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Q: How do I ask homework questions on Physics Stack Exchange?

1:02 AM

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What is the policy on asking homework questions on Physics Stack Exchange? What kinds of questions are considered homework questions? Are homework questions allowed? What should I include in a homework question? Why don't you provide a complete answer to homework questions?

The short version is that we don't want to hear about your homework trouble as such, but will be helpful if you can ask about the conceptual sticking point.

Oh

Nice

I've got a lot of conceptual questions about normal and friction that come up in exercises

So not "I got stuck after this step. Now what do I do?", but "A problem I am working on seems to rely on some difference between static and kinetic friction, but I don't understand why static friction can have any value up to a limit."

How fortuitous that I mentioned the same thing. But lots of people get stuck there.

Like, in what conditions would a block of mass $m$, friction coefficient $\mu$, gravity $g$ and tension $T$ at an angle $\theta$ stay in the surface and don't "fly arround"?

So, you want the book to remains still, so you'll be doing static friction, right?

@dmckee yup!

1:07 AM

Static friction has the property that it acts to prevent motion, but only up to some maximum force.

yup.

That maximum is $\mu_s F_N$ (for $\mu_s$ the coefficient of static friction and $F_N$ the normal force).

I just don't get how a parallel force would act on a perpendicular force

Ah. Got it.

The book probably has some equation like $F_f = \mu_s F_N$, right?

@dmckee yes.

1:09 AM

But those are not vector forces (it's not $\vec{F}_f = \mu_s \vec{F}_N$), so this isn't a vector relationship and direction is not involved.

It just says "the harder things are pressed together the more friction there is".

@dmckee isn't the friction force always parallel to the surface?

Yes, friction is parallel to the surface, but it's maximum magnitude is proportional to the normal force. If the objects were just barely touching, there couldn't be much friction and if you squeeze them together with a hydraulic press there can be huge amounts of friction.

BTW, if you don't already know about ChatJax following this link will start MathJax on chat at let you see rendered math.

Most people bookmark the link.

yes, I do have it

I'm just not too good with it :p

@dmckee how they are related then?

I mean, how do I calculate the force vectors?

The normal force is always normal (perpendicular) to the surface of contact and friction is always parallel to it. Also friction always acts to prevent or resist relative motion.

And you have $F_f=\mu F_N$ to tell you the relationship between their magnitudes (except that with static friction that gives you the maximum magnitude.

1:25 AM

@dmckee then, how would friction prevent perpendicular motion?

like, when we pull something with an angle < 90º, we have partial parallel and perpendicular forces

Friction doesn't affect perpendicular motion unless the objects are actually sticky (not the case for exercises in a introductory course).

It's normal force that prevents objects from moving into one another and in generally you have to figure out what (if anything) s preventing them from losing contact with one another.

then every single force that is not acting with 0º with make things go up?

If the net force points away from the surface they separate. But just because I lift up on a heavy barbell at the gym doesn't mean I supply enough force to pull it off the ground. I still have to counteract it's weight.

my actual exercise was this way: you have a block with weight w on a plane with surface at alpha radians/degrees. we apply a tension force with magnitude T and angle theta in relation to the plane

we have to find the maximum value for the acc (derivate = 0, blah blah blah)

but my problem is understanding the actual physical significance

The usual place to start is with a picture and a free-body diagram (this case is probably simple enough to combine these if you want).

1:30 AM

the perpendicular resultant force is something like Tcos or Tsen of something. which could (or not) be greater than 0

Then you are going to look at the components perpendicular to the surface and find the normal force by requiring that the object not sink into the surface.

@dmckee yup. did this. the problem is what i got (last message)

Once you have the normal force you compute the forces along the surface without friction so see how things would slide.

Then you add friction pointing in a direction that resists that sliding.

@Chinatsu-creepy-chan Yep. It can. Do you have specific numbers for this case?

Or are you doing it all symbolically.

@dmckee This took me a long time to understand.

@dmckee literal solution for any problem of the same kind. but I can plug in some values how the system would work

1:33 AM

@0celo7 It takes most people a long time to understand. And then when you get it is seems so easy that you don't understand why the next person struggles.

it makes sense that if it was an absurd tension the object would really fly. but smalls tensions can even make sense: the gravity center makes the object keep in contact

@Chinatsu-creepy-chan OK. You can now say "If the tension exceeds [some formula], the object is pulled off the surface. If not, we proceed to compute the horizontal acceleration..."

@dmckee the problem is that the resultant force is that one

x.x

@dmckee I meant the part that with static it's the max friction

Where you find the formula for the limiting case by assuming zero perpendicular force without normal forces.

1:35 AM

I finally got it when I wondered why blocks don't slide up inclines from friction

I think

@0celo7 That's a question I sometimes use for students who are stuck on that. But it's risky: sometime it just confuses them more.

but sometimes the friction coefficient is greater than 1

the thing is that the normal isn't greater than the external forces

(i think, lol)

@Chinatsu-creepy-chan never.

kevin Tah N.

When theoretical physicist mostly QG and QFT types talk about memory, and stored information, what do they mean. I am not working on anything, I just thought gosh, I hear it all the time, and what do they mean? It confuses me more because at a computer science level I have a strong understanding of what memory means

@lucas yup, it can be.

