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.
r??
1
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