Relative Equilibrium of Fluids

RELATIVE EQUILIBRIUM OF FLUIDS The treatment will center on a uniformly accelerated body of fluid wherein each particle is in equilibrium and at rest relative to every other particle in the liquid and to the containing vessel, is neither in equilibrium nor at rest relative to the earth. Such condition is called Relative Equilibrium. Mass of liquid in relative equilibrium is free from any shear stress due to the absence of any motion between each particle and also between the liquid mass itself and the container. Two cases of relative equilibrium will be treated here: (1) the translation of liquid mass horizontally and vertically, and (2) the rotation of liquid mass. A. HORIZONTAL DIRECTION To determine the position of the free surface and the pressure distribution or variation in the translated mass of liquid, we proceed as follows: i. First, we consider an elemental mass of liquid on the free surface at point B. the forces which act on this element are the gravity force W which is directed vertically downward and the pressure force P directed away from and normal to the surface (since there is no shear stress). In accordance with the D’Alembert’s principle, an imaginary force, (W/g) a, called the reversed effective force or inertia force, may be added to the system to reduce into one in which the laws of statics are applicable. And so,

Simultaneously solution of this resulting equation gives

ii. Second, let us consider the prism (figure 1) having the length h and a uniform cross-sectional area dA with end-points at O and 1. The forces acting on its prism are its weight WP and the pressure force F1 at the base of the prism. Since there is no vertical motion, and

where p1 is the intensity of pressure at the base h units from the free surface. Hence, as in static fluids, the pressure in a horizontally accelerated liquid mass is constant in a plane parallel to the free surface. Hence, as in static fluids, the pressure in a horizontally accelerated liquid mass is constant in a plane parallel to the free surface. (see figure 2) REMARKS: i. If the horizontal translation does not cause any spillage of the liquid from the container, the new surface M’N’ will exactly bisect the original surface MN, that is, the volume of liquid originally occupying

NAN’ will be displaced to occupy an equal space MAM’. ii. Since the pressure variation still follows that of static fluids, or p = wh, the forces F1 and F2 (see figure 2) on the ends of the container may be calculated by making use of equation F = Awh . The difference between F1 and F2 is the unbalanced force that produces the acceleration of the entire mass and so

B. VERTICAL DIRECTION In the open container (figure 3) which is shown to be subjected to a vertical upward acceleration a, the pressures in the contained liquid will obviously be greater than those existing in the same mass of liquid when at rest.

Motion Going Up

Motion Going Down

C. INCLINED MOTION

Motion Going Up

Motion Going Down

Example # 1. An open tank 1.82 m. square, weighs 3425 N and contains 0.91 m of water. It is acted by an unbalanced force of 10400 N parallel to a pair of sides. a) Find the acceleration of the tank b) What is the force acting on the side with the greatest depth? c) What is the force acting on the side with the smallest depth? Example # 2. An open horizontal tank 2 m high, 2 m wide and 4 m long is full of water. 2 a) How much water is spilled out when the tank is accelerated horizontally at 2.45 m/sec in a direction parallel with its longest side? b) What is the force acting on the side with the greatest depth? c) Compute the required accelerating force. Example # 3: The tank shown in the figure is accelerated to the right. Width of the tank is 1 m. a) Determine the acceleration needed to cause the free surface to touch point A. b) Determine the pressure at B. c) Determine the total force acting on the bottom of the tank. Example # 4: An open tank moves up an inclined plane as shown with constant acceleration. a) Compute the angle that the water surface in the tank makes with the horizontal line. b) Calculate the acceleration required for the water surface to move to the position indicated. c) Compute the vertical component of the acceleration Example # 5: The container shown has a width of 500 mm and contains water at a depth of 300 mm. a) At what angle will the water surface makes with the horizontal when the tank accelerated to the right so that water will just touch point A. b) Determine the acceleration of the tank to the right at this instant. c) What is the maximum pressure at point B of the tank at this instant.