The question is at least supposed to be simple but yeah, I need some detailed answers. I have tried thinking about it in the lines of surface tension but it seems what is required of me is more than that so if anyone has faced a similar question, your help would be most needed.
[Physics] the difference between pressure in water droplet and that in an air bubble
bubblespressuresurface-tension
Related Solutions
About the surface tension of mixtures, this is quite complex, and the answer is certainly not straight. However, it looks like the surface tension of the mixture of two fluids is always in between the surface tension of the pure fluids somehow (so it is simpler than azeotropes for the boiling temperatures of mixtures). The dependency of the surface tension on the concentration may be linear, but usually is parabolic. You can find some graphs of surface tension as a fonction of the concentration of the mixture in this paper. About the temperature dependence, as far as I know, surface tension always decreases with temperature. It is due to the fact that higher thermal fluctuations give smaller cohesion.
Surface tension is a quite confusing subject, especially viewed from a purely mechanical point of view. It appears whenever you have an interface between a condensed phase say $A$ and another immiscible fluid phase $B$.
Thus the first thing to note is that surface tension has always to do with an interface. The surface tension coefficient often denoted $\gamma_{A,B}$ will tell how "costly" it is, in term of energy, for such an interface to exist.
Now the reason why it is costly to have such an interface is ultimately due to the effective adhesion forces between the molecules in each phase. To simplify a bit, there are two principles at play:
(1) In a quite good approximation, molecules interact with van der Waals (vdW) interactions which are always attractive (in vacuum). Furthermore, the vdW forces are the strongest with molecules of the same kind.
(2) In a dense phase of certain molecules, the cohesive energy density is higher than for the same molecules in a more dilute phase.
These two rules have two implications:
If phases $A$ and $B$ comprises the same molecules but have very different densities (e.g. liquid water/water vapour interface), then by the rule (2) there is a big loss in cohesive energy density for each piece of interface created between the two phases. From a mechanical point of view, it is fine to say that molecules in the liquid phase are simply pulled stronger towards the liquid phase than the gas phase.
If phases $A$ and $B$ are two condensed phases comprising different molecules, then by the rule (1), it is also costly to generate an interface between $A$ and $B$.
This leads to the property that the surface tension coefficient $\gamma_{AB}$ is always positive.
Now, in most real cases, multiple interfaces are involved at the same time. Most of the time three interfaces. This is the case for the meniscus you mention but also for the insects walking on water.
To discuss the insect example, one needs to guess whether its legs are wetting or not. If they were, then it is likely that it could not walk on water as it would be preferable for it to actually sink in water. It must have quite a lot of short straight hairs on the legs to induce a hydrophobic effect effectively "repelling" water and inducing only a single contact point with water and then one only needs to care about the deformation of the water/air interface.
Now, regarding the direction of the force, one needs to discriminate two things:
the total interaction between a phase $A$ and a phase $B$
the surface tension interaction between phase $A$ and $B$
While the former accounts for all possible forces between the phases, the latter is only concerned with the shape of an interface and acts by definition tangentially to the interface.
For example, in the first example you mention, this is a mixture of both:
First, the liquid wets the rope which more or less implies a strong adhesion with it, second the liquid exerts a tension related to the $\gamma_{air/soap}$ interface which acts along the interface air/soap but perpendicular to the interface rope/soap; that's mainly because we consider ourselves in a case of ultra-ideal wetting. Thus what it says is that Nature prefers gaining a bit of energy by extending the interface air/soap a bit rather than gaining a much bigger amount of energy by detaching the rope or whatever object you might use from the soap film.
Try the same experiment with a tube made of GoreTex, I am not sure you would get the same outcome.
Best Answer
Let us solve this problem by considering the free energy $F$ of the whole system in thermal equilibrium (temperature and particle number constant, so $d F = - p \; dV$). In fact the type of system does not matter as far as we are considering a bubble of some gaseous or liquid substance inside another substance (can be the same one). Let us for now just look at the case of an air bubble (label $a$) with a spherical surface ($s$) in water (liquid $l$).
The total free energy (differential) of the system is $$dF = dF_a + dF_l + dF_s.$$ The individual free energies of air and liquid are $dF_a = -p_a\; dV$ ($V = V_a$) and $dF_l = -p_l\; dV_l = p_l\; dV$ since an increase in volume of the liquid is equivalent to the decrease in volume of the air. The free energy of the surface can be described as $dF = \Gamma dA$ where $\Gamma$ is the surface tension and $A$ is the surface area of the bubble. Inserting this into the equation above gives
$$0 = -p_a dV + p_l dV + \Gamma dA \qquad \text{or} \qquad p_a = p_l + \frac{dA}{dV} \Gamma.$$
Assuming a spherical geometry of the bubble we can write $dA/dV = 4 \pi \; d(r^2)/(4\pi \; d(r^3)/3) = 3 \times 1/(3 r^2) \times d(r^2)/dr = 2/r$ and
$$p_a = p_l + \frac{2}{r} \Gamma.$$
The calculation is equivalent for different combinations of substance (air bubble in air (soap), water droplet in air, air bubble in water).