←back to thread

896 points tux3 | 9 comments | | HN request time: 0.911s | source | bottom
1. BeetleB ◴[] No.43548232[source]
Oh, BTW, the whole "Friction is directly proportional to the normal force": My Ass!

I could never reproduce it well in the lab, because it's really not true. Take a heavy cube the shape of a book. Orient it so that the spine is on the floor. It's a lot more friction to move it in one direction than in the transverse direction. Yet the normal force is the same. Any kid knows this, and I feel dumb it never occurred to me till someone pointed it out to me.

replies(5): >>43548343 #>>43548487 #>>43548702 #>>43549443 #>>43549561 #
2. mizzao ◴[] No.43548343[source]
Is this possibly because you need to use additional force to horizontally stabilize it in one direction (perpendicular to the spine) but not the other?
replies(1): >>43548995 #
3. ◴[] No.43548487[source]
4. FacelessJim ◴[] No.43548702[source]
The “proportionality constant” is doing a lot of work in that claim. A lot of “constant” parameters are swept under the rug. If you fix enough stuff that claim is indeed correct, although I agree a bit simplistic
5. mercutio2 ◴[] No.43548995[source]
I was about to say exactly this.

Applying force directly to the center of gravity with one finger is hard.

You end up applying torque plus adjustments in response to that torque. And that is heavily dependent on your moment of inertia, unlike the normal force.

But I do agree that explanations of friction are right up there with “how do airfoils work” where poor instructors are liable to get long past the edge of their knowledge and just make shit up.

6. emmelaich ◴[] No.43549443[source]
Yep, cars can accelerate at over 1g.
7. dragonwriter ◴[] No.43549561[source]
Friction is proportional to the normal force, more specifically, it is the normal force times the coefficient of friction.

What you are describing (if the normal force is actually the same) is a contact situation where the coefficient of friction is different in different directions (anisotropic friction.)

replies(2): >>43551150 #>>43556399 #
8. klysm ◴[] No.43551150[source]
Not true in practice for a lot of materials
9. immibis ◴[] No.43556399[source]
When you define a new concept called "coefficient of friction" to be the friction force divided by the normal force, then yes, the friction force is the normal force times the coefficient of friction.

We can do this with any pair of values, such as current and voltage, but it's useful when the graph between current and voltage is close enough to linear, which means the corresponding coefficient is approximately constant. Well, is it? You have to show that with experiment. Once your data shows a line then you can calculate the slope of the line. If your data shows a parabola, you can calculate the quadratic equivalent of slope - don't calculate the actual slope and then declare that to be the result.

Sometimes you see people trying to measure the resistance of diodes, or worse, incorporating the resistance of diodes into calculations. What's the voltage across the diode? The current times the resistance, of course...