PHYS 1441 – Section 004
Lecture #21
Monday, Apr. 19, 2004
Dr. Jaehoon Yu
•
•
•
•
Buoyant Force and Archimedes’ Principle
Flow Rate and Equation of Continuity
Bernoulli’s Equation
Simple Harmonic Motion
Quiz next Wednesday, Apr. 28!
Monday, Apr. 19, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
1
Announcements
• Quiz stats
– Average : 48.4/90  53.8/100
– Top score: 80/90
– Previous quiz scores: 38.2, 41, 57.9
• Next quiz on Wednesday, Apr. 28
– Covers: ch. 10.4 – ch.11, including all that are
covered in the lecture
• Mid-term grade one-on-one discussion
– 11 have not done this yet
– One more opportunity today
Monday, Apr. 19, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
Buoyant Forces and Archimedes’ Principle
Why is it so hard to put a beach ball under water while a piece of small
steel sinks in the water?
The water exerts force on an object immersed in the water.
This force is called Buoyant force.
How does the
The magnitude of the buoyant force always equals the weight of
Buoyant force work? the fluid in the volume displaced by the submerged object.
This is called, Archimedes’ principle. What does this mean?
Let‘s consider a cube whose height is h and is filled with fluid and at its
equilibrium. Then the weight Mg is balanced by the buoyant force B.
B  Fg  Mg And the pressure at the bottom of the
cube is larger than the top by rgh.
h
Mg B
Monday, Apr. 19, 2004
Therefore, P  B / A  rgh
Where Mg is the
B  PA  rghA  rVg
weight of the fluid.
B  Fg  rVg  Mg
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
3
More Archimedes’ Principle
Let’s consider buoyant forces in two special cases.
Case 1: Totally submerged object Let’s consider an object of mass M, with density r0, is
immersed in the fluid with density rf .
The magnitude of the buoyant force is
B  r f Vg
The weight of the object is Fg  Mg  r 0Vg
h
Mg B


Therefore total force of the system is F  B  Fg  r f  r 0 Vg
What does this tell you?
Monday, Apr. 19, 2004
The total force applies to different directions depending on the
difference of the density between the object and the fluid.
1. If the density of the object is smaller than the density of
the fluid, the buoyant force will push the object up to the
surface.
2. If the density of the object is larger that the fluid’s, the
object will sink to the bottom of the fluid.
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
4
More Archimedes’ Principle
Case 2: Floating object
h
Mg B
Let’s consider an object of mass M, with density r0, is in
static equilibrium floating on the surface of the fluid with
density rf , and the volume submerged in the fluid is Vf.
The magnitude of the buoyant force is
The weight of the object is
Therefore total force of the system is
Since the system is in static equilibrium
B  r fVf g
Fg  Mg  r 0V0 g
F  B  Fg  r f V f g  r 0V0 g
0
r f V f g  r0V0 g
r0 V f

rf
V0
What does this tell you?
Monday, Apr. 19, 2004
Since the object is floating its density is always smaller than
that of the fluid.
The ratio of the densities between the fluid and the object
determines the submerged volume under the surface.
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
5
Example for Archimedes’ Principle
Archimedes was asked to determine the purity of the gold used in the crown.
The legend says that he solved this problem by weighing the crown in air and
in water. Suppose the scale read 7.84N in air and 6.86N in water. What
should he have to tell the king about the purity of the gold in the crown?
In the air the tension exerted by the scale on
the object is the weight of the crown
Tair  mg  7.84 N
Twater  mg  B  6.86 N
In the water the tension exerted
by the scale on the object is
B  Tair  Twater  0.98N
Therefore the buoyant force B is
Since the buoyant force B is
B  r wVw g  r wVc g  0.98N
The volume of the displaced
0.98 N
0.98
4
3

V

V


1
.
0

10
m
c
w
water by the crown is
r w g 1000  9.8
Therefore the density of
mc mc g 7.84
7.84
3
3





8
.
3

10
kg
/
m
r

4
the crown is
c V
Vc g Vc g 1.0 10  9.8
c
3kg/m
3, this Spring
Monday,the
Apr.density
19, 2004of pure gold is 19.3x10
PHYS
1441-004,
6
Since
crown2004
is either not made of pure gold or hollow.
Dr. Jaehoon Yu
Example for Buoyant Force
What fraction of an iceberg is submerged in the sea water?
Let’s assume that the total volume of the iceberg is Vi.
Then the weight of the iceberg Fgi is
Fgi  riVi g
Let’s then assume that the volume of the iceberg
submerged in the sea water is Vw. The buoyant force B
caused by the displaced water becomes
B  r wVw g
Since the whole system is at its
static equilibrium, we obtain
Therefore the fraction of the
volume of the iceberg
submerged under the surface of
the sea water is
riVi g  r wVw g
Vw r i 917kg / m3


