Question

practice weight and balance quiz. question 24 of 55
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.

a +30.61.
b +31.61.
c +32.69.

Explanation:

Step1: Calculate the total weight change

• Original empty weight: \(1800\) pounds.
• Weight removed: \(2\times15 = 30\) pounds.
• Weight added: \(14 + 20+30=64\) pounds.
• New total weight \(W=1800 - 30+64=1834\) pounds.

Step2: Calculate the total moment change

• Original moment \(M_0=1800\times31.5 = 56700\) pound - inches.
• Moment from removed seats: \(30\times72=2160\) pound - inches (subtractive).
• Moment from structural modifications: \(14\times76 = 1064\) pound - inches (additive).
• Moment from seat and safety belt: \(20\times73.5=1470\) pound - inches (additive).
• Moment from radio equipment: \(30\times30 = 900\) pound - inches (additive).
• New total moment \(M=56700-2160 + 1064+1470+900=58, 974\) pound - inches.

Step3: Calculate the new CG

• Use the formula \(CG=\frac{M}{W}\).
• \(CG=\frac{58974}{1834}\approx32.15\) (Wait, let's re - calculate the moment and weight accurately.
• Recalculate moment:
• \(M_0 = 1800\times31.5=56700\)
• Removed moment: \(2\times15\times72 = 2160\)
• Added moments: \(14\times76+20\times73.5 + 30\times30=1064 + 1470+900=3434\)
• \(M=56700-2160 + 3434=57974\)
• \(W=1800-30 + 64=1834\)
• \(CG=\frac{57974}{1834}\approx31.61\)

Answer:

B. +31.61