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So, my question might seem silly. I know in real life when we apply a force with our hand and push on lets say a cylinder , we know the force will be distributed over the cross section of the area, so if we had a wider area, we need more force, and if we had smaller area, then we need less force to push the cylinder a certain distance.

So its intuitive. The stress will be the force divided by the given area.

But why? like what happens at the micro-scale, and what makes the force be divided?

Thanks.

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  • $\begingroup$ Maybe I am misreading, but it is not true that you need more force to move a cylinder a certain distance if your area of force application is larger. If you choose to use your finger you need the same force as using your hand. $\endgroup$
    – lalala
    Oct 21, 2021 at 8:13

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Actually as @trula said the external force you apply acts only at the contact point but since all atoms are connected to each other via "interatomic forces" , your external force gets distributed all along the surface and so we need to define force per unit area viz. Stress.

The spring model of atomic structure is quite self explanatory about the interatomic force distribution.

enter image description here

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    $\begingroup$ Thanks. But shouldn't the stress in this case depend on the type of material as well? since its related to the interatomic forces $\endgroup$
    – user134613
    Oct 20, 2021 at 15:30
  • $\begingroup$ @user134613 - it's not the stress itself that depends on the type of material, it's the material's reaction to the stress that depends on the type of material. $\endgroup$
    – Glen O
    Oct 21, 2021 at 2:04
  • $\begingroup$ @GlenO But the stress is the material's reaction to the applied force, isn't it? $\endgroup$
    – user134613
    Oct 21, 2021 at 8:33
  • $\begingroup$ @user134613 - no, that's called the "strain". Stress is the force being applied, strain is the result of the stress. Indeed, stress and strain play comparable roles to force and acceleration, in that $F=ma$ and, for elastic materials, $\sigma=E\epsilon$, where $\sigma$ is the stress and $\epsilon$ is the strain. The $E$, here, is Young's modulus, which depends on the material being deformed. $\endgroup$
    – Glen O
    Oct 21, 2021 at 21:44
  • $\begingroup$ @GlenO Thanks, but I'm still confused because I read different perspectives. for example here quora.com/… . Someone says //External force is the cause, and the internal force is the response of the system. The internal force acts on a give area tells us the stress experienced by the body of the system. In contrast, the external force acts on a given area tells us the load or pressure applied to the system. If both forces are equal and opposite then the system will be in mechanical equilibrium// $\endgroup$
    – user134613
    Oct 22, 2021 at 8:56
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You have a misunderstanding: the force is only on the area your hand touches, if you do it with your fingertip only the area of the fingertip counts, not the area of the cylinder. Thats why you can get a nail with very small area in wood, but not your finger.If you push wit your fingertip you need more force than with your flat hand.

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  • $\begingroup$ So the force is wholly/completely transmitted into the area of contact. While the stress, which is produced by the "nail" in your example, will be of magnitude equals to force/area_of_contact since its produced from the internal material of the nail. Right? $\endgroup$
    – user134613
    Oct 20, 2021 at 14:07
  • $\begingroup$ Did I get it right? $\endgroup$
    – user134613
    Oct 20, 2021 at 14:21
  • $\begingroup$ @trula In the last sentence I thinking you're mixing up force and pressure. The same force is needed to get the cylinder to move, but the pressure will be greater when pushing with your finger, because it is the same force over a smaller area. $\endgroup$
    – zucculent
    Oct 20, 2021 at 16:29
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The answer comes from the fact that every part of an object must be in force equilibrium regardless of the location or shape considered. Additionally, the internal deformation due to these internal forces must be smooth and continuous inside a homogeneous material.

So you cannot have a single line of molecules loaded under the contact all the way to the support because the internal deformation will not be continuous. Depending on the material properties this spreading out of the applied force happens more or less on all materials.

There is a concept of load path, where the most stressed internal molecules lie in the path between the applied load and the support, but this is just a tool for explaining the observed effects, and not a mathematical description of what is going on.

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