Not in this first case, where the energy of the air-glass interface is larger than the combined air-water and glass-water energy - this gives a zero contact angle. For contact angle, see video 11 of this chapter: 2 5 1 5 Contact angle effects on capillary pressure.
I find it hard to believe this is right. In fact that equation only works considering a spherical surface. My confusion comes when you have to think about how would you derive this equation considering a cylindrical surface :( Also is this pressure the one that the liquids exerts on the air or the other way around?
This does take a lot of thinking to understand. Locally at all points on the air-water interface the surface fits a sphere (so long as the curved surface is much smaller than 2 mm). So though the water touches along a line to the cylindrical tube, it forms a hemisphere.
I think the equation is lack of cos(contact angle). The meaning of capillary pressure (when be used in the tube) is as same as the Laplase-pressure. The difference is the R (effective radius of the interface) changes into r/cos(contact angle) (r is the radius of the tube).
Keep in mind that the cos-gamma representation is just looking at the force balance at the three-phase contact, which is really a very stochastic system, depending on local energetics. The first thing to think about is the pressure on an arbitrary "saddle point" on an air-water interface. We derive this in the book, but it is a bit too much for a short video.
@@daveyboy4715 Since the pressures are negative, when unsaturated they won't burst in this case. Also, if the soil becomes saturated and the pressures are positive, then the water is all under compression, so again, no bursting!
Pressure actually does not have direction - it is a scalar, not a vector. So the pressure at any point simply has a single value. The pressure is zero at the surface of the water in the tray, at the bottom, but by hydrostatic principles, pressure in non-moving liquid is always greater going downward (think of diving underwater), and the pressure reduces as you go up in the water column.
@@selkerjss Thank you for your answer. What I was trying to find out is how the parts of the force balance equation (in 4:11) are put together. Which part of the system does it describe? Is it the surface of the meniscus? Is P the pressure as we would measure it in the liquid just under the surface of the water?
@@tomaszmasternak2225 Since the system is symmetrical, forces in the lateral directions are balanced, and net zero. The force balance is for the forces acting in the vertical direction, balanced at the height of the meniscus. You can compute the downward force in two ways. One is to compute the mass of the water being held up, which is 2* pi* r^2 *h*density, where h is the height of the water column. The other way to compute it is to compute the vertical component of the pressure. The easiest way to think of this is to draw a flat circle right below the curved interface. The component of force upward due to the negative pressure is simply that pressure multiplied by the surface area, pi*r^2. It might be easier to visualize by thinking of a cylindrical bucket full of water. What is the downward force on the bottom surface? It is the pressure (density*g*h) times the area of the bottom of the bucket. Here we are just looking at the bucket upside-down!
