Two items weighing 11kg and 8kg are placed 2m and 1m respectively aft of the C of G of an aircraft. How far forward of the C of G must a weight of 30kg be placed so as not to change the C of G?
- 2m
- 3m
- 1m
Explanation
Balance of moments about the C of G requires the forward moment to equal the total aft moment. The aft moment is (11 x 2) + (8 x 1) = 30 kg.m. The 30 kg weight must give the same moment forward, so 30 x d = 30, giving d = 1 m.
ranz130 asking:
still not get it,can somebody explain it.
Community Comments (3)
D
dimky Posts: 514 19.08.2013 / 15:50
Hey.
((11 * 2) + (8 * 1)) / 30 = 1
((11 * 2) + (8 * 1)) / 30 = 1
5
5rain Posts: 1 05.09.2013 / 15:34
WHAT FORMULA DID U USED?
D
dimky Posts: 514 05.09.2013 / 15:39
Hey.
It is just logic...
in this case: sum of (Force * Distance) = sum of (Force * Distance)
(11 * 2) + (8 * 1) = 30 * X
To find X we must divide left part of equation to known value 30.
So result is 1.
It is just logic...
in this case: sum of (Force * Distance) = sum of (Force * Distance)
(11 * 2) + (8 * 1) = 30 * X
To find X we must divide left part of equation to known value 30.
So result is 1.
Please Sign In to post a comment.