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The surface tension at the point of contact for water is inwards (as written in my book) so that would mean it vertical component is downwards but why is that vertical component considered upwards? I found this on google even here I don't understand why isn't the vertical component of $T$ cancelling out vertical component of $R$. What happens to $T\cos{x}$?

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    $\begingroup$ Does this answer your question? Surface tension and capillary rise $\endgroup$ – JansthcirlU Apr 21 '20 at 11:32
  • $\begingroup$ Nope...😔 They dont talk about the forces, im having trouble with the forces $\endgroup$ – Meow Apr 21 '20 at 11:52
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Each water molecule clings to the surface, and to surrounding water molecules, somewhat like a climber clinging to a rock (and to other climbers if there is a whole heap of climbers). The weight of the climber's body (and of anything else dangling below him on a rope) is balanced by the force between his hand and the rock face. It is similar with the water molecules and the capillary tube.

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A glass tube of very small diameter is called capillary

If we dip the capillary tube in water, due to the concave surface, pressure just below the surface becomes $P-\frac{2T}{r}$,

$T$ - surface tension of the liquid. $r$ - radius of curvature of the water surface.

while on the other points at the same horizontal level, pressure is $P$. Due to this less pressure, water level in the tube rises up, till pressure becomes equal at the same horizontal level (i.e. at points A and B)

Let $h$ be the rise in height in the capillary tube.

$P - \frac{2T}{r} + dgh = P$

$h = \frac{2T}{dgr}$

$d$ - density of the liquid.

Although we have derived it for water, it works for other liquids too which forms a conacve meniscus. In case of liquids having convex meniscus it becomes $P + \frac{2T}{R}$ instead of $P - \frac{2T}{R}$.

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Actually the normal force is always considered perpendicular to surface, so that any object in this case , water molecules doesn't enter the surface of capillary tube i.e the horizontal component force due to surface tension $T$ which is $2 \pi RT sin \theta$ gets cancelled by opposite normal force which is acting perpendicular to surface of capillary in horizontal direction, so there is only vertical component of surface tension which is $2 \pi R T \cos \theta$ which is acting and is getting balanced by whole weight of liquid column , so conclusion is normal force is perpendicular to surface of capillary tube there is no component of it in vertical direction.

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  • $\begingroup$ Isnt 2πrTcosx acting downwards then? The surface tension is acting inwards right? So its vertical component should be downwards $\endgroup$ – Meow Apr 21 '20 at 13:43
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    $\begingroup$ @Meow how can 2$\pi rT\cos x$ act downwards if the weight of whole column is to be balanced, the force vector due to surface tension $2\pi r T $ is having 2 components along vertical upward direction($2\pi r \cos x$) and along horizontal direction towards surface of capillary ($2\pi r \sin x$) where $x$ is contact angle $\endgroup$ – maverick Apr 21 '20 at 13:49
  • $\begingroup$ @Meow whole surface tension force not only act towards right but there is vertical component as well , its direction depends on contact angle, usually a concave surface is formed , and forces due to surface tension is along the boundary of liquid towards outside $\endgroup$ – maverick Apr 21 '20 at 13:55
  • $\begingroup$ Thats the part i didnt understand. why is the vertical component of the surface tension upwards? Its written in my book that the surface tension at the point of contact for water is inwards so if thats inwards its vertical component would be downwards. Is there a mistake in my book? $\endgroup$ – Meow Apr 21 '20 at 14:06
  • $\begingroup$ Yes i understood that the vertical component has to be upwards to balance the weight but it doesnt make sense why would it be upwards if the surface tension is acting inwards $\endgroup$ – Meow Apr 21 '20 at 14:07

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