F=μmg. How was μ calculated? Obviously experiments were

[luckis11]

F=μmg. How was μ calculated? Obviously experiments were needed. So, done the experiments, what's the exact conclusionology that μ is that much for that type of ground and that much for the other type of ground?

 

[mibiz1]

F=μmg. How was μ calculated? Obviously experiments were needed. So, done the experiments, what's the exact conclusionology that μ is that much for that type of ground and that much for the other type of ground?

If you're referring the friction of the road vs. tire, I'd say that's μ becomes more of a 'best guess' because weather conditions will always be different, not to mention the way each road and tire is made and the composition of materials used for both. For the same road and tire, friction (both static and kinetic) is different for hot dry weather vs rainy weather. Even for non rainy condition but different temperatures, the friction is different. Thus auto racing is more of an art form than science, IMO. If you ever push your car to its limit, you'll understand why I say it's more of an art than science. It could be a science but who has the time to do the experiments for each and every (unlimited) possible combination of conditions. And how practical is that? And how accurate do you think the experiments will be? Consider that tire doesn't have much grip (read friction) at beginning of the drive for the day, as you drive, the tires heat up and provides a better grip. But extended driving on dry weather will over heat the tire to beyond its optimal performance. That's why in professional auto racing, they switch out the tire while refueling.

/mib

 

[Naty1]

Lots here:

http://en.wikipedia.org/wiki/Force_of_friction

Note that the text says "u" is "empirical"...

that means we have to measure it because we don't understand it well enough to make predictions based on mathematical models.
Also note there are two basic types of friction..static and dynamic...if your question was about auto racing, both will probably be of interest...when dynamic friction drops to a cetain level, rolling friction ensues...

 

[luckis11]

Kinetic friction. Ι found the answer. They could had found the different values for μ by trying different materials solving the example 7.7 at H.Young's physics, setting F=(sin60?)μmg.

 

[luckis11]

Is H.Young wrong at example 7.7 or am I confused? Shouldn't the upward speed be (sin30)5=2.5? That gives that the box cannot rise above 0.32metres whereas he says it goes up to 0.8metres having faced friction too?

 

[luckis11]

It's because of the free body diagram which is based e.g. that when it falls it does not cover the height with the same acceleration... etc. Forget it, if there's some mistake there's also at the free-body diagram (besides one mistake at the example 7.7 assuming the free-body diagramm is correct, but there my question can only be answered by serious doubters).

 

[Doc Al]

Is H.Young wrong at example 7.7 or am I confused? Shouldn't the upward speed be (sin30)5=2.5? That gives that the box cannot rise above 0.32metres whereas he says it goes up to 0.8metres having faced friction too?

Please describe the problem. Sounds like example 7.6, not 7.7, where there's a crate being slid up a ramp with friction. If that's the problem you mean, then you are told that the initial velocity is 5 m/s up the ramp. (Don't take the vertical component of that velocity.)

 

[luckis11]

Forget it. Do you know where I can find the proofs that the free body diagram is true?

 

[Doc Al]

Forget it. Do you know where I can find the proofs that the free body diagram is true?

What free body diagram?

 

[luckis11]

The analysis of forces at H.Young's examples 5-9, 5-17, 5-18.

 

[Doc Al]

The analysis of forces at H.Young's examples 5-9, 5-17, 5-18.

I don't see any problem with those examples. (Assuming we have the same book, the 12th edition.) Did you have a specific issue?

 

[luckis11]

No "problem". I asked for the proof.

 

[Subductionzon]

The empirical data is the proof. Friction is not well enough understood to develop the sort of proof that you want. What they did was to run a series of experiments and base their value on those experiments. What has been found is that once you find a value for μ it is fairly constant.

 

[Doc Al]

No "problem". I asked for the proof.

What do you mean by "proof"? I thought you had a question about how to draw a free body diagram? Is there something that you think is done wrong? Be specific.

 

[luckis11]

Forget about the constant μ. Regarding H.Young's examples 5-9, 5-17, 5-18, alone:

By proof I mean that instead of wondering WHY this force analysis is true (if it is true), the BECAUSE should be ready somewhere as either logical proof or experimental evidence with no logical explanation.

 

[Doc Al]

By proof I mean that instead of wondering why this force analysis is true (if it is true), the because should be ready somewhere as either logical proof or unexplained experimental evidence.

Sorry, but I have no idea what you're looking for. To do a force analysis one identifies all the forces acting, draws a free body diagram, and applies Newton's laws. You cannot tell whether a particular analysis is correct without actually looking at the details.

Do you think there's something wrong with some free body diagram you saw in the text? Or some force analysis. Please be specific.

