Chapter 9 Fluid Mechanics Fluids A nonsolid state

Chapter 9 Fluid Mechanics

Fluids “A nonsolid state of matter in which the atoms or molecules are free to move past each other, as in a gas or liquid. ” (p. 318) Solids: definite shape and volume. Liquids: definite volume but no definite shape. Gases: no definite shape or volume; has volume and shape of container.

Mass Density “The mass per unit volume of a substance. ” (p. 319) Mass density = mass / volume ρ=m/V For mass, we will use grams (g) or kilograms (kg) and for volume we will use cm 3, liters (L) or milliliters (m. L). 1 m 3 = 1 x 106 cm 3 1 L = 1000 m. L 1 cm 3 = 1 m. L Common units of density are g / cm 3, g / m. L, and kg / m 3 Solids and liquids are almost incompressible, which means their densities do not change. Gases are compressible; so their densities depend on temperature and pressure.

Buoyant Force “A force that acts upward on an object submerged in a liquid or floating on a liquid’s surface. ” (p. 319) Archimedes Principle: “any object submerged in a fluid experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the object. ” (p. 320) Buoyant force = weight of displaced fluid FB = m f g (mf = mass of fluid displaced)

Floating Objects If an object floats, then its density is less than the density of the fluid (ρo < ρf ; ρo is the density of object and ρf is the density of the fluid) Recall that the weight of an object is Fg = mg. We will let mo = mass of the floating object. So, the weight of the object in air is Fg (object) = mog For a floating object, FB = Fg (object) mf g = mo g And, since m = ρV, ρf. Vf g = ρo. Vog See problem 3, p. 324, Practice 9 A

Sinking Objects… If an object sinks, then its density is greater than the density of the fluid (ρo > ρf ). An object submerged in a fluid has an “apparent weight” that is less then it’s normal weight: Fnet = FB – Fg (object) If an object sinks below the surface of a fluid, then the volume of the fluid displaced equals the volume of the object. So, Vf = Vo and we can replace both with just V: Fnet = FB – Fg (object) = ρf. Vf g– ρo. Vog = ρf. Vg– ρo. Vg A simple relationship results from the above equation Fg (object) / FB = ρo / ρf See problem #1, p. 324, Practice 9 A

Fluid Pressure and Temperature Pressure is “the magnitude of the force on a surface per unit of area. ” (p. 325) Pressure = Force / Area = F / A The SI (International Standard) unit for pressure is the Pascal (Pa), which is 1 N / m 2 Atmospheric pressure is 101, 000 Pa, which is equal to 1 atmosphere (atm). 1 atm = 14. 7 psi (pounds per square inch)

Pascal’s Principle Pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid and to the walls of the container. If pressure is increased at point 1 in the fluid, it increases by the same amount at point 2. Pressure increase = F 1 / A 1 = F 2 / A 2

Pressure as a function of depth Pressure increases with depth: Pressure = F / A = mg / A = ρVg / A = ρAhg / A = ρgh (gauge pressure) Absolute pressure = atmospheric pressure + ρgh = P 0 + ρgh