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Question 90-1 : Refer to figure 031 04..given.force fa 100 n.distance a 6 m.distance b 3 m.calculate force fa to obtain equilibrium ? [ Exam pilot ]
50 n.
Question 90-2 : 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 90-3 : The fuel index is ?
Used to calculate the correct position of the cg due to different locations of the fuel tanks.
Question 90-4 : 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.
Question 90-5 : 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.
Question 90-6 : Refer to figure 031 55..find the cg position as %mac for the take off ?
24%.
Question 90-7 : 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
Question 90-8 : Refer to figure 031 00..if the aft galley is loaded with 102 kg, dry operating index doi will change by ?
1.52592.
Question 90-9 : Refer to figure 031 02..given.distance a 2 m.force fa 300 n.calculate force fc to obtain equilibrium ?
100 n.
Question 90-10 : 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
Question 90-11 : 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.
Question 90-12 : 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.
Question 90-13 : Refer to figure 031 46..how much fuel could be loaded at reference station 4.575, whan all thanks were fitted in this aircraft ?
806 l.
Calculate the values from the tanks at station 4.575.236 litre.324 litre.246 litre. .806 litre.
Question 90-14 : 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.
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 = 1.9 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 + 1.9 = 23.9 in
Question 90-15 : 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.
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 c.step 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 m/m = d/dm/3500 = 30/210m = 500 kgby removing 500 kg from hold c, the cg will move to the forward limit.
Question 90-16 : 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 ?
0.33 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 room.method 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 m/m = d/d400/2400 = /2cg movement = 0.33 mby adding 400 kg into this aft positioned hold, the cg has moved backwards by 0.33 metres.method 2 alternatively, if we assume the datum is on the original cg mass kg arm m moments kg.m. original200000added400+2+800new total2400+0.33+800
Question 90-17 : 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.
The cg is currently in limits but needs to move aft to the aft limit. this distance is 80 in to 80.5 in = 0.5 in..in order to move the cg on to the aft limit, we will need to load the mass into a hold at +150 in...step 1. establish the parameters..m = mass to be added ..m = total mass 5000 lb..d = cg movement required 0.5..d = relevant distance, between the target cg position and the hold being loaded distance between +150 and +80.5 = 69.5..step 2. time for the formula..m/m = d/d..m/5000 = 0.5/69.5..m = 35.97 lb..by adding 35.97 lb into the rear hold, the cg is moved back on to the aft limit.
Question 90-18 : 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 other...the allowable tom is the lower of... regulated take off mass rtom. maximum zero fuel mass mzfm + take off fuel. regulated landing mass rlm + trip fuel...this 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 answer....underload 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 used...underload = allowable tom dom useful load
Question 90-19 : 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 +2.08 m..forward cg limit +2.14 m..aft cg limit +2.41 ?
41 kg into compartment 2.
.the cg is currently in front of the fwd limit and needs to move aft. this distance is +2.08 m to +2.14m = 0.06 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 already.step 1. establish the parameters m = mass to be added m = total mass 1785 kg d = cg movement required 0.06 d = relevant distance, between the target cg position and the compartment being loaded distance between +4.76 and +2.14 = 2.62 step 2. time for the formula m/m = d/dm/1785 = 0.06/2.62m = 40.9 kgby adding 41 kg into the rear compartment number 2 the cg is moved back into limits.
Question 90-20 : 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 ?
44.86 inches forward of the datum.
A few different techniques available here mass x arm = moment massarmmomentoriginal2966 48 142 368added mass128283584new3094 44.85 138 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 = 44.85 inches forward of the datum
Question 90-21 : 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 lb...three 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 forwards...in order to move the cg into limits, we will need to load the ballast into the front seat...step 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..m/m = d/d..m/2750 = 3/82..m =100.6 lbs..by adding 100.6 lbs of ballast on the front seats, the cg is moved back into limits....alternatively cg is 3'' behind the aft limit. therefore, the cg is 150'' aft of datum.... old total moment = 2 750 x 150 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 = 2 750 + m...this will result in a new total moment... new total moment = 2 750 + m x 147 lb.in...now 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 = 100.6 lb
Question 90-22 : Given total mass 7500 kg..centre of gravity cg location station 80.5..aft cg limit station 79.5..how 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 ?
