,$ the liquid in the tube will reach a height $h=2 \gamma / \rho g r$ above the level of the liquid outside the tube, where $g$ is the gravitational acceleration. Assume that the liquid "wets" the capillary (the liquid surface is vertical at the contact with the inside of the tube).
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$h=2 \gamma / \rho g r$
Physics 101 Mechanics
Rutgers, The State University of New Jersey
University of Sheffield
University of Winnipeg
In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.
A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas.
Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion.
Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids.
Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.
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So here we know that the mass of the liquid that rises in the tube will have a force of gravity acting down on it. And then we're going to assume that the surface tension is acting perfectly vertically, upwards. And so we're going to apply essentially Newton's second law. And we know that the two forces must be equal for the liquid to be an equilibrium. And so we can say that the mass of the risen liquid is the density times the volume. So we can say that the force tension first is equaling mg. And so this becomes that surface tension times two pi r. This is going to be equal the density times pi r squared times, h times gi And then we're simply solving for H so H would be equaling 22 times the surface tension divided by the density times the acceleration due to gravity times are the radius. This would be us essentially proving that equation. That is the end of the solution. Thank you for watching
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