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Question 86-1 : Refer to figure 031 05 .given .force fa 100 n.distance a 6 m.distance b 3 m.calculate force fb to obtain equilibrium ? [ Result topography ]

200 n

exemple 186 200 n.

Question 86-2 : Refer to figure 031 34 .given.bem 1200 kg.bem cg 3 00.pilot and front pax 200 kg.cargo on pax floor 50 kg.fuel 250 kg.find the loaded centre of gravity cg by using the attached graph ?

2 96 m


Question 86-3 : Refer to figure 031 04 .given .force fa 100 n.distance a 6 m.distance b 3 m.calculate force fa to obtain equilibrium ?

50 n

exemple 194 50 n.

Question 86-4 : Select the correct statement for the cg safe range ?

The safe range falls between the front and rear cg limits and includes both limits

exemple 198 The safe range falls between the front and rear cg limits and includes both limits.

Question 86-5 : The fuel index is ?

Used to calculate the correct position of the cg due to different locations of the fuel tanks

exemple 202 Used to calculate the correct position of the cg due to different locations of the fuel tanks.

Question 86-6 : Refer to figure 031 03 .given .force fa 300 n.force fb 150 n.distance a 3 m.calculate distance b to obtain equilibrium ?

6 m

exemple 206 6 m.

Question 86-7 : Refer to figure 031 60 .given .weight 110800 kg.mac 31 6%.from the given values and information given by the attached sheet find the dry operating index if an additional pilot 85 kg will join the flight in zone e ?

120 4

Note de moi this question also exists at examination with .weight 112600 kg mac 30 7% and an additional pilot of 75 kg in zone g answer is 119 3 exemple 210 120.4.

Question 86-8 : Refer to figure 031 55 .find the cg position as %mac for the take off ?

24%

exemple 214 24%.

Question 86-9 : A mass of 600 kg is loaded at a station which is located 12 metres behind the present centre of gravity and 18 metres behind the datum the moment for that mass used in the loading manifest is assume g=10 m/s2 ?

108000 n m

exemple 218 108000 n.m

Question 86-10 : Refer to figure 031 00 .if the aft galley is loaded with 102 kg dry operating index doi will change by ?

1 52592

exemple 222 1.52592.

Question 86-11 : Refer to figure 031 02 .given .distance a 2 m.force fa 300 n.calculate force fc to obtain equilibrium ?

100 n

exemple 226 100 n.

Question 86-12 : For this question use annex ecqb 031 048 v2015 07 calculate the centre of gravity if the mass of the baggage is increased to 50 kg ?

85 79 mm

exemple 230 85.79 mm

Question 86-13 : A 5 m long plank is on a pivot located at 3 m of the left side a user applies a 600 n force on the end of the left side to be in balance what force must be apply on the opposite side ?

900 n

Ecqb04 dec 2018 . .3m x 600 n = 2m x n.1800 nm = 2m x n.1800 / 2 = 900 n exemple 234 900 n.

Question 86-14 : Refer to figure 031 52 .what is the forward cg limit and what is the aft cg limit for an aircraft having a gross mass of 4300 lbs ?

87 3 and 94 6

exemple 238 87.3 and 94.6.

Question 86-15 : Refer to figure 031 46 .how much fuel could be loaded at reference station 4 575 whan all thanks were fitted in this aircraft ?

806 l

Calculate the values from the tanks at station 4 575 .236 litre.324 litre.246 litre. .806 litre

Question 86-16 : Length of the mean aerodynamic chord 1 m.moment arm of the forward cargo 0 50 m.moment arm of the aft cargo + 2 50 m.the aircraft mass is 2 200 kg and its centre of gravity is at 25% mac.to move the centre of gravity to 40% which mass has to be transferred from the forward to the aft cargo hold ?

110 kg

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 and 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 exemple 246 110 kg.

Question 86-17 : For the transport aeroplane the moment balance arm for the forward hold centroid is . 210 ?

367 9 inches

Img149 exemple 250 367.9 inches.

Question 86-18 : Referring to the loading manual for the transport aeroplane the maximum load intensity for the lower forward cargo compartment is . 211 ?

68 kg per square foot

Img150 exemple 254 68 kg per square foot.

Question 86-19 : The maximum intensity floor loading for an aeroplane is given in the flight manual as 650 kg per square metre .what is the maximum mass of a package which can be safely supported on a pallet with dimensions of 80 cm by 80 cm ?

416 kg

The max floor loading is 650 kg per square metre .the area of the pallet is 0 8m x 0 8m = 0 64 m² .650 kg x 0 64 = 416 kg exemple 258 416 kg

Question 86-20 : A pallet having a freight platform which measures 200 cm x 250 cm has a total mass of 300 kg .the pallet is carried on two ground supports each measuring 20 cm x 200 cm .using the loading manual for the transport aeroplane calculate how much mass may be added to or must be off loaded from the ?

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 the current pallet mass 285 5 kg may be added exemple 262 285.5 kg may be added.

Question 86-21 : From the loading manual for the jet transport aeroplane the maximum floor loading intensity for the aft cargo compartment is . 213 ?

68 kg per square foot

On the line of the aft cargo compartment table we read 'maximum distribution load intensity kg per ft ² ' 68 exemple 266 68 kg per square foot.

Question 86-22 : From the loading manual for the transport aeroplane the aft cargo compartment has a maximum total load of . 214 ?

4187 kg

Img151 exemple 270 4187 kg.

