Mass & Balance questions

[HF13]

Hi guys!!!

Can you help me to answer to the following questions?
19 days to go and still i lot of unanswered questions


1) Basic mass: 1 200 kg ; Basic balance arm: 3.00 m
Under these conditions the Basic centre of gravity is at 25% of the mean aerodynamic chord (MAC). The length of MAC is 2m.
In the mass and balance section of the flight manual the following information is given:
Position Arm
front seats: 2.5 m
rear seats: 3.5 m
rear hold: 4.5 m
fuel tanks: 3.0 m
The pilot and one passenger embark; each weighs 80 kg. Fuel tanks contain 140 litres of petrol with a density of 0.714. The rear seats are not occupied. Taxi fuel is negligible.
The position of the centre of gravity at take-off (as % MAC) is:

a)34
b)29
c)22
d)17


2 ) 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?

a)110
b)183
c)165
d)104

3)An aeroplane with a two wheel nose gear and four main wheels rests on the ground with a single nose wheel load of 500 kg and a single main wheel load of 6000 kg. The distance between the nose wheels and the main wheels is 10 meter.
How far is the centre of gravity in front of the main wheels?
a) 4 meters
b) 40 cm
c) 25 cm
d) 41.6
thank you all!!!

[yellowsubmarine]

Answer to no.1????

I've had a go at no.1......the answer i get is 22.5% mac although i expect to be corrected.

Is tough to explain without drawing the diagram but here goes......

Balane arm = the dist. from the datum to C og G

This makes your C of G (working in cm) 300 aft of datum. It is also 25% mac so Le mac = 250 and Te mac = 450

your are adding 160 Kg onto the Le mac position (50 cm fore of C of G). I then pressumed that because the fuel is added onto origional C of G position it makes no difference to final position except in overall weight of the a/c.

So....using little m over big M = little d over big D:

M = 1,200 (bem) + 160 (pilot and pax) + 95 (fuel, this may be inaccurate)

m = 160 (pax)

D = 50 (cm)

d = ? 160/1455 x 50 = 5.5cm subtract this from origional C of G because the weight is added fore of origional.

Finally 45cm of 200 cm = 22.5%

This maybe horribly wrong and i stand to be correted but i hope it helps. Would have a look at others but should really be revising Met...(very easily side-tracked). good luck, YS

[Quazimodo007]

For first one is 22

[yellowsubmarine]

Screw met... No.2

Screw the Met....No.2 looks much easier.

Again need a diagram but answer should be 110 I think.

M = 2200 (mass of a/c)

D = 300cm (fore to aft cargo compartments)

d = 15cm (old C of G to new C of G)

m = ? = 15/300 x 2200 = 110

Sorry for any mistakes! YS

[HF13]

Right answers!!!

YS, thanks a lot, i don't know if how you proceed is the correct way but definitely it takes me to the right answer so it works for me
thanks!!!


....any idea for the question number 3...anyone???

[Neptunus Rex]

HF 13

Are you perchance sitting an exam with a Blackberry?
I think we should be told!!!
Neppie

[Quazimodo007]

3

answer is 40cm

[Nozwaldo]

Just had a look at questions 2 & 3 and I agree with the answers given previously.

Q2. Using m/M = d/D m=? M=2200 d=0.15m D=3m

m = Md/D

m = 2200 x 0.15 / 3 = 330/3 = 110Kg to move from FWD to AFT cargo hold

Q3. Total weight through nose gear = 1000Kg. Total weight through main gear = 24000Kg. Therefore total aircraft weight is 25000Kg.

Using the main gear position as a datum, the nose gear turning moment is 1000Kg x -10m = -10000Kgm.

CG position = moment/weight = -10000/25000 = -0.40m = -40cm i.e 40cm forward of main gear. Answer b.

When doing these remember to use the correct signs for positive (clockwise) & negative (counterclockwise) turning moments with respect to the datum.

Noz

[Nozwaldo]

Just had a crack at Q1 using a different method to YS by calculating turning moments about the datum:
Firstly I calculate the fuel weight to be 100Kg (140 x 0.714).
LEMAC is at 2.5m from datum.

Turning moment for pilot & passenger = 160Kg x 2.5m = 400Kgm

Turning moment for the basic mass + fuel at same arm = 1300Kg x 3m =3900Kgm
Total moment = 3900Kgm + 400Kgm = 4300Kgm
Total weight = 1200kg(basic) + 100Kg(fuel) + 160Kg(pilot & PAX) = 1460Kg

CG position = 4300Kgm/1460Kg = 2.945m from datum, which is 0.445m from LEMAC, which is 22.25% MAC. Close enough to answer c.

Slightly long winded, but another way to get there.

Noz

[HF13]

Quazimodo,Noz thanks so much for the answers and Noz i really appreciate your detail answer. for question number 1 i guess i have only to decide which one of the ways work easyer for me but they both seem correct.

For Neppie

I'm sorry but i didn't understand your question???

bye you all