Discussion
Fuel of \( 4000 \ \mathrm{kg} \) burns from a tank \( 1.5 \ \mathrm{m} \) aft of the centre of gravity on an aeroplane of \( 40\,000 \ \mathrm{kg} \). The centre of gravity moves forward by approximately
Answer: D
Correct answer: 0.167 m
Removing a mass \( m \) at a distance \( d \) from the centre of gravity shifts it by \( m d \) divided by the mass that remains:
\[ \Delta x = \frac{4000 \times 1.5}{40\,000 - 4000} = \frac{6000}{36\,000} = 0.167 \text{ m} \]The fuel was aft, so removing it draws the centre of gravity forward and the static margin increases. Dividing by the original 40,000 kg rather than the final 36,000 kg gives 0.15 m, which is the usual slip: the moment must be taken about the mass that is actually left.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development