## HYDROSTATIC FORCES ON SURFACES

**Introduction: Remember the second law of Pascal**: In a container, pressure acts perpendicular to the boundary \in this lecture we will investigate how forces act on surfaces:

**Nature of plane or curved surface **

**Total force **

**Center of force**

**Hydrostatic Force on a Plane Surface:** When a surface is submerged in a fluid, forces develop on the surface due to the fluid. The determination of these forces is important in the design of storage tanks, ships, dams, and other hydraulic structures. For fluids at rest we know that the force must be perpendicular to the surface since there are no shearing stresses present. We also know that the pressure will vary linearly with depth if the fluid is incompressible. For a horizontal surface, such as the bottom of a liquid-filled tank, the magnitude of the resultant force is simply FR = pA, where p is the uniform pressure on the bottom and A is the area of the bottom. For the open tank shown, p = yh. Note that if atmospheric pressure acts on both sides of the bottom, as is illustrated, the resultant force on the bottom is simply due to the liquid in the tank. Since the pressure is constant and uniformly distributed over the bottom, the resultant force acts through the centroid of the area.

In equation form:

Where h = y sin e. For constant 'Y and θ(1)

The integral appearing in Eq. (1) is the first moment of the area with respect to the x axis, So we can write

Where Yc is the Y coordinate of the centroid measured from the x axis which passes through O. Equation (1) can thus be written as ;

**F**_{R}**= y**_{A}**y**_{c}**sin θ**

**Hydrostatic Force on a Curved Surface:** The equations developed for the magnitude and location of the resultant force acting on a submerged surface only apply to plane surfaces. However, many surfaces of interest (such as those associated with dams, pipes, and tanks) are no planar. Although the resultant fluid force can be determined by integration, as was done for the plane surfaces, this is generally a rather tedious process and no simple, general formulas can be developed. As an alternative approach we will consider the equilibrium of the fluid volume enclosed by the curved surface of interest and the horizontal and vertical projections of this surface.

For example, consider the curved section BC of the open tank of Fig. We wish to find the resultant fluid force acting on this section, which has a unit length perpendicular to the plane of the paper. We first isolate a volume of fluid that is bounded by the surface of interest, in this instance section BC, the horizontal plane surface AB, and the vertical plane surface AC. The free-body diagram for this volume is shown in Fig. The magnitude and location of forces F_{l} and F_{2}can be determined from the relationships for planar surfaces. The weight, OW, is simply the specific weight of the fluid times the enclosed volume and acts through the center of gravity (CO) of the mass of fluid contained within the volume. The forces F_{H} and F_{v} represent the components of the force that the tank exerts on the fluid.

In order for this force system to be in equilibrium, the horizontal component F_{H} must is equal in magnitude and collinear with F_{2}, and the vertical component F_{v }equal in magnitude and collinear with the resultant of the vertical forces F_{l} and OW. This follows since the three forces acting on the fluid mass (F_{2}, the resultant of F_{l} and OW, and the resultant force that the tank exerts on the mass) must form a concurrent force system. That is, from the principles of statics, it is known that when a body is held in equilibrium by three nonparallel forces they must be concurrent (their lines of action intersect at a common point), and coplanar.

**Thus, F**_{H}**= F**_{2}

** F**_{v}**= F**_{l}** W**