Question on forces acting on an aircraft in climb
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Ahh, so what you are saying is the upward pull of the downward thrust has increased so in order for the aircraft to remain in equilibrium the lift must be decreased. But what about the downward pull of the increased drag?
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Ahh, so what you are saying is the upward pull of the downward thrust has increased so in order for the aircraft to remain in equilibrium the lift must be decreased. But what about the downward pull of the increased drag?
I just searched on google and came to this thread. I just wanted to lite a spark to this!
It makes things more clear if we say weight is equal to the F in F=MG. It won't change one bit while in a turn or while being accelerated at 2G. F=MG is your mass, which is constant, and the force of gravity exerted by the earth, which again is constant! The only time it will change is when you change your distance from the center of the body which is exerting the acceleration on you (earth). As you get higher in altitude your "weight" will change, gravity decreases. It is very slight however for civilian aircraft. We can't think of weight and apparent weight while analyzing the same set of forces because you are switching back and forth between frames of reference to try and explain the actual force you are feeling. So for all this I'm going to talk in the frame of reference of weight, which is an inertial frame of reference.
I think the definition of weight should be clear. Now, looking at the forces in the turn we should agree that the weight doesn't change. The aircraft has an acceleration force applied to it. We have to think in 3 dimensions here. Let's take this to a 0G environment so that we can isolate the force in the turn, space. If we fix a vertical pole in space, and space meaning deep space away from ANYTHING that could affect the results of this. Then take a string and attach it to small box with a small rocket engine on the back. As we fire the rocket it will speed up the box on a circular track around the pole, assuming it doesn't wind around the pole. At a constant speed around the pole, the box will have "1" force on it. Since we are in the inertial frame of reference, the box has a centripetal force acting on it toward the center. Centrifugal force does not exist in the inertial frame of reference. (If I were to sit in the box I would be in a non-inertial reference frame and I would feel the centrifugal force (a pseudo-force), which would translate to me feeling an "apparent weight" that is greater than my actual weight"). So that 1 force, since it is obviously unbalanced, will be 'accelerating' the box inwards. The box's weight HAS NOT changed!!
In a turn, the aircraft's weight DOES NOT change!! Once you understand that, you'll be able to clearly understand the forces in a turn.
The concept that in a turn you have a reduced effective wingspan is still correct when talking about lift according to the y axis. If you enter a level gentle turn and do not increase power, you will have to trade off airspeed for some extra lift to counteract the reduced effective wingspan by adding up elevator.
Now adding up elevator can't be looked at as adding "apparent weight" to the aircraft. We still need to remain in an inertial reference frame to analyze this. The horizontal stabilizer and elevator are a separate airfoil, just like the wing, that happens to be attached to an extremity of an aircraft. It will create it's own lift and it's own drag. The most correct way of looking at it is that drag will be the "apparent weight" in the non-inertial reference frame. It won't be the lift of the horizontal stabilizer!! The lift, or lack of lift, will only "point" the aircraft in an attitude. There is a range of lift it will produce from positive to zero to negative lift. It will have induced drag if it's producing lift at all and that's what will be the "apparent weight". If it's at zero lift, it will only be experiencing form/skin friction drag.
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So let's stir this up a little.
There you are, standing on the surface of the earth. You have your mass, and you have your weight. You can tell your weight because you are standing on a scale.
Suppose the earth were to be spinning twice as fast. Your mass is unchanged. Are you heavier? (That is, does the scale show more?)
PBL
There you are, standing on the surface of the earth. You have your mass, and you have your weight. You can tell your weight because you are standing on a scale.
Suppose the earth were to be spinning twice as fast. Your mass is unchanged. Are you heavier? (That is, does the scale show more?)
PBL
No, your apparent weight would actually be less, as the angular velocity of the Earth would impart a force in the opposite direction to gravity. It does that now, but if you increased it's rate of rotation, that force would also increase.
Suppose the earth were to be spinning twice as fast. Your mass is unchanged. Are you heavier? (That is, does the scale show more?)
I think the definition of weight should be clear.
From Wiki:-
The ISO standard ISO 31-3 (1992) defines weight as follows:
The weight of a body in a specified reference system is that force which, when applied to the body, would give it an acceleration equal to the local acceleration of free fall in that reference system.[9]
The weight of a body in a specified reference system is that force which, when applied to the body, would give it an acceleration equal to the local acceleration of free fall in that reference system.[9]
If you are using a NON-inertial frame of reference- such as an aircraft undergoing an acceleration- your weight will be greater than your mass.
I actually want to renew my dis-agreement with Checkboard and Captain Pit Bull on this point- by the ISO definition weight is not defined soley as force due to gravitational pull from a mass- or rather, the force felt due to acceleration IS force due to gravity, and force due to gravity IS force due to acceleration- they are actually the same thing. As I said previouly, there is no experiment that would allow you to differentiate between the two.
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I asked
Wizofoz replied
Thanks for biting! I'm not -yet- saying what I think the answer is, because I don't want to spoil the UTC-Monday-morning fun However, I don't think your reasoning works:
Capt Pit Bull has already pointed out that the appropriate way to think of things is as forces causing accelerations, and not motions causing (or, as you say, imparting) forces.
