1. STABILITY
PROBLEMS
NO. 1

2. 1. You are loading a cargo of canned goods with a stowage factor of 65
1. You are loading a cargo of canned goods with a stowage factor of 65. If you allow 15% for broken stowage, how many tons can be loaded in a space of 55,000 cubic feet?
Ans Tons

3. 1. Solution; Weight = Volume x Allowance Stowage Factor65
Weight = tons

4. 2. What is the stowage factor of cargo of canned goods packed in cartons weighing 340 pounds and 36 cubic feet each carton?
Ans ft³/ton

5. 2. Solution:
SF = Volume
Weight / 2240 lbs.
SF = ft³
340 tons / 2240 lbs
SF = 237 ft³/ton

6. 3. If the stowage factor of cotton is 56, allowing a stowage loss of 15%, it will stow at a factor of ___.
Ans

7. 3. Solution:
S. Factor = Stowage Factor of Cotton
Allowance
S. Factor =
0.85
S. Factor =

8. 4. A ship of 11,000 tons displacement has a moment of statical stability of 500 tons meters when heeled 5°. Find the initial metacentric height?
Ans m

9. 4. Solution;
MSS = ∆ x GM x Sin θ
∆ x GM x Sin θ = MSS
GM = MSS
∆ x Sin θ
= tons meter
11,000 tons x Sin 5º
GM = m

10. 5. A vessel displacing 18,000 tons has a KG of 50 feet
5. A vessel displacing 18,000 tons has a KG of 50 feet. A crane is used to lift cargo weighing 20 tons from a supply vessel. When lifting, the head of the crane boom is 150 feet above the keel. What is the change in KG?
Ans foot

11. 5. Solution:
GG’ = W x D ∆
GG’ = 20 t x ( 150 ft – 50 ft )
18,000 t
GG’ = 20 t x 100 ft.
GG’ = foot

12. 6. A weight of 500 tons is loaded into a ship so that its centre of gravity is 10 meters from that of the ship. Find the shift of G if the ship's original displacement was 3,000 tons.
Ans m

13. 6. Solution:
GG’ = W x D ∆
GG’ = 500 t x 10 m
3,500 tons
GG’ = m

14. 7. What will be the reduction of GM on a 5,000 tons displacement ship with 6.3 KG in sea water after half of the sea water ballast is discharged from the double bottom ballast tank 22 m x 10 m x 2 m with 1.3 m KG?
Ans m.

15. 7. Solution: Wt. of Ballast = L x B x D x R. Den.
= 22 x 10 x 1 x 1.025
Wt. of Ballast = tons
Old ∆ = 5,000 t t
New ∆ = tons
Dist = KG of Vsl.– KG of Tank
= 6.3 m – 1.3 m
Dist = 5.0 m
GG’ = W x D ∆
= t x 5 m t
GG’ = m

16. 8. The original displacement of a ship was 4,285 tons and her KG was 6
8. The original displacement of a ship was 4,285 tons and her KG was 6.00 meters. Find her new KG after she has loaded the following weights; 800 tons at 3.6 meters above the keel, 440 tons at 7.0 meters above the keel tons at 5.8 meters above the keel, 630 tons at 3.0 meters above the keel.
Ans m

17. 8. Solution:
Weight KG Moments
4, m ,710
m ,880
m ,080 m
m ,890
6, ,198
New KG = Total Moments / Total Weights
= 34,198 / 6265
New KG = m

18. 9. A vessel displacing 18,000 short tons and has a VCG of 70 ft
9. A vessel displacing 18,000 short tons and has a VCG of 70 ft. You dump 100 short tons of water, which had a VCG of 88 ft., what is the new KG?
Ans feet

19. 9. Solution:
Weight VCG Moments
18, ft ,260,000
( - ) ft ,800 ( - )
17, ,251,200
New KG = Total Moments
Total Weights
New KG = 1,251,000 / 17,900
New KG = ft

20. 10. A rectangular shaped vessel 200 m in length and 32 m in breadth, floats at an even keel draft of 9.0 m. The KG is 10.0 m, KB is 4.737, BM is m. What is her GM?
Ans m

21. 10. Solution:
KB = m
BM = m ( + )
KM = m
KG = m ( - )
GM = m

22. 11. At a given draft, a ship of 120 meters
length and 15 meters beam has a
coefficient of fineness of the
waterplane of What is her TPC
at this draft?
Ans tons

23. 11. Solution:
TPC = L x B x Cw x Rel. Density
100
TPC = 120 m x 15 m x x 1.025
TPC = tons

24. 12. What is the TPC of a box-shaped vessel 120 m x 15 m x 10 m floating at even keel draft of 4.5 m?
Ans tons

25. 12. Solution:
TPC = x Area of Waterplane
100
TPC = x 120 m x 15 m
TPC =
TPC = tons

26. 13. A rectangular shaped vessel 100 m in length, 20 m breadth is floating in salt water at an even keel draft of 6.0 m. A forward hold 10.0 m long was bilged. What is the TPC?
Ans tons

27. 13. Solution:
TPC = x Area
100
TPC = x ( 100m – 10m ) x 20 m
TPC = x 90m x 20m
TPC = tons

28. 14. A drum which is 1. 5 m long and 60 cm
14. A drum which is 1.5 m long and 60 cm. in diameter has mass of 20 kgs when empty. Find what will be her draft in water of density 1,024 kg/m³ If it contains 200 liters of paraffin of relative density 0.6.
Ans m

29. 14. Solution:
Den. of Paraffin = RD of Paraffin x FW Density
= 0.6 x 1,000 kgs./m³
Den. Of Paraffin = 600 kgs./m³
Mass of the Paraffin = Volume x Density
= 0.2 m³ x 600 kgs./m³
Mass of the Paraffin = 120 kgs.
Mass of the Drum = kgs. ( + )
Total Mass = kgs.

30. 14. Solution:
Vol. of water displaced = Mass / Density
=140 kgs. / 1,024 kgs./m³
Vol. of water displaced = m³
Vol. of water displaced (V) = π²d
d = V / π²
d = m³
x .09
Draft = m

31. 15. A box-shaped barge 16m x 6m x 5 m is floating alongside a ship in fresh water at a mean draft of 3.5 m. The barge is to be lifted out of the water and loaded on to the ship with a heavy lift derrick. Find the load in tons borne by the purchase when the draft of the barge is reduced to 2 m?
Ans tons

32. 15. Solution:
Mass of the Barge = L x W x Dr x Rel. Den.
= 16 x 6 x 3.5 x 1.0
Mass of the Barge = 336 tons
Mass of water = 16 x 6 x 2 x 1.0
Displaced at 2m = tons
Mass of the Barge = 336 tons
Mass of Barge at 2 m = 192 tons ( - )
Load Borne by the = 144 tons
Purchase