Question 90-1 : Refer to figure 031 34givenbem 1200 kgbem cg 300pilot and front pax 200 kgcargo on pax floor 50 kgfuel 250 kgfind the loaded centre of gravity cg by using the attached graph ? [ Exam pilot ]

Question 90-2 : Refer to figure 031 04given force fa 100 ndistance a 6 mdistance b 3 mcalculate force fa to obtain equilibrium ?

50 n.

Question 90-3 : Select the correct statement for the cg safe range ?

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

Question 90-4 : The fuel index is ?

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

Question 90-5 : Refer to figure 031 03given force fa 300 nforce fb 150 ndistance a 3 mcalculate distance b to obtain equilibrium ?

6 m.

Question 90-6 : Refer to figure 031 60given weight 110800 kgmac 316%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 ?

1204.

Note de moi this question also exists at examination with weight 112600 kg mac 307% and an additional pilot of 75 kg in zone g answer is 1193

Question 90-7 : Refer to figure 031 55find the cg position as %mac for the take off ?

24%.

Question 90-8 : 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 ms2 ?

108000 nm.

Question 90-9 : Refer to figure 031 00if the aft galley is loaded with 102 kg dry operating index doi will change by ?

152592.

Question 90-10 : Refer to figure 031 02given distance a 2 mforce fa 300 ncalculate force fc to obtain equilibrium ?

100 n.

Question 90-11 : 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 ?

8579 mm.

Question 90-12 : 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 n1800 nm = 2m x n1800 2 = 900 n

Question 90-13 : Refer to figure 031 52what is the forward cg limit and what is the aft cg limit for an aircraft having a gross mass of 4300 lbs ?

873 and 946.

Question 90-14 : Refer to figure 031 46how much fuel could be loaded at reference station 4575 whan all thanks were fitted in this aircraft ?

Question 90-15 : An aircraft has a loaded mass of 5500 lbs the cg is 22 inches aft of the datuma passenger mass 150 lbs moves aft from row 1 to row 3 a distance of 70 incheswhat will be the new position of the cg assuming all dimensions aft of the datum ?

239 inches.

The new centre of gravity position can be calculated using the following formula m x d = m x d where m stands for the total mass of the aircraft d is the distance between the previous and new cg m is the moved mass and d stands for the distance between the old and new position of the moved mass d = m x d m d = 150 lbs x 70 in 5 500 lbs d = 19 into determine the new cg position we must the add distance between the previous and new cg d to the original cg remember that since the passenger has moved aft the cg will follow this direction 22 + 19 = 239 in

Question 90-16 : An aircraft has three holds situated 10 inches 100 inches and 250 inches aft of the datum identified as holds a b and c respectivelythe total aircraft mass is 3500 kg and the cg is 70 inches aft of the datumthe cg limits are from 40 inches to 70 inches aft of the datum how much load must be removed ?

500 kg.

The cg is currently in limits but needs to move to the forward limit why we do not know this distance is +70 to +40 = 30in order to move the cg onto the forward limit we will need to remove load from hold cstep 1 establish the parameters m = mass to be removed m = total mass 3500 kg d = cg movement required 30 d = relevant distance between the target cg position and the hold being unloaded distance between +40 and +250 = 210 step 2 time for the formula mm = ddm3500 = 30210m = 500 kgby removing 500 kg from hold c the cg will move to the forward limit

Question 90-17 : The mass of an aircraft is 2000 kg and 400 kg of freight is added to a hold 2 m aft of the present cg position the movement of the cg is ?

033 m aft.

Whenever we have to add or subtract a known amount we can simply use mass x arm = moments as shown by method 2 as an arguably simpler method for the exam roommethod 1 step 1 establish the parameters m = mass to be added 400 kg m = new total mass 2400 kg d = cg movement = d = relevant distance between the original cg position and the hold being loaded distance of 2 metres step 2 time for the formula mm = dd4002400 = 2cg movement = 033 mby adding 400 kg into this aft positioned hold the cg has moved backwards by 033 metresmethod 2 alternatively if we assume the datum is on the original cg mass kg arm m moments kgm original200000added400+2+800new total2400+033+800

Question 90-18 : An aircraft has a mass of 5000 lbs and the cg is located at 80 inches aft of the datumthe aft cg limit is at 805 inches aft of the datumwhat is the maximum mass that can be loaded into a hold situated 150 inches aft of the datum without exceeding the limit ?

