(^722) A Textbook of Engineering Mechanics
Divide the whole surface into a no. of small parallel strips as shown in the figure. Now let
us consider a strip of thickness dx, width b and at a depth x from the free surface of the liquid as
shown in Fig. 35.4.
We know that the intensity of pressure on the strip
=wx
and area of the strip =b d x
∴ Pressure on the strip p= Intensity of pressure × Area
=w x b dx
Now* total pressure on the surface,
P=∫ wxb dx
=wxbdx∫
But ∫ xb dx = Moment of the surface area about the liquid level.
=Ax
∴ P=wA x ...(Same as in Art. 35.7)
Example 35.4. Find the total pressure on a rectangular plate 2m × 4m vertically immersed
in water, such that 2 metre side is parallel to the water surface, and 2.5 metres below it.
Take specific weight of water as 9.8 kN/m^3.
Solution. Given: Width of plate (b) = 2 m; Depth of plate (d) = 4 m and specific weight
of water (w) = 9.8 kN/m^3
We know that area of the rectangular plate,
A= 2 × 4 = 8 m^2
and depth of centre of gravity of the plate from the water surface,
x =
4
2.5 4.5 m
2
+=
∴ Total pressure on the plate,
P=wA x=××=9.8 8 4.5 352.8 kN Ans.
- Total pressure may also be found out by dividing the whole surface into a no. of small parallel
strips.
Let a 1 , a 2 , a 3 ... = Areas of the strips,
x 1 , x 2 , x 3 ... = Depths of the centres of gravity of the corresponding strips
from the liquid surface.
∴ Pressure on the first strip =wa 1 x 1
Similarly, pressure on the second strip
=wa 2 x 2
and pressure on the third strip =wa 3 x 3 and so on
∴ Total pressure on the surface,
P=wa 1 x 1 + wa 2 x 2 + wa 3 x 3 + ....
=w (a 1 x 1 + a 2 x 2 + a 3 x 3 + ...)
= wAx
Fig. 35.5.