Fluids - StaticsLevel 1 Physics

Essential Questions and Objectives

Essential Questions What are the physical

properties of fluid states of matter?

What is pressure? Is the term “suction” real? What is an ideal fluid? What happens to velocity

when the area of an opening inc/dec?

Objectives Define atmospheric pressure, gauge

pressure and absolute pressure  and the relationship among these terms

Define and apply the concept of fluid pressure

State and apply Pascal's principle in practical situations such as  hydraulic lifts

State and apply Archimedes' principle to calculate the buoyant force

Demonstrate proficiency in accurately drawing and labeling free-body  diagrams involving buoyant force and other forces

State the characteristics of an ideal fluid

Apply the equation of continuity in solving problems

Understand that Bernoulli's equation is a statement of conservation of energy

Demonstrate proficiency in solving problems involving changes in  depth and/or changes in pressure and/or changes in velocity

Primary States of Matter

3 Primary States of Matter Solid – Definite

shape & volume Liquid – Takes shape

of container, still has definite volume

Gas – Takes shape and volume of container

The “fourth state” of matter is considered plasma. Matter at very high temperatures and pressures which typically occur on the Sun, or during re-entry from space

DensityDensity is an important factor that determines the behavior of a fluid

Density is the mass per unit volume of a substance. The variable given to density is the Greek letter ρ

€

ρ =m

V

ρ =kg

m3⇒ Units

PressureWhy should we be concerned about fluids?

Pressure is transmitted through liquid. Since liquid is effectively incompressible, pressure applied to a liquid is transmitted without loss throughout the liquid.

PressureAn important application in fluids would be the pressure of a fluid.

Pressure is defined as the amount of force per unit area

€

P =F

A⇒ Units

N

m2

ExampleA water bed is 2.0 m on a side an 30.0 cm deep.

a. Find its weight if the density of water is 1000 kg/m3.

b. Find the pressure the that the water bed exerts on the floor. Assume that the entire lower surface of the bed makes contact with the floor.

mgWV

m

V

m

Va

1000

30.022)

ρ

1.2 m3

1200 kg

11760 N

24

11760)

m

N

A

mg

A

FPb 2940 N/m2

Pressure continuedTwo important pieces of information

A fluid exerts a pressure in all directionsPressure acts perpendicularly to any surface.

Pressure and DepthAll 3 objects have pressureexerted on them that isperpendicular to the surface.

But notice that at the bottom, the magnitudeof the pressure is greater(as shown in the lengthof the arrow).

Pressure vs. Depth – Submerged Object

mg

Fatm

Fwater

An object is submerged just under the surface. In the FBD at the right, what are the 3 forces acting on the object?

1. The weight (mg) the object

2. The force of the atmosphere pressing down (Fatm)

3. The force of the water pushing up (Fwater)

€

Fwater = Fatm + mg

Pressure vs. Depth Equation

Remember the pressure equation

€

P =F

A

€

Fwater = Fatm + mg

€

PA = PoA + mg

ρ =m

Vm = ρV

PA = PoA + ρVg

V = Ah

PA = PoA + ρAhg

P = Po + ρgh

*Po is atmospheric pressure.It has a constant value of 1.013 x 105 Pa

An in depth look

€

P = Po + ρghDepth belowsurface

Initial Pressure – may be atmospheric1.013 x 105 Pa

ABSOLUTE PRESSURE

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ΔP = ρghGAUGE PRESSURE– change in pressure

Examplea) Calculate the absolute pressure at an ocean depth of 1000

m. Assume that the density of water is 1000 kg/m3 and that Po= 1.01 x 105 Pa (N/m2).

b) Calculate the total force exerted on the outside of a 30.0 cm diameter circular submarine window at this depth.

P

xP

ghPP o

)1000)(8.9)(1000(101 5

ρ 22 )15.0(

F

r

F

A

FP

9.9x106 N/m2 7.0 x 105 N

A closed systemIf you take a liquid and place it

in a system that is CLOSED like plumbing for example or a car’s brake line, the PRESSURE is the same everywhere.

Since this is true, if you apply a force at one part of the system the pressure is the same at the other end of the system. The force, on the other hand MAY or MAY NOT equal the initial force applied. It depends on the AREA.

You can take advantage of the fact that the pressure is the same in a closed system as it has MANY applications.

The idea behind this is called PASCAL’S PRINCIPLE

Pascal’s Principle

BuoyancyWhen an object is immersed in a fluid, such as a liquid, it is buoyed UPWARD by a force called the BUOYANT FORCE.

When the object is placed in fluid is DISPLACES a certain amount of fluid. If the object is completely submerged, the VOLUME of the OBJECT is EQUAL to the VOLUME of FLUID it displaces.

Archimedes’ Principle" An object is buoyed up by a force equal to the weight of the fluid displaced."

In the figure, we see that the difference between the weight in AIR and the weight in WATER is 3 lbs. This is the buoyant force that acts upward to cancel out part of the force. If you were to weight the water displaced it also would weigh 3 lbs.

Archimedes’ Principle

Fluidobject

FluidB

FLUIDB

VV

VgF

VmmgF

)(

)(

ρρ

ExampleA bargain hunter purchases a "gold" crown at a flea market. After

she gets home, she hangs it from a scale and finds its weight in air to be 7.84 N. She then weighs the crown while it is immersed in water (density of water is 1000 kg/m3) and now the scale reads 6.86 N. Is the crown made of pure gold if the density of gold is 19.3 x 103 kg/m3?

object

objectobject

object

object

fluid

fluidfluidFluidB

B

buoyantwaterobjectairobject

V

m

mass

V

V

gVmgF

F

FFF

ρ

ρ)(

86.684.7

)()(

0.98 N

0.0001 m3

0.0001 m3

0.80 kg

8000 kg/m3

NO! This is NOT gold as 8000<19300