Contact angle of liquid drop on solid surface.

AI Thread Summary
Point 44 of the referenced PDF document discusses the concept of surface tension and its directional properties. Surface tension acts tangentially to the surface of a liquid, influencing the shape of droplets and the behavior of interfaces. The direction of surface tension is determined by the forces acting on the molecules at the surface, which are unbalanced compared to those in the bulk of the liquid. Additionally, surface tension at solid-air interfaces is considered in force balance calculations for liquids to account for interactions at the boundary. Understanding these principles is crucial for analyzing fluid behavior in various applications.
amk_dbz
Messages
41
Reaction score
0
Can anyone explain point 44 of the attached pdf document on surface tension.

(Here's the link in case attachment doesn't work:
http://www.sakshieducation.com/EAMCET/QR/Physics/Jr Phy/12Surface tension _198-208_.pdf)

How is the surface tension direction found out?
Also why is surface tension for Solid-Air Surface considered when balancing forces on liquid?
Any help will be appreciated
Thank you.
 
Physics news on Phys.org
direction will be tangential to both interfaces.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top