jfuh
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the infamous Jet Question
- Thread starter PolySciGuy
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Thinker
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The laws were first published in Newton's "Philosophiae Naturalis Principia
Mathematica (1687)".
The first law can be stated as "An object at rest or moving at constant velocity
will continue in that state unless acted on by an external force".
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That is correct. But it has nothing to do to my question.
In the context of the question Newton only says: if there is no external force acting on object it means the object is at rest or moving at constant velocity.
That is correct. It is my understanding.
But the question was:
Who says: if the object is at rest or moving at constant velocity it means there is no external force acting on it ?
Can you answer the direct and simple question?
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The main point has been that I had decided to let you think there was somebody who could say so. I'd decided to let you think that the velocity I was using was constant.
So, I have been asking the direct and simple question:
Is the covered ground distance still = 0, if
accelerations of the planes are equal?
Yes or No?
So, I have been asking the direct and simple question:
Is the covered ground distance still = 0, if
accelerations of the planes are equal?
Yes or No?
Thinker
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Your stupidity is frightening. Not only do you have scant knowledge of
physics, you do not possess the minuscule amount of logic it needs to
understand that the two statements you see as being different are in fact
equivalent. Either that or you are feigning stupidity for your own amusement.
I suspect the former.
The answer to your question is that Newton's first law leads to that conclusion.
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Ok. You can say it in either way. No problem to understand.
So it is all still the same --- you still are choosing only one out of 2 Qs to reply.
And I cannot understand it yet: how do you always find time to make a long reply to any kind of stupidity or to a missing symble but still cannot answer "Yes" or "no''?
So, no it is the 4th (or5th?) post where I have been asking the direct and simple question:
Is the covered ground distance still = 0, if
accelerations of the planes are equal?
Yes or No?
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So, no it is the 4th (or5th?) post where I have been asking the direct and simple question:
Is the covered ground distance still = 0, if
accelerations of the planes are equal?
Yes or No?
Well, Thinker, I have given you more than enough time to say “No, it is not 0’’ and to burst into another tirade about my stupidity, interpretations, etc…
For some reason you have not done this, even when you know that such an answer would only cause applauds from some fools. It seems applauds of the fools wouldn’t make it for you.
I hope you see it is still = 0. And the plane on the belt (on floor of the big plane) is at rest, - it has 0 velocity in relation to the ground. That proves that both you and I have had our moments of being wrong. As far as you remember, at first, I myself said that as long as it had speed it would take off…The problem looked so simple that I did a mistake. Even after endorsing star####’s post, I still was able to re-post my mistake. It will not take off. As our friend jfuh would say: ''no matter what’s going between the belt and the wheels,'' - the number is still ‘’0’’.
As you can see all your forces are back in action. And you see I did not lie when I said that I agreed with all your forces (even when the belt had the same ‘’speed’’ as my ‘’velocity’’- it was not a real mistake of yours to apply forces). I didn’t lie when I said that you were correct when you saw the plane was moving; and when I said that you could see it moving only when you were standing on the belt, the floor of the big plane.
As to inertial frame reference, - you can see the little plane has acceleration = 0, which means it is at rest or v=constant, but you still can see the force applied to it, thus constant velocity does not mean absence of force. And it is just one example. There are others. You cannot read the 1st law in either way. Mathematical equivalents of the first law are:
1: If ∑F = 0, then v = a constant.
2: If ∑F = 0, then a = 0.
But they never write it opposite way. You can try to say that in F=ma it is in either way. Unfortunately one must not deduct the 1st law from the 2nd, these are different laws. It is well known that velocity itself does not establish an inertial frame.
So the bottom line is – potentially there is no problem for the plane running on the belt to take off, but at the specs of OP it does not take off.
The answer for 1) is NO.
When velocities are matched (which means accelerations are matched too) it does not move in relation to the ground/ air. You yourself did not have to take a pencil in order to calculate “0”. You did not have to watch TV, as well as Newton would not have to…
The answer for 2) is NO.
You may have to decide what they mean as ‘’the speed of the wheels.’’ Since it is unreasonable to think that it is speed of drugging, it must be only speed of rolling. Then you don’t even have to look at the plane. It does not matter what is going with plane as our friend jfuh would say. Look at my original drawing and you would see the speed of the plane is 0. If you cannot see ‘’0’’ looking at wheels and the belt, you still can measure the speed as the distance that a point of a wheel travels along the belt (ground, floor) per a minute thus making it the same as speed of the plane in 1). Thus, whatever is you approach you will get the same number 0.Thus, the answer for 2) is: NO.
“’NO”” for both.
You have seen the numbers, you have calculated them yourself. Numbers do not have an opinion. TV has to follow numbers, but numbers follow no TV.
Thus, again, the main problem in the solving the OP has been that ppl has made it unreasonably complicated. As I have been saying – it is explicitly simple as long as you understand what it is about, - about relative velocities.
