Staility – I : Chapter 2

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 – 2468.4)
                                                      = 440.4t