**1. Find the thrust exprienced by a flat keel plate 10m x 2m when the draft is 8m in SW.**

__Solution :__

**Area of the flat keel plate** = (10 x 2)
= 20m2
**Pressure = (depth x density)**
=(8 x 1.025)
=8.2t /m2
** Thrust = (pressure x area)**
= 8.2 x 20 =164 t .

**2. A box-shaped vessel 150m x20m x 12m is floating in a dock of RD 1.010 at an even keel draft of 10m. Find the total water pressure experienced by the hull.**

__ Solution:__

** The total water pressure exerted on the hull to the water = (Thrust of keel plate which acting horizontally + Thrust of forward and aft which act vertically +Thrust of port and stbd side which act vertically)**
**Thrust of Keel plate = (pressure x area) **
** Pressure = (depth x density)**
=10 x1.010 t/m2

Area = 150m x 20m = 3000m2 Thrust =(10 x1.010 x 3000) = 30, 300 t

**Thrust of forward and Aft =(pressure x area) **

** Pressure = depth x density**
=( 5 x1.010)t/m2

**Area =L X B **
=(20 x10)
= 200m2

**Thrust = (P x A)**
= (5Ã—1.010 x200)
= 1010t

**Thrust acting both forward and aft** = (1010 x2)
= 2020 t

**Thrust of port and stbd side = (pressure x area)**

** Pressure =( depth x density)**
=( 5Ã—1.010)

** Area = (L X B) **
= 150m x 10 = 1500m2

** Thrust = ( P X A )**
= 5 x 1.010 x1500
= 7575 t

**Thrust acting on Both sides** = 7575 x 2
=15150 t

Hence , Total pressure on the hull = (30300 +2020 + 15150) =47,470 tonnes.

**3. A submarine has a surface area of 650m2 and can with stand a total water pressure of 1332500 t . Find at what approximate depth in SW she would collapse.**

__Solution:__

Area = 650m2 Thrust = 1332500t RD = 1.025 Depth =?

** Thrust = ( P x A )**
1332500 = (P x 650)
P = 2050 t/m2

**Now, P =(depth x density)**
2050 = (depth x 1.025)

Hence , Depth = 2000m.

**4. A rectangular lock gate 40m wide and 20m high has water of RD 1.010 12m deep on one sideand water of RD 1.020 11m deep on the other. Find the resultant thrust experienced and direction in which it acts.**

__Solution: __

Water of RD =1.010 Depth =12 m C = (12/2) =6m

**Pressure = (depth x density)**
=(6 x1.010) t/m2

**Thrust =(P x A) **=( 6 x 1.010 x40 x12)
= 2908.8t

Water of RD = 1.020 Depth = 11m C = (11/2) =5.5m

**Pressure =(depth x density)**
= (5.5 x 1.020 ) t/m2

Area =( L X B) =(40 x 11) = 440m2

**Thrust =( P X A)** = (5.5 x 1.020 x 440)
= 2468.4 t

**Thus we can Resultant thrust** =(2908.8 â€“ 2468.4)
= 440.4t

**5. A rectangular lock gate 36m wide and 20m high has FW on one side to a depth of 16m. Find what depth of SW on the other side will equalize the thrust.**

Solution :

Inside the lock gate area =( 36 x 16)m2

**Pressure = depth x density** =(16/2 x 1)
= 8 t/m2

**Thrust inside the lock gate = (pressure x area)**
Thrust =(8 x 36 x 16)
= 4608 t

**Let â€˜Xâ€™ depth of salt water on other side will equalize the thrust .**

**Pressure = depth x density **=( X/2 x 1.025)
Area =(36 x X )
Thrust = (P x A)

4608 = (X/2 x 1.025 x 36X ) 4608 = (1.025 x 36X x X/2) 4608 = (1.025 x 36X2/2) = ( 1.025 x 18X2) X2 = (4608 /1.025 x 18)

X = 15.8 m

Hence, Depth of other side of lockgate =15.8m

**6. A collision bulkhead is triangular in shape. Its maximum breadth is 12m and its high 15m. Find the thrust experienced by it if the fore peak tank is pressed up to a head of 3m of SW.**

Solution :

