Refer to figure 031 04 Given Force Fa 100 NDistance A 6 mDistance B 3 mCalculate force Fa to obtain equilibrium ?
Exam > pilot
Select the correct statement for the CG safe range ?
The fuel index is ?
Refer to figure 031 03 given force fa 300 nforce fb 150 ndistance a 3 mcalculate distance b to obtain equilibrium Used to calculate correct position of cg due to different locations of fuel tanks.. 
Refer to figure 031 60 given weight 110800 kgmac 31 6%from given values and information given the attached sheet find dry operating index if an additional pilot 85 kg will join flight in zone e Used to calculate correct position of cg due to different locations of fuel tanks. note moi this question also exists at examination with weight 112600 kg mac 30 7% an additional pilot of 75 kg in zone g answer 119 3. 
Refer to figure 031 55 find cg position as %mac the take off Used to calculate correct position of cg due to different locations of fuel tanks. note moi this question also exists at examination with weight 112600 kg mac 30 7% an additional pilot of 75 kg in zone g answer 119 3. 
A mass of 600 kg loaded at a station which located 12 metres behind present centre of gravity and 18 metres behind datum the moment that mass used in loading manifest assume g=10 m/s2 Used to calculate correct position of cg due to different locations of fuel tanks.. Question 70-8
Refer to figure 031 00 if aft galley loaded with 102 kg dry operating index doi will change Used to calculate correct position of cg due to different locations of fuel tanks.. Question 70-9
Refer to figure 031 02 given distance a 2 mforce fa 300 ncalculate force fc to obtain equilibrium Used to calculate correct position of cg due to different locations of fuel tanks.. Question 70-10
for this question use annex ecqb 031 048 v2015 07 calculate centre of gravity if mass of baggage increased to 50 kg Used to calculate correct position of cg due to different locations of fuel tanks.. Question 70-11
A 5 m long plank on a pivot located at 3 m of left side a user applies a 600 n force on end of left side to be in balance what force must be apply on opposite side Used to calculate correct position of cg due to different locations of fuel tanks. Ecqb04 dec 2018 3m x 600 n = 2m x ? n 1800 nm = 2m x ? n 1800 / 2 = 900 n. Question 70-12
Refer to figure 031 52 what the forward cg limit and what the aft cg limit an aircraft having a gross mass of 4300 lbs Used to calculate correct position of cg due to different locations of fuel tanks. Ecqb04 dec 2018 3m x 600 n = 2m x ? n 1800 nm = 2m x ? n 1800 / 2 = 900 n. Question 70-13
Refer to figure 031 46 how much fuel could be loaded at reference station 4 575 whan all thanks were fitted in this aircraft Used to calculate correct position of cg due to different locations of fuel tanks. Calculate values from tanks at station 4 575 236 litre 324 litre 246 litre 806 litre. Question 70-14
Length of mean aerodynamic chord 1 mmoment arm of forward cargo 0 50 mmoment arm of aft cargo + 2 50 mthe aircraft mass 2 200 kg and its centre of gravity at 25% macto move centre of gravity to 40% which mass has to be transferred from forward to aft cargo hold Used to calculate correct position of cg due to different locations of fuel tanks. Change in mass / total mass = change in cg / total distance moved change in cg = 0 15 m (25%mac to 40%mac of 1 metre) total distance moved = distance between front forward cargo aft cargo = 0 5 m to 2 5 m = 3 m change in mass = total mass x change in cg / total distance moved change in mass = 2200 x 0 15 /3 = 110 kg. Question 70-15
For transport aeroplane moment balance arm the forward hold centroid 210 Used to calculate correct position of cg due to different locations of fuel tanks. Change in mass / total mass = change in cg / total distance moved change in cg = 0 15 m (25%mac to 40%mac of 1 metre) total distance moved = distance between front forward cargo aft cargo = 0 5 m to 2 5 m = 3 m change in mass = total mass x change in cg / total distance moved change in mass = 2200 x 0 15 /3 = 110 kg. Question 70-16
Referring to loading manual the transport aeroplane maximum load intensity the lower forward cargo compartment 211 68 kg per square foot. Change in mass / total mass = change in cg / total distance moved change in cg = 0 15 m (25%mac to 40%mac of 1 metre) total distance moved = distance between front forward cargo aft cargo = 0 5 m to 2 5 m = 3 m change in mass = total mass x change in cg / total distance moved change in mass = 2200 x 0 15 /3 = 110 kg. Question 70-17
The maximum intensity floor loading an aeroplane given in flight manual as 650 kg per square metre what the maximum mass of a package which can be safely supported on a pallet with dimensions of 80 cm 80 cm 68 kg per square foot. The max floor loading 650 kg per square metre the area of pallet 0 8m x 0 8m = 0 64 m² 650 kg x 0 64 = 416 kg. Question 70-18
