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 from PART66Online
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.
R
ranz130 asked · 18 Aug 2013
still not get it,can somebody explain it.
Community Comments (3)
P
PART66Online
19 Aug 2013
Hey.
((11 * 2) + (8 * 1)) / 30 = 1
((11 * 2) + (8 * 1)) / 30 = 1
5
5rain
5 Sep 2013
WHAT FORMULA DID U USED?
P
PART66Online
5 Sep 2013
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.
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