(^736) „„„„„ A Textbook of Engineering Mechanics
Thus the intensity of pressure on the wall per unit length,
= Area of triangle ABC

1 2
22
wH
HwH×=
∴Total pressure on the wall of length (l),
P =
2
2
wH
l×
This total pressure will act at the centre of gravity of the triangle, i.e., at a depth of
2
3
H
from
the liquid surface, or at a height of
3
H
from the bottom of the liquid.
Example 35.14. A water storage tank has one of its vertical side 10 m long. Find the
pressure exerted on this wall, when the water is 1.5 m deep. Also find the point where the pressure
acts.
Solution. Given: Length of vertical side of tank (l) = 10 m and depth of water (H) = 1.5 m
Total pressure exerted on the wall.
We know that the pressure BC
=w H = 9.8 × 1.5 = 14.7 kN/m^2
∴ Total pressure per metre length of the storage tank
= Area of triangle ABC

1
1.5 14.7 11.03 kN
2
×× =
and total pressure exerted on the 10 m long wall,
P = 10 × 11.03 = 110.3 kN Ans.
Point where the pressure acts
We know that the point, where the pressure acts, is the centre of gravity of the triangle ABC.
This point is at depth of h i.e.
21.5
3
×
= 1 m from A (or at a height of
1.5
3
= 0.5 m from the bottom
of the water). Ans.
35.16.PRESSURE DIAGRAM DUE TO ONE KIND OF LIQUID OVER ANOTHER
ON ONE SIDE
Consider a vertical wall, subjected to pressure due to
one kind of liquid, over another, on one side as shown in Fig.
35.21. This will happen, when one liquid is insoluble into
the other.
Let w 1 = Specific weight of liquid 1
H 1 = Height of liquid 1, and
w 2 , H 2 = Corresponding values for liquid 2,
We know that the pressure in such a case will be zero
at the upper liquid surface, and will increase by a straight
line law to (w 1 H 1 ) up to a depth of H 1. It will further
increase, by a straight line law, to (w 1 H 1 + w 2 H 2 ) as shown in
Fig. 35.18.
Fig. 35.20.
Fig. 35.21. Pressure diagram due to one
kind of liquid over another.