Is the friction on a parallelepiped with mass m equal to the friction on a dice with the same mass? Because the overhang of the parallelepiped causes an irregular distribution of the normal force on the ground.

  • $\begingroup$ How does "friction on an object" make sense? There is only friction between sliding surfaces. Are all faces of the objects in contact with a surface that is sliding? Or should we imagine the object lying on a surface with on of its sides and sliding over it? $\endgroup$ – Steeven Nov 12 at 14:19
  • $\begingroup$ It’s a parallelepiped. It is stationary and is laying on a surface with its biggest face completely touching the ground. $\endgroup$ – Hell stormer Nov 12 at 16:03

As outlined by OP in a comment, we are dealing with an object lying on a surface with one side fully in contact with that surface.

Kinetic friction doesn't depend on area. In only depends on normal force (the pressure on the surface):*

$$f_k=\mu_k n$$

$\mu_k$ is the friction coefficient and can be thought of as a constant (depends on the two materials, on their interaction, roughnesses etc.).

Two objects with equal masses have the same weight. When lying freely on a surface (no extra pushes or burdens), that weight equals the normal force (due to Newton's 1st law).

So, it doesn't matter if the object is a parallelepipedum, a cube, a box or other - meaning if the contacting surface is triangular, square, rectangular or other. If their masses are the same, then the normal force is the same and then kinetic friction is the same.**

* This kinetic-friction model applies for not-too-large normal forces per area (called normal pressure), so it is useful in many daily-life scenarios, and I'm assuming also in yours. If you are dealing with, say, cutting tools in machining or turning or similar, then we shouldn't use this model.

** When saying this, I am referring to the total kinetic friction. While the total kinetic frictions are equal for any objects of the same mass, regardless of contact area, the distributions of the frictions may not be equal. More weight in one side of an object (like in the parallelepipedum you mention) will cause a larger normal force in this side and thus, according to the friction model above, a larger kinetic friction on this area, whereas an object with symmetric mass (like a cube) will give a different and symmetric kinetic-friction distribution.

  • $\begingroup$ But isn’t the infitesimal normal force distributed irregularly on the surface. Meaning that if we took an parallelepipedium, and the edge E which is touching the ground with the overhang being above it being a pivot. The overhang causes rotational momentum. So the points which are at the opposing edge apply a smaller normal force to the ground than the ones directly next to E. So the normal force isn’t distributed regularly. $\endgroup$ – Hell stormer Nov 13 at 12:51
  • $\begingroup$ @Hellstormer Sure, the normal force is a distribution over the contact area. If this distribution is uneven/irregular, then the kinetic friction is different at different parts of the contact area. The total kinetic friction is still the same, because the total normal force is still the same. $\endgroup$ – Steeven Nov 14 at 10:37
  • $\begingroup$ @Hellstormer Your original question was if the two kinetic frictions on two differently shaped objects with the same mass are equal, and you are correct that they are not equal in their distribution - but they are equal in total magnitude. If you draw a force arrow representing kinetic friction on each object, those arrows will be equal. That's what I assumed you were asking about. If you draw the distribution on each, it will not be equal for both. I have added a note about this to the answer. $\endgroup$ – Steeven Nov 14 at 10:48

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