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I have a structure colored in blue which we are looking directly downwards on (birdsview). Its high up in the air, and the only thing keeping it from falling down is the friction surface (red) between structure (blue) and the steel plate (black line). The reason there is enough friction to keep the entire thing together is due to bolts (grey lines) pulling the whole thing together with a strong force, which creates friction.

What I am wondering is which of these two cases, given that the force pulling them together is equal, would see the highest friction force. Case 1 only has 2 friction surfaces, but the surfaces are larger. Case 2 has 4 friction surfaces, but the surfaces are much smaller. Since friction is F= N*μ the friction surfaces might not even matter and it only depends on the force pulling it all together?

Case 1 enter image description here

Case 2 enter image description here

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    $\begingroup$ I would be hesitant to use theoretical friction calculations if this involved a beam than might fall on someone's head. There is Contact Theory which I don't claim to understand, but it points that there are differences in friction based on contact area. $\endgroup$
    – Tiger Guy
    Nov 10, 2021 at 22:43

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You are correct, for the same jamming force (N), regardless of the size of the contact area, the magnitude of the shear friction is the same for both cases, however, the effectiveness is different:

  1. If intended to prevent the beam from deflection, none of the scenarios works, as the plates and bolts will follow the deflection of the beam.

  2. If intended to prevent footing from slipping off the narrow-slippery surface, case 1 works.

enter image description here

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The case with four contact surfaces can have initial equal friction but ultimately it can resist much more force because the flanges will deflect and rotate creating an indentation on the steel plates which is going to act as a bracket.

Some car jacks take advantage of this mechanism.

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Assuming you consider rigid bodies (i.e. no deformation) then --like you said-- the friction would be the same because it depends on the reaction force.

However, if you are considering deformable bodies then it's a lot different.

Update: The reason that deformable bodies have an effect, is that then the contact of the two points can change, and instead of friction coupling you can have shape coupling (I guess that needs a bit more explaining).

In any case, the deformation can be either

  1. on the beam elements that carry the pads and are used to apply the forces externally
  2. on the pads themselves, due to Hertzian pressure.

Regarding the latter, I guess its easier to explain it with an example

enter image description here

In the picture above is a set of ice skates. If a person is standing on ice with the ice skates then its much more difficult to push him laterally, than if he were standing on shoes. The reason --IMHO-- is that the ice skates due to the small area develop larger stresses and "dig into" the ice. As a result, its not just friction any more, that governs the resistance but a "shape" interlocking.

This can also happens with any other material if you apply enough force. The contact stresses might exceed the contact strength and a dent might develop.

The following image is from Wikipedia for contact mechanics and exaggerates the local deformation of a ball and a contact surface.

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  • $\begingroup$ How so? - if we have deforming bodies $\endgroup$
    – greatoak763
    Nov 10, 2021 at 20:54

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