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It's been a long time since I did problems on friction but here's one...

Consider a cuboidal block of dimensions 123 meters with a total mass of 1 kg. There are 3 cases that we need to look at:

  1. Its base of area 1*2 sq. meters is in contact with the surface
  2. Its base of area 1*3 sq. meters is in contact with the surface
  3. Its base of area 2*3 sq. meters is in contact with the surface

for all the above cases, the coefficient of friction ($\mu$) is a constant. the normal force ($N$) is also the same since the weight of the block is the same and $N$ has to equal the downward acting force mg

since $f = \mu N$, that means the friction force is the same in each of the three cases.

That means the friction force is independent of the orientation of the block!? But that doesn't make sense... with a larger contact area friction should intuitively increase. If friction doesn't, what does? And does the contact area even matter then?

Maddy
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1 Answers1

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The normal force is the same, regardless of which side of the box is contacting the ground. It will always be equal and opposite to $mg$ for an earthly system in equilibrium - but I'm certain you understand that.

On a more intuitive level, as the weight force is equally distributed on the bottom surface of the box, think of dividing this surface into a collection of minuscule areas $da$. Each of these little areas exerts some normal force - in the case of the contact surface with the largest area, this normal force will be of course be smaller, but there are more $da$'s composing the whole surface. For the smallest surface, there will be fewer $da$'s, but each $da$ will exert a greater normal force. All of these normal forces necessarily sum to the same net force ($-mg$), and all of their respective tiny frictional forces will therefore necessarily sum to the same net frictional force.

Editing: Something else that may be tripping you up. The geometry and quirkiness of the contact surface does matter, but $N$ is completely independent of this - $\mu$, however, is absolutely not.

dnxtn
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  • yeah well that's what I said... The normal force, as it seems, doesn't depend on the orientation. So WHAT DIFFERENCE DOES THE ORIENTATION MAKE? does the force of friction change (since you say $\mu$ changes)? – Maddy Nov 13 '23 at 10:24
  • The orientation does not make a difference. The frictional force does not change. $\mu$ is determined by the surface characteristics of the material - how bumpy or pointy it is at a microscopic level. If all sides of the ideal cube are made of the same material, $\mu$ will be the same and so will the frictional force. – dnxtn Nov 13 '23 at 19:56
  • Woah! so you mean, if you drag (not roll) a tire down while it's on its edge, it will experience the same force of friction as when the tyre is dragged down on its side?? since the weight of the tire does not chance... – Maddy Nov 16 '23 at 16:45
  • That's a great question - you'd think so, considering the equation, right? Actually, in that case, the tire is experiencing two different kinds of friction. When dragged, it feels static friction (as we're used to), and when rotating normally, it feels rolling friction (which is completely different). Here's a good resource on rolling friction from Thomas Moore's physics page: http://physics.thomasmore.edu/ConnectedPhysics/ss-friction-rolling.html – dnxtn Nov 16 '23 at 22:55
  • Yeah, I know that... that is why I said "drag (not roll)". So if you DRAG it down at a single point as opposed to by the side, would the forces be the same?

    Take a different example... taking a block which seems like a trapezium from the side, 1 of its base areas is huge while the opposite one is almost point-sized. What you suggest is $F_{friction}$ is the same for both cases, i.e. if we drag it from either side (large surface and a near point-sized surface), regardless of the area of contact. yeah?

     – Maddy Nov 17 '23 at 10:37