Hydrostatic Force on a Curved Surface

We encounter many arbitrarily shaped bodies immersed in liquids such as pipes and walls of containers. Forces on these may be calculated in the same manner as in the previous section. But the required integration of involved terms becomes very tedious. A more simplistic approach is to consider the forces resolved in the three coordinate directions separately. It may be noted that each component of force acts upon a projected area of the body. For example, force in x-direction will act normally on the area projected upon the y-z plane.

Figure 2.15 : Hydrostatic forces on a curved surface

 

Consider a curved surface as shown in Fig.2.15, immersed in a liquid. The resultant force $F_R$ can be resolved into two components - $F_H$ in the horizontal direction and $F_V$ in the vertical direction. We are considering a thin body which is two-dimensional and as such there is no force in the direction normal to the paper. The configuration can be split into two parts for discussion purposes- one, the part between the body and the free stream , $pqrs$ and two, the body itself, $poq$. Consider each part separately.

Block of fluid $pqrs$ is in equilibrium. The horizontal forces acting on it cancel out. Thus,

$$F_{HL1}~=~F_{HR1}$$ (2.43)

Similarly on the block $poq$ we have,

$$F_{HL2}~=~F_{HR2}~=~F_H$$ (2.44)

The vertical forces acting upon the fluid are (1) $F_A$ due to atmosphere, (2) $W_{pqrs}$, the weight of block $pqrs$ and (3) $W_{poq}$, the weight of block $poq$. Consequently,

$$F_V~=~F_A~+~W_{pqrs}~+~W_{poq}$$ (2.45)

Force $F_A$ is given by the atmospheric pressure times the projected area normal to it. The resultant force, $F_R$ is given by,

$$F_R~= \sqrt{F^2_H~+~F^2_V}$$ (2.46)

(c) Aerospace, Mechanical & Mechatronic Engg. 2005
University of Sydney