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The formula for calculating pressure would be:
$$P=\frac{F}{A}$$
i.e. force per unit area.

But, I'm confused, shouldn't the $A$ factor also consider the thickness of the surface? E.g. a sheet of paper $200\ \mathrm{m}$ in length and width may not be able to reduce the pressure exerted by the feet of the elephant as would be reduce by a piece of rubber(or even a paper) $50\ \mathrm{m}\times 50\ \mathrm{m}\times 2\ \mathrm{m}$ in length, breadth and height respectively.

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4 Answers 4

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The pressure is the same, whether it is a sheet of paper or piece of rubber. The object doesn't alter the applied pressure (stress). What is different is the effect of the pressure (stress) on the object. That is, the resulting strain (change in thickness) that occurs due to the applied pressure (stress).

Can you add the formula for strain?

Stress σ is another term for pressure in mechanics of materials. So

$$σ=\frac{F}{A}$$

Where $F$ is the vertical load

Longitudinal strain ε is the change in length or deformation δ divided by the original length $L$, or

$$ε=\frac{δ}{L}$$

If the deformation is linear elastic, then the relationship between stress and strain is

$$σ=Eε$$

Where $E$ is Young's Modulus, or the modulus of elasticity of the material.

Putting all the above together,

$$δ=\frac{FL}{AE}$$

Hope this helps.

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  • $\begingroup$ thanks a lot! :-) But, if I understand correctly, do you mean the pressure at the bottom of the 0.1mm paper and the 2m(2 meter) thick paper , stays the same? I re-read your answer, and it seems you told me about the stress exterted on the paper and not at the bottom of it , e.g. a hand but at the bottom of the paper( in the first case, it'd be crushed by the elephant, in the second, only some force) $\endgroup$
    – juztcode
    Commented Jul 18, 2020 at 18:01
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    $\begingroup$ @juztcode I assume the paper and the rubber are each resting on a solid surface beneath. That surface than exerts an upward normal force equal to the downward force $F$. So both the paper and rubber are subjected to compressive stress. The amount each squeezes (deforms) will depend on the value of E and the original thickness L for the paper and the rubber. $\endgroup$
    – Bob D
    Commented Jul 18, 2020 at 18:37
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Neither paper nor rubber would significantly reduce the pressure on the ground due to the elephant's feet. This is because the paper or rubber would bend easily, so not spreading the force exerted by each foot over the 'nominal' area of the paper. But if a rigid board or sheet of metal were used instead of paper, you could then use your formula. Provided the board were rigid it wouldn't matter how thick it was, except that its own weight would add slightly to the force on the ground.

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The downward pressure of the elephant foot is done on the contact surface and is $$p_d = \frac{F_e}{A}$$

For a thin sheet of paper, the upward reaction of the ground on the other paper surface is equal, because the weight of the paper itself is very small compared to the elephant force.

But if the thickness is too big, the weight of the upper layers of paper adds more force to the layers close to the ground. In this case, the upward pressure to the paper block is $$p_u = \frac{F_e + w_p}{A}$$.

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Pressure is only the measure of the force an area experiences. How it responds to has nothing to do with the pressure. It's response is measured by the strain:

$$strain=(L-L_0)/(L_0)$$ or the elongation/ the original length.

Here is the difference: An elephant steps on a square of fabric, so the fabric is stretched quite a bit. Measure the amount it's length changed on the x axis and divide and you have the strain in the x direction. Measure the length change for the y axis and divide and you have the strain in the y direction.

If and elephant stepped on a metal sheet of the same thickness and area, there will be the same pressure but considerably less stretching so less strain.

Thickness has nothing to do with pressure exerted, it only depends on how much surface area there is. Strain doesn't either, it only depends on the material. To see which of two objects experiences less strain, we need the same thickness for it to be a fair comparison.

Stress is the cause which was the same in both cases. Strain is the effect.

Note: technically it seems there is a little more nuance to how I used strain direction in this example, but the idea is hopefully clearer this way.

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