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Announcements
Physics 131: Lecture 23


Today’s Agenda

Description of Fluids at Rest


 Pressure vs Depth

Pascal’s Principle: hydraulic forces

Archimedes Principle: objects in a fluid

Bernoulli’s equation



Today: Last new material
Wednesday in class (Dr. Harlow and I will
review the semester)
Final Exam: Thursday 12/13 at 2:00pm
Review Session:
 Thursday 12/6 10:00am
10:00am-12pm
12pm in SS2118
(Dr. Harlow and I) Exam Jam
My extra office hours:
 12/11 9am – 12 and 2 – 5pm
 12/12 9am – 12 and 2 – 5pm
Drop in center: Check link on portal
Conceptual Questions (mastering physics)
Physics 201: Lecture 1, Pg 1
Physics 201: Lecture 1, Pg 2
Fluids
States of Matter
What parameters do we use to describe fluids?

Fluids


Solid
 Hold Volume
 Hold Shape
Liquid
 Hold
H ld V
Volume
l
 Adapt Shape
Gas
 Adapt Volume
 Adapt Shape
Density

Pressure
m
V
P
F
A
These letters look similar! Be careful.
Physics 201: Lecture 1, Pg 3
Physics 201: Lecture 1, Pg 4
You are my density
Pressure

The density of a material is its
mass per unit volume:


Physics 201: Lecture 1, Pg 5
Pressure = Force/Area
Units SI
Pascal = Newtons/m2

If I apply 30 Pa of pressure over an area of
20 m2, how much force do I apply?

F = P  A = 30 N/m2  20 m2 = 600 Newtons
Physics 201: Lecture 1, Pg 6
Page 1
Pressure

Pressure
The same force applied over a smaller area
results in greater pressure – think of poking a
balloon with your finger and then with a
needle.
Physics 201: Lecture 1, Pg 7
Physics 201: Lecture 1, Pg 8
Pressure vs. Depth
Atmospheric Pressure



Basically weight of atmosphere!
Air molecules are colliding with you right now!
Pressure = 1x105 N/m2 = 14.7 lbs/in2!
Example: Magdeburg sphere w/ r =
0.1 m
A = 4 p r2 = 0.125 m2
F = 12,000 Newtons
(over 2,500 lbs)!
Physics 201: Lecture 1, Pg 9
Physics 201: Lecture 1, Pg 10
Gauge Pressure
Clicker Question 1:
Many pressure gauges, such as tire gauges and the
gauges on air tanks, measure not the actual or absolute
pressure p but what is called gauge pressure pg.
What can you say about the
pressures at points
1, 2, and 3?
A. p1 = p2 = p3.
B. p1 = p2 > p3.
C p3 > p1 = p2.
C.
D. p3 > p1 > p2.
E. p1 = p3 > p2.
where 1 atm = 101.3 kPa.
Physics 201: Lecture 1, Pg 11
Physics 201: Lecture 1, Pg 12
Page 2
Pascal’s Principle
Pascal’s Principle
Pascal’s principle:
An external pressure applied to an enclosed fluid is transmitted
unchanged to every point within the fluid.

Consider the system shown:
 A downward force F1 is applied to
the piston of area A1.
 This force is transmitted through the
liquid to create an upward force F2.
 Pascal’s Principle says that
increased pressure from F1 (F1/A1) is
transmitted throughout the liquid.
F1
F
 2
A1 A 2
F1
F2
d2
d1
A1
A2
A
F2  F1 2
A1
Check that F•d is the same on both sides.
Energy is conserved!
Physics 201: Lecture 1, Pg 13
Physics 201: Lecture 1, Pg 14
Clicker Question 2:
Clicker Question 3:
The small piston of a hydraulic lift has a cross sectional
area of 3 cm2 . You need to lift a 15,000 N weight by
applying a force of only 200 N.
What is the area of the large piston, assuming both
pistons are at the same height, as shown in the
drawing?
If you push the small piston down a distance of 25 cm.
How far up does the weight move?
(a) 0.3 cm
(b) 8.2 cm
(c) 25 cm
(a) 75 cm2
(b) 125 cm2
(c) 225 cm2
Physics 201: Lecture 1, Pg 15
Physics 201: Lecture 1, Pg 16

When the cylinder is lowered into the water,
how will the scale readings for the cylinder and
the water change?
Suppose we weigh an object in air and
in water.
Since the pressure at the bottom of the object is
greater than that at the top of the object, the water
exerts a net upward force, the buoyant force, on the
object.
The buoyant force is equal to the difference in
th pressures times
the
ti
the
th area.


a)
b)
c)
d)
e)
W2?
W1
Archimedes’ Principle
Clicker Question 4:
cylinder’s scale will decrease, but water’s scale will increase
cylinder’s scale will increase, but water’s scale will decrease
cylinder’s scale will decrease, but water’s scale will not change
both scales will decrease
W1
W2?
both scales will increase
FB  F2  F1
FB  ( p2  p1 )  A  ghA
FB  liquid gVliquid  M liquid  g  Wliquid
y1
F
1
y2
p
1
A
p
2
F
2
Therefore, the buoyant force is equal to
the weight of the fluid displaced.
Physics 201: Lecture 1, Pg 17
Physics 201: Lecture 1, Pg 18
Page 3
Clicker Question 5:
Archimedes’ Principle
Imagine holding two identical bricks in place underwater. Brick 1 is
just beneath the surface of the water, and brick 2 is held about 2
feet down. The force needed to hold brick 2 in place is:

