A&P weight and balance: moments, CG and adverse loading

Weight and balance questions on the General test are arithmetic dressed in terminology. If you can compute a moment, a loaded CG, the effect of removing equipment and a percent of MAC, and you know what tare, datum and basic empty weight mean, the subject is free points.

Questions and answers, explained

All 12 questions above, with the correct answer and why it is correct. Everything here is on a&p weight and balance: moments, cg and adverse loading.

1. A 170-pound passenger sits at an arm of +45 inches. What moment does the passenger create?

   • A3.78 inch-pounds
   • B215 inch-pounds
   • C7,650 inch-poundsCorrect
   • D76,500 inch-pounds

   Why: Moment = weight × arm = 170 lb × 45 in = 7,650 in-lb. Moments are always weight times distance from the datum, expressed in inch-pounds.

   Reference AM.I.C.K4 · FAA-H-8083-30B, Ch. 6 (Moment)

2. An airplane's empty weight is 1,800 lb at +30.0 in. A 180-lb pilot is added at +40.0 in and 240 lb of fuel at +48.0 in. Where is the loaded CG?

   • A+32.76 inCorrect
   • B+39.33 in
   • C+30.00 in
   • D+36.80 in

   Why: Moments: 1,800 × 30 = 54,000; 180 × 40 = 7,200; 240 × 48 = 11,520. Total moment 72,720 in-lb ÷ total weight 2,220 lb = +32.76 in. Averaging the three arms (39.33) ignores how much each item weighs.

   Reference AM.I.C.K4 · FAA-H-8083-30B, Ch. 6 (Center of Gravity)

3. In weight and balance, the datum is

   • Aalways the tip of the propeller spinner
   • Ban imaginary vertical plane set by the manufacturerCorrect
   • Cthe point where the aircraft actually balances
   • Dthe main landing gear axle centerline

   Why: The datum is an imaginary vertical plane from which all horizontal measurements are taken, with the aircraft in level flight attitude. The manufacturer chooses it and it is identified in the aircraft specifications or TCDS. The balance point is the CG, not the datum.

   Reference AM.I.C.K1 · FAA-H-8083-30B, Ch. 6 (Datum)

4. The term basic empty weight applies when the empty weight includes

   • Athe pilot and a full load of usable fuel
   • Bonly the residual (undrainable) engine oil
   • Conly the equipment the manufacturer installed as standard
   • Dthe full capacity of the engine oil systemCorrect

   Why: The handbook: basic empty weight includes the full engine oil capacity, standard for most aircraft certified after 1978. Licensed empty weight includes only residual oil (older aircraft). Standard empty weight is the manufacturer's figure without optional equipment.

   Reference AM.I.C.K1 · FAA-H-8083-30B, Ch. 6 (Empty Weight)

5. An airplane has an empty weight of 3,100 lb and a maximum allowable gross weight of 3,950 lb. What is its useful load?

   • A7,050 lb
   • B1,100 lb
   • C850 lbCorrect
   • D3,950 lb

   Why: Useful load = maximum allowable gross weight − empty weight = 3,950 − 3,100 = 850 lb. It covers fuel, crew, passengers, baggage and any fluids not in the empty weight. This exact example appears in the FAA handbook.

   Reference AM.I.C.K1 · FAA-H-8083-30B, Ch. 6 (Useful Load)

6. A main-wheel scale reads 1,260 lb, but wheel chocks weighing 12 lb were left on the scale. What net weight should be recorded for that wheel?

   • A1,248 lbCorrect
   • B1,272 lb
   • C1,260 lb
   • D1,236 lb

   Why: Anything on the scale that is not part of the aircraft, such as chocks, blocks or a jack used to level a tailwheel airplane, is tare weight and must be subtracted: 1,260 − 12 = 1,248 lb.

   Reference AM.I.C.K3 · FAA-H-8083-30B, Ch. 6 (Tare Weight)

7. For an adverse-loading check on an airplane whose piston engine is rated at 230 METO horsepower, how much minimum fuel is used?

