Fluids that are static ( at rest )

Figure shows a tank of water or other liquid open to atmosphere. As every diver knows, the pressure increases with depth below the air – water interface. The diver’s depth gauge, in fact , is a pressure sensore much like that of figure. As every mountaineer knows, the pressure decreases with altitude as one ascends into the atmosphere . The pressures encountered by the diver and the mountaineer are usually called hydrostatic pressures, because they are due to fluids that are static ( at rest ). Here we want to find an expression for hydrostatic pressure as a function of depth or altitude.

Let us look first at the increase in pressure with depth below the water’s surface. We set up vertical y axis in the tank, with its origin at the air – water ontreface and positive direction upward. We next consider a water sample contained in a imaginary right circular cylinder of horizontal base ( or face ) area A, such thay y₁ and y₂ ( both of which are negative numbers ) are the depths below the surface of the upper and lower cylinder faces, respectively.

Figure

shows a free- body diagram for the water in cylinder. The water is static equilibrium ; that is, it is stationary and the forces on it balance. Three forces act on it vertically: Force F₁ acts at the top surface of the cylinder and is due to the water above the cylinder. Similarly, force F₂ acts at the bottom surface of the cylinder and is due to the water just below the cylinder . The gravitational force on the water in the cylinder is represented by mg , where m is the mass

of the water in the cylinder . The balance of these forces is wriiten as

We want to transform equation into an equation involving pressures. From equation we know that

The mass m of the water in the cylinder is from

Where the cylinders volume V is the product of its face area A and its height y₁- y₂ . Thus, m is equal

Substituting this , we find

Figure

This equation can be used to find pressure both in liquid ( as a fuction of depth ) and in the atmosphere ( as a fuctin of altitude or height ). For the former , suppose we seek the pressure p at a depth h below the liquid surface . then we choose level 1 to be surface , level 2 to be a distance h below it, and p₀ to the represent the atmosphere pressure on the surface . We then substitute

Which becomes

Sample Problem

Balancing of pressure in U – tube

The U-tube in figure contains two liquids in static equilibrium: Water of density ( = 998 kg/m³ ) is in the right arm, and oil of unknown density is in the left . Measurement gives l = 135 mm and d = 12.3 mm. What is the density of the oil ?

The pressure p_int at the level of the oil – water interface in left arm depends on the density of oil and height of the oil above the interface . The water in the right arm at the same level must be at the same pressure p_int. The reason is that, because the water is in static equilibrium , pressure at points in the water at the same level must be the same even if points are separated horizontally.

Calculation : in the right arm, the interface is distance l below the free surface of the water, and we have