# Why is there a $1/2$ factor in the surface tension for a thin film?

According to Wikipedia on the surface tension of a thin film: $$\gamma = \frac{1}{2} \frac{F}{L}$$ Where $\gamma$ is the surface tension, $L$ is the length of the movable side and $F$ is the force required to stop it from sliding. (Please refer to the picture of a thin film below, also from Wikipedia)

Why is there a $1/2$ factor included in the equation? Some books said it's because there are two layers of the liquid. But in the case of a thin film, shouldn't there be only one layer? (i.e. one molecule thick). My current understanding is that there are supposedly two layers and each layer exerts an equal amount of force and thus the $1/2$ factor.

EDIT

Here is a picture to clarify what I am thinking (although it doesn't exactly depicts a "thin" film.) It shows a huge magnification of the cross-section around the boundary between the movable rod and the liquid.

$ABCH$ is the cross-section of the movable rod and $DEFG$ is the cross-section of the liquid. The circles in the liquid are liquid molecules, named as lower case letters for the sake of conversation.

Now lets consider molecule $f$ for instance. It stays static because there is a balance of cohesive forces in every direction. Molecules on the liquid surfaces such as $a$ and $d$, however, experience a net downward liquid-cohesive force and therefore must be balanced by the adhesive force from the movable rod, resulting in surface tension. Each line of molecules (along $a$ and $d$) exerts a pulling force of $F=\gamma L$. There are two lines of molecules so $F = 2 \gamma L$.

However, molecules $b$ and $c$ also experience net downward liquid-cohesive force and therefore must be balanced by the adhesive force from the movable rod, resulting in a pulling force exerted by molecules $b$ and $c$ on the movable rod. Therefore $F$ no longer equals $2 \gamma L$ but equals $4 \gamma L$ in this case.

In case of a thin film, there are two layers or rather surfaces, I would say. 2 surfaces, that is, the one that is visible in the given diagram and the one on the opposite side of this surface that cannot be seen in the picture.
The net surface tension is due to the 2 surfaces pulling the slider back with equal force. So the factor $\frac{1}{2}$ is present in the equation.

EDIT: According to your picture (I am re-posting your picture for clarification), there are certain flaws in your understanding which I think I have understood now. I mention them as points:

1. First of all, in case of a THIN FILM, there will not exist molecules like $b$ and $c$. That is what is meant by a thin film and hence this formula is so simple without any complex mathematics.
2. Secondly, I must admit that a thin film is rare and that your case can be roughly approximated to be a thin film. Even then you must know that there is something called an angle of contact that exists due to the difference in adhesive forces and cohesive forces. In this case, the water film will bend outwards to prepare a somewhat concave meniscus and hence those $b$ and $c$ molecules will spread to the surface and be a part of it.

I hope I am clear with my answer and it will clear your doubts. Ask me if it still hurts.

• I understand that surface tension is caused by the imbalance of attractive intermolecular forces. The amount of force shouldn't depend on the number of surfaces, but rather the number of layers. The molecules on the edge of each layer will experience imbalance force. So a higher number of layers (i.e. a thicker edge) should exert more force, thus a higher surface tension, disagreeing with the equation. – krismath Oct 17 '15 at 7:48
• You are partly correct. The amount of force truly depends on the number of layers. But the name surface tension suggests that it has something to do with surfaces.The molecules in the layers exert force on the slider but these forces cancel out. The net force is due to the 2 surfaces since they have no layer of liquid molecules on the outer side. – SchrodingersCat Oct 17 '15 at 15:59
• Yes, the molecules on the surface have no layer of liquid molecules on the other side, but in fact all the molecules adjacent to the movable side have no liquid molecules adjacent on the other side either. Thus, it shouldn't be only the molecules on the surface, adjacent the the movable rod, but it should be all of the molecule adjacent to the movable rod (including ones that are not on the surface.) – krismath Oct 24 '15 at 13:13
• Can you supply a picture showing what you are trying to say? It is becoming a bit difficult to frame everything from your description. – SchrodingersCat Oct 24 '15 at 15:23
• I have added a picture and further descriptions in my original post. Hope you could understand it. I am kinda desparate to get an answer. This has been bothering me and my friends for a while now. – krismath Oct 24 '15 at 17:07