I like that! So, what's the answer?
PBL
Originally Posted by PBL
Suppose the earth were to be spinning twice as fast. Your mass is unchanged. Are you heavier? (That is, does the scale show more?)
Originally Posted by Wizofoz
No, your apparent weight would actually be less
However, I don't think your reasoning works:
Originally Posted by Wizofoz
as the angular velocity of the Earth would impart a force in the opposite direction to gravity
Originally Posted by bookworm
After the groundspeed thread, I thought for a moment you were going to come up with another definitional question: Is the "acceleration of free fall" changed?
PBL
Capt Pit Bull has already pointed out that the appropriate way to think of things is as forces causing accelerations, and not motions causing (or, as you say, imparting) forces.
An EQUATION means the two sides are EQUAL, meaning they are the same thing. One side doesn't CAUSE the other, both sides ARE each other.
If it is correct to say F=M*A it is equally correct to say A=F/M.
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Originally Posted by Wizofoz
[PBL: Capt Pit Bull has already pointed out that the appropriate way to think of things is as forces causing accelerations, and not motions causing ... forces.]
And I think he's wrong.
An EQUATION means the two sides are EQUAL, meaning they are the same thing. One side doesn't CAUSE the other, both sides ARE each other.
And I think he's wrong.
An EQUATION means the two sides are EQUAL, meaning they are the same thing. One side doesn't CAUSE the other, both sides ARE each other.
For example, suppose I drop a ball and it hits the floor. This entire interaction is well described by Newtonian mechanical principles, all expressed in equations as you note. But the action of gravity along with my releasing the ball are causes of its impact with the ground. Similarly, its impact with the ground is not a cause either of gravity or of my dropping it.
Back to the original problem. Say I weigh 80 kgs on the scale. You think I will weigh less (that is, the scale will read less) if the earth spins faster. So you probably think I will weigh more if the earth spins slower. How much would you think I would weigh if the earth just stops rotating? What would be the cause of my weighing that much? Indeed, what would be the cause of my weighing anything at all in any of these scenarios?
PBL
Last edited by PBL; 28th Jun 2010 at 09:16. Reason: Dropping a bigger hint
Don't know what you mean about causation. I'm saying it is wrong to say "Force causes a mass to accelerate". Force IS the acceleration of a mass. Thus, the acceleration of a mass IS force. If you have one, you can calculate the other. If my aircraft is accelerating, there is force. If I know my Mass, I can derive my Force.
As to the problem:-
Trick question perhaps?
Are you at the pole or the equator?
Assuming you are at the Equator:-
F=MV^2/R
= 80*465^2/6 378 000
= 2.7N
Normally, you weigh 80 kg ( at the equator) = 784N
Twice the rotational speed, you weigh 781.3N
If the Earth stopped, 786.7
See Eötvös effect.
Are we sticking to Newton, or bringing Einstein in off the bench?
Newton- because Masses attract each other.
Einstein- Because Mass causes Space/Tme to curve.
Interesting discusstion, but please don't keep me in suspence if I'm missing something.
As to the problem:-
Trick question perhaps?
Are you at the pole or the equator?
Assuming you are at the Equator:-
F=MV^2/R
= 80*465^2/6 378 000
= 2.7N
Normally, you weigh 80 kg ( at the equator) = 784N
Twice the rotational speed, you weigh 781.3N
If the Earth stopped, 786.7
See Eötvös effect.
Indeed, what would be the cause of my weighing anything at all in any of these scenarios?
Newton- because Masses attract each other.
Einstein- Because Mass causes Space/Tme to curve.
Interesting discusstion, but please don't keep me in suspence if I'm missing something.
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Originally Posted by Wizofoz
I'm saying it is wrong to say "Force causes a mass to accelerate". Force IS the acceleration of a mass. Thus, the acceleration of a mass IS force.
You asked not to be kept in suspense. OK, how does the following strike you?
Originally Posted by Wizofoz
[PBL: ....what would be the cause of my weighing anything at all in any of these scenarios?] ....
Newton- because Masses attract each other. Einstein- .....
Newton- because Masses attract each other. Einstein- .....
Earth is big and close, so as you say it attracts me and gives me my weight. Sun is big but far away; we can ignore its contribution to my weight. Other planets are smaller and far away; we can ignore their contribution also. To a good approximation, we can throw them all away! And the rest of the universe beyond our planetary system as well. To a good approximation, there is just me and the earth giving me my weight, according to you.
But wait a minute: you said rotation of the earth had something to do with my weight. Rotation? What rotation? Rotation relative to what? There isn't anything any more except the earth and me........
Conclusion?
PBL
Physicists consider mass and energy to be the same,
If force and acceleration are the same, why are the units different?
I didn't say they were the same. I said force was the same as mass TIME acceleration. What you are saying is that 2=2*2.
But wait a minute: you said rotation of the earth had something to do with my weight. Rotation? What rotation? Rotation relative to what? There isn't anything any more except the earth and me........
Last edited by Wizofoz; 28th Jun 2010 at 11:21.
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Hi PBL
Then please explain the tides.
