1. CHS
PROBLEM SOLVING
1 - 13

2. 1. A ship has a displacement of 2,400 tons and KG 10. 8 meters
1. A ship has a displacement of 2,400 tons and KG 10.8 meters. Find the new KG if a weight of 50 tons mass already on board is raised 12 meters vertically.
Answer: meters

3. M G G G G G G B K THE POS. OF K IS FIXED THE POS OF M VARIES
W/ THE DFT OF VSL. G G G
G MOVES TOWARDS G
WT. LOADED G G
G MOVES AWAY FROM
WT. DISCHARGE B
G MOVES PARALLEL
TO THE DIR. OF THE WT
BEING SHIFTED K
B MOVES TOWARDS
THE LOW SIDE OF
INCLINED V/L

4. M G K

5. 1. Solution:
GG’ = Wt. x Dist. Δ
= 50 tons x 12m
2,400 tons
GG’ = m
Old KG = m ( + )
New KG = m

6. 2. A ship has displacement of 2,000 tons and KG 10. 5 meters
2. A ship has displacement of 2,000 tons and KG 10.5 meters. Find the new KG if a weight of 40 tons mass already on board is shifted from the ‘tween deck to lower hold trough a distance of 4.5 meters vertically.
Ans m

7. M G K

8. 2. Solution:
GG’ = Wt x Dist Δ
= 40 tons x 4.5 m
2,000 tons
GG’ = m
Old KG = m ( - )
New KG = m

9. 3. A ship of 2,000 tons displacement has KG 4. 5 meters
3. A ship of 2,000 tons displacement has KG 4.5 meters. A heavy lift of 20 tons mass is in the lower hold and has KG 2 meters. The weight is then raised 0.5 meter clear of the tank top by a derrick whose head is 14 meters above the keel. Find the new KG of the ship?
Ans m

10. M G K

11. 3. Solution:
Height of Derrick fr. Keel = 14 m
KG. of Cargo = 2.0 m ( - )
Distance = 12.0 m
GG’ = Wt x Dist. Δ
= 20 tons x 12.0 m
2,000 tons
GG’ = m
Old KG = m ( + )
New KG = 4.62 m

12. 4. A ship has a displacement of 7,000 tons and KG 6 meters
4. A ship has a displacement of 7,000 tons and KG 6 meters. A heavy lift in the lower hold has KG 3 meters and mass 40 tons. Find the new KG when this weight is raised through 4.5 meters vertically and is suspended by a derrick whose head is 17 meters above the keel.
Ans m

13. 4. Solution:
Ht of Derrick = m
KG of Cargo = m ( - )
Distance = m
GG’ = Wt. x Dist. Δ
= tons x 14.0 m
7,000 tons
GG’ = m
Old KG = m ( + )
New KG = m

14. 5. Find the ship in the center of gravity of a ship of 1,500 tons displacement when a weight of 25 tons mass is shifted from the starboard side of the lower hold to the port side on deck through a distance of 15 meters.
Ans m

15. 5. Solution:
GG’ = Wt. x Dist. Δ
= 25 tons x 15 m
1,500 tons
GG’ = m

16. 6. A tank holds 120 tons when full of fresh water
6. A tank holds 120 tons when full of fresh water. Find how many tons of oil of relative density 0.84 it will hold, allowing 2% of the volume of the tank for expansion in the oil.
Ans tons

17. 6. Solution:
Capacity = Wt. of FW x Expansion
= tons x 2 % or ( .02 )
Capacity =
(-)
Capacity = tons
Wt. of Oil = Capacity x R.D. of Oil
117.6 tons x 0.84
Wt. of oil = tons

18. 7. A tank when full will hold 130 tons of salt water
7. A tank when full will hold 130 tons of salt water. Find how many tons of oil relative density it will hold allowing 1% of the volume of the tank for expansion.
Ans tons

19. 7. Solution:
Capacity = Capacity of FW x Expansion
= x 1% or .01
Capacity =
(-) tons
Capacity = tons
R.D. Oil = ( x )
Wt. of Oil = tons
Actual Wt. = Weight of Oil / Density of SW
= tons / 1.025
Actual Wt. of Oil = tons