 

[Doc Al]

Forget about the constant μ. Regarding H.Young's examples 5-9, 5-17, 5-18, alone:

By proof I mean that instead of wondering WHY this force analysis is true (if it is true), the BECAUSE should be ready somewhere as either logical proof or experimental evidence with no logical explanation.

Again: Be specific. What exactly in those examples is troubling you?

 

[luckis11]

Let me remind you that Force is the mdu/dt that happens to the mass. Thus, e.g. why the mdu/dt because of gravity should be equal to mgsin30? Another example: The force n goes upwards, whereas no mass goes toward that direction.

Thus it's most probably all correct, but...why?

 

[Doc Al]

Let me remind you that Force is the mdu/dt that happens to the mass.

Let me remind you that that's only useful if the force in question is the only force acting on the mass. When multiple forces are involved, mdu/dt gives the net force.

Thus, e.g. why the mdu/dt because of gravity should be equal to mgsin30?

The force of gravity is mg acting down. If gravity were the only force acting, the object would accelerate downward with du/dt = g. But gravity isn't the only force acting on an object on an incline.

The force of gravity, like any vector, can be broken into components. mgsin30 is the component of gravity parallel to the incline.

Another example: The force n goes upwards, whereas no mass goes toward that direction.

The normal force is perpendicular to the surface. There's no net force in that direction, so there's no acceleration in that direction. (The normal force and the perpendicular component of gravity add to zero.)

 

[luckis11]

If mdu/dt is only the net force, how are the non-net forces defined? Vectors of what?

 

[Doc Al]

If mdu/dt is only the net force, how are the non-net forces defined?

You can use mdu/dt if you isolate the force so that it's the only one acting.

 

[luckis11]

n=mgcos30 can "act" (happen) only when mgsin30 happens. But when mgsin30 happens it's the only one which happens as a mdu/dt. Then how can I isolate n as a mdu/dt?

 

[deadstar33]

Haha, "conclusionology". I may steal that word...

 

[Doc Al]

n=mgcos30 can "act" (happen) only when mgsin30 happens.

The components 'act' whenever gravity acts.

But when mgsin30 happens it's the only one which happens as a mdu/dt.

That's because mgsin30 is the net force.

Then how can I isolate n as a mdu/dt?

Since the normal force is a 'reactive' force, you can't (at least not in this scenario). But why would you want to?

 

[luckis11]

It seems you changed your mind and now you say that mg and n cannot be mdu/dt. Then, what are they, since the definition of force is mdu/dt?

 

[Doc Al]

It seems you changed your mind and now you say that mg and n cannot be mdu/dt.

Huh? Recall that I said that it's the net force on an object that equals mdu/dt. If you can arrange for a given force to be the only force acting, then you can directly apply mdu/dt.

Then, what are they, since the definition of force is mdu/dt?

That's a 'generic' definition of force. Not very useful here when multiple forces act. You can obviously have forces acting without a mass accelerating.

 

[luckis11]

Which are "obviously", what?

 

[Doc Al]

Which are "obviously", what?

Huh?

 

[luckis11]

n and mg, what are they? They cannot be mdu/dt as you said, therefore they are vectors of what?

 

[Doc Al]

n and mg, what are they?

They are forces, specifically electromagnetic and gravitational forces.

They cannot be mdu/dt as you said, therefore they are vectors of what?

You are getting hung up on mdu/dt as some sort of 'definition' of force; much better to think of mdu/dt (Newton's 2nd law) as a property of force in general.

 

[luckis11]

No comment.

 

[_PJ_]

You can obviously have forces acting without a mass accelerating.

What?
No you can't.

The sum of acceleration due to all forces may equate to 0, but that just implies at least two equal and opposite accelerations.

 

[Doc Al]

The sum of acceleration due to all forces may equate to 0, but that just implies at least two equal and opposite accelerations.

Nah. There's only one acceleration. There are multiple forces acting, not multiple accelerations.

 

[_PJ_]

Nah. There's only one acceleration. There are multiple forces acting, not multiple accelerations.

If two equal and opposite forces act on an entity, they each provide an acceleration.

Imagine a car driving along a road. It has the engine providing acceleration forwards, plus gravity providing acceleration downwards. Two forces, two accelerations.


A force is only a force because it accelerates (confers energy to) an entity.
Acceleration is only a name for "result of action of a force"

So the two are mutually interchangeable in this sense.

 

[Doc Al]

If two equal and opposite forces act on an entity, they each provide an acceleration.

While you can imagine that if you like, it's misleading. Luckily, since net force and acceleration are proportional you'll 'get the right answer' despite a confusing physical picture.

