4 Basics

darrell said:
I really dislike the "sum must be zero" idea, it's either wrong or a tremendous oversimplification.
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It's one of those things where someone reads the words, but doesn't understand them.

It's obviously wrong that there are no net forces ever.

However, an airplane moving in a straight line -- even if it is climbing or descending uniformly -- has all the forces adding to zero or it would be accelerating.

It doesn't work for turns. Turning requires an acceleration, always.
 
brywd said:
A plane carrying hundreds of birds in cages takes off. In flight the door to the cages opens somehow and all of the birds escape and are flying around the cabin. Does the weight of the plane go down?
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Why does a tank of compressed helium weigh so much?
 
Henning said:
Yep, however we still reference those accelerations in units of gravity. My point is it doesn't apply to this thread in any meaningful form. In our general society here on earth, kilograms are used, correctly or incorrectly, as a unit of weight. Find me a bathroom scale calibrated in Newtons rather than kilograms if you would like to refute that.
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Sac's right…if you have acceleration other than gravity (or even including it if you're into GR), there is a net force that is producing that acceleration.

It's much more than a semantic argument.
 
MAKG1 said:
The "stone."

Yes, some Brits do actually use it. I've never heard it said in the US except as the nerd equivalent of a bar bet.
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Aussies use it even more frequently, although I believe stone is a unit of weight.
 
Sac Arrow said:
Okay now what's the English unit of mass?
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Depends on what you mean by "English unit".

One system is pounds mass / pounds force, another is pounds force / slug, and another is pounds mass / poundel. The official US legal definition of the "pound" is 0.45359237 Kg exactly.


"Weight" is a concept that predates the differentiation between "mass" and "force" and depending on context (and the technology used to measure it) can refer to either mass or force.
 
brywd said:
A plane carrying hundreds of birds in cages takes off. In flight the door to the cages opens somehow and all of the birds escape and are flying around the cabin. Does the weight of the plane go down?
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My answer to that is YES because the in-flight birds must burn calories in order to remain in flight and in so doing their own weight is converted to heat. After half an hour of constant flight not only will the birds themselves weigh less (even the exhausted and dead ones laying on the floor) but the heat trapped in the cabin will....

Okay never mind. How about this: Does a pressurized aircraft weigh more than an unpressurized one at 30,000 feet? :rolleyes:
 
skidoo said:
Its all a matter of interpreting the OP question. You are correct above except the answer is yes, if you interpret the OP question literally. He asks "So as an airplane's wing generates lift, does it weigh less?" A scale measures "weight" or "force on an object due to gravity". If you were able to place a scale under the wheels of an airplane and keep it there through the takeoff roll (to measure the "weight" relative to the ground), as it traveled down the runway, it would measure less and less as the wing generates lift.

Now, once it leaves the ground, there is no weight (to measure the force relative to.) But, if you measure the weight at a sea level field and then measure it at a 5,000 ft field at the same latitude, it will "weigh" less simply because the "force on an object due to gravity" will be less. This is because the object if further away from the center of the earth.

With highly sensitive instruments, I have measured weights of objects that differ when measured on the floor as compared to the weight on a table 4 ft higher, all due the increased radius from the center of the earth.
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That's true, I assumed the weight difference with elevation to be small, but that's right it will be less, plus it can even vary with geographic location. The "standard" acceleration due to gravity is just like mean sea level, it's an arbitrary datum approximating average conditions that is used for calculations. Your actual weight is what it is, based on local gravity.
 
MAKG1 said:
snip...

It doesn't work for turns. Turning requires an acceleration, always.
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Always?

Say you stand still for 24 hours, you have effectively done a 360 degree turn (earth rotation). You have turned at the same rate for that 24 hours. What acceleration was involved?
 
Sac Arrow said:
That's true, I assumed the weight difference with elevation to be small, but that's right it will be less, plus it can even vary with geographic location. The "standard" acceleration due to gravity is just like mean sea level, it's an arbitrary datum approximating average conditions that is used for calculations. Your actual weight is what it is, based on local gravity.
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Correct. And, as you say, geographic location makes a difference too. For example, the same mass will weigh less at the equator than it will north or south of there both due to elongation of the globe (further from the center of the earth) and due to centripetal force of the rotating earth.
 
skidoo said:
Always?

Say you stand still for 24 hours, you have effectively done a 360 degree turn (earth rotation). You have turned at the same rate for that 24 hours. What acceleration was involved?
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Centripetal acceleration.
 
I think you could see it as a problem like a buoyancy problem, so the apparent weight of the plane is the mass times the gravity (mg) minus the lifting force (Fl). So:

Wapp = mg - Fl

So the apparent weight changes depending on the difference between those forces, so while you are climbing you could actually have a negative apparent weight. At least that's my take on it. Of course "weight" is just mg. So that will be a constant (at low altitudes), although if you climb high enough you can see a difference in the magnitude of the gravitational force defined by g = G mM/r^2
 
Apparent weight? Operational weight? Gravitational weight? Which definition?

