Engineering Mechanics

Chapter 35 : Hydrostatics „„„„„ 723

Example 35.5. A circular door of 1 m diameter closes on an opening in the vertical side of
a bulkhead, which retains water. The centre of the opening is at a depth of 2 m from the water level.
Determine the total pressure on the door.

Take specific gravity of sea water as 1.03.

Solution. Given: Diameter of the door (d) = 1 m; Depth of centre of opening from water

level ()x = 2 m and specific gravity of water = 1.03

We know that area of the circular door,

A=

()^22 (1) 0.7854m^2

44

d

ππ

==

∴ Total pressure, P = wAx=×(9.8 1.03) 0.7854× 2 = 15.86 kN Ans.
Example 35.6. A triangular lamina ABC is immersed in water with the side AB coinciding
with the water surface as shown in Fig. 35.6.

Fig. 35.6.
A point D is taken in AC, such that the water pressure on the two areas ABD and DBC
are equal. Show that AD : AC = 1 : 2.

Solution. Given: Water pressure on ABD = Water pressure on DBC (i.e. PABD = PDBC )

Let w = Specific weight of the liquid,

a = Length of side AB,

x 1 = Depth of vertex D from the water surface, and

x 2 = Depth of vertex C from the water surface.

We know that the pressure on ABD

PABD =

2

11 1

236

ax x wax

w

×

××=

∴ x 1 =

6 PABD

wa

...(i)

Similarly, PABC =

2

22 2

236

ax x wax

w

×

××=

∴ x 2 =

6 PABC

wa

But PDBC =PABD =

2

2

212

PABC =wax

∴ x 2 =

12 PABD

wa

...(ii)