ranklein
At the risk of flogging a dead horse ………………
If you are looking to understand the answer to your CG question rather than be buried in formulae, you might like to try the following:
Fig. 1 below shows an aircraft balanced on a fulcrum positioned under its CG. The fulcrum is on a load cell which can measure the weight of the aircraft. The balance is unstable, so that if the aircraft tips slightly forward it will tend to fall over forward more, and similarly if it tips slightly clockwise it will tend to fall back onto its tail. But for the moment its in perfect balance. However, this is not (yet) the aircraft in the problem.
Fig. 1
Inside the aircraft is a level floor as shown below, and on this floor are two heavy steel cylinders, “A” and “B“, each of mass m and therefore weight mg. “A” is placed at Station 1030 and “B” is an equal distance on the far side of the fulcrum. The cylinder “A” produces an anti-clockwise turning moment tending to cause the aircraft to fall on its nose, while “B” produces an equal and opposite clockwise turning moment tending to cause the aircraft to fall on its tail. These two moments are equal and opposite, so the aircraft is still in balance. The magnitude of each of these moments is numerically equal to mg times (the distance of the cylinder from the fulcrum). The load cell shows that the aircraft including the cylinders weighs a total of 285,000 lbs. The aircraft including the cylinders represents the aircraft in the problem.
Fig. 2
If we now enlist the aid of a ghostly FA (who is naturally weightless) to roll “A” from Station 1030 to Station 600, the anti-clockwise moment produced by “A” will INCREASE to mg times (the new distance of “A” from the fulcrum), and the aircraft will tend to fall over forwards (cylinder “B“ is not moved). The CHANGE in the moment produced by “A” will be equal to the DIFFERENCE between the initial and final moments due to “A”, which turns out to be mg times (the distance from Station 1030 to Station 600), or mg x 430 inch-lbs.
Fig. 3
Now, if you were outside the aircraft, with no knowledge of what was going on inside, you could observe the following sequence:
Initially the aircraft would be in balance and the load cell would show the weight to be 285,000 lbs as in Fig. 1 above.
At a later time the aircraft would be seen to have an unchanged weight of 285,000 lbs but it would be trying to fall over forwards as shown below, and the obvious explanation would be that something had happened internally to move the CG forward. In the “external” view, the anti-clockwise moment causing this would be equal to 285,000 times (the distance that the CG had moved forward) in inch.lbs. We are told that this distance is 2.3 inches.
Fig. 4
Thus, we have two different descriptions of why the aircraft is tending to fall over forwards:
(a) the “internal” view where we know that the tipping moment is due to moving the cylinder “A” from Station 1030 to Station 600, and is equal to mg x 430, and
(b) the “external” view where we know only that the CG has moved 2.3 inches involving a tipping moment of 285,000 x 2.3 inch.lbs.
Since these are two descriptions of the SAME THING, these two moments must be equal.
That is, the forward-tipping moment = mg x 430, and is also = 285,000 x 2.3 inch.lbs
Therefore, mg = (285,000 x 2.3)/430 = 1,524 lbs.
Cheers,