In skidoo's first experiment (bathroom scale/overhead bar) You need to pull on a spring scale attached to the bar and sum that scales reading with the bathroom scale. You'll find that your weight doesn't change no matter how hard you pull.

[So, does the weight of the airplane change]
No, and let's just assume a constant g for the intent of the OPs question, the weight of the airplane does not change.
 
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PaulS said:
The sum of the forces must equal zero.
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This is only true when there is no acceleration.

F = ma

If a is not 0 and m is not 0 then F is not 0. We can assume that an airplane has a mass greater than 0.
 
MarleyW said:
I think you could see it as a problem like a buoyancy problem, so the apparent weight of the plane is the mass times the gravity (mg) minus the lifting force (Fl). So:

Wapp = mg - Fl

So the apparent weight changes depending on the difference between those forces, so while you are climbing you could actually have a negative apparent weight. At least that's my take on it. Of course "weight" is just mg. So that will be a constant (at low altitudes), although if you climb high enough you can see a difference in the magnitude of the gravitational force defined by g = G mM/r^2
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I'm having a little bit of a hard time following. Ignoring the effects of gravity on altitude, if you climb at a constant rate your weight will not change because there is no net acceleration. In the transitional period when you go from level flight to a constant rate climb, there will be a net acceleration and your weight will increase during that time. The converse will happen in a decent.
 
Sac Arrow said:
I'm having a little bit of a hard time following. Ignoring the effects of gravity on altitude, if you climb at a constant rate your weight will not change because there is no net acceleration. In the transitional period when you go from level flight to a constant rate climb, there will be a net acceleration and your weight will increase during that time. The converse will happen in a decent.
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Yah I am talking about apparent weight, like, the weight of something while submerged in a fluid. If you have a ball, that completely sinks in water, that means that the force of gravity on the ball (mg) is greater than the buoyant force of the water. If they ball floats, it means the buoyant force is greater than the mg. So if you were to weigh a ball that is sinking, it would show as less than it would if it were not in the water, due to the effect of the buoyant force. It obviously is not the same thing because the buoyant force is only determined by the volume of the displaced liquid, while lift is another force entirely. But I was thinking that the apparent weight of the plane would in fact change (between negative and positive value and also zero). Dunno if that is a valid way of looking at it though.
 
JohnF said:
Apparent weight? Operational weight? Gravitational weight? Which definition?

In skidoo's first experiment (bathroom scale/overhead bar) You need to pull on a spring scale attached to the bar and sum that scales reading with the bathroom scale. You'll find that your weight doesn't change no matter how hard you pull.

[So, does the weight of the airplane change]
No, and let's just assume a constant g for the intent of the OPs question, the weight of the airplane does not change.
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That's a good point. My assumption was that weight (i.e. force relative to the ground) does change. So, if we go back to the definition of weight, it is W=mg, where m=the mass and g=the force on the object due to gravity.

If the mass does not change, and the force on the object due to gravity does not change, then the Weight does not change. Your modified example demonstrates this. So, I concede and change my answer to No.
 
MAKG1 said:
No. There are two forces. They may oppose each other, but they are still there. It makes a difference to your wing roots.
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^^^ this.

All matter has mass and, when acted up by gravity, weight. Period. What one may choose to define as "the system" by which to discuss said mass and weight may vary, but the mass and weight of objects is still present.
 
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MarleyW said:
Yah I am talking about apparent weight, like, the weight of something while submerged in a fluid. If you have a ball, that completely sinks in water, that means that the force of gravity on the ball (mg) is greater than the buoyant force of the water. If they ball floats, it means the buoyant force is greater than the mg. So if you were to weigh a ball that is sinking, it would show as less than it would if it were not in the water, due to the effect of the buoyant force. It obviously is not the same thing because the buoyant force is only determined by the volume of the displaced liquid, while lift is another force entirely. But I was thinking that the apparent weight of the plane would in fact change (between negative and positive value and also zero). Dunno if that is a valid way of looking at it though.
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The bottom line is that whether it's aerodynamic lift or buoyancy, it's still an external force propping up your airplane or ship.

Consider two airships. They might be "weightless" as a unit, but what happens when they whack in to each other?
 
brywd said:
A plane carrying hundreds of birds in cages takes off. In flight the door to the cages opens somehow and all of the birds escape and are flying around the cabin. Does the weight of the plane go down?
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This made my day
 
skidoo said:
Always?

Say you stand still for 24 hours, you have effectively done a 360 degree turn (earth rotation). You have turned at the same rate for that 24 hours. What acceleration was involved?
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Yes, always.

There is a small centrifugal acceleration.
 
