4r98 NICHOLLS'S SEAMANSHIP AND NAUTICAL KNOWLEDGE

Moment about K =(2x4)+(5x6)+(7xlO)
=108 foot-lbs

_. „ ~ Moment 108 .

Distance K Or, =--------------- = —=5*4 leer

1 Total weight 20

The initial distance K G was 4*0 ft.
The new distance K Gl is 54 ft

GQi = 14ft.

The addition of the weights has shifted the C. of G. 14 leet to the
right.

(ii) Redistribution of Weights.—Suppose the 10 tbs. weight were now
shifted 3 feet to the left, find the position of the new centre ot gravity.

1

r??

10 6 Li*
Fig. 18.
Draw Figure 18 for the new condition.
Moment about K =(2 x4)+(4= XlO)+(6 X6)
= 78 foot-lbs.
_, _ Moment 78
Distance K (?« = • . ^^ =-~=3'9 fe«t
2 Weight 20
K GI was 54 feet.
KG± is 3-9 „
G&te 1-5 „
Shifting the weight 3 feet to the left has shifted the C. of G. 1 -5 feet
to the left.
(iii) Lifting off a Weight.—Lift off the 4 Ibs. weight and tmd the
of the centre of gravity 6?2.
Draw Figure 19 for new condition