Understanding surface tension in liquids

Here we discuss two different (but equivalent) ways of understanding surface tension in liquids, in terms of both forces and energy. We finish by explaining how surface tension causes liquid droplets to take on a spherical shape in the absence of external forces.
To support the channel: ko-fi.com/beny...
About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
My website: benyelverton.com/
#physics #mathematics #surfacetension #forces #energy #intermolecularforces #workdone #surfacearea #area #interface #liquid #fluid #particles #pressure #molecules #maths #math #science #education

[@MathwithMing]

Thanks so much for taking my suggestion!!

[@DrBenYelverton]

I've encountered a lot of sources that seem to suggest surface tension acts inwards so it was definitely worth making a video to clarify that it doesn't!

[@MathwithMing]

@@DrBenYelverton Perhaps the authors of those sources mean the *net* force due to surface tension on a curved surface pointing inwards to explain surface area minimization? But ultimately this is due to attracting forces among molecules so in that sense the tangential components pointing outward

[@alphadek]

Is it really necessarily attractive forces ? If the molecules are all identical, including their charge, shouldn't the force be repulsive ?

[@DrBenYelverton]

@@alphadek The molecules are neutral overall, but they can still attract each other. Fluctuations in the charge density of the electron cloud around one molecule will cause that molecule to gain a temporary dipole moment. This will then induce dipole moments in other nearby molecules, and the molecules attract through dipole-dipole interactions.

[@alphadek]

@@DrBenYelvertonok I get it !

[@hellonepp]

Cleared all my doubts. Thanks a lot❤

[@DrBenYelverton]

Glad to hear that!

[@JNVbudgam-p8t]

thank you sooo much sir ❤❤

[@DrBenYelverton]

Thanks for watching, I'm glad it was helpful!

[@JNVbudgam-p8t]

@DrBenYelverton sir kindly explain it through the ncert of class 11

[@AmeerHamza-bc4fh]

So theoretically there should be a vertical reaction R on the surface element because of net downward inward forces of attraction. How that vertical R is not there and only the surface tension which is a horizontal force comes into the surface? I'm confused about this situation and my mind is not clear. Thanks for your reply.

[@ayaan_maan]

Will you be making a series on properties of solids like elasticity, stress and strain? I really can't get my head around these concepts or the use of tensors in them (I tried reading Landau-Lifschitz but that got me nowhere).

[@DrBenYelverton]

Will try to cover this some day, it's on my to-do list. I have never read the Landau & Lifshitz books but my impression is that they're probably not the best resource for gaining intuition!

[@asharani759]

Imagine a person has compressed a massive spring with his finger and thumb, (let the total compression be x). Now the person releases the spring without any external jerk (Note that the spring has mass(M) and it is in air)... I am required to find the time in which the spring will be back in its natural length.. Sir cn u pls try to solve this question... my initial approach was finding the acceleration of topmost and bottomost points of spring, in order to find their respective displacements (wrt COM) (let be x1 and x2), thus to get back in natural length, x1+x2= x... But i am unable to draw FBD of points, and how do i consider their weight... If u have any other approach, or cn continue this method further, kindly tell. Thanks 😊

[@DrBenYelverton]

That's an interesting question, I'll think about this and see if I can figure it out!

[@supramayro434]

Solution is kinda simple: Since there are no forces acting on a body,its center of mass stays on the same place. If we consider all the situation with respect to center of mass,it would just be two springs with k'=2k (since k is proportional to 1/L and we consider half of a spring). Then we can just write a harmonic oscillator equation with frequency of oscillations equal to w=√(k/2m) and the time you need is actually 1/4 of the period,which is equal to 2π/w. Then,t=0.5*π√(2m/k)