3597 lbs.

The cg is currently in limits but needs to move aft to the aft limit this distance is 80 in to 805 in = 05 inin order to move the cg on to the aft limit we will need to load the mass into a hold at +150 instep 1 establish the parameters m = mass to be added m = total mass 5000 lb d = cg movement required 05 d = relevant distance between the target cg position and the hold being loaded distance between +150 and +805 = 695 step 2 time for the formula mm = ddm5000 = 05695m = 3597 lbby adding 3597 lb into the rear hold the cg is moved back on to the aft limit

Question 90-19 : Which of the following is the correct method to determine the 'underload' ?

Allowable tom – dom – useful load = underload.

Another question that tests your knowledge of the various definitions used in this subject and also their relationship to each otherthe allowable tom is the lower of regulated take off mass rtom maximum zero fuel mass mzfm + take off fuel regulated landing mass rlm + trip fuelthis is demonstrated on the attached load sheet in the red box looking at the answers allowable tom – dom – useful load = underload > correct the maximum tom with the dom and useful load removed equals the underload remember that the useful load is the take off fuel and actual traffic load as shown in the attached load sheet actual tom – dom – useful load = underload > incorrect actual tom with the dom and useful load removed would equal 0 allowable lm + trip fuel – dom = underload > incorrect a nonsense answer allowable tom + useful load – dom = underload > incorrect another nonsense answerunderload is the weight that still is available until the first limiting maximum weight is reached maximum mass and balance limits for zero fuel take off or landing limitation of any compartment that is intended to be usedunderload = allowable tom dom useful load

Question 90-20 : Given the following information what is the minimum additional load in whole kilograms that must be loaded into either compartment 1 or 2 to bring the cg into limits for take off aircraft take off mass 1785 kg initial cg position +208 m forward cg limit +214 m aft cg limit +241 m compartment 1 ?

41 kg into compartment 2.

The cg is currently in front of the fwd limit and needs to move aft this distance is +208 m to +214m = 006 min order to move the cg into limits we will need to load this cargo into compartment 2 compartment 1 is further forward and would give the opposite effect this removes one answer option alreadystep 1 establish the parameters m = mass to be added m = total mass 1785 kg d = cg movement required 006 d = relevant distance between the target cg position and the compartment being loaded distance between +476 and +214 = 262 step 2 time for the formula mm = ddm1785 = 006262m = 409 kgby adding 41 kg into the rear compartment number 2 the cg is moved back into limits

Question 90-21 : A loaded aircraft has its centre of gravity cg 48 inches forward of the datum and its mass is 2 966 lb if a last minute load of 128 lb is added 28 inches aft of the datum what is the new position of the cg ?

4486 inches forward of the datum.

A few different techniques available here mass x arm = moment massarmmomentoriginal2966 48 142 368added mass128283584new3094 4485 138 784new total moment = old total moment + additional load m 2966 + 128 x d = 2966 x 48 + 128 x 28 3094 x d = 142368 + 3584 3094 x d = 138784 d = 138784 3094 = 4485 inches forward of the datum

Question 90-22 : 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 145 inches aft the aft limit is 147 inches aft the current mass is 2750 lbthree passengers are in the rear seats with the baggage ?

101 lb.