A pallet having a freight platform which measures 200 cm x 250 cm has a total mass of 300 kg the pallet carried on two ground supports each measuring 20 cm x 200 cm using loading manual the transport aeroplane calculate how much mass may be added to or must be off loaded from pallet in order the load intensity to match maximum permitted distribution load intensity lower deck forward cargo compartment 212 285 5 kg may be added. Surface contact area = 0 2 m x 2 m x 2 ground supports = 0 8m² maximum permitted distribution load intensity 1m² converted in feet = 3 28 x 3 28 = 10 76 ft² 0 8 x 10 76 = 8 61 ft² reference states that we can load 68 kg per ft² thus 68 x 8 61 = 585 3 kg 585 3 300 kg of current pallet mass 285 5 kg may be added. Question 70-19
From loading manual the jet transport aeroplane maximum floor loading intensity the aft cargo compartment 213 68 kg per square foot. On line of aft cargo compartment table we read 'maximum distribution load intensity (kg per ft ²)' 68. Question 70-20
From loading manual the transport aeroplane aft cargo compartment has a maximum total load of 214 68 kg per square foot. On line of aft cargo compartment table we read 'maximum distribution load intensity (kg per ft ²)' 68. Question 70-21
From loading manual the transport aeroplane maximum load that can be carried in that section of aft cargo compartment which has a balance arm centroid at 215 835 5 inches 3 62 kg. On line of aft cargo compartment table we read 'maximum distribution load intensity (kg per ft ²)' 68. Question 70-22
An aeroplane whose specific data shown in annex has a planned take off mass of 200 000 kg with its centre of gravity c g located at 15 38 m rearward of reference point representing a c g location at 30% mac mean aerodynamic cord the current cargo load distribution front cargo 6 500 kgrear cargo 4 000 kg for performance purposes captain decides to reset value of centre of gravity location to 33% mac the front and rear cargo compartments are located at a distance of 15 m and 25 m from reference point respectively after transfer operation new cargo load distribution 219 Front cargo 3 74 kg rear cargo 6 76 kg. Mean aerodynamic cord lenght = 18 6 m 14 m = 4 6 m centre of gravity modification = 30% to 33% = 3% 3% of mac lenght = 3% of 4 6 m = 0 138 m mass change / total mass = change of cg / total distance moved mass change = change of cg x total mass / total distance moved mass change = 0 138 m x 200000 kg / 10 m mass change = 2760 kg we have to transfer 2760 kg from front cargo to rear cargo in order to move cg aft front cargo = 6500 kg 2760 kg = 3740 kg rear cargo = 4000 kg + 2760 kg = 6760 kg. Question 70-23
The floor limit of an aircraft cargo hold 5 000 n/m2 it planned to load up a cubic container measuring 0 4 m of side it's maximum gross mass must not exceed assume g=10m/s2 Front cargo 3 74 kg rear cargo 6 76 kg. Weight in kg = 5000 n/m² / 10 = 500 kg 500 x (0 4 x 0 4) = 80 kg. Question 70-24
The floor of main cargo hold limited to 4000 n/m2 it planned to load a cubic container each side of which measures 0 5 m its maximum gross mass must not exceed assume g = 10m/s2 Front cargo 3 74 kg rear cargo 6 76 kg. Footprint of one cubic container is 0 5 x 0 5 = 0 25 m² 4000 n/m² x 0 25 m² = 1000 n 10 n = 1 kg maximum gross mass must not exceed 100 kg per container. Question 70-25
Given actual mass 116 500 lbs original cg station 435 0 compartment a station 285 5 compartment b station 792 5if 390 lbs of cargo moved from compartment b aft to compartment a forward what the station number of new cg Front cargo 3 74 kg rear cargo 6 76 kg. Note that this the only answer that moves center of gravity forward! change in mass / total mass = change in cg / total distance moved change in mass = 390 lbs change in cg = ? total distance moved = distance between a b = 792 5 285 5 = 507 change in mass = total mass x change in cg / total distance moved 390 = 116500 x 507 / change in cg change in cg = 507 x 390 / 116 500 = 1 70 station number of new cg 435 0 1 7 = 433 3. Question 70-26
Pallet ground base 1 44 m² the pallet carried on two ground supports each measuring 1 2 m x 0 2 m each using maximum floor loading intensity the cargo compartment of 732 kg/m² calculate maximum mass which can be loaded onto pallet Front cargo 3 74 kg rear cargo 6 76 kg. Area in contact with surface = 2 x 1 2 x 0 2 = 0 48 m² 732 x 0 48 = 351 kg. Question 70-27
The maximum floor loading a cargo compartment in an aircraft given as 750 kg per square metre a package with a mass of 600 kg to be loaded assuming pallet base entirely in contact with floor which of following the minimum size pallet that can be used Front cargo 3 74 kg rear cargo 6 76 kg. The minimum size that can be used 600 / (0 4 x 2) = 750 kg/m² or 600 kg = 80% of 750 kg 80% of 1 m² = 0 8 m² 40 x 200 = 800 cm² (0 8 m²). Question 70-28