   • A230 lb
   • B38 lb
   • C460 lb
   • D115 lbCorrect

   Why: Minimum fuel is 1/12 gallon per METO horsepower, which works out to METO hp ÷ 2 in pounds: 230 ÷ 2 = 115 lb. It represents half an hour of operation and is the fuel you cannot leave out when checking CG limits.

   Reference AM.I.C.K7 · FAA-H-8083-30B, Ch. 6 (Adverse-Loaded CG Checks)

8. In a forward adverse-loaded CG check, how is the useful load arranged?

   • AEverything at maximum, wherever it is located
   • BLoad forward of the forward limit at maximum; load behind it empty, except minimum fuelCorrect
   • COnly the pilot and full fuel, with nothing else aboard
   • DLoad behind the forward limit at maximum and forward items empty

   Why: The check deliberately tries to push the CG past the forward limit: every item of useful load forward of that limit is loaded and everything behind it is left empty. The exception is fuel behind the limit, which is entered as minimum fuel because the aircraft cannot fly without it.

   Reference AM.I.C.K7 · FAA-H-8083-30B, Ch. 6 (Adverse-Loaded CG Checks)

9. An airplane weighs 1,600 lb empty with its CG at +36.5 in. The aft CG limit is +36.0 in, and ballast can be installed at −14.0 in. Using the FAA handbook formula, how much ballast is needed?

   • A8 lb
   • B22 lb
   • C16 lbCorrect
   • D36 lb

   Why: Ballast = aircraft weight × distance out of limits ÷ distance between the ballast and the desired CG = 1,600 × 0.5 ÷ (36.0 − (−14.0)) = 800 ÷ 50 = 16 lb, placed forward to pull the CG back within the aft limit.

   Reference AM.I.C.K9 · FAA-H-8083-30B, Ch. 6 (Ballast)

10. An airplane's CG is at +70.0 in. The leading edge of the mean aerodynamic chord (LEMAC) is at +58.0 in and the MAC is 60 in long. Where is the CG in percent of MAC?

    • A12 percent
    • B83 percent
    • C117 percent
    • D20 percentCorrect

    Why: %MAC = (CG − LEMAC) ÷ MAC × 100 = (70.0 − 58.0) ÷ 60 × 100 = 20 percent. The CG is 12 inches behind the leading edge of a 60-inch chord.

    Reference AM.I.C.K4 · FAA-H-8083-30B, Ch. 6 (Mean Aerodynamic Chord)

11. An airplane weighs 2,000 lb empty with its CG at +35.0 in. A 20-lb radio located at +60.0 in is removed. What is the new empty-weight CG?

    • A+35.25 in
    • B+34.75 inCorrect
    • C+35.00 in
    • D+34.40 in

    Why: Original moment 2,000 × 35 = 70,000 in-lb. Removing the radio subtracts 20 lb and 20 × 60 = 1,200 in-lb: 68,800 ÷ 1,980 = +34.75 in. Removing weight from behind the CG moves the CG forward.

    Reference AM.I.C.K4 · FAA-H-8083-30B, Ch. 6 (Equipment Change and Aircraft Alteration)

12. When raising an airplane with two wing jacks for weighing, what should the technician do as the jacks go up?

    • ARaise one side fully first, then the other
    • BLeave the safety collars loose so the jacks move freely
    • CRaise them together and keep the safety collars run downCorrect
    • DHave a person sit in the cockpit to steady it

    Why: The handbook: raise two wing jacks simultaneously so the aircraft does not slip off, and keep the safety collars screwed down against the cylinder so the aircraft cannot tilt if one jack loses hydraulic pressure. Follow the manufacturer's jacking instructions in detail.

    Reference AM.I.C.K10 · FAA-H-8083-30B, Ch. 6 (Jacking the Aircraft)

Drill the whole domain

• Weight and Balance (6%)

Other topics

• A&P General: Ohm's law, power and series-parallel circuits
• Aircraft corrosion types for the A&P General: how to tell them apart
• A&P mechanic privileges and limitations (14 CFR part 65)
• Maintenance records, return to service and FAA Form 337
• Aircraft bolts, nuts, rivets and safety wire for the A&P General
• Aviation fuel identification, fueling safety and fire classes
• The Dirty Dozen and maintenance human factors (A&P General)

See all topics · Study Guides · Full practice exam