FAA Question Chapter 6, Section B

6-36 AMG002

An aircraft with an empty weight of 2,100 pounds and an empty weight CG of +32.5 was altered as follows:

1. two 18-pound passenger seats located +73 were removed;

2. structural modifications were made at +77 increasing weight by 17 pounds;

3. a seat and safety belt weighing 25 pounds were installed at +74.5; and

4. radio equipment weighing 35 pounds was installed at +95.

What is the new empty weight CG?

A— +34.01

B— +33.68

C— +34.65

6-36. Answer B. JSGT 6B, FGH

It is typically easier to solve this type of problem if you enter the information into a table as seen below.

Item	Weight	Arm	Moment
Aircraft	2,100	32.5	68,250.0
2 Pass seats	-36	73.0	-2,628.0
Modification	17	77.0	1,862.5
Seat/belt	25	74.5	3,325.0
Radio	35	95.0	3,325.0
Total	2,141		72,118.5

First, multiply all weights by their arms to obtain the moments.

Watch the negative sign on removed items.

Total the weight and moment columns, then divide the new moment by the new weight.

The new CG is 33.68 inches (72,118.5 ÷ 2,141 = 33.68).

6-37 AMG002

An aircraft as loaded weighs 4,954 pounds at a CG of +30.5 inches. The CG range is +32.0 inches to +42.1 inches. Find the minimum weight of the ballast necessary to bring the CG within the CG range. The ballast arm is +162 inches.

A— 61.98 pounds

B— 30.58 pounds

C— 57.16 pounds

6-37. Answer C. JSGT 6B, FGH

Apply the ballast formula to the values given in the problem:

In the above formula, the aircraft weight is entered as 4,954 pounds.

The current CG is +30.5 inches, and the desired CG is the forward limit of +32.0 inches.

The arm of the ballast is +162.0 inches. A positive number requires addition of weight at the ballast point, and a negative number requires removal of weight.

6-38 AMG002

An aircraft with an empty weight of 1,800 pounds and an empty weight CG of +31.5 was altered as follows:

1. two 15-pound passenger seats located at +72 were removed;

2. structural modifications increasing the weight 14 pounds were made at +76;

3. a seat and safety belt weighing 20 pounds were installed at +73.5; and

4. radio equipment weighing 30 pounds was installed at +30.

What is the new empty weight CG?

A— +30.61

B— +31.61

C— +32.3

6-38. Answer B. JSGT 6B, FGH

It is typically easier to solve this type of problem if you enter the information into a table.

Item	Weight	Arm	Moment
Aircraft	1,800	31.5	56,700
Seats	-30	72.0	-2,160
Mod	14	76.0	1,064
Seat/belt	20	73.5	1,470
Radio	30	30.0	900
Total	1,834		57,974

First multiply all weights by their arms to obtain the moments.

Watch the negative sign on removed items.

Total the weight and moment columns and divide the new moment by the new weight.

The new CG is 31.61 inches (57,974 ÷ 1,834 = 31.61).

6-39 AMG002

An aircraft had an empty weight of 2,886 pounds with a moment of 101,673.78 before several alterations were made. The alterations included:

1. removing two passengers seats (15 pounds each) at +71;

2. installing a cabinet (97 pounds) at +71;

3. installing a seat and safety belt (20 pounds) at +71; and

4. installing radio equipment (30 pounds) at +94.

The alterations caused the new empty weight CG to move

A— 1.62 inches aft of the original empty weight CG.

B— 2.03 inches forward of the original empty weight CG.

C— 2.03 inches aft of the original empty weight CG.

6-39. Answer A. JSGT 6B, FGH

It is typically easier to solve this type of problem if you enter the information into a table.

Item	Weight	Arm	Moment
Aircraft	2,886	35.23	101,673.78
Seats	–30	71.00	–2,130
Cabinet	97	71.00	6,887
Seat/belt	20	71.00	1,420
Radio	30	94.00	2,820
Total	3,003		110,670.78

The old CG is calculated by dividing the original moment (101,673.78) by the original weight (2,886 lbs).

The original CG was 35.23 (101,673.78 ÷ 2,886 = 35.23).

The new CG is calculated by dividing the new moment (110,670.78) by the new weight (3,003 lbs).

Because the new CG (36.85) is a larger number than the original (35.23), you know that the CG has shifted aft of the original.

