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Question 92-1 : The cg centre of gravity of an aicraft is 58 inches forward of the datum, with a total mass of 2609 lb. if a last minute load of 128 lb occurs and it is placed 32 inches aft of the datum, where is the new cg located ? [ Diploma registration ]
53.79 inches forward of the datum.
1st method..mass change / new total mass = change of cg / distance from mass to old cg... mass change 128 lb. new total mass 2 609 lb + 128 lb = 2 737 lb. change of cg . distance from mass to old cg 58 in + 32 in = 90 in...128 lb / 2 737 lb = change of cg / 90 in..change of cg = 128 / 2 737 x 90 = 4.208 in..since the initial cg position is forward of the datum and the last minute load is added aft of the datum, the new position of cg will be shifted closer to the datum and will be..new cg position = old cg position + change of cg = 58 inches 4.2 in = 53.79 inches forward of datum...2nd method..moment. the turning force created by the mass over a distance or lever arm...moment = mass x arm..remember that, by convention, the items aft of the datum have a positive arm, and consequently a positive moment, and those which are forward of the datum have a negative arm and moment...generally, the centre of gravity cg can be found by adding all the moments and then dividing by the total mass..cg = total moment / total mass... initially, the aircraft's moment was 2 609 lb x 58 in. = 151 322 lb in.. after adding the last minute load, the aircraft's moment became 151 322 lb in. + 128 lb x 32 in = 147 226 lb in...thus, the new position of the cg, after loading the last minute load will be..cg = total moment / total mass = 147 226 lb in / 2 609 lb + 128 lb = 53.79 in...the minus symbol implies that the new cg position is 53.79 in. forward of the datum.
Question 92-2 : An aircraft has its datum at the nose. the front seats are 65 inches aft of datum, the passenger seats are 105 inches aft and the separate baggage compartment is 137 inches aft. the aft limit is 139 inches aft. the current mass is 2530 lb...three passengers are in the rear seats, with the baggage ?
103 lb
Learning objective 031.05.03.02.01 calculate the amount of additional load or ballast to be loaded at or removed from a given position or compartment to establish a defined cg position....current situation..the aft cg limit is 139 inches aft of the datum...however, due to the current arrangement of passengers, the cg exceeds the aft cg limits by 3 inches making the current cg at 142 inches aft of the datum...how may we bring the cg forward by 3 inches to meet the cg aft limit without re arranging the passengers we will use an extra ballast mass forward of the aircraft front seats to bring the cg to the aft limit...how much extra ballast..cg is 3'' behind the aft limit. therefore, the cg is 142'' aft of datum.... old total moment = 2530 x 142 lb.in...to bring the cg forward to within limits, a ballast 'm' must be added to the front seat. the front seat is located at 65'' aft of datum.... ballast moment = m x 65 lb.in...the new total mass will be... new total mass = 2530 + m...this will result in a new total moment... new total moment = 2530 + m x 139 lb.in...now apply the formula old total moment + moment of the ballast = new total moment....2530 x 142 + m x 65 = 2530 + m x 139..2530 x 142 + m x 65 = 2530 x 139 + m x 139..m = 102.6 lb = 103 lb...alternatively, simply apply the following formula..new mass x new cg = old mass x old cg ± mass x arm.. 2530 + m x 139 = 2530 x 142 + 65 x m..351 670 + 139 m = 359 260 + 65 m..74 m = 7590..m = 102.6 lb 103 lb
Question 92-3 : You are departing for a flight with a planned landing fuel of 3000 kg. according to aircraft limitations, the centre tank fuel must be used prior to the wing tank fuel...given the following information, the location of the cg on landing will be how many metres aft of the datum..maximum take off mass ?
15.95
These are the formulas we will need to solve this exercise..moment = mass x arm...cg = total moment / total mass...for this question, we are looking for the cg on landing. thus, the moment and mass of the burnt fuel must be subtracted from the take off moment and take off mass correspondingly, so as to find the landing moment and landing mass...the cg on landing is given by the formula landing cg = landing moment / landing mass.... 1 determine the landing moment..landing moment = take off moment centre tank fuel moment wings tanks fuel burnt moment... take off moment = actual take off mass x take off cg location.... . . take off moment = 59 500 kg x 16.26 m = 967 470 kgm. . .... centre tank fuel moment = fuel in centre tank x centre fuel tank centroid.... . . centre tank fuel moment = 16 000 kg x 15.38 m = 246 080 kgm. . .... wing tanks fuel burnt moment = wing tanks fuel burnt at landing x wing fuel tank centroid.... . . wing tanks fuel burnt moment = 18 000 3000 kg x 17.79 m = 266 850 kgm. . .....landing moment = 967 470 kgm 246 080 kgm 266 850 kgm = 454 540 kgm... 2 determine the landing mass... landing mass = actual take off mass fuel centre tank burnt wing tanks fuel burnt.... . . landing mass = 59 500 kg 16 000 kg 18 000 3000 kg = 28 500 kg. . .... 3 determine the landing cg... landing cg = landing moment / landing mass.... . . landing cg = 454 540 kgm / 28 500 kg.. landing cg = 15.95 m
Question 92-4 : Prior to departure, the load controller informs the pilot about a lmc of 4763 kg loaded in the aft cargo compartment. can the pilot accept it take off fuel 8000 kg.take off cg 44%.take off mass 60 000 kg.underload 5687 kg bigger than the lmc.cg limit 16% 46%.operating mass 55 000 kgmac length 4 ?
