Given that the total mass of an aeroplane is 112 000 kg with a centre of gravity position at 22.62 m aft of the datum. The centre of gravity limits are between 18 m and 22 m.
How much mass must be removed from the rear hold (30 m aft of the datum) to move the centre of gravity to the middle of the limits?
The Center of Gravity (CG) is where the entire mass is evenly distributed. Remember the formula for CG calculation:
- CG = Total Moment ÷ Total Mass
Where Total Moment = Total Mass x CG.
To determine the mass, we'll use a simple equation. By calculating the total moment in the current configuration and subtracting the moment of the load removal, we can ascertain the new total moment. Given the provided data - old and new CG, station of load removal, and total mass, we may find the mass "m" of the load removal:
New Total Moment = Old Total Moment - Load Removal
(112 000 - m) × 20 = 112 000 × 22.62 - m × 30
112 000 × 20 - m × 20 = 112 000 × 22.62 - m × 30
2 240 000 - m × 20 = 2 533 440 - m × 30
-m × 20 = 293 440 - m × 30
m × 10 = 293 440
m = 29 344 kg
Alternatively, for a more straightforward approach, you may use the following formula:
Change in Mass / Old Total Mass = Change in CG / Distance from mass to new CG
Change in Mass = (Change in CG / distance from mass to new CG) x Old total mass
Change in Mass = (2.62 / 10) x 112 000
Change in Mass = 29 344 kg
Your Notes (not visible to others)
This question has appeared on the real examination, you can find the related countries below.
-
Austro Control1
-
Lithuania1
-
Romania1
Use Keyboard Shortcuts
