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Classical Physics
How Does Surface Tension Balance Small Objects on Water Surface?
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[QUOTE="ergospherical, post: 6550232, member: 691758"] Again as is described in the paper, to separate a unit area of liquid-liquid interface (##E_1 = 2\gamma_{LV}##) and a unit area of solid-solid (##E_2 = 2\gamma_{SV}##) interface before recombining them into two unit area solid-liquid interfaces has an energy cost\begin{align*} 2\gamma_{LV} + 2\gamma_{SV} - 2A_{SL} \equiv 2\gamma_{SL} \end{align*}where ##A_{SL}## is the energy reduction upon forming a solid-liquid interface. From Young's law,\begin{align*} A_{SL} = (\gamma_{SV} - \gamma_{SL}) + \gamma_{LV} = \gamma_{LV} (\cos{\theta} + 1) \end{align*}which is the force per unit length on the solid due to liquid-solid adhesion: this depends on ##\gamma_{LV}##. Finally, to obtain the full picture for the capillary action, one must also take into account the normal Laplace forces acting on the pin, due to its curvature; these are of magnitude ##\kappa \gamma_{LV}## per unit length, where ##\kappa## is the curvature. [/QUOTE]
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How Does Surface Tension Balance Small Objects on Water Surface?
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