The cg is currently 3 inches behind the aft limit and needs to move forwardsin order to move the cg into limits we will need to load the ballast into the front seatstep 1 establish the parameters m = mass to be added m = total mass 2750 lb d = cg movement required 3 d = relevant distance between the target cg position and the front seat distance between +147 and +65 = 82 step 2 time for the formula mm = ddm2750 = 382m =1006 lbsby adding 1006 lbs of ballast on the front seats the cg is moved back into limitsalternatively cg is 3'' behind the aft limit therefore the cg is 150'' aft of datum old total moment = 2 750 x 150 lbinto 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 lbinthe new total mass will be new total mass = 2 750 + mthis will result in a new total moment new total moment = 2 750 + m x 147 lbinnow apply the formula old total moment + moment of the ballast = new total moment 2 750 x 150 + m x 65 = 2 750 + m x 147 2 750 x 150 + m x 65 = 2 750 x 147 + m x 147 m = 1006 lb

Question 90-23 : Given total mass 7500 kg centre of gravity cg location station 805 aft cg limit station 795how much cargo must be shifted from the aft cargo compartment at station 150 to the forward cargo compartment at station 30 in order to move the cg location to the aft limit ?

625 kg.

The cg is currently to the rear of the aft limit and needs to move forwards this distance is 805 to 795 = 1 in order to move the cg into limits we will need to move cargo from the aft compartment to the forward compartmentstep 1 establish the parameters m = mass to be moved m = total mass 7500 kg d = cg movement required 1 d = relevant distance between the compartments distance between +150 and +30 = 120 step 2 time for the formula mm = ddm7500 = 1120m = 625 kgby moving 625 kg into the forward compartment the cg is moved into limits

Question 90-24 : As for the mass and balance calculations the index is ?

A figure without unit of measurement which represents a moment.

When completing load sheets particularly for large aircraft it is convenient to use an index to represent the large numbers involved and to simplify the calculationsgenerally loading index is a non dimensional figure ie 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 usea 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 90-25 : Length of the mean aerodynamic chord = 1 m moment arm of the forward cargo – 050 m moment arm of the aft cargo + 250 m the aircraft mass is 2 200 kg and its centre of gravity is at 25% macto move the centre of gravity to 40 % which mass has to be transferred from the forward to the aft cargo ?

110 kg.

To simplify calculations we advise converting centre of gravity cg positions from %mac into metres the question gives a mac length of 1 m therefore 25% mac = 025 m40% mac = 04 mas we will move cargo from the forward to the aft cargo compartment the airplane mass will remain the same 2200 kgthe difference in cg position in metres is 04 – 025 = 015 mthe distance between the forward and aft cargo holds is 05 + 25 = 3 mthe mass to be transferred can be calculated using the formula total mass x d = mass to be moved x d where d distance between the previous and new cg position 015 m d distance between the forward and aft cargo compartments 3 m mass to be moved = total mass x d ÷ dmass to be moved = 2 200 x 015 ÷ 3mass to be moved = 110 kgnote the datum is located between the forward and aft cargo compartments therefore take care with the negative value which means that the compartment is located forward of the datum

Question 90-26 : The dry operating index is ?

The reduced moment of the dry operating mass.

The word index in mass and balance is a moment divided by a constantit’s used to simplify the calculation process instead of using big numbers for calculations we use the index to get faster resultsdry operating index is reduced moment for dry operating mass

Question 90-27 : What are the advantages of using the index method to determine moments it ?

Reduces the magnitude of moments making it less time consuming to compute.

When completing load sheets particularly for large aircraft it is convenient to use an index to represent the large numbers involved and to simplify the calculationsgenerally loading index is a non dimensional figure ie 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 usea '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 90-28 : The index method in mass and balance calculations is used for ?

Reducing the magnitude of the moment.

When completing load sheets particularly for large aircraft it is convenient to use an index to represent the large numbers involved and to simplify the calculationsgenerally loading index is a non dimensional figure ie 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 usea '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 90-29 : Calculate the centre of gravity location for the take off mass given bem 2 635 lb bem moment 204 7935 inlb mass on front seat 90 lb arm on front seat 78 in mass on aft seat 186 lb arm on aft seat 117 in block fuel 850 lb taxi fuel 85 lb fuel arm 86 in ?

8144 inches aft of datum.

