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In the case of equilibrium of hemispherical drop, what i know is that the surface tension is responsible for holding the hemispherical drop with the other hemisphere thus forming a spherical drop and the surface tension also holds the liquid on the curved surface thus giving it the curved shape. Why is force due to surface tension measured only along periphery(T2πr) whereas the surface tension exists also on the surface(2πr²) ?

Equilibrium condition: Atmospheric pressure pushing the curved surface along with the surface tension from the outside and pressure inside the liquid acting in the opposite direction to both these forces thus maintaining an equilibrium

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  • $\begingroup$ Please provide more information about "the case of equilibrium of hemispherical drop." $\endgroup$
    – sammy gerbil
    Commented Jan 10, 2020 at 10:13
  • $\begingroup$ Atmospheric pressure pushing the curved surface along with the surface tension from the outside and pressure inside the liquid acting in the opposite direction to both this maintaining an equilibrium. Now I'll add this to my post $\endgroup$
    – Vivek karunakaran
    Commented Jan 10, 2020 at 10:18
  • $\begingroup$ Please can you post the problem which you have been asked to solve. $\endgroup$
    – sammy gerbil
    Commented Jan 11, 2020 at 3:20
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    $\begingroup$ That's not a problem actually. It's the derivation for finding the excess pressure inside a liquid drop. It's there all over the internet but no one tells the reason behind this circumference approach $\endgroup$
    – Vivek karunakaran
    Commented Jan 11, 2020 at 5:33
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    $\begingroup$ No. If the tension is completely across the surface(curved part) then why are we calculate it only along the base circumference? That's my question $\endgroup$
    – Vivek karunakaran
    Commented Jan 11, 2020 at 19:11

3 Answers 3

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I think I understand your question. You wish to ask why the surface tension force acts only on the periphery (circumference) of the hemispherical water droplet. Well, it doesn't. It acts all over the surface, at every point. To understand the answer, draw any given line of a very small length (say $dx$) anywhere on the surface. The force $Tdl$ acts on it due to tension by its neighbouring molecules, say in a given direction (right, say). But it too exerts a net equal and opposite surface tension force $Tdl$ (towards the left). So the net force on the small line is zero. This argument can be extended to the whole surface. The only region where this line experiences a net non-zero force is the periphery. So the net force due to surface tension is exerted on the periphery as shown and is given by (assuming surface tension to be constant) $$F= \int{Tdl} = T \cdot 2\pi R$$ enter image description here

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  • $\begingroup$ In that case shouldn't it be experiencing a net force by the molecules below the periphery as well as above since the hemisphere is an assumption from the whole Shperical drop despite shown isolated as a hemisphere here?? If so net force in the periphery case also would be zero $\endgroup$
    – Vivek karunakaran
    Commented Jan 11, 2020 at 12:06
  • $\begingroup$ Yes, when dealing with the whole sphere, the net force at this periphery will obviously also be zero. I have shown net non-zero force on periphery assuming only a hemispherical drop. When trying to find excess pressure inside water droplet, we deal with hemisphere because the assumption that the force by the left hemisphere on the right one is the same as the force by the right hemisphere on the left one. So we get the equation $$P_0 (\pi r^2) + T(2\pi R) = P(\pi r^2)$$ when we apply it for a hemisphere, not the whole sphere. $\endgroup$
    – kushal
    Commented Jan 11, 2020 at 12:14
  • $\begingroup$ Is it clear now? $\endgroup$
    – kushal
    Commented Jan 12, 2020 at 17:33
  • $\begingroup$ I understood that both the hemispheres hold each other through attraction along the surface but why do we have to find it in upwards direction towards the other half when the considered hemisphere contains molecules only below the periphery? Isn't that the molecules which are in the bottom surface responsible for excess pressure by contracting the surface? Hence shouldn't we be finding the force due to surface tension pointing in downward direction $\endgroup$
    – Vivek karunakaran
    Commented Jan 12, 2020 at 18:11
  • $\begingroup$ Got it. We are taking only the external forces from things like atmosphere, pressure by other half and surface tension by other half. Right? $\endgroup$
    – Vivek karunakaran
    Commented Jan 12, 2020 at 18:14
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Excess pressure inside a liquid drop

In the calculation you consider the balance of forces on a hemisphere of liquid. The only forces which are relevant are external forces which other objects exert on this hemisphere. These include the pull of surface tension $\sigma$ from the other hemisphere at the rim where they join, and the pressure $p$ from the liquid inside the other hemisphere acting across the flat face of this hemisphere (the yellow area in the diagram). (Also the atmospheric pressure across the blue surface, which I have not shown. So in effect in my diagram $p$ is the excess pressure.)

It is not necessary to consider the surface tension forces at other points in the surface of this hemisphere (the blue area in the diagram) because these are internal forces. They always occur in equal and opposite pairs between adjacent elements of water which are both in this hemisphere. It is only along the circular rim of the hemisphere that the opposite paired surface tension forces are missing because they act on water which is in the other hemisphere not this one.

enter image description here

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  • $\begingroup$ I think i understood it. Calculating the surface tension without something to experience the force in a resultant direction doesn't make sense eventhough that object is under stress $\endgroup$
    – Vivek karunakaran
    Commented Jan 11, 2020 at 19:08
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The surface acts like a stretched membrane. The attachment point of that membrane is the triple junction between the solid surface, the air, and the liquid surface. The summation of the forces involved at that triple point is what results in the contact angle of the liquid on the surface.

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