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Old Dec 21, 2012, 01:59 PM
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Originally Posted by richard hanson View Post
Sarcastic -
not really
prove my staement wrong.
What statement do you want me to prove wrong?
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Old Dec 21, 2012, 02:41 PM
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What statement do you want me to prove wrong?
Pick one any one -
Your choice --
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Old Dec 21, 2012, 02:59 PM
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Great - so therefore, the chopper will react with an equal and opposing force called lift.

Do you agree with the above statement?

If not - you clearly don't understand physics, particularly Newton's third law.
So you really believe that if you place a piece of plywood on the floor and stand on it, it will accelerate downward at a rate equal to W/m where W is you weight and m is the mass of the plywood?
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Old Dec 21, 2012, 04:26 PM
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Do I understand Newton's thrird law ?
example:
You make a silly comment - I will reply with a silly comment
How's that?
Seriously--
Newton's three laws - are important in that it was the first time we saw explanation of physical laws which had been applied by engineers , for thousands of years.
Enough already--
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Old Dec 22, 2012, 08:27 AM
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Originally Posted by ShoeDLG View Post
For an object that consists of multiple elements that can move relative to one another (like the air, for example), the momentum is defined as the sum of the momentum of the constituent elements. Momentum (like force) is a vector quantity with a magnitude and direction. When you consider the momentum of all the elements of an object, it is entirely possible for their sum to be zero (even if the individual elements are in motion).

This isn't my definition for the momentum, it's the one you'll find in any Classical Mechanics textbook.
Is it possible, in the case of a disturbance in a fluid, for the net momentum in that fluid to always be zero?
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Old Dec 22, 2012, 11:11 AM
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Is it possible, in the case of a disturbance in a fluid, for the net momentum in that fluid to always be zero?
It depends. If the density of the fluid is uniform and the fluid is bounded by fixed walls, then its momentum will always be zero. Otherwise a disturbance can change its net momentum.
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Old Dec 23, 2012, 05:33 AM
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I’m not sure if this will resonate with anyone, but it might help illustrate a key concept in the momentum discussion. In order to perform this “thought experiment” you will need to imagine:

1. An accurate scale with a large footprint. By “large footprint” I mean the scale can measure the downward force exerted on it over a large area.

2. A small electric helicopter that has a weight equal to W

3. A rigid foam block with a weight equal to w (mass equal to w/g)

In the first part of the thought experiment you place the foam block on the scale and adjust the scale’s dial to read zero. You then place the helicopter on top of the foam block. The scale’s dial should show a reading of W. Newton’s Second Law says that the foam block’s momentum will change a rate equal to the unbalanced force acting on it. What are the vertical forces acting on the foam block?

1. The helicopter pushes down on it with a force equal to W

2. The scale pushes up on the air with a force equal the reading on its dial, W

3. Gravity pulls down on it with a force equal to w

4. The scale pushes up on it with an additional force equal to w (this is the force we “zeroed out” when we set the dial to zero)

The net vertical force acting on the block is: (W + w) – (W + w) = 0, suggesting the block’s vertical momentum remains the same. This result is entirely consistent with our experience. If someone invoked Newton’s Second Law to tell you the block must be accelerating downward at a rate equal to Wg/w (by rearranging F=ma), it would be an obvious misapplication of Newton’s Laws.

You now remove the helicopter and foam block from the scale and set its dial to read zero. You then hover the helicopter above the scale (within a rotor diameter or two of the scale’s surface). In this case, the helicopter is supported not by the block, but by the air. What are the vertical forces acting on the air?

1. The helicopter pushes down on it with a force equal to W

2. The scale pushes up on the air with a force equal the reading on its dial. You can convince yourself (by actually doing the experiment or finding a video of it), that as long as the helicopter’s rotor area is small compared to the scale’s footprint, the reading on the scale’s dial will be very close to W

3. Gravity pulls down on the air with a force equal to its weight (w_air)

4. The scale pushes up on the air with an additional force equal to w_air (this is the force we “zeroed out” when we set the dial to zero)

The net vertical force acting on the air in this case is: (W + w_air) – (W + w_air) = 0, suggesting that the air’s vertical momentum remains the same. Unlike the block where we could look at it and verify that it wasn’t accelerating, it’s not as obvious with the air. Some of it is clearly being accelerated downward, but some of it is also being accelerated upward. However, just as the net vertical force acting on the foam block was zero, the net vertical force acting on the air is also zero (or very close to zero). Invoking Newton’s Second Law to suggest the air’s downward momentum is increasing at a rate equal to the force exerted on it by the helicopter is no different from suggesting that the foam block had to be accelerating downward at a rate equal to Wg/w. The fact that the air is in put in motion does not change the fact that the net vertical force acting on it is zero (or very close to zero). According to Newton's Second Law, it therefore experiences no change in its vertical momentum.

