# Staility – I : Chapter 2

1. 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.

1. ###### 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 – 2468.4)
= 440.4t #### Amit Sharma

Graduated from M.E.R.I. Mumbai (Mumbai University), After a brief sailing founded this website with the idea to bring the maritime education online which must be free and available for all at all times and to find basic solutions that are of extreme importance to a seafarer by our innovative ideas.

### 1 Comment

• rasel says:

sir i need stability chapter 12 and 13 situation

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