Sun is big but far away; we can ignore its contribution to my weight. Other planets are smaller and far away; we can ignore their contribution also.
rudder,
The sun exerts about 0.0006N at the radius of the Earths orbit. The Moon somewhat more. It certainly is enough to cause the tides, but isn't that significant when it comes to weight.
Interesting question came up elsewhere, though- why is there also a high tide on the OPPOSITE side of the Earth to the moon? (Hint, you can ignore the suns influence for this example).
The sun exerts about 0.0006N at the radius of the Earths orbit. The Moon somewhat more. It certainly is enough to cause the tides, but isn't that significant when it comes to weight.
Interesting question came up elsewhere, though- why is there also a high tide on the OPPOSITE side of the Earth to the moon? (Hint, you can ignore the suns influence for this example).
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Originally Posted by rudderrudderrat
[PBL: Sun is big but far away; we can ignore its contribution to my weight. Other planets are smaller and far away; we can ignore their contribution also.] Then please explain the tides.
PBL
Last edited by PBL; 28th Jun 2010 at 11:24.
PBL,
Sorry, offending comment removed. I thought YOU were being condescending, whilst also being mistaken.
Anyway, as I pointed out, the rotation of the Earth DOES change your apparent weight as, as I pointed out, acceleration is absolute, not relative.
Do you dis-agree with that?
Sorry, offending comment removed. I thought YOU were being condescending, whilst also being mistaken.
Anyway, as I pointed out, the rotation of the Earth DOES change your apparent weight as, as I pointed out, acceleration is absolute, not relative.
Do you dis-agree with that?
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Hi Wizofoz,
Would you agree it's because the earth and moon orbit one another about their common c of g? Since that point is not at the centre of the earth (must be somewhere say about 1/10? distance to the moon), then there is a high tide on earth's moon side (due closer to moon) and another on the far side because it's "centrifuged" out.
why is there also a high tide on the OPPOSITE side of the Earth to the moon?
Hi Rudder,
No, that's not it. Consider that the orbital period is 28 days, the centre of the rotation is very close to the earth, and you'll see any centripetal force due to the Earths orbit of the moon (YES, it orbits the Moon just as surley as the Moon orbits the earth- they are a non-symetriacl binary system) is very small.
No, it's because the far side of the Earth (from the Moons POV) is sufficiently further away from the Moon than the near side for the Moons gravity to be have appreciably less effect there. This Gravity Gradient means on the near side it pulls the WATER away from the EARTH, on the far side, it pulls the EARTH away from the WATER!
PBL,
Hope bridges are mended.
Now, do you see what I mean about the Earths rotation being an absolute, as it is an acceleration?
No, that's not it. Consider that the orbital period is 28 days, the centre of the rotation is very close to the earth, and you'll see any centripetal force due to the Earths orbit of the moon (YES, it orbits the Moon just as surley as the Moon orbits the earth- they are a non-symetriacl binary system) is very small.
No, it's because the far side of the Earth (from the Moons POV) is sufficiently further away from the Moon than the near side for the Moons gravity to be have appreciably less effect there. This Gravity Gradient means on the near side it pulls the WATER away from the EARTH, on the far side, it pulls the EARTH away from the WATER!
PBL,
Hope bridges are mended.
Now, do you see what I mean about the Earths rotation being an absolute, as it is an acceleration?
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what i find interesting is why PBL implied that you might feel heavier if the earth spun faster, opposite to what i would think a normal mind would think. even more surprising is that many high school students think that if the world stopped spinning, gravity would be 0G.
You are quite confused!.. regarding that quote and what you said after. Does this clear it up: gravity is not a force, it is an acceleration, 9.8m/s2.
Just come out clean to what you think it is!
wiz... suppose you are standing still and i run up and crash right into you. im applying a force to you, to make you accelerate backwards. to accelerate backwards, a force has to be applied, correct? so without that force, you won't accelerate. you can't have acceleration without a force, but you can have a force without acceleration!!!
yes F=MA balances out, but an equation doesn't explain everything.
This discussion is filled with ridiculous banter... going back to PBLs dilemma, no one has mentioned one thing about the gravity equation while trying to explain gravity.
I'm waiting for the Lord PBL to explain his views because I think you and I will have some interesting points!
I actually want to renew my dis-agreement with Checkboard and Captain Pit Bull on this point
Thanks for biting! I'm not -yet- saying what I think the answer is, because I don't want to spoil...blah blah blah
I'm saying it is wrong to say "Force causes a mass to accelerate".
yes F=MA balances out, but an equation doesn't explain everything.
This discussion is filled with ridiculous banter... going back to PBLs dilemma, no one has mentioned one thing about the gravity equation while trying to explain gravity.
I'm waiting for the Lord PBL to explain his views because I think you and I will have some interesting points!
You are quite confused!.. regarding that quote and what you said after. Does this clear it up: gravity is not a force, it is an acceleration, 9.8m/s2.
This discussion is filled with ridiculous banter... going back to PBLs dilemma, no one has mentioned one thing about the gravity equation while trying to explain gravity.
As to this-
but you can have a force without acceleration!!!