Velocity Needed For Water To Overcome Adhesion Force To A Rounded Steel Inclined Plane

Question

Context: I work at a fast food restaurant with a 3-module sink. The sides are slightly angled down to allow the water to trickle down into the compartments. I found that water clings to the side of sink due to adhesion and cohesion of the water while running down instead of flying off the edge.

Question: what velocity does cold water have to reach while trickling down a steel round edged incline plane to overcome its adhesion to the surface material of the incline plane and fly off the edge at any arbitrary angle?

(I hope the diagram helps)

Answer 1

First let's get a few things straight,

• When it flies off both the normal and adhesive forces will be 0.

• Actually in this question there is no need to talk about adhesive forces because normal is a self adjusting force, if there were no adhesive forces normal would have been lowered a bit because,

$$\ F_{adhesive} - \ N_{Along Radius Of Curvature} +\ mg_{Along Radius Of Curvature} = 0$$

Please note that since the curve is not a circular one, normal force will also be taken along the radius of curvature of the point the water flies off, also $ \ g_{Along Radius Of Curvature} $ is not gravitational acceleration but the component of gravitational force along radius of curvature divided by mass of water that flies off.

In case there were no adhesive forces,

$$ \ -N_{Along Radius Of Curvature} +\ mg_{Along Radius Of Curvature} = 0$$

We can see that in both cases the resultant force by the surface is actually equal and it is equal to $\ mg_{Along Radius Of Curvature}$.

• For the sake of convenience, I am assuming that the ledge is making a cylindrical curve with the sink, as in the curve looked from the side view is a part of a circle (the view the diagram provides us with). We could still solve it otherwise it is just that the radius we would take would be radius of curvature.

Now we can just apply conservation of energy from a point on the ledge till the point on the curve where the water will fly off.

We would get a relation between initial velocity and v_escape.

Then you can say that at the point of escape $$\ mg_{Along Radius Of Curvature} = \frac {m(v_{escape})^2}{R}$$

With this you would get a neat relationship in between the initial velocity required to have the water flow off at angle $\theta$, to find minimum angle you can differentiate the expression with respect to the angle to find minimum velocity required.