Physics 231 Lecture 24
•
•
Main points of today’s lecture:
Buoyancy
y y
FB  Wdisplaced   f Vdisplacedg
•
Bernouli’s equation
1 2
v f  gy f
2
1
 P0  v02  gy0
2
• Viscous flow
Av
F 
d
Pf 
•
Poisseuille’s law
V R 4 P1  P2 

8L
t
Bulk modulus example
•
A steel block with a volume of 8 cm3 is dropped in the ocean and
comes to rest at the bottom, which is 3 km below the surface. 1) What
is the pressure at the bottom of the ocean? 2) By what amount V is
the volume decreased. ((The Bulk modulus for steel is 1.6x1011 N/m2.
The density of water is 1000kg/m3. 1 atm. =1.01x105 N/m2)
1) Pbot  Ptop  gh
– a) 2x105 N/m2
– b) 4x105 N/m2
Pbot  1.0x10
1 0x105 Pa  1000kg / m 3 99.8m
8m / s 2 3000m
– c) 2x106 N/m2
7
2
P

3.0x10
N
/
m
7
2
bot
– d) 3x10 N/m

V
2) P  B
V
V
V
Pbot  Ptop
V   P 
B
B




8cm3
7
2
5
2


3x10
N
/
m
1x10
N
/
m
1.6x1011 N / m 2
 1.5x10 3 cm 3

Reading Quiz
2. Gauge pressure is
A.
B.
C.
D
D.
larger than
smaller than
the same as
None of the above
true pressure. Explain.
Gauge pressure
Comes from the definition
Pgauge  P  Patmosphere
Slide 13-8
Reading Quiz
3. The buoyant force on an object submerged in a liquid depends on
A.
B.
C.
D
D.
the object’s mass.
the object’s volume.
the density of the liquid.
both B and C.
C
Slide 13-10
Buoyancy
•
Consider the small volume in the vessel at the right. If
the volume is filled with the same fluid in the vessel, it
will be in equilibrium. Thus:
FB  W  0 where W  water Vin _ box g,
Vin _ box  Vdisplaced  therefore FB  water Vdisplaced g
•
•
T
FB
Now, replace the volume of fluid with the same volume
of some other material. There will still be the same
buoyant force pushing up on the block of material.
Example: A kidney weighs 5.7 N in air. If the kidney
is completely submerged in water, its apparent
weight is 1.6 N. Determine the specific gravity of
the kidney. (specific gravity = kidney/water)

water Vkidney g 
Apparent weight  T  W  FB W(1  FB / W)  W 1 

 kidney Vkidney g 
Apparent weight  W 1  water / kidneyy  1  water / kidneyy  Apparent weight/W


water / kidney  1  Apparent weight/W  1-1.6/5.7  .71
Specific gravity  kidney / water  1/ .71  1.39
Example
•
A bl
blockk off wood
d fl
floats
t on the
th surface
f
off a lake
l k with
ith 60% off its
it volume
l
below the surface of the water. What is the density of the wood?
FB  water Vdisplaced g
W  wood Vwood g
FB  W
 water Vdisplaced g  wood Vwood g
also; Vdisplaced  0.6Vwood
\ wood g\
 water  0.6  V\ wood  g\  wood V
wood  0.6  water  600kg / m 3
Quiz
•
A metall block
bl k (
( steel=7.86x10
7 86 103 kg/m
k / 3) floats
fl
on the
h top off a pooll off
mercury (mercury=1.35x104 kg/m3). The fraction of the volume of the
block which lies below the surface is
– a)) 0.0
00
Vdisplaced  f below Vsteel
– b) 0.25
– c) 0.58
W  FB ; W  steel Vsteel g; FB  mercury f below Vsteel g
– d) 0.42
 steel Vsteel g  mercury f below Vsteel g  steel / mercury  f below
– e) 0.75
 7.86x10
7 86x103 kg / m 3 / 11.35x10
35x104 kg / m 3  .58
58  f below
Hint: mercury plays the role of water. The Buoyancy force
comes from displacing the mercury. Equate the weight of
the steel to the buoyancey force from the mercury.
Does the water level go up or down when the anchor
is lowered to the bottom of the lake?
When resting on the bottom
Vdisplaced=Vanchor
When floating in the boat
w Vdisplacedg= anchor Vanchorg
 Vdisplaced= anchor/ w Vanchor
– a) The water level goes up.
– b) The water level goes down.
– c) The water level goes neither up nor down.
Conceptual question
•
A lead
l d weight
i ht is
i ffastened
t
d on top
t off a large
l
solid
lid piece
i
off
Styrofoam that floats in a container of water. Because of the
weight of the lead, the water line is flush with the top surface of
the Styrofoam
Styrofoam. If the piece of Styrofoam is turned upside down
so that the weight is now suspended underneath it,
– a) the arrangement sinks.
– b) the
th water
t line
li is
i below
b l
th
the ttop surface
f
off th
the St
Styrofoam.
f
– c) the water line is still flush with the top surface of the
Styrofoam.
Same displaced water in both figures!
Conceptual quiz
•
A lead
l d weight
i ht is
i ffastened
t
d to
t a large
l
solid
lid piece
i
off St
Styrofoam
f
th
thatt
floats in a container of water. Because of the weight of the lead,
the water line is flush with the top surface of the Styrofoam. If
the piece of Styrofoam is turned upside down
down, so that the weight
is now suspended underneath it, the water level in the container
– a) rises.
– b) d
drops.
– c) remains the same.
If an object floats:
FB=wgVdisplaced=weight of object.
object
The weight doesn’t change.
Vdisplaced doesn’t change.
The level doesn’t change.