If your goal is to demonstrate that the air's rate of downward momentum change is equal to the force exerted on it by the helicopter, then you have to show that the scale's dial continues to read zero with the helicopter hovering over it. I haven't seen this yet.
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Old Dec 23, 2012, 07:12 AM
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If your goal is to demonstrate that the air's rate of downward momentum change is equal to the force exerted on it by the helicopter, then you have to show that the scale's dial continues to read zero with the helicopter hovering over it. I haven't seen this yet.
Is it not the fact that force equals the rate of change of momentum?

F=d(m.v)/dt

The rate of change of momentum of the air being caused to move by the rotor blades must be equal to the helicopters weight.
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Old Dec 23, 2012, 10:19 AM
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Is it not the fact that force equals the rate of change of momentum?
Newton's Second Law does not say that the air's rate of momentum change is equal to the force exerted on it by a helicopter. Netwon's Second Law says the air's rate of momentum change is equal to the sum of the forces exerted on it. If the scale in the above helicopter example exerts any upward force on the air at all, then the sum of the downward forces exerted on the air will be less than the helicopter's weight, and the air's rate of downward momentum change will also be less than the helicopter's weight. It turns out (from doing the experiment) that the scale exerts an upward force on the air that is almost exactly equal to the helicopter's weight. This means that the sum of the downward forces exerted on the air, and the air's rate of downward momentum change are almost exactly zero.

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Originally Posted by TangoKilo View Post
The rate of change of momentum of the air being caused to move by the rotor blades must be equal to the helicopters weight.
If this were true then the downward rate of change of the foam block's momentum would also be equal to the helicopter's weight. It's obvious that the momentum of the foam block isn't changing because it doesn't move. It isn't as obvious that the momentum of the air isn't changing, because the rotor puts it in motion. Just because the air is in motion doesn't mean that it has any net momentum. Your statement is a misapplication of Newton's Second Law (because you are not considering all the forces acting on the air).
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Old Dec 23, 2012, 11:02 AM
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if the foam block is placed on the scale - then the scale is zeroed -- whyis the block even considered
Are logic and laws independant?
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Old Dec 23, 2012, 11:41 AM
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The foam block just illustrates that if you support the helicopter above the scale with a solid object, the vertical forces on the object sum to zero and the object experiences no vertical momentum change (no acceleration). This is readily apparent when the object is solid.

If you replace the foam block with the air as the supporting object, the scale shows that the vertical forces on the supporting object (the air in this case) still sum to zero. For some reason, the motion of the air seems to lead some to the conclusion that its momentum must be changing. The conclusion that the air experiences net acceleration (momentum change) is just as wrong as the conclusion that the foam block is accelerated by the weight of the helicopter.

It isn't necessarily obvious that whenever you weigh something with a scale, you have effectively "zeroed out" the weight of the air acting on the scale. The reason for zeroing out the weight of the foam block in the above example was to follow the exact same procedure in both cases... in order to keep the logic and laws aligned.
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Old Dec 23, 2012, 12:02 PM
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OK- but the logic also says: if you zero the scale - the block is now simply part of the scale and as such, not involved.
I am of course a bit surprised at how the newtonian laws are construed (and misconstrued )
The three laws are pretty clear -but the attempts to apply them is much like the attempts to apply US tax laws .
The tax laws are of course are "interpretive" laws -
newtonian laws were meant to be absolute - unless I misunderstand them.
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Old Dec 23, 2012, 12:13 PM
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The trouble arises when Newton's Laws are misapplied. If you calculate an object's rate of momentum change without considering all of the forces acting on it, you will get the wrong answer. This isn't a problem with the laws, it's usually a problem associated with oversimplifying their application.
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Old Dec 23, 2012, 12:34 PM
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the block is now simply part of the scale and as such, not involved.
The block's involvement is that it is keeping the helicopter at a fixed height above the scale. It is performing exactly the same role that the air plays, and you have accounted for its weight in exactly the same way that you accounted for the weight of the air.
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Old Dec 23, 2012, 12:50 PM
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The block's involvement is that it is keeping the helicopter at a fixed height above the scale. It is performing exactly the same role that the air plays, and you have accounted for its weight in exactly the same way that you accounted for the weight of the air.
-OK- It seems to me that the size of the foam block is of no consequence in your test .
Y/N
therefore the block could have been HD foam (very tiny) or LD foam (quite large )
and your results would be the same?
Just muddying the water --- but then ----- I do like to simplify
PS how does one oversimplify?
I Know how to over complicate (even tho that is a nonsense term)
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