MarleyW said:
...But I was thinking that the apparent weight of the plane would in fact change (between negative and positive value and also zero). Dunno if that is a valid way of looking at it though.
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When you toss the word "apparent" in there it kind of muddies things up. Apparent to who or what? If an airplane weighs 2,000 pounds and is sitting on the ground then the weight is on the wheels and can be measured with the use of three scales. When the airplane is in steady straight and level flight there must be a lifting force on the wings and fuselage equal to 2,000 pounds to keep it there. It might be difficult or impossible to measure that with a set of scales but it's there.

Same thing with a boat. A 2,000 pound boat floats because it has displaced 2,000 pounds of water which is trying to reclaim it's territory and push the boat out.
 
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Jaybird180 said:
As a plane goes through the air, its thrust offsets or overpowers the drag, hence forward movement. The wing, based on the speed of the airflow at a certain angle of attack generates lift which offsets or overcomes weight.

So as an airplane's wing generates lift, does it weigh less?
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Thrust overpowers drags and hence the airplane accelerates forward. However, a plane in steady state flight is acted upon equally by both thrust and drag.
 
cirrusmx said:
Thrust overpowers drags and hence the airplane accelerates forward. However, a plane in steady state flight is acted upon equally by both thrust and drag.
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Still a simplification.

Thrust is not necessarily parallel to direction of motion. Drag is.

Not even in straight and level flight, especially slow flight. There is a vertical component of thrust and a horizontal component of lift.

The correct statement is that the vector sum of lift, drag, thrust, and weight is zero for steady flight. They don't separate as nicely as the cartoons suggest.
 
The other force acting on an airplane in straight and level flight is the Coriolis effect. Well it's an "effect" if you shoot a projectile over some distance. It will land on some point that is different from its straight line projection due to the rotation of the earth under it, to the left or right, or forward or rearward depending on the direction of travel.

For an airplane, it is an actual force, since it will be necessary to apply a correction factor in order to fly a straight line in any other than an east-west heading. The effect is very slight and unnoticeable (WCA is much larger) but nevertheless it does represent a very wide radius turn and consequent lateral acceleration.
 
Sac Arrow said:
The other force acting on an airplane in straight and level flight is the Coriolis effect. Well it's an "effect" if you shoot a projectile over some distance. It will land on some point that is different from its straight line projection due to the rotation of the earth under it, to the left or right, or forward or rearward depending on the direction of travel.

For an airplane, it is an actual force, since it will be necessary to apply a correction factor in order to fly a straight line in any other than an east-west heading. The effect is very slight and unnoticeable (WCA is much larger) but nevertheless it does represent a very wide radius turn and consequent lateral acceleration.
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Is the atmosphere also turning with the earth?
 
Jaybird180 said:
As a plane goes through the air, its thrust offsets or overpowers the drag, hence forward movement. The wing, based on the speed of the airflow at a certain angle of attack generates lift which offsets or overcomes weight.

So as an airplane's wing generates lift, does it weigh less?
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The answer is no. The air passing over/under the wing has nothing to do with the planes weight.
 
MAKG1 said:
For those of us who don't live in Florida, prevailing winds are in the opposite direction of earth rotation.
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Right, because they are directed by the larger, faster, more powerful band nearer the equator. The water on one side of your drain is flowing the opposite direction of the water on the other side of your drain.That's what the Coriolis force is all about. The difference in the speed of rotation a different latitudes.
 
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Henning said:
Yes, but not at the same speed, that's what causes weather patterns to be circular and move.
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The question was to give Sac some food for thought. Now go to the principal's office for not raising your hand. :D
 
MAKG1 said:
Still a simplification.

Thrust is not necessarily parallel to direction of motion. Drag is.

Not even in straight and level flight, especially slow flight. There is a vertical component of thrust and a horizontal component of lift.

The correct statement is that the vector sum of lift, drag, thrust, and weight is zero for steady flight. They don't separate as nicely as the cartoons suggest.
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sure. in other words, the important thing to know is that in steady flight the sum of the forces that act upon the airplane is equal to zero.
 
Jaybird180 said:
The question was to give Sac some food for thought. Now go to the principal's office for not raising your hand. :D
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Henning was generally correct so I didn't really have anything to add. My point is that regardless of the actions of the winds, a straight ground track actually follows a curved path through space.
 
cirrusmx said:
sure. in other words, the important thing to know is that in steady flight the sum of the forces that act upon the airplane is equal to zero.
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Yes, and it's widely misunderstood that this includes steady climbs and descents as well. Even slips. You might need an excess force to initiate a climb, but you do not need it to maintain.
 
Jaybird180 said:
As a plane goes through the air, its thrust offsets or overpowers the drag, hence forward movement. The wing, based on the speed of the airflow at a certain angle of attack generates lift which offsets or overcomes weight.

So as an airplane's wing generates lift, does it weigh less?
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I know I'm late to the discussion, but IIRC, mass increases with velocity in quantum physics. IMO, the plane would actually weigh more as its speed increases. :lol:
 
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