62.5 kg.
The cg is currently to the rear of the aft limit and needs to move forwards. this distance is 80.5 to 79.5 = 1..in order to move the cg into limits, we will need to move cargo from the aft compartment to the forward compartment...step 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..m/m = d/d..m/7500 = 1/120..m = 62.5 kg..by moving 62.5 kg into the forward compartment, the cg is moved into limits.
Question 90-23 : 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 calculations...generally, loading 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 90-24 : 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 ?
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 = 0.25 m..40% mac = 0.4 m..as we will move cargo from the forward to the aft cargo compartment, the airplane mass will remain the same 2200 kg..the difference in cg position in metres is 0.4 – 0.25 = 0.15 m...the distance between the forward and aft cargo holds is 0.5 + 2.5 = 3 m...the 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 0.15 m.. d distance between the forward and aft cargo compartments 3 m. . .....mass to be moved = total mass x d ÷ d..mass to be moved = 2 200 x 0.15 ÷ 3..mass to be moved = 110 kg..note 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 90-25 : 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 constant.it’s used to simplify the calculation process, instead of using big numbers for calculations we use the index to get faster results...dry operating index is reduced moment for dry operating mass.
Question 90-26 : 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 calculations...generally, loading 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 90-27 : 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 calculations...generally, loading 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 90-28 : Calculate the centre of gravity location for the take off mass, given..bem 2 635 lb..bem moment 204 793.5 in.lb..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 ?
81.44 inches aft of datum.
Formulas needed..center of gravity = total moment / total mass..moment = mass x arm. 1 calculate total mass..total mass = bem + front seat + aft seat + take off fuel..total mass = 2 635 lb + 90 lb + 186 lb + 765 lb = 3 676 lb.. 2 calculate the moments... bem = 204 793.5 in.lb. front seat = 90 lb x 78 in = 7 020 in.lb. aft seat = 186 lb x 117 in = 21 762 in.lb. take off fuel = 765 lb x 86 in = 65 790 in.lb... 3 calculate total moment..total moment = 204 793.5 + 7 020 + 21 762 + 65 790 = 299 365.5 lb.in... 4 calculate cg position..centre of gravity = total moment / total mass..centre of gravity = 299 365.5 / 3676 = 81.44 in aft of datum..note 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 90-29 : A twin engine aeroplane has an actual mass of 6 063 kg. its cg is located at 2.7 m aft of the datum. how much mass must be transferred from a cargo compartment 1.2 m aft of the datum to compartment 4.1 m aft of the datum to establish a cg 2.8 aft of the datum ?
209 kg.
The cg is currently in front of the fwd limit and needs to move aft. this distance is +2.7 m to +2.8m = 0.1 min order to move the cg into limits, we will need to move cargo from the forward compartment to the aft compartment.step 1. establish the parameters m = mass to be moved m = total mass 6063 kg d = cg movement required 0.1 d = relevant distance, between the compartments distance between +1.2 and +4.1 = 2.9 step 2. time for the formula m/m = d/dm/6063 = 0.1/2.9m = 209 kgby moving 209 kg into the rear compartment, the cg is moved back into limits.alternatively calculate the current total moment.moment = total mass x arm.moment = 6063 kg x 2.7 m = 16 370.1 kgmcalculate the final moment after the mass has been moved .moment = 6063 kg x 2.8 m = 16976.4 kgmdetermine the additional moment required difference between final and initial moment .16976.4 kgm 16 370.1 kgm = 606.3 kgmtotal distance moved.4.1 m – 1.2 m = 2.9 mcalculate the required mass to increase the total moment by 606.3 kgm.606.3 kgm 2.9 m = 209 kg
Question 90-30 : An aircraft has three holds situated at 20 in, 120 in and 240 in aft of the datum, identified as holds a,b 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 ?
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 = 15..in 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..m/m = d/d..m/4900 = 15/205..m = 358.5 kg..by removing 359 kg from compartment c, the cg will move to the forward limit. 360 kg is the nearest answer.
Question 90-31 : 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 he/she should move a box with a mass of 362 kg to change the cg to within the take off limits ?