20. 8. A tank measuring 8m x 6m x 7m is being filled with oil of relative density 0.9. Find how many tons of oil in the tank when the ullage is 3 meters.
Ans tons

21. 8. Solution:
Wt. of Oil = L x W x (H – Ullage) x R.D of Oil
= 8m x 6m x ( 7m – 3m ) x 0.9
= 8m x 6m x 4m x 0.9
Wt. of Oil = tons

22. 9. Oil of relative density 0
9. Oil of relative density 0.75 is run into a tank measuring 6m x 4m x 8m until the ullage is 2 meters. Calculate the number of tons of oil the tank then contains.
Ans tons

23. 9. Solution:
Weight of Oil = L x W x ( H – Ullage ) x R.D. Oil
= 6m x 4m x (8m – 2m) x 0.75
= 6m x 4m x 6m x 0.75
Weight of Oil = 108 tons

24. 10. A tank will hold 100 tons when full of fresh water
10. A tank will hold 100 tons when full of fresh water. Find how many tons of oil of relative density 0.85 may be loaded if 2% of the volume of the oil loaded is to be allowed for expansion.
Ans tons

25. 10. Solution:
Capacity = Wt. of FW x Expansion
= 100 t x 2% or .02 =
(-)100
Capacity = 98 tons
Weight of Oil = Capacity x R.D. of Oil
= 98 tons x 0.85
Weight of Oil = tons

26. 11. A box-shaped vessel 120m x 6m x 2. 5 m floats at a draft of 1
11. A box-shaped vessel 120m x 6m x 2.5 m floats at a draft of 1.5 m in water of density 1,013 kgs/m³. Find the displacement in tons, and the height of the centre of buoyancy above the keel.
Ans tons / 0.75 m

27. 11. Solution:
Δ = L x B x Draft x Cb x R. Density
= 20m x 6m x 1.5m x 1 x tons/m³
Δ = 1, tons
KB = ½ x Draft
= ½ x 1.5 m
KB = m

28. 12. A box-shaped barge 55m x 10m x 6m is floating in fresh water on an even keel at 1.5 m draft. If 1,800 tons of cargo is now loaded, find the difference in the height of the centre buoyancy above the keel?
Ans m

29. 12. Solution:
Δ = L x W x Draft x Cb x R. Density
= 55m x 10m x 1.5m x 1 x 1.000ton/m³
Δ = 825 tons
KB = ½ x Draft
KB = ½ x 1.5m
KB = 0.75m

30. If 1,800 is now loaded
Δ = L x W x Draft x Cb x R. Density
1,800 tons = 55m x 10m x Draft x 1 x 1.000
Draft = tons
55m x 10m x 1 x ton/m³
Draft =
550
Draft = 3.27m
KB = ½ x Draft
= ½ x 3.27m
KB = m

31. 12. Solution:
New Draft New Δ
Old Draft Old Δ
New Draft 1,800 tons
1.5m tons
New Draft = 1.5m x 1,800 tons
825 tons
New Draft = m
KB = ½ x Draft
KB = ½ x 3.27m
KB = m

32. 13. A box-shaped barge 75m x 6m x 4m displaces 180 tons when light
13. A box-shaped barge 75m x 6m x 4m displaces 180 tons when light. If 360 tons of iron are loaded while the barge is floating in fresh water, find her final draft reserve buoyancy.
Ans m / 70%

33. 13. Solution:
Light Δ = L x W x Draft x Cb x R. Density
180 tons = 75m x 6 m x Draft x 1 x 1.0
Light Draft = tons
75m x 6m x 1 x 1.000
Light Draft = tons
450
Light Draft = 0.4m

34. New Draft New Δ
Old Draft Old Δ
New Draft tons
0.4m tons
New Draft = 0.4m x 360 tons
180 tons
New Draft = 0.8m
Light Draft = 0.4m ( + )
Final Draft = 1.2m

35. Reserve Buoyancy = Final Draft/ Depth = 1.2m / 4m
Intact Buoyancy = or 30 %
( - ) %
Reserve Buoyancy = %
RESERVE BUOYANCY 4m
Depth
1.2m Draft