No, because it will be outside of the cg limits.
. 1 find old cg.the old cg position is 44% relative to mac and it should be converted into distance meters from the datum.cg distance from leading edge = cg position % relative to mac x mac length cg distance from leading edge = 44% x 4 m = 1.76 mdistance of mac trailing edge 16 m aft of the datumlength of mac 4 mdistance of mac leading edge 16 m 4 m = 12 m aft of the datumold cg is 12 m + 1.76 m = 13.76 m aft of datum. 2 find new cg.1st method.mass change / new total mass = change of cg / distance from mass to old cg mass change 4763 kgnew total mass 60 000 kg + 4763 kg = 64 763 kgchange of cg distance from mass to old cg 15.5 m 13.76 m = 1.74 m4763 kg / 64 763 kg = change of cg / 1.74 m.change of cg = 4763 / 64 763 x 1.74 = 0.13 msince the initial cg position is aft of the datum and the lmc is added aft of the datum, the new position of cg will be shifted away from the datum and will be new cg position = old cg position + change of cg = 13.76 m + 0.13 m = 13.89 m aft of datum2nd method.cg = total moment / total mass.moment = mass x armtotal moment = old moment + lmc moment = 60 000 kg x 13.76 m + 4763 kg x 15.5 m = 825 600 + 73 827 = 899 427 kg mtotal mass = old mass + lmc mass = 60 000 kg + 4763 kg = 64 763 kgcg = 899 427 kg m / 64 763 kg = 13.89 msince the initial cg position is aft of the datum and the lmc is added aft of the datum, the new position of cg will be shifted away from the datum and will be 13.89 m aft of datum. 3 check if the new cg is within limits.cg limit 16% 46%.new cg position 13.89 m aft of datum.the new cg position needs to be converted into % mac to check whether it is within the cg limitations..cg position % relative to mac = cg distance from leading edge / mac lengthnew cg distance from leading edge = 13.89 m 12 m = 1.89 mmac lenth = 4 mnew cg position % relative to mac = 1.89 m / 4 m = 47.25 %the cg limit is 16% 46% and the new cg position is at 47.25% which means that the new cg is out of limits and the lmc cannot be loaded.
Question 92-5 : After the weighing procedure of an aircraft with the results given below, what is the cg position of the bem with reference to the main landing gear nose landing gear jack point 4350 kg.each main landing gear jack point 15 505 kg.longitudinal distance between the nose landing gear and the main ?
1.5 m forward of datum
Learning objective 031.05.01.01.01 calculate the cg position of an aircraft by using the formula cg position = sum of moments / total mass.moment is the turning force created by the mass over a distance or lever arm moment = mass x armremember that, by convention, the items aft of the datum have a positive arm, and consequently a positive moment, and those which are forward of the datum have a negative arm and moment.thus, since the nose landing gear jack point is forward of the datum main landing gears jack points , both its arm and moment will be negative.generally, the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment / total mass.the total mass of the aircraft is 4350 kg + 2 x 15 505 kg = 35 360 kgand its total moment 4350 kg x 12.2 m + 2 x 15 505 kg x 0 = 53 070 kg.m.therefore, the cg position of the bem will be 53 070 kg.m / 35 360 kg = 1.5 m.the minus symbol implies that the cg position is 1.5 m forward of the main landing gears.
Question 92-6 : The pilot of an aircraft is planning a flight carrying a bridal pair, both sitting on the rear seats. the calculated cg is out of limits and the pilots decides to load some ballast at the front seats to move the cg within limits. the aircraft total mass is 2570 lb. given the following data, how much ?
336 lb
The cg is currently rear of the aft limit and needs to move forwards. this distance is 160 in to 149 in = 11 inin order to move the cg into limits, the pilot will load ballast on to the front seat. the arms for the rear seats and the compartment are not required in this question.step 1. establish the parameters m = mass to be added m = total mass 2570 lbs d = cg movement required 11 in d = relevant distance, between the target cg position and the front seat being loaded distance between 149 in and 65 in = 84 in step 2. time for the formula m/m = d/dm/2570 = 11/84m = 336.5 lbsby adding 336.5 lbs on the front seat, the cg is moved into limits.