 0.890
3
Vi r w 1030kg / m
About 90% of the entire iceberg is submerged in the water!!!
Monday, Apr. 19, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
7
Flow Rate and the Equation of Continuity
Study of fluid in motion: Fluid Dynamics
If the fluid is water: Water dynamics?? Hydro-dynamics
•Streamline or Laminar flow: Each particle of the fluid
Two main
types of flow follows a smooth path, a streamline
•Turbulent flow: Erratic, small, whirlpool-like circles called
eddy current or eddies which absorbs a lot of energy
Flow rate: the mass of fluid that passes a given point per unit time m / t
m1
r1V1 r1 A1l1


 r1 A1v1
t
t
t
since the total flow must be conserved
m1 m2
r1 A1v1  r 2 A2 v2

t
t
Equation of Continuity
Monday, Apr. 19, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
8
Example for Equation of Continuity
How large must a heating duct be if air moving at 3.0m/s along it can
replenish the air every 15 minutes, in a room of 300m3 volume?
Assume the air’s density remains constant.
Using equation of continuity
r1 A1v1 r 2 A2 v2
Since the air density is constant
A1v1  A2 v2
Now let’s imagine the room as
the large section of the duct
A2l2 / t
V2
A2 v2
300



A1 
 0.11m 2
v1
v1  t
v1
3.0  900
Monday, Apr. 19, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
9
Bernoulli’s Equation
Bernoulli’s Principle: Where the velocity of fluid is high, the
pressure is low, and where the velocity is low, the pressure is high.
Amount of work done by the force, F1,
that exerts pressure, P1, at point 1
W1  F1l1  P1 A1l1
Amount of work done on the other
section of the fluid is
W2   P2 A2 l2
Work done by the gravitational force to
move the fluid mass, m, from y1 to y2 is
W3  mg  y2  y1 
Monday, Apr. 19, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
10
Bernoulli’s Equation cont’d
The net work done on the fluid is
W  W1 W2 W3  P1 A1l1  P2 A2 l2 mgy2  mgy1
From the work-energy principle
1
1
2
mv2  mv12  P1 A1l1  P2 A2 l2 mgy2  mgy1
2
2
Since mass, m, is contained in the volume that flowed in the motion
A1l1  A2 l2
Thus,
and
m  r A1l1  r A2 l2
1
1
2
2
r A2 l2 v2  r A1l1v1
2
2
 P1 A1l1  P2 A2 l2  r A2 l2 gy2  r A1l1 gy1
Monday, Apr. 19, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
11
Bernoulli’s Equation cont’d
Since
1
1
r A2 l2 v22  r A1l1v12  P1 A1l1  P2 A2 l2  r A2 l2 gy2  r A1l1 gy1
2
2
We
obtain
Reorganize P1 
1 2 1 2
r v2  r v1  P1  P2  r gy2  r gy1
2
2
1 2
1 2
Bernoulli’s
r v1  r gy1  P2  r v2  r gy2 Equation
2
2
Thus, for any two
points in the flow
1 2
P1  r v1  r gy1  const.
2
For static fluid P2  P1  r g  y1  y2   P1  r gh
1
2
2
P

P

r
v

v

2
1
1
2
For the same heights
2
Pascal’s
Law
The pressure at the faster section of the fluid is smaller than slower section.
Monday, Apr. 19, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
12
Example for Bernoulli’s Equation
Water circulates throughout a house in a hot-water heating system. If the water is
pumped at a speed of 0.5m/s through a 4.0cm diameter pipe in the basement under
a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter
pipe on the second 5.0m above? Assume the pipes do not divide into branches.
Using the equation of continuity, flow speed on the second floor is
2
A1v1  r12 v1
 0.020 

v2 

0.5

 1.2m / s
2


A2
 r2
 0.013 
Using Bernoulli’s equation, the pressure in the pipe on the second floor is


1
P2  P1  r v12  v22  r g  y1  y2 
2
1
5
 3.0 10  1103 0.52  1.22  1103  9.8   5 
2
 2.5 105 N / m 2

Monday, Apr. 19, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu

13