Question 86-23 : From the loading manual for the transport aeroplane the maximum load that can be carried in that section of the aft cargo compartment which has a balance arm centroid at . 215 ?

835 5 inches is 3062 kg

Img152 exemple 274 835.5 inches is 3062 kg.

Question 86-24 : An aeroplane whose specific data is shown in the 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 the reference point representing a c g location at 30% mac mean aerodynamic cord .the current cargo load distribution is .front cargo 6 500 ?

Front cargo 3 740 kg rear cargo 6 760 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 the front cargo to the rear cargo in order to move the cg aft .front cargo = 6500 kg 2760 kg = 3740 kg .rear cargo = 4000 kg + 2760 kg = 6760 kg exemple 278 Front cargo: 3 740 kg, rear cargo: 6 760 kg

Question 86-25 : The floor limit of an aircraft cargo hold is 5 000 n/m2 .it is 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 ?

80 kg

Weight in kg = 5000 n/m² / 10 = 500 kg .500 x 0 4 x 0 4 = 80 kg exemple 282 80 kg.

Question 86-26 : The floor of the main cargo hold is limited to 4000 n/m2 .it is 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 ?

100 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 exemple 286 100 kg.

Question 86-27 : Given . actual mass 116 500 lbs. original cg station 435 0. compartment a station 285 5. compartment b station 792 5.if 390 lbs of cargo is moved from compartment b aft to compartment a forward what is the station number of the new cg ?

433 3

Note that this is the only answer that moves the 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 and 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 the new cg 435 0 1 7 = 433 3 exemple 290 433.3

Question 86-28 : Pallet ground base 1 44 m² .the pallet is carried on two ground supports each measuring 1 2 m x 0 2 m each .using the maximum floor loading intensity for the cargo compartment of 732 kg/m² calculate the maximum mass which can be loaded onto the pallet ?

351 kg

Area in contact with surface = 2 x 1 2 x 0 2 = 0 48 m².732 x 0 48 = 351 kg exemple 294 351 kg.

Question 86-29 : The maximum floor loading for a cargo compartment in an aircraft is given as 750 kg per square metre .a package with a mass of 600 kg is to be loaded .assuming the pallet base is entirely in contact with the floor which of the following is the minimum size pallet that can be used ?

40 cm by 200 cm

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 exemple 298 40 cm by 200 cm.

Question 86-30 : Given the following data how much cargo must be moved from the forward hold to the aft hold to achieve a cg at 33% mac .take off mass 200000 kg.forward hold cargo 6500 kg.aft hold cargo 4000 kg.distance between holds 10 m.current cg 30%mac.mac 4 6 m ?

2760 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 exemple 302 2760 kg.

Question 86-31 : What is the maximum running load in the aft section of the forward lower compartment . 232 ?

13 12 kg/in

Img174 exemple 306 13.12 kg/in.

Question 86-32 : Palletised cargo ?

Consists of different cargo box on pallets stored in the cargo holds

exemple 310 Consists of different cargo box on pallets stored in the cargo holds.

Question 86-33 : Bulk cargo ?

Can be loaded without specific loading equipment

Bulk cargo loose unpackaged non containerized cargo such as cement grains ores exemple 314 Can be loaded without specific loading equipment.

Question 86-34 : Containerised cargo ?

Consists of baggage and cargo loaded into standard size containers stored in the cargo holds

Containerised cargo baggage and cargo can be loaded into standard size containers designed to fit and lock into the cargo compartment the containers have an individual maximum mass limit and an individual floor loading limit mass per unit aera exemple 318 Consists of baggage and cargo loaded into standard size containers stored in the cargo holds.

Question 86-35 : Bulk cargo ?

Consists of cargo box baggage loosely loaded in the cargo holds

Bulk cargo can be loaded without specific loading equipment exemple 322 Consists of cargo (box, baggage) loosely loaded in the cargo holds.

Question 86-36 : Define the maximum load distribution ?

Load divided by smallest area

exemple 326 Load divided by smallest area.

Question 86-37 : A pallet having a freight platform which measures 100 cm x 150 cm has a total mass of 300 kg .the pallet is carried on two ground supports each measuring 20 cm x 100 cm using the maximum floor loading intensity for the cargo compartment of 800 kg/m² .calculate how much mass may be added to or must ?

20 kg may be added

Area in contact with surface = 0 2 m x 1 m x 2 = 0 4m².maximum floor loading = 800 kg/m².therefore max load for 0 4m² is .800 x 0 4 = 320 kg.20 kg may be added to a 300 kg pallet exemple 330 20 kg may be added.

Question 86-38 : A container that measure 1 42m² is to be loaded the maximum floor loading is 720 kg/m² the maximum load that can be put in the container is ?

1022 kg

1 42 x 720 = 1022 4 kg exemple 334 1022 kg.

Question 86-39 : An aircraft has a mass of 5000 lbs and the cg is located at 80 inches aft of the datum .the aft cg limit is at 80 5 inches aft of the datum .what is the maximum mass that can be loaded into a hold situated 150 inches aft of the datum without exceeding the limit ?

35 97 lbs

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 exemple 338 35.97 lbs.

Question 86-40 : An aircraft has a loaded mass of 5500 lbs the cg is 22 inches aft of the datum .a passenger mass 150 lbs moves aft from row 1 to row 3 a distance of 70 inches .what will be the new position of the cg assuming all dimensions aft of the datum ?

23 9 inches

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 exemple 342 23.9 inches.


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