Transcript Fluids - Statics

Fluids - Statics
Level 1 Physics
Essential Questions and
Objectives
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Essential Questions
 What are the physical properties of
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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?
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Objectives
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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
Density
Density 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
m

V
kg
  3  Units
m
Pressure
Why 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.
Pressure
An important application in fluids would be the pressure of a fluid.
Pressure is defined as the amount of force per unit area
F
N
P   Units 2
A
m
Example
A 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.
a) V  2  2  0.30 
m
m
   1000  
V
V
W  mg  11760 N
1.2 m3
1200 kg
F mg 11760 N
b) P  


2
A
A
4m
2940 N/m2
Pressure continued
 Two important pieces of information
 A fluid exerts a pressure in all directions
 Pressure acts perpendicularly to any surface.
Pressure and Depth
All 3 objects have pressure
exerted on them that is
perpendicular to the
surface.
But notice that at the
bottom, the magnitude
of the pressure is greater
(as shown in the length
of the arrow).
Pressure vs. Depth –
Submerged Object
Fatm
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)
mg
Fwater
3. The force of the water pushing
up (Fwater)
Fwater  Fatm  mg
Pressure vs. Depth Equation
Remember the pressure equation
Fwater  Fatm  mg
F
P
A
PA  Po A  m g
m

m  V
V
PA  Po A  Vg
V  Ah
PA  Po A  Ahg

P  Po  gh
*Po is atmospheric pressure.
It has a constant value of
1.013 x 105 Pa

An in depth look
P  Po  gh
Depth below
surface
Initial Pressure – may be atmospheric
1.013 x 105 Pa
ABSOLUTE PRESSURE
P  gh
GAUGE PRESSURE– change
in pressure
Example
a) 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  Po  gh
P  1x105  (1000)(9.8)(1000)
P
9.9x106 N/m2
F
F
F
P  2 

2
A r
 (0.15)
7.0 x 105 N
A closed system
If 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
Buoyancy
When 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
FB  (mg ) FLUID m  V
FB  ( Vg ) Fluid
Vobject  VFluid
Example
A 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?
Fobject ( air)  Fobject ( water )  Fbuoyant
7.84  6.86  FB 
0.98 N
FB  (mg ) Fluid   fluidV fluid g
V fluid 
0.0001 m3
Vobject 
0.0001 m3
massobject 
 object 
0.80 kg
mobject
Vobject

8000 kg/m3
NO! This is NOT gold as 8000<19300