# Would a flattened coin float or sink?

I came across this problem which mentioned that a coin which would otherwise sink in water, would float if it was flattened by a hammer. What i dont understand is how that is possible, since the volume would not change and that implies that the density of the coin doesn't change either. If we use archimed principle or buoyancy, why do you think that the coin should float in the second case?

• Surface area increases in the second case which is why it floats. It helps distribute the weight across and displace the water. In the case where it sinks, it cannot displace enough water before it starts to sink Oct 6 '16 at 12:33
• Why must weight distribution affect buoyancy of an object? Oct 6 '16 at 12:50

If you take a pin and place it into water very carefully with the pin pointed down, no matter how careful you are, the pin will sink. However, you can easily do the experiment at home where you carefully lay a pin on its side on the surface of the water and presto, it floats. The volume doesn't change, but the surface area in contact with the water does.

The equation relevant to this situation is as follows:

$$F_W=2\gamma L\cos\theta$$

where $F_W$ is the weight of what you want to float, $\gamma$ is the surface tension, $L$ is the length of the object (let's use the pin example), and $\theta$ is the angle of contact with the surface (which changes to make the equation work, if possible). IF this equation is satisfied, the object will float on the surface. Given that $\gamma$ is a fixed value and $\cos\theta$ can only vary between $0$ and $1$ (negative numbers wouldn't make any sense whatsoever in this case), it's pretty obvious that there's a minimum value of $L$ for which this equation will hold.

For our pin of mass $m$, the minimum length contacting the surface of the water would be:

$$L\ge\frac{1}{2}\frac{mg}{\gamma}$$

This corresponds to $\cos\theta=1$. If $L$ is any smaller than this value, then the equation at the top cannot be satisfied no matter what and the pin sinks. From this, it's plain to see that when you try to float the pin on its tip, the length contacting the surface, $L$, is way too small. But it turns out that on its side, $L$ is long enough that some angle $\theta$ allows this equation to be satisfied. Thus, the pin can float.

If we go back to the coin flattened by a hammer, before flattening, the case likely is that $L$ is too small. However, the coin can be flattened to increase $L$ enough that it satisfies the inequality above, and thus, can float on water.

• If one uses a strict definition of "floats", doesn't this otherwise correct argument become irrelevant? Oct 6 '16 at 13:27
• @DJohnM if by "strict definition" you mean "only due to density-related buoyancy", then yes. But I say as long as the object doesn't sink, it can be considered as floating
– Jim
Oct 6 '16 at 13:30

If a coin is flattened it doesn't lose any volume, which is why it is counter intuitive. If however you would flatten a coin, and then place it on still water (carefully on top), then the surface area is larger, and the surface tension of the water could (if the coin is sufficiently thin) hold the coin up.

The coin would not float though, if you put it in vertically (as it would break the surface layer of water).

The coin will sink in most situations though (if it is flat). If it has a bough in it made through the strikes of the hammer, then it would float even easier (but only if it doesnt get water in the indented side of the coin).

• That would mean that it is the surface tension holding the coin up and not buoyant force? Oct 6 '16 at 12:42
• Yes. Unless you change the shape of the coin, it definetly would be the surface tension, assuming its surface area has been increased sufficiently. Oct 6 '16 at 12:48
• Does surafce tension have a relation with increased surface area? Oct 6 '16 at 12:50
• This is mroe exact than I could say it: en.wikipedia.org/wiki/Surface_tension#Floating_objects Oct 6 '16 at 12:54

If it is thin enough and the water is calm enough its mass per unit area will be small enough to be supported by surface tension. Why can water flies run across the surface of a pond?

I think this is related to the surface tension of the water. For the coin to sink, the water surface has at some point to be cuten, for the water to wet the coin.

If the coin is flattened, its weight stays the same, but its surface increase a lot, especially its circumference. Then if the flat coin is placed carefully on the flat water surface, surface tension is able to work against the gravity, and the peace of metal floats.

• What do you mean by the fact that 'For the coin to sink, the water surface has at some point to be cuten, for the water to wet the coin.' i couldnt understand this. Oct 6 '16 at 12:52
• Take the flat water surface and try to insert the coin starting from the edge. When the coin penetrates the water, then the water surface has been cut; the water surface is no more continuous. Surface tension is the force which acts against that. See answer from Dirk Bruere for the example of water flies. Oct 6 '16 at 13:57

A non-wetted object placed on water makes a kind of 'dent' in the surface, which increases the water surface as it sinks down. So, a sufficiently large object will make a stronger 'dent', and the buoyant effect of surface tension (to flatten the water surface) applies more strongly to the large object.

It isn't the area of the flattened coin that matters, though (the water in contact with the flat area does not change surface area with coin elevation), but the length of the circumference of the coin. It's the negative meniscus around the circumference that is the stretched-surface source of upward force on that coin.

The circumference-to-weight ratio determines if the coin can be supported by surface tension. Water-strider insects have long feet that are applied parallel to the water surface, maximizing their likelihood of staying dry.