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If a one cent coin is set carefully onto the surface of water horizontally (flat side parallel to the water surface). What are the relevant forces to determine if it would sink.

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  • $\begingroup$ en.wikipedia.org/wiki/Surface_tension#Floating_objects. $\endgroup$ – valerio Jun 23 '16 at 12:40
  • $\begingroup$ I saw that but the shape is what bugs me. Everywher ee you look for the surface tension you see sphere taken into consideration. what if that shape is flat on the water. $\endgroup$ – user3699627 Jun 23 '16 at 12:42
  • $\begingroup$ The situation is the same, but you will have to measure $\cos \theta$ experimentally, I guess. $\endgroup$ – valerio Jun 23 '16 at 12:43
  • $\begingroup$ When there's a corner, here of 90 degrees, the contact line stays pinned on it within the corresponding range of angles. This makes the situation much easier than in the case of a sphere. $\endgroup$ – Joce Jul 7 '16 at 9:57
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This foto is the first such a foto ever made;

Timo Suvanto, floating coin

The "secret" is told on his blog in Finnish. Including the whole calculations;

Mass of the coin; $0.45 g$
Diameter; $15.8mm$ So the force produced by surface tension should be more than the force produced by the weight of the coin. $T=\frac{mxg}{l}=\frac{0.00045 kgx9.81m/s^2}{\pi x0.0158m}=89\frac{mN}{m}$

But the known value for surface tension of water is $72.8\frac{mN}{m}$

Thus, the coin should NOT float.

The secret of the trick is told in the blog; - The water is cold, and has thus more surface tension. - The coin is treated with grease, nd this turns the contact angle of water around, causing the effective radius to increase. (This was the explanation from the blog.)

But if we calculate this with +10% radius; $T=\frac{0.00045 kgx9.81m/s^2}{\pi x0.0174m}=80.8\frac{mN}{m}$

And we use the surface tension in $0\unicode{xb0}C;75.6$ $mN$

So we still don't have mathematically solid explanation! Solving the diameter $l=\frac{mxg}{T}=\frac{0.00045 kgx9.81m/s^2}{\pi x0.0756N/m}=18.6$ $mm$ Which gives a factor $1.176$ for "effective" diameter.

this picture with simple scale shows that this is not plausible; floating penny

With the distance of $1\frac{1}{3}$ the surface is allready complete neutral.

There is one thing still missing; Buoyancy. The penny is clearly below the surface.

This penni from year 1979 is aluminium. So it's density is $2.7$ $g/cm^3$ and volume $0.45x2.7=1.1215$ $g/cm^3$ Or it's "weight on water" is $0.45/2.7=16.7$ $g$

$T=\frac{0.000167 kgx9.81m/s^2}{\pi x0.0158m}=32.9\frac{mN}{m}$

So now we are in the correct range, as It should not be forgotten, that this penny has kinetic energy when it's placed on water.

Normally this is done so, that the penni is attached on finger with the adhesion of the grease, which carries it's own weight in air. This adhesion must be released against the tensions of water, which means that at least factor 2 is needed to place the coin in the water.

Most probaply the person who has done this, has not calculated it before, he has just tried it and succeed. And the calculations are done afterwards.

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