Pressure of a sphere against the ground

Question

So we all know that the pressure of a force applied to an area is defined like this: $$P:=\frac{F}{A}$$ Where $F$ is the magnitude of the component of force applied orthogonal to the Area $A$

Now this relation is convenient for calculating the pressure of the weight of a something like cube on an area on the ground because the cube (if being stable) necessarily in all possible orientations have a contact area greater than zero.

But now consider a sphere being stable on an area on the ground. Suddenly the contact area shrinks to a "Point" which means the amount of area is zero. Does that mean pressure of the sphere exerted by its mass on the ground is infinity. If so why the ground won't tear apart.

Answer 1

The situation you are describing is well-studied: the stress distribution is known as Hertzian Stress.

Because the substrate (the surface that the sphere is resting on) and the sphere itself both have a finite elastic modulus, the surfaces will deform and the actual contact area will be a circle of finite size; the stress distribution (as a function of radius) will follow a quadratic law (as a function of distance from the center of the contact patch).

From the link above, a few key results. First, the radius $a$ of the contact area for a sphere (radius R, elastic modulus $E_1$, Poisson ratio $\nu_1$) on a plane with elastic modulus $E_2$ and Poisson ratio $\nu_2$ when subject to a force $F$ (which may be just the weight of the sphere) is given by

$$a = \sqrt[3]{\frac34 FR\left(\frac{1-\nu_1^2}{E_1}+\frac{1-\nu_2^2}{E_2}{}\right)}$$

And the maximum pressure is

$$P_0 = \frac{3F}{2\pi a^2}$$

The distribution of pressure with radial distance is then given by

$$P(r) = P_0 \sqrt{1-\frac{r^2}{a^2}}$$

as given here

Edit: Fixed formula.

Answer 2

A waving hands explanation:

It will momentarily be infinity, and then the ground will give away until the area of contact will be large enough such that the pressure is less the than the tensile strength of the ground.

P.S. This is a common explanation for taking ideal settings and thinking what would happen in the real world.

Answer 3

The contact area shrinking to zero is not a useful concept.
If the area became very small the the pressure would certainly increase and the ground in contact with the body will compress thus producing an upward force in opposition to the weight of the object.
The compression will continue until the upward force due to the compressed ground equals the weight of the body.
Note that during this process the body will also suffer a compression.
If as a result of the compression the ground has deformed permanently then what happens next depends on the body which has been compressed.
Some of the deformation of the body may have been elastic and so the body will then rebound ie start moving upwards.
On the other hand if it was a permanent compression of the body it would eventually just stop moving.

Answer 4

Both the ball and the ground are deformable media. When the ball is at rest on the ground, the contact is not at a geometric point. The deformation of both the ball and the ground result in a small but finite contact patch and an associated stress typically much smaller than the failure stress of either material.

Answer 5

Actually you have it backwards. The sphere has a finite contact patch area and well defined pressure values, but the cube has zero pressure on most of the sides and infinite at the edges.

In the diagram above the two forces $F$ are the same, and the area under pressure curves $P(x)$ is the same. But for the cube the pressure is concentrated at the edges and in the sphere is centered and more spread out.

The theory behind this is called Hertzian Contact theory. The reasoning is that materials are never perfectly rigid and they deform according to the pressure applied on their surface. This deformation forces the two sides of a contact to have no overlap.

As soon as load is applied on the contact the contact are spreads as more material deforms (squeezes). The contact area $A$ changes with the load $F$. For a sphere the area changes by $A \propto F^\frac{2}{3}$ and the peak pressure by $P_{max} \propto F^\frac{1}{3}$.

For the cube the situation is actually more complex, but in essence the contact spreads quickly from the center and it runs out of room when it reaches the edges. There in concentrates the pressure as in theory the peak pressure on a sharp edge is infinite. In real life, local deformations bring the peak pressure down from infinite, but it is still a really high value compared to the sphere case.

_{References: http://www.mech.utah.edu/~me7960/lectures/Topic7-ContactStressesAndDeformations.pdf}

Answer 6

When you get close enough to the surface of a perfect sphere is becomes effectively flat. Think about the earth. It appears not to have any curvature at all from out perspective.

But no physical sphere is perfect. Get close enough and you would see that the surface is not smooth. There are a veritable plethora of deformities on the surface.

On the other side, look at the flat surface. It only appears to be flat. Get close enough and you will see that the actual surface is erratic.