Fluids 101Fluids 101Chapter 10Chapter 10

FluidsFluids• Any material that flows and offers little resistance to changing its shape. –Liquids

–Gases

–Plasma?

DensityDensity• An important property of matter is density.

• Density is defined as mass per unit volume.

• The symbol is . m

V

Density of waterDensity of water

•1.0 g/ml •1.0 g/cm3

•1,000 kg/m3

PressurePressure

•Force per unit area.

ForcePressure

Area F

PA

PressurePressure•Measured in atm or Pascals (Pa)

21 1

NPa

m

PressurePressure•Standard atmospheric pressure = force per 1 m2 of air at sea level.

1.01x105 Pa or 101 kPa

Pressure Under Pressure Under WaterWater

•The pressure due to fresh water increases about 1 atm or 100 kPa for every 10 m.

Liquid PressureLiquid Pressure• Pressure of a liquid at rest

depends only on density and depth.

Pressure = density x depth

Water pressureWater pressure

•Water pressure is greater in the deeper lake.

Pascal’s PrinciplePascal’s Principle•The force exerted by a fluid on the walls of its container always acts perpendicular to the walls.

Pascal’s PrinciplePascal’s PrincipleWhen you are under water the water’s pressure pushes in on you from all sides. The force is perpendicular to your body.

Pascal’s PrinciplePascal’s Principle

F orces exerted by flu id on w a ll o f con ta iner a re p erpend icu la r a t every po in t.

Pascal VasesPascal Vases• The pressure of

a liquid is the same at any depth regardless of the shape of its container.

Pressure in a ColumnPressure in a Column

• The pressure in a fluid column can be found using this equation:

P gh

Pressure in a ColumnPressure in a Column

• AP test version of the same equation:

0p p gh 0p - this stands for the initial pressure.

ProblemProblem• What is the pressure

exerted by water at a depth of 45.0 m?

SolutionSolution0p p gh

P ghP = (1000 kg/m3)(9.8 m/s2)(45 m)

P = 4.41 x105 Pa or 441 kPa441 kPa

BuoyancyBuoyancy• Buoyancy is an upward force

that a fluid exerts on an object that is immersed in it. This is called the buoyant force.

buoyF Vg

Buoyant Buoyant ForcesForces

Archimedes PrincipleArchimedes Principle

• An object is buoyed up by a force equal to the weight of the fluid it displaces

• A brick in water weighs less than when in air. The difference (buoyant force) is equal to the weight of the water displaced.

Archimedes’ PrincipleArchimedes’ Principle

Apparent WeightApparent Weight• The apparent weight

(FA) is the difference between its actual weight and the buoyant force.

A BF w F

ProblemProblem• A cube of steel that measures 5.0

cm on each side is immersed in fresh water. The density of steel is 9.0 x 103 kg/m3. What is the buoyant force acting on the cube? What is the apparent weight of the steel?

Answer Answer pt 1pt 1

buoyF Vg

FB = (1000 kg/m3)(0.05 m)3(9.8m/s2)

FB = 1.23 N

Answer Answer pt.2

FW = (9.0 x 103 kg/m3)(0.05 m)3(9.8m/s2)

FA = W – FB

FA = 11.03 N – 1.23 N = 9.8 N9.8 N

m

V m = ρV

“Bob, do you think I’m sinking? Be honest”