All general physics textbooks include a chapter regarding hydrostatics. Archimedes’ and Pascal’s principles are commonly covered, but surface tension is merely named. Few people knows that surface tension contributes to the buoyancy of any object, with more or less relevance.1 I think that including surface tension in general physics courses is important to understand all phenomena regarding hydrostatics, and this is why every year, before starting the lesson regarding hydrostatics, I pose a question to the students: Can 1 kg of iron float on water? Usually, some of them answer instinctively, saying, “no, that is not possible.” Others, knowing that I am used to showing them physics “tricks” in the classroom, remain thoughtful. Anyway, it is a question that would be answered negatively by around 90% of questioned people. However, as I will show, there are at least two ways to make 1 kg of iron float. When we think about 1 kg of iron or other metal, we often think about it as a compact cubic-shaped block. In that case, the answer is no, since the weight overcomes the other forces, such as the buoyancy force, among others.2–5

However, what would happen if the kilogram of iron was an extremely thin sheet? If we place it on the water, will it float? To answer this, we need to look at the forces that would act on the iron sheet. Supposing that it does not penetrate into the water, the only forces acting on it are its weight and the surface tension exerted by the water surface (Fig. 1). If the iron sheet is circular, the force exerted by the surface tension6,7 is given by
$F=2πrγ,$
(1)
Fig. 1.

Forces acting on the iron sheet on the water.

Fig. 1.

Forces acting on the iron sheet on the water.

Close modal
where r is the radius of the sheet and γ is the surface tension (γ = 72.5 × 10−3 N/m for the water). This force should be enough to counteract the weight, given by w = mg, where m is the mass of the sheet and g = 9.81 m/s2 is the gravitational acceleration. By applying the static condition, we can obtain the necessary surface to counteract the weight action, whose radius is
$r=mg2πγ,$
(2)
resulting in r = 21.5 m. From the radius, the surface area is calculated as S = 1460 m2. Considering that the density of iron is 7874 kg/m3, the thickness of the sheet should be t ≅ 87 nm. This would be an extremely thin film, but in theory, it would be possible. The other option is making the buoyancy force equal to the weight810  (Fig. 2). This force is given by
$B=ρ1gV,$
(3)
Fig. 2.

Forces acting on the iron boat on the water.

Fig. 2.

Forces acting on the iron boat on the water.

Close modal
where ρl = 1000 kg/m3 is the density of water and V is the volume of water displaced by the iron. The displaced volume needs to be V = m/ρl, resulting in V = 10−3 m3. In this case, the force due to the surface tension will be negligible.
There are many ways to achieve the necessary displaced volume, but we will assume a hemispherical shape. The volume of a hemisphere is given by $2πre3/3$. Thus, the radius of the hemisphere can be obtained as
$re=3m2πρ1,3$
(4)
resulting in re = 7.8 cm. To achieve this exterior radius with 1 kg of iron, the hemisphere must be hollow. Then, we may obtain the interior radius as
$ri=re3−3m2πρm,3$
(5)
where ρm = 7874 kg/m3, resulting in ri = 7.4 cm. Supposing a shell with hemispherical shape, another valid expression for the inner radius results:
$ri=re1−ρ1ρm.3$
(6)

The necessary thickness of the iron boat is t = ∼4 mm, and, in this way, the kilogram of iron will float.

Finally, since it is difficult to manage 1 kg of iron, we may perform a didactic experiment using aluminum foil. First, we cut a circular piece of foil and fill a basin with water. As we may observe in Fig. 3(a), the piece of aluminum remains on the surface of the water. We have measured the radius and mass of the piece of aluminum, resulting in r = 6.5 cm and m = 0.45 g, respectively. Then the surface tension, obtained with Eq. (1), and the weight are F = 2.96 × 10−2 N and w = 4.4 × 10−3 N, respectively. As the surface tension is higher than the weight, the foil floats. Having a closer look at Fig. 3(a), we observe the effect of the surface tension in the light shadows and reflections around the aluminum piece. Next, we may decrease the surface tension of water by adding liquid soap. Thus, the aluminum piece does not float but falls to the bottom of the basin [Fig. 3(b)]. Finally, making a shell with it, we can make it float again [Fig. 3(c)]. This time, it floats because of the buoyancy force, since the surface tension is negligible.

Here, I have answered the question, can 1 kg of iron float on water? It is a question that cannot be answered without some knowledge about hydrostatics. I show that there are at least two ways to achieve it: one by using the surface tension of water and the other by using the buoyancy force or Archimedes’ theorem. In both cases, the kilogram of iron surprisingly floats. Finally, we prove it experimentally by using aluminum foil. This exercise and experimental demonstration could be interesting as examples of both phenomena in introductory physics courses at all levels.

I thank my son Ahren and my daugther Nell for their help with the experiments. This work has been partially supported by Gobierno de Aragón-Fondo Social Europeo (Grupo de Tecnologías Ópticas Láser, E44_23R) and Universidad de Zaragoza (PRAUZ_23_4628 and PRAUZ_23_4652).

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Francisco Jose Torcal-Milla is an associate professor of physics in the Applied Physics Department of the University of Zaragoza (Spain). His areas of research include optics and applied physics. He is also passionate about teaching and classroom experimentation.