So the force the water pushes up with is equal to
weight of the water displaced by the object.
FB = Wfluid

FB = mfluidg = (Vg)fluid

a) greater
b) the same
1
c) smaller
2
Physics 201: Lecture 1, Pg 19
Physics 201: Lecture 1, Pg 20
Clicker Question 5.5:
Will it Float?
Object is in equilibrium
Two blocks are of identical size. One is made of
lead and sits on the bottom of a pond; the other is
of wood and floats on top. Upon which is the
buoyant force greater?
y
FB  (mg ) object
FBmg
liquid  g  Vdispl.  object  g  Vobject
Vdispl.
displ
Vobject
A. On the lead block.
B. On the wood block.
C. They both experience the same buoyant force.

object
liquid
The Tip of The Iceberg: What fraction of an iceberg is submerged?
Vwater displ.
Vice
Physics 201: Lecture 1, Pg 21

ice
917 kg/m3

 90%
 water 1024 kg/m3
Physics 201: Lecture 1, Pg 22
Archimedes’ Principle

Clicker Question 6:
An object made of material that is denser than
water can float only if it has indentations or
pockets of air that make its average density less
than that of water.
A plastic cube (ρ = 780 kg/m3, V = 0.71 m3) is suspended
by a string such that half the volume is submersed in
water (ρ = 1000 kg/m3) as shown in the figure.
What is the tension T in the string?
(a)
(b)
(c)
(d)
(e)
Physics 201: Lecture 1, Pg 23
372 N
428 N
931 N
1950 N
2380 N
Physics 201: Lecture 1, Pg 24
Page 4
Clicker Question 7:
Clicker Question 8:
Suppose you float a large ice-cube in a glass of water,
and that after you place the ice in the glass the level of
the water is at the very brim. When the ice melts, the
level of the water in the glass will:
Which weighs more:
A. A large bathtub filled to the brim with water.
B. A large bathtub filled to the brim with water with a
battle-ship floating in it.
C. They will weigh the same.
A. Go up,
A
up causing the water to spill out of the glass
glass.
B. Go down.
C. Stay the same.
Tub of water
Overflowed water
Physics 201: Lecture 1, Pg 25
Physics 201: Lecture 1, Pg 26
Fluids in motion
Example






Water comes out of a hose which has a radius of
1 cm. You are holding your finger over the hose
and the area the hose can flow through is 2.01 
10-4 m2. If water in hose has speed 1m/s, what is
speed
p
when water p
passes yyou finger?
g
Assume
the water is incompressible.
1A1v1 = 2A2v2
1 = 2
A1v1 = A2v2
(.01 m)2(1 m/s) = (2.01  10-4 m2) v2
v2 = 1.56 m/s
Physics 201: Lecture 1, Pg 27
Physics 201: Lecture 1, Pg 28
Bernoulli’s equation
Bernoulli’s equation
When a fluid moves from a wider area of a pipe to a narrower one, its
speed increases; therefore, work has been done on it.


W = Fd = P(Ad) = (P)V

W =½mvf2 - ½mvi2
W = Fd = P(Ad) = (P)V
So:
 ((P2 – P1))V = ½mvf2 - ½mvi2
 m= V
 (P2 – P1)V = ½ Vvf2 - ½ Vvi2

(P2 – P1) = ½ v12 - ½ v22

P2 + ½ v22 = P1 + ½ v12
P2+gy2 + ½v22 = P1+gy1 + ½ v12
Physics 201: Lecture 1, Pg 29
Physics 201: Lecture 1, Pg 30
Page 5
Bernoulli’s equation
Bernoulli’s equation
The general case, where both height and speed may change, is
described by Bernoulli’s equation:
The kinetic energy of a fluid element is:
Equating the work done to the increase in kinetic energy gives:
This equation is essentially a statement of conservation of energy in a
fluid.
Physics 201: Lecture 1, Pg 31
Physics 201: Lecture 1, Pg 32
Applications of Bernoulli’s equation

Bernoulli’s equation (at constant height)




P2 + ½ v22 = P1 + ½ v12
P + ½ v2
 This quantity must always remain the same
So fluids with faster velocity will exert less
pressure!
This has many interesting applications!
The Bernoulli effect is simple to demonstrate – all
you need is a sheet of paper. Hold it by its end,
so that it would be horizontal if it were stiff, and
blow across the top. The paper will rise, due to
the higher speed, and therefore lower pressure,
above the sheet.
Demos!
Physics 201: Lecture 1, Pg 33
Physics 201: Lecture 1, Pg 34
Applications of Bernoulli’s equation
Applications of Bernoulli’s equation:
Airplane wing

Physics 201: Lecture 1, Pg 35
This lower pressure at high speeds is what rips
roofs off houses in hurricanes and tornadoes,
and causes the roof of a convertible to expand
upward. It even helps prairie dogs with air
circulation!
Physics 201: Lecture 1, Pg 36
Page 6
Clicker Question 9:
Oil (ρo = 920 kg/m3) flows through a horizontal cylindrical
pipe of radius R1 = 12 cm with a speed of v1 = 2 m/s.
Along the flow direction the pipe widens to a radius R2
= 24 cm (as shown in the figure), the speed of the oil
is then equal to v2.
What is the pressure difference ∆P = P2 – P1 between
the wide and narrow parts of the pipe?
(a) ∆P = 660 Pa
(b) ∆P = 1185 Pa
(c) ∆P = 1520 Pa
(d) ∆P = 1725 Pa
(e) ∆P = 1950 Pa
Physics 201: Lecture 1, Pg 37
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