Question 92-7 : The cg limits of an aircraft are between 4.08 m and 4.32 m aft of the datum. the take off mass of an aircraft is 3195 kg and the calculated cg is at 4.35 m aft of the datum. if a mass of 84 kg can be moved, what is the distance the mass should be moved in order to get a cg at the aft limit ?
114 cm forward
Use the formula mass change / total mass = change of cg / distancemass change 84 kgtotal mass 3195 kgchange of cg 4.35 m 4.32 m = 0.03 m = 3 cmdistance 84 kg / 3195 kg = 3 / distance.distance = 3195 kg x 3 cm / 84 kg.distance = 114 cmthe cg has to be moved forward from 4.35 m aft of datum to 4.32 m aft of datum , so the mass must be moved 114 cm forward.
Question 92-8 : The calculated cg before take off is 8 cm beyond the cg limit. the aircraft has a total mass of 7132 kg. in order to have a cg within limits, a mass of 320 kg can be shifted within the hold. how much should you move this mass ?
1.783 m
Use the formula mass change / total mass = change of cg / distancemass change 320 kgtotal mass 7132 kgchange of cg 8 cm = 0.08 mdistance 320 kg / 7132 kg = 0.08 m / distance.distance = 7132 kg / 320 kg x 0.08 m.distance = 1.783 mthe mass of 320 kg has to be moved by 1.783 m.
Question 92-9 : A pallet of 1 500 kg needs to be removed from the cargo hold situated at the position of 1 650 cm. the new total mass after removal of the pallet will be 19 340 kg and the new cg at 1 270 cm. the cg forward cg limit is at 860 cm and the aft limit is at 1 360 cm. by mistake another package of 1 500 ?
15 cm forward of the aft cg limit.
1 calculate the old cg before removing the pallet from the position of 1 650 cm...mass change / old total mass = change of cg / distance from mass to new cg... mass change 1 500 kg. old total mass 19 340 kg + 1 500 kg = 20 840 kg. change of cg . distance from mass to new cg 1 650 cm 1 270 cm = 380 cm...1 500 kg / 20 840 kg = change of cg / 380 cm..change of cg = 1500 kg / 20 840 kg x 380 cm..change of cg = 27 cm..the pallet would be removed from a position of 1 650 cm and the new cg after removal is 1 270 cm. this means that the old cg before removing the pallet would be further backwards... old cg = 1270 cm + 27 cm = 1 297 cm.... 2 calculate the final cg after removing the wrong pallet from a position of 680 cm...mass change / new total mass = change cg position / distance from mass to old cg... mass change = 1 500 kg. new total mass = 20 840 kg 1 500 kg = 19 340 kg. change of cg =. distance from mass to old cg = 1 297 cm 680 cm = 617 cm...1 500 kg / 19 340 kg = change of cg / 617 cm..change of cg = 1 500 kg / 19 340 kg x 617 cm..change of cg = 48 cm..the pallet is removed from a position of 680 cm and the old cg before removal is 1 297 cm. this means that the new cg after removing the pallet would be further backwards... new cg = 1 297 cm + 48 cm = 1 345 cm.... 3 determine the distance between the final cg and the aft cg limit...the aft cg limit is at 1 360 cm. the cg of the aircraft after removing the wrong pallet is 1 345 cm. this means that the final cg is located... 1 360 cm 1 345 cm = 15 cm forward of the aft cg limit.....note both formulas mass change / old total mass = change of cg / distance from mass to new cg and mass change / new total mass = change cg position / distance from mass to old cg can be used for these calculations. the choice between the two formulas depends on the available data every time.
Question 92-10 : Given total mass 2900 kg.centre of gravity location station 115.aft cg limit station 116the maximum mass that can be added at station 130 is... ?
207 kg.
In this question we want to add mass at an aft location station 130 without taking the cg out of limits. the cg is currently at station 115 and the limit is station 116...use the formula..added mass / total mass = cg movement distance/ relevant distance... added mass . total mass 2900 kg. cg movement distance from 115 location station to aft limit 116 = 1. relevant distance mass being added at station 130 to aft limit 116 = 14...added mass / 2900 = 1 / 14..added mass = 2900 x 1 / 14 = 207 kg..by adding 207 kg at station 130, the cg will move from station 115 to 116 but will still be in limits...note the distances do not need units.
Question 92-11 : You are tasked with determining the takeoff mass cg with the following data provided..takeoff fuel 9400 lbs..actual takeoff mass 56 800 lbs..fuel index chart..for 9360 lbs 0.3..for 9600 lbs 0.9..for 9850 lbs 1.1..instruction if the fuel mass does not match the chart values, use the next higher ?
