1. Solids and Fluids Chapter 9

2. Phases of Matter  Solid – definite shape and volume  Liquid – definite volume but assumes the shape of its container  Gas – assumes the shape and volume of its container.  Liquids and Gases are collectively referred to as FLUIDS.  Fluid – A substance that can flow.

3. Fluid Pressure  Pressure = Force / Area  Units of [Newtons/m 2 ] or [pounds/in 2 ]  p= F / A to find pressure, use the component of force normal to the surface area  Pressure is commonly measured in [Pascals] = [Pa] = [N/m 2 ]

4. Density  Density = Mass / Volume  Density is measured in [kg/m 3 ] or [g/cm 3 ]  Density of Water = 1 g/cm 3 = 1000 kg/m 3  Density is symbolized with ρ

5. Pressure and Depth  Water pressure increases with depth.  Water Pressure = F/A mg/A (ρhA)g/A  Water pressure at height, h isρgh [N/m 2 ] height, h isρgh [N/m 2 ]

6. Water Pressure and Depth  Pressure of a liquid at a depth of h is p = p o + ρgh where p o is the pressure at the surface of the liquid.

7. Water Pressure and Depth  p o is pressure at the surface  Often surface pressure is air pressure, p a  Standard air pressure at sea level is [1 atmosphere] = [1 atm] = [1.013 X 10 5 N/m 2 ] = [101.3 kPa]

8. Scuba Diver Example  What is the total pressure on the back of a scuba diver in a lake at a depth of 8.0 meters?  What is the force on the divers back due to the water alone, taking the surface of the back to be a rectangle 60.0 cm by 50.0 cm?

9. Pascal’s Principle  Pressure which is applied to an enclosed incompressible fluid is transmitted to every point in the fluid and to the walls of the container.

10. Pascal’s Principle  Pressure applied to an incompressible fluid is transmitted instantaneously throughout the fluid.  Pressure is the same throughout.

11. Pascal’s Principle  Hydraulic systems make use of Pascal’s Principle.  An applied force can be multiplied and made to lift a large load.  (F/A) remains constant

12. Example  A garage lift has input and lift (output) pistons with diameters of 10cm and 30 cm. The lift is used to hold up a car with a weight of 1.4 X 10 4 N. a) what is the magnitude of the input force on the piston? b) what pressure is applied to the input piston?

13. Archimedes Principle  A body immersed wholly or partially in a fluid experiences a buoyant force equal in magnitude to the weight of the volume of fluid that is displaced.

14. Examples  A spherical helium filled balloon was a radius of 1.10m. Does the buoyant force on the balloon depend on the density of 1) helium 2) air or 3) the weight of the rubber skin?  Compute the magnitude of the buoyant force on the balloon. ρ air = 1.29 kg/m 3 and ρ He = 0.18 kg/m 3.  If the rubber skin of the balloon has mass of 1.2 kg, find the balloon’s initial acceleration when released if it carries a load of 3.52 kg.

15. Fluid Flow Conditions of molecules in a flowing fluid may be unpredictable - difficult to quantify. Therefore it is helpful to identify several conditions of an ideal fluid: Condition 1: Steady flow means that all particles have the same velocity as they pass a given point. Condition 2: Irrotational flow means that a fluid element has not angular velocity. (no whirlpools) Condition 3: Nonviscous flow means viscosity is negligible. Condition 4: Incompressible flow means the fluid’s density is constant.

16. Continuity of Fluid Flow Consider fluid flowing in a tube with different diameters…

17. Equation of Continuity  Mass of fluid flowing into the tube in a given time must equal the mass flowing out of the tube…  Δm 1 = ρV 1 = ρA 1 Δx 1 = ρA 1 v 1 Δt  Δm 2 = ρV 2 = ρA 2 Δx 2 = ρA 2 v 2 Δt  Since Δm 1 = Δm 2, A 1 v 1 = A 2 v 2  Equation of Continuity: A 1 v 1 = A 2 v 2

18. Bernoulli’s Equation  Work – Energy Formula become Bernoulli’s Equation:  W net = ΔKE + ΔU  p 1 + ½ρv 1 2 + ρgy 1 = p 2 + ½ρv 2 2 + ρgy 2

19. Bernoulli’s Equation Special Cases  p 1 + ½ρv 1 2 + ρgy 1 = p 2 + ½ρv 2 2 + ρgy 2  p + ½ρv 2 + ρgy = constant  If the fluid is at rest, then the formula becomes the pressure depth relationship: p 2 – p 1 = ρg(y 2 – y 1 ) p 2 – p 1 = ρg(y 2 – y 1 )  If y 1 = y 2, then p 1 + ½ρv 1 2 = p 2 + ½ρv 2 2 which says if the velocity of a fluid increases, the pressure it exerts decreases.

20. Examples  A cylindrical tank containing water has a small hole punched in its side below the water level, and water runs out. What is the approximate initial flow rate of water out of the tank?