Breadth of collision bulkhead = 12m
Height = 15m
**Pressure = (depth x density)**
** Thrust = (pressure x area)**
Breadth = 12m
Height of the water inside the tank = 15m
Water pressed up to ahead of = 3m
â€˜Câ€™ inside the tank = (1/3 x15)
= 5 m
Outside the tank = 3m
so total =5 + 3 =8m

**Now pressure =(depth x density)**
=( 8 x 1.025)

Area = (1/2 x 12 x15) = 90m2

**Thrust experienced = (pressure x area)**
= (90 x 8 x 1.025)
=738t.

**7. A collision bulkhead is triangular shaped, having a breadth of 14m at the tank top and a height of 12m. As a result of a collision, the forepeak tank gets ruptured and SW enters the tank to a sounding of 9m. Calculate the thrust on the bulkhead.**

Solution :

B = 14m, H =12m
As we know that :
** Pressure = (depth x density)**
**Thrust = (pressure x area)**

**An right angle triangle PBC , PB =7m**
** Since, Height divide AB equally into two parts.**
** Triangle PBC and FEC are similar**
** So by law of similar angle triangle**

**PB/ FE = PC / FC**
7 / FE =12 / 9

Hence, FE = (63 / 12) = 5.25m

**In triangle DEC**
** DF = FE**
DE = (2 X FE)
= 2X 5.25
=10.5m

Area of triangle DEC = (1/2 x10.5Ã—9) =47.25m2

**Pressure = (depth x density)**
= (9 x 1/3 x 1.025)
= 3.075 t/m2

**Hence Thrust experienced = (P X A)**
= (3.075 x 47.25)
=145. 29 t

**8. A tank has a triangular bulkhead, apex upwards. Its base is 14m and its sides, 15m each. It has a circular inspection hole of radius 0.5m the centre of a manhole 0.8m above the base and 1.6 from one corner. Find the thrust on the manhole cover when the tank contain oil of RD 0.95 to a sounding of 10m.(Assume Ï€ to be 3.1416).**

Solution:

Base = 14m , Sides = 15m , Depth of the oil = 10m

**Pressure = (depth x density)**
= (10 â€“ 0.8 x 0.95)
= 8.74 t/m2

**Area of the circle = Ï€ r2**
= (3.1416 x 0.5 x 0.5)
=0.7854m2

**Thrust = (P X A)**
=( 8.74 x 0.7854)
= 6.864 t.

**9. A rectangular deep tank is 22m x 20m x10m. Above the crown of the tank is a rectangular trunkway 0.2m high, 5m long and 4m wide. Find the thrust on the tank lid when the tank is pressed up with SW to a head of 2.64m above the crown of the tank .**

Solution :

Rectangular deep tank = 22m x 20m x10m
Trunk way = (5m x 4m x 0.2m)
**Pressure = (depth x density)**
= (2.64m â€“ 0.2) x1.025

**Since , Depth is (2.64m â€“ 0.2 ) and we have the calculate the thrust act on the trunkway and the water level above the trunkway will act as thrust )**

**If this trunkway would have been placed inside the tank then the depth will be **

(10m + 2.64m â€“ 0.2/2m ).

**Area of the trunkway = (L X B)**
= 4Ã—5
=20 m2

**Thrust experienced = (P X A)**
= (2.501 x 20 )
= 50.02tonnes.

**10. A double bottom tank measures 25m x20m x 2m. Find the thrust on the tank top when pressed upto a head 16m of SW. Also find the resultant thrust on the tank bottom, and the direction that it acts, if the shipâ€™s draft in SW is 10m.**

Solution :

**Volume of tank = (L X B X H)**
= (25m x 20m x 2m)

Area =(25 x20) = 500m2

**Pressure = (depth x density)**
= (16 x 1.025)
=16.4 t/m2

**Thrust on tank top = (PX A)**
= (16.4 X 500)
= 8200 t

**Pressure on the tank bottom = (depth x density)**
= (18 x1.025)

= 8.45t/m2

**Thrust act on the tank bottom = (P X A)**
= (500 x 18 .45)
= 9225t

**Pressure act on outside of the tank bottom = (depth x density)**
= (10 x 1.025)
= 10.25t

**Hence, Thrust acting = (P X A)**
= (10.25 X500)
= 5125 tonnes.

Finally, Resultant thrust = (9225 -5125) = 4100 tonnes.

Explain why triangle's pressure point where will ?