Given following data how much cargo must be moved from forward hold to aft hold to achieve a cg at 33% mac take off mass 200000 kgforward hold cargo 6500 kgaft hold cargo 4000 kgdistance between holds 10 mcurrent cg 30%macmac 4 6 m Front cargo 3 74 kg rear cargo 6 76 kg. Mean aerodynamic cord lenght = 18 6 m 14 m = 4 6 m centre of gravity modification = 30% to 33% = 3% 3% of mac lenght = 3% of 4 6 m = 0 138 m mass change / total mass = change of cg / total distance moved mass change = change of cg x total mass / total distance moved mass change = 0 138 m x 200000 kg / 10 m mass change = 2760 kg. Question 70-29
What the maximum running load in aft section of forward lower compartment 232 Front cargo 3 74 kg rear cargo 6 76 kg. Mean aerodynamic cord lenght = 18 6 m 14 m = 4 6 m centre of gravity modification = 30% to 33% = 3% 3% of mac lenght = 3% of 4 6 m = 0 138 m mass change / total mass = change of cg / total distance moved mass change = change of cg x total mass / total distance moved mass change = 0 138 m x 200000 kg / 10 m mass change = 2760 kg. Question 70-30
Palletised cargo Consists of different cargo box on pallets stored in cargo holds. Mean aerodynamic cord lenght = 18 6 m 14 m = 4 6 m centre of gravity modification = 30% to 33% = 3% 3% of mac lenght = 3% of 4 6 m = 0 138 m mass change / total mass = change of cg / total distance moved mass change = change of cg x total mass / total distance moved mass change = 0 138 m x 200000 kg / 10 m mass change = 2760 kg. Question 70-31
Bulk cargo Can be loaded without specific loading equipment. Bulk cargo loose unpackaged non containerized cargo (such as cement grains ores). Question 70-32
Containerised cargo Consists of baggage cargo loaded into standard size containers stored in cargo holds. Containerised cargo baggage cargo can be loaded into standard size containers designed to fit lock into cargo compartment the containers have an individual maximum mass limit an individual floor loading limit (mass per unit aera). Question 70-33
Bulk cargo Consists of cargo (box baggage) loosely loaded in cargo holds. Bulk cargo can be loaded without specific loading equipment. Question 70-34
Define maximum load distribution Load divided smallest area. Bulk cargo can be loaded without specific loading equipment. Question 70-35
A pallet having a freight platform which measures 100 cm x 150 cm has a total mass of 300 kg the pallet carried on two ground supports each measuring 20 cm x 100 cm using maximum floor loading intensity the cargo compartment of 800 kg/m² calculate how much mass may be added to or must be off loaded from pallet in order the load intensity to match maximum permitted distribution load intensity the cargo compartment Load divided smallest area. Area in contact with surface = 0 2 m x 1 m x 2 = 0 4m² maximum floor loading = 800 kg/m² therefore max load 0 4m² is 800 x 0 4 = 320 kg 20 kg may be added to a 300 kg pallet. Question 70-36
A container that measure 1 42m² to be loaded the maximum floor loading 720 kg/m² the maximum load that can be put in container Load divided smallest area. 1 42 x 720 = 1022 4 kg. Question 70-37
An aircraft has a mass of 5000 lbs and cg located at 80 inches aft of datum the aft cg limit at 80 5 inches aft of datum what the maximum mass that can be loaded into a hold situated 150 inches aft of datum without exceeding limit Load divided smallest area. Mass added / old total mass = change of cg / distance from hold to new cg mass added = (change of cg / distance from hold to new cg) x old total mass mass added = (0 5 / (150 80 5)) x 5000 mass added = 35 97 lbs. Question 70-38
An aircraft has a loaded mass of 5500 lbs the cg 22 inches aft of datum a passenger mass 150 lbs moves aft from row 1 to row 3 a distance of 70 inches what will be new position of cg assuming all dimensions aft of datum Load divided smallest area. Mass moved / total mass = change of cg / distance moved change of cg = mass moved x distance moved / total mass change of cg = 150 x 70 / 5500 = 1 9 inches new cg location 22 + 1 9 = 23 9 inches. Question 70-39
The cg limits of an aircraft are from 83 inches to 93 inches aft of datum the cg as loaded found to be at 81 inches aft of datum the loaded mass 3240 lbs how much mass must be moved from forward hold 25 inches aft of datum to aft hold 142 inches aft of datum to bring cg onto forward limit Load divided smallest area. Mass moved / total mass = change of cg / distance moved mass moved = (change of cg x total mass) / distance moved mass moved = (2 x 3240) / (142 25) mass moved = 55 38 lbs. Question 70-40
An aircraft has three holds situated 10 inches 100 inches and 250 inches aft of datum identified as holds a b and c respectively the total aircraft mass 3500 kg and cg 70 inches aft of datum the cg limits are from 40 inches to 70 inches aft of datum how much load must be removed from hold c to ensure that cg positioned on forward limit Load divided smallest area. Mass change / total mass = change of cg / total distance moved mass change = change of cg x total mass / total distance moved mass change = ((70 40) x 3500 / (250 40) mass change = 30 x 3500 / 210 mass change = 500 kg.
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