Basics of Fluid Mechanics, 2014a

110 CHAPTER 4. FLUIDS STATICS
dA, outward as positive. The total
force on the area will be the
integral of the unit force
∫
F = − P ˆn dA (4.139)
A
The result of the integral is a vector. So, if the y component of the force is needed,
only a dot product is needed as
dF y = dF • ĵ (4.140)
From this analysis (equation (4.140)) it can be observed that the force in the direction
of y, for example, is simply the integral of the area perpendicular to y as
∫
F y =
The same can be said for the x direction.
The force in the z direction is
∫
F z =
The force which acting on the z direction
is the weight of the liquid above the
projected area plus the atmospheric pressure.
This force component can be combined
with the other components in the
other directions to be
√
F total = F 2 z + F 2 2
x + F y (4.143)
And the angle in “xz” plane is
tan θ xz = F z
F x
(4.144)
A
A
PdA y (4.141)
h g ρdA z (4.142)
only the
liquid above
the body
affecting
the body
and the angle in the other plane, “yz”is
tan θ zy = F z
F y
(4.145)
Fig. -4.27. Schematic of Net Force on floating
body.
The moment due to the curved surface require integration to obtain the value. There
are no readily made expressions for these 3–dimensional geometries. However, for some
geometries there are readily calculated center of mass and when combined with two
other components provide the moment (force with direction line).