Formulas needed center of gravity = total moment total massmoment = mass x arm 1 calculate total mass total mass = bem + front seat + aft seat + take off fueltotal mass = 2 635 lb + 90 lb + 186 lb + 765 lb = 3 676 lb 2 calculate the moments bem = 204 7935 inlb front seat = 90 lb x 78 in = 7 020 inlb aft seat = 186 lb x 117 in = 21 762 inlb take off fuel = 765 lb x 86 in = 65 790 inlb 3 calculate total moment total moment = 204 7935 + 7 020 + 21 762 + 65 790 = 299 3655 lbin 4 calculate cg position centre of gravity = total moment total masscentre of gravity = 299 3655 3676 = 8144 in aft of datumnote question asks for centre of gravity at take off taxi fuel will be consumed before take off therefore is must be subtracted from our calculations

Question 90-30 : A twin engine aeroplane has an actual mass of 6 063 kg its cg is located at 27 m aft of the datum how much mass must be transferred from a cargo compartment 12 m aft of the datum to compartment 41 m aft of the datum to establish a cg 28 aft of the datum ?

209 kg.

The cg is currently in front of the fwd limit and needs to move aft this distance is +27 m to +28m = 01 min order to move the cg into limits we will need to move cargo from the forward compartment to the aft compartmentstep 1 establish the parameters m = mass to be moved m = total mass 6063 kg d = cg movement required 01 d = relevant distance between the compartments distance between +12 and +41 = 29 step 2 time for the formula mm = ddm6063 = 0129m = 209 kgby moving 209 kg into the rear compartment the cg is moved back into limitsalternatively calculate the current total moment moment = total mass x arm moment = 6063 kg x 27 m = 16 3701 kgmcalculate the final moment after the mass has been moved moment = 6063 kg x 28 m = 169764 kgmdetermine the additional moment required difference between final and initial moment 169764 kgm 16 3701 kgm = 6063 kgmtotal distance moved 41 m – 12 m = 29 mcalculate the required mass to increase the total moment by 6063 kgm 6063 kgm 29 m = 209 kg

Question 90-31 : An aircraft has three holds situated at 20 in 120 in and 240 in aft of the datum identified as holds ab and c respectively the total aircraft mass is 4 900 kg and the cg is 50 in aft of the datum the cg limits are from 35 in to 65 in aft of the datum how much load must be removed from hold c to ?

360 kg.

The cg is currently in limits but needs to move to the forward limit why we do not know this distance is +50 to +35 = 15in order to move the cg onto the forward limit we will need to remove load from compartment 3 step 1 establish the parameters m = mass to be removed m = total mass 4900 kg d = cg movement required 15 d = relevant distance between the target cg position and the compartment being unloaded distance between +35 and +240 = 205 step 2 time for the formula mm = ddm4900 = 15205m = 3585 kgby removing 359 kg from compartment c the cg will move to the forward limit 360 kg is the nearest answer

Question 90-32 : During pre flight calculations the pilot finds out that the take off cg position is 8 cm aft of the take off cg aft limit the tom is 7 312 kg determine the minimum distance forwards that heshe should move a box with a mass of 362 kg to change the cg to within the take off limits ?

1616 m.

Mass change total mass = change of cg distance mass change 362 kg total mass 7312 kg change of cg 8 cm distance 362 kg 7 312 kg = 8 distance distance = 7 312 kg x 8 cm 362 kg distance = 1616 cm = 1616 m

Question 90-33 : An aircraft has a total mass of 2 222 kg with a cg of 613 m aft of the datum line after calculations to return the cg within limits an additional ballast of 136 kg was loaded the new cg is at the aft limit of 628 m determine where the additional ballast was loaded ?

873 m aft of the datum line.

Mass change new total mass = change of cg distance from mass to old cgmass change 136 kgnew total mass 2 222 kg + 136 kg = 2 358 kgchange of cg 628 – 613 = 015 mdistance from mass to old cg 136 kg 2 358 kg = 015 m distance from mass to old cgdistance from mass to old cg = 015 m x 2 358 kg 136 kgdistance from mass to old cg = 26 m=> to move the cg aft we must add the mass 136 kg further back from the original position therefore 613 m + 26 m = 873 m

Question 90-34 : During pre flight calculations you discover that the take off cg position is 8 cm aft of the take off cg limit the aircraft's take off mass is 7231 kg determine the minimum distance forwards that you should move a box with a mass of 302 kg to change the cg to within the take off limits ?

