How to calculate the force holding a leaf floating on water? 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. 
 A: 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.
A: Any object that 'floats' on a liquid surface has a static equilibrium between the downward body force (due to downward pull of gravity), a buoyant force due to the displacement of liquid volume by the body, and a surface tension force in the direction of the contact angle (the meniscus angle).
The pin floating in water does not float solely by means of the surface tension, although indeed it is the dominant force. If you look carefully part of the pin is below the surface of the water, and so there is a buoyant force as well as surface tension.
The body forces act on the entire body, although one can resolve all forces at the body's center of mass as a single downward force. Likewise the buoyant forces act on the entire body and can be resolved as a single upward force at the center of buoyancy which is the centroid of the displaced liquid volume.
The surface tension force is a bit different. Think of the surface of the liquid as a net made of small mesh of fibers. An object placed on the net either stretches the fibers around the object or if too heavy, rip the fibers and allow the object to fall through the net. The surface of water for example acts like this net by the stretching of hydrogen bonds which are more plentiful at the surface than within the body of water below. That's because of the air interface above. The surface tension acts along a line, the line that surrounds the body at the water-air interface for example. That's the location where stretching of the bonds takes place and it's why the units are indeed in terms of length not area. It's the force per unit length around this peripheral contact area.
So to complete the vertical equilibrium of the floating object you need to know not only the surface tension force, but also the peripheral boundary (length) and the net contact angle (due to tension and buoyancy). Using this angle to resolve a vertical component. This vertical component can be resolved to a single vector acting upwards at the centroid of the area defined by the peripheral contact line.
If the centers do not line up, there will be moments that cause rotation of the object - until equilibrium is reached.
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