Mechanics of Fluids

which has different depths of water on the two sides. The surface may extend
up through the liquid and into the atmosphere; only the part below the free
surface then has a net hydrostatic thrust exerted on it.
The pressures we have considered have been expressed as gauge pressures.
It is unnecessary to use absolute pressures because the effect of atmospheric
pressure at the free surface is to provide a uniform addition to the gauge
pressure throughout the liquid, and therefore to the force on any surface
in contact with the liquid. Normally atmospheric pressure also provides a
uniform force on the other face of the plane, and so it has no effect on either
the magnitude or position of the resultant net force.
It should be particularly noted that, although the total force acts at the
centre of pressure, its magnitude is given by the product of the area and the
pressure at the centroid.
Example 2.2 A cylindrical tank 2 m diameter and 4 m long, with
its axis horizontal, is half filled with water and half filled with oil of
density 880 kg · m −3 . Determine the magnitude and position of the net
hydrostatic force on one end of the tank.
Solution
We assume that the tank is only just filled, that is, the pressure in
the fluids is due only to their weight, and thus the (gauge) pressure
at the top is zero. Since two immiscible fluids are involved we must
consider each separately. In equilibrium conditions the oil covers the
upper semicircular half of the end wall.
Since the centroid of a semicircle of radius a is on the central
radius and 4a/3π from the bounding diameter, the centroid Co of
the upper semicircle is 4(1 m)/3π = 0.4244 m above the centre of
the tank, that is, (1 − 0.4244) m = 0.5756 m from the top. The
pressure of the oil at this point is
ϱgh = (880 kg · m −3 )(9.81 N · kg −1 )(0.5756 m) = 4969 Pa
Hydrostatic thrusts on submerged surfaces 63