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?
[Physics] Would a flattened coin float or sink
buoyancysurface-tensionwater
Best Answer
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.