1.616 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 = 161.6 cm = 1.616 m
Question 90-32 : An aircraft has a total mass of 2 222 kg with a cg of 6.13 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 6.28 m...determine where the additional ballast was loaded ?
8.73 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 6.28 – 6.13 = 0.15 mdistance from mass to old cg 136 kg / 2 358 kg = 0.15 m / distance from mass to old cgdistance from mass to old cg = 0.15 m x 2 358 kg / 136 kgdistance from mass to old cg = 2.6 m=> to move the cg aft, we must add the mass 136 kg further back from the original position. therefore, 6.13 m + 2.6 m = 8.73 m.
Question 90-33 : 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 ?
1.915 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 = 191.5 cm = 1.915 m
Question 90-34 : An aircraft had a total mass of 5 430 kg with a cg of 5.43 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 5.56 m aft of the datum.. determine where the additional ballast was loaded ?
8.92 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 5.56 – 5.43 = 0.13 m distance moved 210 kg / 5 640 kg = 0.13 m / distance moved. distance moved = 0.13 m x 5 640 kg / 210 kg. distance moved = 3.49 m=> to move the cg aft, we must add the mass 210 kg further back from the original position. therefore, 5.43 m + 3.49 m = 8.92 m
Question 90-35 : An aircraft has a total mass of 3 927 kg with a cg of 4.28 m aft of the datum line.. determine the amount of additional load to be loaded at station 5.8 m aft of the datum line, to establish a new cg of 4.37 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 4.37 m – 4.28 m =0.09 m distance from mass to new cg 5.8 m – 4.37 m = 1.43 mmass change / 3 927 kg = 0.09 m / 1.43 m. mass change = 3 927 kg x 0.09 m / 1.43 m = 247 kg
Question 90-36 : 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 performance...to answer this question we need to tefer to the attached figures from cap 696 figures 4.4 and 4.13, 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 90-37 : 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 90-38 : 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 datum...determine the new position of the cg ?
31.2 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 value.old moment = 4 500 lb x 35 in = 157 500 lb.inload moment = 334 lb x 20 in = 6 680 lb.innew moment = 157 500 lb.in 6 680 lb.in = 150 820 ln.in. new mass = 4 834 lbcg = total moment / total mass. cg = 150 820/ 4 834 = 31.2 in aft of the datum
Question 90-39 : Determine the dry operating index doi of the aircraft, given bem 3 208 kg. cg at bem 129.7 inches aft of datum. 2 crew members 160 kg. crew seats 47.5 inches aft of datumdoi= mass kg x cg inches / 5000 30 ?
54.74
Let's start by calculating the dom mass and cg..dom mass = 3 208 kg + 160 kg = 3 368 kg..moment = cg x mass..cg = moment / mass..individual moments... bem moment = 3 208 kg x 129.7 in = 416 078 kg.in. crew moment = 160 kg x 47.5 in = 7 600 kg.in...total moment = 416 078 kg.in + 7 600 kg.in = 423 678 kg.in..cg = 423 678 kg.in / 3 368 kg = 125.795 in..the dry operating index doi is the reduced moment of the dry operating mass dom...doi= mass kg x cg inches /5000 30..doi = 3 368 kg x 125.795 / 5 000 30 = 54.74..note 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.
Question 90-40 : An aircraft has 100 lb of ballast installed in the forward cargo hold located 22.5 inches aft of the datum. the load and balance calculation gives a tom of 4 000 lb, and the cg at 85 inches aft of the datum...what should the pilot do to the ballast to move the cg to 86 inches aft of the datum ?
Remove 63 lb.
The cg is currently at +85 inches but needs to move aft to +86 inches. this distance is +85 to +86 = 1 inch.in order to move the cg aft, we will need to remove some of the ballast installed in the forward cargo hold located 22.5 inches aft of the datum. this hold is forward of the cg, so removing mass from it will move the cg aft. this eliminates 2 of the answers already.step 1. establish the parameters m = mass to be removed m = total mass 4000 lb d = cg movement required 1 d = relevant distance, between the target cg position and the compartment being unloaded distance between +86 and +22.5 = 63.5 step 2. time for the formula m/m = d/dm/4000 = 1/63.5m = 63 lbsby removing 63 lbs of ballast from the forward cargo hold, the cg will move to the desired position of +86 inches.
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