$\tan \theta = \mu_s$

1:39 AM

reddit.com/r/askscience/comments/214foz/…

@Chinatsu-creepy-chan What is the $\theta$?

@Chinatsu-creepy-chan Not usually in intro courses, but such materials do exist.

@lucas inclination of plane

@lucas probably the angle of the inclined plane

@dmckee like rubber on rubber!

according to reddit

@Chinatsu-creepy-chan NO it isn't $theta$

$\theta =\tan^{-1} \frac{F_f}{N}$

1:41 AM

cast iron on copper has $\mu_s=1.05$

(P = w. Weight in portuguese is 'Peso' :p)

@0celo7 Is that Coulomb friction?

@lucas yeah, i got that wrong :p

@0celo7 You right. I saw that here. "https://en.wikipedia.org/wiki/Friction#Coefficient_of_friction" But what is the problem with $\mu \gt 1$?

@Chinatsu-creepy-chan What?

@Chinatsu-creepy-chan I cannot see your uploaded image above.

@lucas what's the problem?

1:53 AM

@Chinatsu-creepy-chan It says for me, "Image not found"

@lucas Oh. Try reloading the page, IDK e.e

@Chinatsu-creepy-chan OK. I saw.

@Chinatsu-creepy-chan So, what is the problem?

32 mins ago, by Chinatsu-creepy-chan

@dmckee then, how would friction prevent perpendicular motion?

@Chinatsu-creepy-chan perpendicular with respect what?

some plane

1:58 AM

Incline plane?

Or horizontal?

yup. you know, not only. it could be an inclined plane to make stuff easier

or you just say the body is a particle and balances the forces on dS (infinitesimal part of parametrized surface)

@Chinatsu-creepy-chan Friction force doesn't prevent the perpendicular motion.

mathematically not. but that's not the deal in real life

i think i've got things clear

In your case $mg\cos \alpha$ prevents perpendicular motion.

@lucas but we have tension at angle theta. with makes perpendicular motion

supose you have a sandpaper body over a rubber surface. really big friction coefficients.

2:03 AM

@Chinatsu-creepy-chan $T\sin\theta-mg\cos \alpha=0$

if you make tension at angle theta, it will lift up. but the new center of gravity would still make the body have a contact surface and friction would not let body just fly from the plane

@lucas nope. normal makes mgcos alpha just disappear

@Chinatsu-creepy-chan "if you make tension at angle theta, it will lift up" Are you sure? If $T \sin \theta \lt mg \cos \alpha$ then contact will be maintained.

@lucas told'ya: the normal perpendicular force is $mg\cos\alpha$, so the only force acting is $T \sin\theta$

@Chinatsu-creepy-chan I cannot understand what is your problem with perpendicular motion. In a horizontal plane, when you exert a force to lift something, does it go up for any amount of your force?

If your force is little than its weight, then it won't move.

@lucas what happens is that the perpendicular force is greater than 0, then it will lift. i was just wondering why in real life some things do not look like they're lifting

2:13 AM

What do you mean by "some things do not look like they're lifting"?

"the normal perpendicular force is $mg \cos α$" Not necessarily.

Remember that the normal force takes on the value needed to prevent the two object from interpenetrating.

@lucas when the tension is just not "enough"

$mg \cos a$ is the value it has when the object is resting on it with no other forces acting. But as soon as you start pulling away from the surface it will drop.

@Chinatsu-creepy-chan In real life if $T\sin \theta \gt mg \cos \alpha$ then it will lift.

@dmckee e-e. I'm so confused. my teacher already told me this - "saying that normal is the weight is a bad habit of ours. normal is just the contact perpendicular force"

now I understand the concept but not the math

2:18 AM

@dmckee I agree with you "the normal perpendicular force is $mg \cos α$ Not necessarily". So, what?

@Chinatsu-creepy-chan They way to work is long and slow and always right is ...

1. Work out the sum of the forces perpendicular to the surface without taking the normal force into account.

2. If that total would cause the two object to move through each other, then (and only then) give the normal force the magnitude of the net you computed in step (1) pushing the two objects apart.

The normal force is just whatever you need it to be so that an object does not go into another object. In addition, it must be normal (perpendicular) to the surface.

The result of step (2) will be the two objects at rest with respect to each other in the perpendicular direction.

But after a time you will come to the kind of understanding that @Benzene just gave and not bother with the long way except in particularly difficult cases.

@Benzene this got so much easier xD

lol

2:22 AM

I have another really important question

kk

normal normally (yay, puns) has a perpendicular and parallel component

when we talk about friction, we're talking about the entire magnitude or just the magnitude of the perpendicular component?

Contact forces normally have perpendicular and parallel components. The normal force is the perpendicular component by definition. It is always and only perpendicular to the surface of contact.

"Normal" means "perpendicular".

You talk about the entire normal force. The normal force is a measure of how much two objects are pushing against eachother, and friction is proportional to how hard two things are pushing against eachother Ff=u*N

kevin Tah N.

my family keeps asking me about my school, life etc, but nobody supports me

what a bunch of arses

2:27 AM

=(

@Benzene now that @dmckee said that about contact forces, friction is $\mu$ times normal or contact force magnitude?

it is $\mu$ times normal