Thus:
1. Forces are utterly irrelevant.
1. Frictions are utterly irrelevant
2. Bearings are utterly irrelevant
3. Jerk start is utterly irrelevant
4. Drag,
5. CoF,
6. angle of attack
7. giant fans
8. straps and anchors
9. fluid dynamics
10. tours into other chapters of physics
11. ..
12. …
13. ….
….
…
100 posts are utterly irrelevant to the simplest problem.
As you can see I have not been having a problem talking about all those things, pointing each time that they have been irrelevant. No need to try to catch me on lack of knowledge about planes, and for me to demonstrate my knowledge, because it is utterly irrelevant - you may to grow gray hair before you exhaust my knowledge, and thus, - before you solve the simplest problem. You yourself saw how many exercises had been irrelevant … and you fell into the same pit.
At the same time you have been like a partner to me, because in spite of your nasty personal attacks you still have been able talk physics and do math, even you have been talking out of the pit and rejecting my hand. I hope now, when you have numbers, you need no hand and you are out of the pit.
You remember, aren't you – distance is a number. It is “0”.
Is the covered ground distance still = 0, if
accelerations of the planes are equal?
Yes or No?
Well, Thinker, I have given you more than enough time to say “No, it is not 0’’ and to burst into another tirade about my stupidity, interpretations, etc…
For some reason you have not done this, even when you know that such an answer would only cause applauds from some fools. It seems applauds of the fools wouldn’t make it for you.
I hope you see it is still = 0. And the plane on the belt (on floor of the big plane) is at rest, - it has 0 velocity in relation to the ground. That proves that both you and I have had our moments of being wrong. As far as you remember, at first, I myself said that as long as it had speed it would take off…The problem looked so simple that I did a mistake. Even after endorsing star####’s post, I still was able to re-post my mistake. It will not take off. As our friend jfuh would say: ''no matter what’s going between the belt and the wheels,'' - the number is still ‘’0’’.
As you can see all your forces are back in action. And you see I did not lie when I said that I agreed with all your forces (even when the belt had the same ‘’speed’’ as my ‘’velocity’’- it was not a real mistake of yours to apply forces). I didn’t lie when I said that you were correct when you saw the plane was moving; and when I said that you could see it moving only when you were standing on the belt, the floor of the big plane.
As to inertial frame reference, - you can see the little plane has acceleration = 0, which means it is at rest or v=constant, but you still can see the force applied to it, thus constant velocity does not mean absence of force. And it is just one example. There are others. You cannot read the 1st law in either way. Mathematical equivalents of the first law are:
1: If ∑F = 0, then v = a constant.
2: If ∑F = 0, then a = 0.
But they never write it opposite way. You can try to say that in F=ma it is in either way. Unfortunately one must not deduct the 1st law from the 2nd, these are different laws. It is well known that velocity itself does not establish an inertial frame.
So the bottom line is – potentially there is no problem for the plane running on the belt to take off, but at the specs of OP it does not take off.
The answer for 1) is NO.
When velocities are matched (which means accelerations are matched too) it does not move in relation to the ground/ air. You yourself did not have to take a pencil in order to calculate “0”. You did not have to watch TV, as well as Newton would not have to…
The answer for 2) is NO.
You may have to decide what they mean as ‘’the speed of the wheels.’’ Since it is unreasonable to think that it is speed of drugging, it must be only speed of rolling. Then you don’t even have to look at the plane. It does not matter what is going with plane as our friend jfuh would say. Look at my original drawing and you would see the speed of the plane is 0. If you cannot see ‘’0’’ looking at wheels and the belt, you still can measure the speed as the distance that a point of a wheel travels along the belt (ground, floor) per a minute thus making it the same as speed of the plane in 1). Thus, whatever is you approach you will get the same number 0.Thus, the answer for 2) is: NO.
“’NO”” for both.
You have seen the numbers, you have calculated them yourself. Numbers do not have an opinion. TV has to follow numbers, but numbers follow no TV.
Thus, again, the main problem in the solving the OP has been that ppl has made it unreasonably complicated. As I have been saying – it is explicitly simple as long as you understand what it is about, - about relative velocities.
Thus:
1. Forces are utterly irrelevant.
1. Frictions are utterly irrelevant
2. Bearings are utterly irrelevant
3. Jerk start is utterly irrelevant
4. Drag,
5. CoF,
6. angle of attack
7. giant fans
8. straps and anchors
9. fluid dynamics
10. tours into other chapters of physics
11. ..
12. …
13. ….
….
…
100 posts are utterly irrelevant to the simplest problem.
As you can see I have not been having a problem talking about all those things, pointing each time that they have been irrelevant. No need to try to catch me on lack of knowledge about planes, and for me to demonstrate my knowledge, because it is utterly irrelevant - you may to grow gray hair before you exhaust my knowledge, and thus, - before you solve the simplest problem. You yourself saw how many exercises had been irrelevant … and you fell into the same pit.
At the same time you have been like a partner to me, because in spite of your nasty personal attacks you still have been able talk physics and do math, even you have been talking out of the pit and rejecting my hand. I hope now, when you have numbers, you need no hand and you are out of the pit.
You remember, aren't you – distance is a number. It is “0”.