4

It is common to see a heap of conical shape formed by a large number of similar size hard spherical objects, for example, a heap of pebbles, sand etc. Suppose we want to model this system as a collection of identical hard balls, with dry friction between the balls described by coefficient $\mu_1$ and the friction between the balls and the floor described by coefficient $\mu_2$. Based on a "toy" problem with three cylinders forming a pyramid one can conjecture that there is some threshold condition for $\mu_1$ and $\mu_2$ that guarantees equilibrium for a heap. What are the conditions for a heap of identical hard spheres to be in a static equilibrium?

BioPhysicist
  • 56,248
Maxim Umansky
  • 4,377
  • And this would be independent of the packing geometry and the slope of the pile? That seems pretty unlikely. – Emilio Pisanty Mar 04 '16 at 06:04
  • Good point. But sand piles do exist, apparently with some irregular packing. What are the conditions for their existence? There are only two important parameters here , $\mu_1$ and $\mu_2$, the rest should not matter - diameter, gravity, number of balls (if it is large) - correct? Without friction no equilibrium here - correct? – Maxim Umansky Mar 04 '16 at 06:19
  • the packing could be random, especially if you don't take spheres of having the same size (which could create artificial strange situations). – Fabrice NEYRET Mar 04 '16 at 06:39
  • You could I would be inclined to solving this computationally using granular dynamics (e.g. here and here). Would make for quite a nice project. – lemon Mar 04 '16 at 07:42
  • @lemon Probably the brute force computational approach is the only option for irregular packing. But for a regular arrangement one can probably solve it with pencil and paper. One can try it first in 2D, like in the three-cylinder arrangement, then add another layer etc. From my preliminary analysis one condition will be $\mu_1 > \tan(15^{\circ})$ – Maxim Umansky Mar 04 '16 at 16:15
  • @MaximUmansky could you augment the question with some of that preliminary analysis? – David Z Mar 05 '16 at 11:28
  • @DavidZ Would not it make more sense to present it as an answer to the question? It is not the full answer but covers an important particular case. – Maxim Umansky Mar 05 '16 at 16:43
  • Yeah, you could do that too. But presumably you have some specific problem that you want answered, and in order to have the best chance of getting it answered, you'd want to give as much relevant information as possible (up to the point where it becomes a distraction). – David Z Mar 05 '16 at 17:32

2 Answers2

4

I don't have a full answer (which would probably require some computer simulations) but one can rather easily analyze a case of a regular pyramid in hexagonal close packing, assuming strong friction on the floor (non-slip condition). This would be something like a bunch of marbles on a rubber mat.

First look at it in 2D, so we have an arrangement like in the Figure. Consider the forces on the lower right cylinder. At the threshold of stability there is no force from the second right cylinder in the bottom row, and the lower right cylinder is at the threshold of slippage at point B (no slippage at point A by the assumption). The force $F_1$ from the top log applied at point B is directed towards point A so that there is no net torque on the lower right cylinder. Therefore we immediately find from the geometry the condition $\mu_1 \geq \tan(15^{\circ})$, otherwise the lower right cylinder would start rolling to the right.

Now, back to 3D; if there is a regular pyramid of identical spheres in the HCP packing then an outermost sphere in the bottom layer would have a single sphere in the layer above pushing on it. Then situation is analogous to the 2D case except instead of 60$^{\circ}$ for the angle DCB we would use $\arccos(\sqrt{2/3})$ (from the regular tetrahedron geometry) which is about 35$^{\circ}$. So in 3D the critical value of $\mu_1$ becomes $\tan(({90^{\circ}-\arccos(\sqrt{2/3})})/2) \approx \tan(26^{\circ})$. For a general packing this condition will probably still be necessary (but not sufficient) for the equilibrium to exist (if we exclude unstable arrangements such as the cubic lattice).

enter image description here

sammy gerbil
  • 27,277
Maxim Umansky
  • 4,377
  • @sammy gerbil I am not sure I understand your comment. Why are you saying it is $\mu_2$ rather than $\mu_1$? As I explained above, this solution corresponds to the limit when $\mu_2$ is large, so $\mu_2$ is completely out of the picture. Also, I am puzzled by that comment about the force $Mg$, my solution uses torques with respect to point A; the gravity force does not create a torque w/respect to that point, so the gravity force is out of the picture. – Maxim Umansky Jun 05 '21 at 17:13
  • Sincere apologies @Maxim. You are correct, and your explanation is clear and simple. My own calculation was much more complicated and has caused me to make a mistake. – sammy gerbil Jun 08 '21 at 10:38
  • As a comment, your result is very different from that suggested by Niels - viz. $\mu = \tan\theta$ where $\theta$ is the angle of repose which is $60^{\circ}$ here. – sammy gerbil Jun 08 '21 at 10:49
2

This is a semi-solved problem in civil engineering, where the angle made by a pile of small objects is called the angle of repose, which depends on the density, size distribution, and sharpness or irregularity of the objects in the pile. If the objects in the pile are randomly stacked, small compared to the size of the pile, and the coefficient of static friction between them is known, then an approximate solution for the angle of repose is

angle of repose = arctan(coefficient of static friction)

niels nielsen
  • 92,630
  • 1
    Good answer, but would be nice to have a link to a good article or source on angle of repose. e.g. https://en.wikipedia.org/wiki/Angle_of_repose – fyodrpetrovich Aug 08 '19 at 14:19