Forces describe interactions whereas acceleration describes motion. You can have multiple forces acting, but only one resultant acceleration. When two equal and opposite forces act there is no acceleration.

Imagine a car driving along a road. It has the engine providing acceleration forwards, plus gravity providing acceleration downwards. Two forces, two accelerations.

Nope. Multiple forces may act, but the car has but a single acceleration.

A force is only a force because it accelerates (confers energy to) an entity.

Nope. Forces don't have to accelerate or confer energy to an object in order to exist.

Acceleration is only a name for "result of action of a force"

Better to say that the acceleration is the result of action of all the forces acting on a body.

 

[luckis11]

Anyway, let's say that "I am confused and there is no question".
However, some of my conclusions so far are:

The definition of Force is
mdu/dt= (mass)(the acceleration of the mass which happens visibly-macrocosmically)
so it's resultant force that sometimes happens.

mg can only be mdu/dt, there's no other proof that the force of gravity is that much, besides the acceleration of a mass at free fall.

Therefore, the definition of all forces besides the resultant, is unknown, and it is upon us to guess what they might be e.g. the mdu/dt that would had happened if the other non-resultant forces were not happening together. Or, that they might be momentums=(mass)(velocities) that strike the m (I call these green momentums).

So, one quess is "if the canted wall did not exist, the mdu/dt would be mg", so n is...?

 

[mibiz1]

Perhaps you should review these:

* http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-3"
* http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-6"
* http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-7"
* http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-8"

 

[RandomMystery]

Anyway, let's say that "I am confused and there is no question".
However, some of my conclusions so far are:

The definition of Force is
mdu/dt= (mass)(the acceleration of the mass which happens visibly-macrocosmically)
so it's resultant force that sometimes happens.

mg can only be mdu/dt, there's no other proof that the force of gravity is that much, besides the acceleration of a mass at free fall.

Therefore, the definition of all forces besides the resultant, is unknown, and it is upon us to guess what they might be e.g. the mdu/dt that would had happened if the other non-resultant forces were not happening together. Or, that they might be momentums=(mass)(velocities) that strike the m (I call these green momentums).

So, one quess is "if the canted wall did not exist, the mdu/dt would be mg", so n is...?

If you ignore the air particles, then the normal force (or what you call n) should be zero (since there isn't any normal force, since we are ignoring internal forces).

If you don't ignore the air particles, then the air particles hit against the falling body (which causes the momentum of the air particles to transfer from themselves to the falling body) we call this "air drag" which eventually gets bigger (because you hit more particles as you speed up due to gravity going down) and eventually the transfers of momentum will stop you from accelerating since so many particles will strike you.

An atom is at rest on Earth's surface and Gravity is pulling down on it. However the atom does not go through the Earth because another atom is under it. The atom under it is one of the atoms of the ground. As the atom on top is pushed down by gravity, the outer electron shells of the top atom and the bottom atom begin to repeal each other, since same charged particles repeal each other.

The force of the "repeal" is the normal force. As the top atom is pushed closer to the bottom atom by gravity, the repellent force of their electron shells grows until it has the same magnitude and direction as the force of gravity. (The atom under the bottom atom does the same thing and so on until you get to the center of the earth.)

Here people deem them to be "still", but since energy is conserved, eventually the electrons outer shell force or pressure (which is called Electron degeneracy pressure) would weaken and then the electrons would collapse into the nucleus of the atom. Hence, electrons are accelerated toward the center of the atom. People tend to ignore this fact because it's hard to observe and usually so small that it is negligible.

However, this only occurs when a lot of joules (energy) is applied to the electrons. Since the force of gravity is relatively weak in this scenario, it would not happen in a very long time.

The same thing occurs with multiple atoms and compounds, the electron shell "fights back" sometimes forming a bonded outer shell (hence compounds) in structures.

Edit: I am ignoring all the atoms under that bottom atom for simplification

mg can only be mdu/dt, there's no other proof that the force of gravity is that much, besides the acceleration of a mass at free fall.

Everything is a guess. But what I think your doing is decomposing every structural logic until nothing else is left of it. Sounds like something a teenager would do. Luckis11, Do you know the difference between force and acceleration?

Wikipedia: In physics, a force is any influence that causes a free body to undergo a change in speed, a change in direction, or a change in shape.

The electrons shape is changing, they are being compressed at an extremely slow rate.

I hope this clears some confusion, and I hope I've guessed right.

Edit: I over simplified some facts in my explanation. I did so because I do not want to write a "thesis" which would be too long for most people to want to read.