Subtracting the original CG from the new CG results in a difference of 1.62 inches (36.85 – 35.23 = 1.62).

Therefore, the CG shifted aft 1.62 inches.

6-40 AMG002

When making a rearward weight and balance check to determine the CG will not exceed the rearward limit during extreme conditions, the items of useful load which should be computed at their minimum weights are those located forward of the

A— forward CG limit.

B— datum.

C— rearward CG limit.

6-40. Answer C. JSGT 6B, FGH

When making a rearward extreme loading check, you must load the items of useful load behind the rearward CG limit to their maximum, and items of useful load ahead of the rearward CG limit to their minimum.

6-41 AMG002

An aircraft with an empty weight of 1,500 pounds and an empty weight CG of +28.4 was altered as follows:

1. two 12-pound seats located at +68.5 were removed;

2. structural modifications weighing +28 pounds were made at +73;

3. a seat and safety belt weighing 30 pounds were installed at +70.5; and

4. radio equipment weighing 25 pounds was installed at +85.

What is the new empty weight CG?

A— +23.51

B— +31.35

C— +30.30

6-41. Answer C. JSGT 6B, FGH

It is typically easier to solve this type of problem if you enter the information into a table.

Item	Weight	Arm	Moment
Aircraft	1,500	28.4	42,600
Seats	-24	68.5	-1,644
Mod	28	73.0	2,044
Seat/belt	30	70.5	2,115
Radio	25	85.0	2,125
Total	1,559		47,240

First, multiply all weights by their arms to obtain the moments.

Watch the negative sign on removed items.

Total the weight and moment columns and divide the new moment by the new weight.

The new empty weight CG is 30.3 (47,240 ÷ 1,559 = 30.3).

6-42 AMG002

The following alteration was performed on an aircraft: A model B engine weighing 175 pounds was replaced by a model D engine weighing 185 pounds at a -62.00-inch station. The aircraft weight and balance records show the previous empty weight to be 998 pounds and an empty weight CG of 13.48 inches. What is the new empty weight CG?

A— 13.96 inches

B— 14.25 inches

C— 12.73 inches

6-42. Answer C. JSGT 6B, FGH

It is typically easier to solve this type of problem if you enter the information into a table.

Item	Weight	Arm	Moment
Aircraft	998	13.48	13,446.04
Engine B	-175	-62.00	10,850.00
Engine D	185	-62.00	-11,470.00
Total	1,008		12,833.04

Calculate the new CG by dividing the total moment by the total weight.

Upgrading to the model D engine increased the aircraft by 10 pounds at station -62.00.

Since more weight is added ahead of the datum, the CG shifts forward to 12.73 inches (12,833.04 ÷ 1,008 = 12.73).

6-43 AMG002

If the empty weight CG of an airplane lies within the empty weight CG limits,

A— it is necessary to calculate CG extremes.

B— it is not necessary to calculate CG extremes.

C— minimum fuel should be used in both forward and rearward CG checks.

6-43. Answer B. JSGT 6B, FGH

Adverse loading checks are a deliberate attempt to load an aircraft in a manner that will create the most critical balance condition and still remain within the aircraft’s design CG limits. If the empty weight CG falls within the empty weight CG range it is unnecessary to perform this check.

6-44 AMG002

When computing the maximum forward loaded CG of an aircraft, minimum weights, arms, and moments should be used for items of useful load that are located aft of the

A— rearward CG limit.

B— forward CG limit.

C— datum.

6-44. Answer B. JSGT 6B, FGH

A forward adverse-loading check is performed to determine if it is possible to load an airplane so that its CG will fall ahead of the forward CG limit. To perform this type of check, use maximum values for items of useful load that are forward of the forward limit, and minimum values for those items located aft of the forward CG limit.

6-45 AMG002

All other things being equal, if an item of useful load located aft of an aircraft’s CG is removed, the aircraft’s CG change will be

A— aft in proportion to the weight of the item and its location in the aircraft.

B— forward in proportion to the weight of the item and its location in the aircraft.

C— forward in proportion to the weight of the item, regardless of its location in the aircraft.

6-45. Answer B. JSGT 6B, FGH

When weight is added or removed to an aircraft, the CG is effected depending on an amount of weight and its location relative to the CG. When weight is added, it moves the CG toward the added weight. When weight is removed, the CG moves away from the location of the removed weight. Only when weight is added or removed at the CG does the CG remain constant.