1915 m.

Mass change total mass = change of cg distance mass change 302 kg total mass 7 231 kg change of cg 8 cm distance 302 kg 7 231 kg = 8 distance distance = 7 231 kg x 8 cm 302 kg distance = 1915 cm = 1915 m

Question 90-35 : An aircraft had a total mass of 5 430 kg with a cg of 543 m aft of the datum line after calculations to return the cg within limits an additional ballast of 210 kg was loaded the aircraft's new cg is 556 m aft of the datum determine where the additional ballast was loaded ?

892 m aft of the datum line.

Mass change new total mass = change of cg distance moved mass change 210 kg new total mass 5 430 kg + 210 kg = 5 640 kg change of cg 556 – 543 = 013 m distance moved 210 kg 5 640 kg = 013 m distance moved distance moved = 013 m x 5 640 kg 210 kg distance moved = 349 m=> to move the cg aft we must add the mass 210 kg further back from the original position therefore 543 m + 349 m = 892 m

Question 90-36 : An aircraft has a total mass of 3 927 kg with a cg of 428 m aft of the datum line determine the amount of additional load to be loaded at station 58 m aft of the datum line to establish a new cg of 437 m aft of the datum line ?

247 kg.

Mass change old total mass = change of cg distance from mass to new cg mass change old total mass 3 927 kg change of cg 437 m – 428 m =009 m distance from mass to new cg 58 m – 437 m = 143 mmass change 3 927 kg = 009 m 143 m mass change = 3 927 kg x 009 m 143 m = 247 kg

Question 90-37 : Select 3 factors which affect the stab trim for take off ?

Tom cg and flap setting.

Stabilizer trim setting is one of the most critical factors affecting our take off performanceto answer this question we need to tefer to the attached figures from cap 696 figures 44 and 413 and see its variables tom is entered into the cg envelope so that we can find the cg stab trim depends on the flap setting

Question 90-38 : For a take off in a large aeroplane what is the relationship between the cg and the stabiliser trim setting ?

With a more aft cg a lower nose up stabiliser trim setting is required.

When the cg is forward of the cp forward cg position there is a natural tendency for the aircraft to want to pitch nose down if the cp is forward of the cg aft cg position a nose up pitching moment is created consequently an aft cg will need less assistance from the stabiliser nose up trim

Question 90-39 : A loaded aircraft has its centre of gravity cg 35 inches aft of the datum and its mass is 4 500 lb a last minute load of 334 lb is placed on board 20 inches forward of the datumdetermine the new position of the cg ?

312 inches aft of the datum.

Note that the items aft of the datum have a positive value items forward of the datum have a negative valueold moment = 4 500 lb x 35 in = 157 500 lbinload moment = 334 lb x 20 in = 6 680 lbinnew moment = 157 500 lbin 6 680 lbin = 150 820 lnin new mass = 4 834 lbcg = total moment total mass cg = 150 820 4 834 = 312 in aft of the datum

Question 90-40 : Determine the dry operating index doi of the aircraft given bem 3 208 kg cg at bem 1297 inches aft of datum 2 crew members 160 kg crew seats 475 inches aft of datumdoi= mass kg x cg inches 5000 30 ?

5474.

Let's start by calculating the dom mass and cg dom mass = 3 208 kg + 160 kg = 3 368 kgmoment = cg x masscg = moment massindividual moments bem moment = 3 208 kg x 1297 in = 416 078 kgin crew moment = 160 kg x 475 in = 7 600 kgintotal moment = 416 078 kgin + 7 600 kgin = 423 678 kgincg = 423 678 kgin 3 368 kg = 125795 inthe dry operating index doi is the reduced moment of the dry operating mass dom doi= mass kg x cg inches 5000 30doi = 3 368 kg x 125795 5 000 30 = 5474note this question has been created based on incomplete feedback please let us know if you come across this question as we don't know for sure if the available options will be the same in official exams