A leaf is denser than water, but floats on top of it because of surface tension. But how to calculate the force holding up the leaf? Surface tension has units of force per length, not area, but clearly the area of the leaf matters. It's not just the edges of the leaf that are carrying the weight.
The standard calculations with water surface tension concerns e.g. a needle floating on water, and the calculation proceeds by considering the surface tension along the two lines of the water-needle interface. If we extend this treatment to a disc, we would calculate the force as the water's surface tension times the circumference of the disc (neglecting the effect of the small contact angle). But this seems wrong to me – intuitively, it seems that the disc/leaf would experience a pressure from the surface, so that the force pushing it up against gravity would be proportional to the area, not the circumference.
Best Answer
First of all, we know in advance that the force holding up the leaf is going to be equal to its weight minus whatever buoyant force is present. That force must be equal to the surface tension effect.
Surface tension effects act only where the liquid involved presents a free surface. It is absent in the absence of any free surface; that is, a surface that is immersed completely in liquid experiences no surface tension forces.
Now note that surface tension measurements have units of force per unit length, or dynes per centimeter. To find the net magnitude of the surface tension force you multiply the surface tension number by the length of the boundary of the liquid with the solid surface touching the liquid. In this example, this is the circumference of the leaf.
So, (circumference in cm x surface tension in dynes/cm) = dynes = surface tension force.