0.9
This question is testing knowledge of the fuel index technique as used with the mrjt on the final page of cap 696.the question states if the fuel mass does not match the chart values, use the next higher mass.takeoff fuel 9400 lbs, which does not match, so the next higher mass is 9600 lbs with an index of 0.9for 9360 lbs 0.3.for 9600 lbs 0.9.for 9850 lbs 1.1when completing load sheets, particularly for large aircraft, it is convenient to use an index to represent the large numbers involved and to simplify the calculations.generally, an index is a non dimensional figure i.e. a figure without unit of measurement that is a scaled down value of a moment and the effect of reducing the magnitude of the moment to one that is much easier to use.a loading index li is simply a moment load mass x cg arm divided by a constant li = load mass x cg arm / constant = load moment / constant.
Question 92-12 : During dispatch, you perform the mass and balance calculations. given the following information, calculate the zero fuel mass cg in inches aft of the datum.basic empty mass 2635 lb.basic empty mass moment 204 397.5 in.lb.front seat occupant's mass 93 lb.front seat occupant's arm 78 in.aft seat ?
80.22
This question is a newer development of question 319614, now asking for the zfm cg rather than the tom cg.a moment is the turning force created by the mass over a distance or lever arm moment = mass x arm.generally, the centre of gravity cg can be found by adding all the moments and then dividing by the total mass cg = total moment / total mass.step 1 calculate the zero fuel mass zfm = bem + front seat occupant's mass + aft seat occupant's mass no fuel zfm = 2635 lb + 93 lb + 186 lb = 2914 lbstep 2 calculate the moments at the zero fuel mass bem = 204 397.5 in.lbfront seat occupant's moment = 93 lb x 78 in = 7254 in.lbaft seat occupant's moment = 186 lb x 119 in = 22 134 in.lbtotal moment = 204 397.5 + 7254 + 22 134 = 233 785.5 in.lbstep 3 calculate cg position at zero fuel mass centre of gravity = total moment / total mass centre of gravity = 233 785.5 lb.in / 2914 = 80.22 in. aft of datum
Question 92-13 : 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.
Question 92-14 : For the transport aeroplane the moment balance arm for the forward hold centroid is.. 210 ?
367.9 inches.
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Question 92-15 : 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.
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Question 92-16 : 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
Question 92-17 : 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.
Question 92-18 : 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.
Question 92-19 : From the loading manual for the transport aeroplane, the aft cargo compartment has a maximum total load of.. 214 ?
4187 kg.
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Question 92-20 : 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.
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Question 92-21 : 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 ?
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.
Question 92-22 : 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.
Question 92-23 : 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.
Question 92-24 : 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.
Question 92-25 : 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.
Question 92-26 : 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.
Question 92-27 : 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.
Question 92-28 : What is the maximum running load in the aft section of the forward lower compartment.. 232 ?
13.12 kg/in.
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Question 92-29 : Palletised cargo... ?
Consists of different cargo box on pallets stored in the cargo holds.
Question 92-30 : Bulk cargo... ?
Can be loaded without specific loading equipment.
Bulk cargo loose, unpackaged, non containerized cargo such as cement, grains, ores.
Question 92-31 : 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.
Question 92-32 : Bulk cargo... ?
Consists of cargo box, baggage loosely loaded in the cargo holds.
Bulk cargo can be loaded without specific loading equipment.
Question 92-33 : Define the maximum load distribution. ?
Load divided by smallest area.
Question 92-34 : 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 ?
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.
Question 92-35 : 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.
Question 92-36 : 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.
Question 92-37 : 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.
Question 92-38 : The cg limits of an aircraft are from 83 inches to 93 inches aft of the datum..the cg as loaded is found to be at 81 inches aft of the datum. the loaded mass is 3240 lbs..how much mass must be moved from the forward hold, 25 inches aft of the datum, to the aft hold, 142 inches aft of the datum, to ?
55.38 lbs.
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 92-39 : An aircraft has three holds situated 10 inches, 100 inches and 250 inches aft of the datum, identified as holds a, b and c respectively..the total aircraft mass is 3500 kg and the cg is 70 inches aft of the datum..the cg limits are from 40 inches to 70 inches aft of the datum..how much load must be ?
500 kg.
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.
Question 92-40 : A 5 kg mass is located at the end of a plank, at 4m of the pivot. on the other end of 8m long, what mass must be put for the plank to be in balance ?
2.5 kg.
Moment = mass x balance arm..5 kg x 4 m = 20 kgm. kg x 8 m = 20 kgm..20 / 8 = 2.5 kg.
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