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General Physics (PHY 2130)
Lecture 23
•  Solids and fluids
  buoyant force
  Archimedes’ principle
  Fluids in motion
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Solids and fluids
  different states of matter; fluids
  density, pressure, etc.
Review Problem: The surface pressure on the planet Venus is 95 atm. How far
below the surface of the ocean on Earth do you need to be to experience the
same pressure? The density of seawater is 1025 kg/m3.
P = Patm + ρgd
95 atm = 1 atm + ρgd
ρgd = 94 atm = 9.5 ×10 6 N/m 2
(1025 kg/m )(9.8 m/s )d = 9.5×10
3
2
d = 950 m
6
N/m 2
Pascal’s Principle
•  A change in pressure applied to
an enclosed fluid is transmitted
undiminished to every point of
the fluid and to the walls of the
container.
•  The hydraulic press is an
important application of
Pascal’s Principle
F1 F2
P=
=
A1 A2
•  Also used in hydraulic brakes,
forklifts, car lifts, etc.
Since A2>A1, then F2>F1 !!!
4
ΔP at point 1 = ΔP at point 2
F1
F2
=
A1 A 2
⎛ A 2 ⎞
⎟⎟ F1
F2 = ⎜⎜
⎝ A1 ⎠
The work done pressing the
smaller piston (#1) equals
the work done by the larger
piston (#2).
F1d1 = F2d2
Using an hydraulic lift reduces the amount of force needed to lift a load, but
the work done is the same.
5
Example: Assume that a force of 500 N (about 110 lbs) is applied to the smaller
piston in the previous figure. For each case, compute the force on the larger
piston if the ratio of the piston areas (A2/A1) are 1, 10, and 100.
Using Pascal’s Principle:
A2 A1
1
10
100
F2
500 N
5000 N
50,000 N
6
Example: In the previous example, for the case A2/A1 = 10, it was found that
F2/F1 = 10. If the larger piston needs to rise by 1 m, how far must the smaller
piston be depressed?
Since the work done by both pistons is the same,
F2
d1 = d 2 = 10 m
F1
7
Example: Depressing the brake pedal in a car pushes on a piston with crosssectional area 3.0 cm2. The piston applies pressure to the brake fluid, which is
connected to two pistons, each with area 12.0 cm2. Each of these pistons presses a
brake pad against one side of a rotor attached to one of the rotating wheels. See
the figure for this problem. (a) When the force applied by the brake pedal to the
small piston is 7.5 N, what is the normal force applied to each side of the rotor?
The pressure in the fluid
Also,
Pr essure P =
P = Fb Ab .
Normal Force N
Area of the brake pad piston A
the normal force applied to each side of the rotor
A
12.0 cm 2
N = PA =
Fb =
(7.5 N) = 30 N
2
Ab
3.0 cm
Measuring Pressure
•  The spring is calibrated by a
known force
•  The force the fluid exerts on
the piston is then measured
 
 
 
One end of the U-shaped tube
is open to the atmosphere
The other end is connected to
the pressure to be measured
Pressure at B is Po+ρgh
 
 
A long closed tube is
filled with mercury and
inverted in a dish of
mercury
Measures atmospheric
pressure as ρgh
9
Manometer
A manometer is a
U-shaped tube that
is partially filled
with liquid.
Both ends of the tube
are open to the
atmosphere.
10
If there is a pressure difference between the gas and the atmosphere, a
force will be exerted on the fluid in the U-tube. This changes the
equilibrium position of the fluid in the tube.
From the figure:
At point C
Also
Pc = Patm
PB = PB'
The pressure at point B is the
pressure of the gas.
PB = PB ' = PC + ρgd
PB − PC = PB − Patm = ρgd
Pgauge = ρgd
Gauge Pressure
Pgauge = Pabs - Patm
11
Barometer
The atmosphere pushes on the container
of mercury which forces mercury up the
closed, inverted tube. The distance d is
called the barometric pressure.
PA = PB = Patm
PA = ρgd
Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall.
How would you measure blood pressure?
Has to be:
(a) accurate
(b) non-invasive
(c) simple
sphygmomanometer
13
Example: An IV is connected to a patient’s vein. The blood pressure in the vein
has a gauge pressure of 12 mm Hg. At least how far above the vein must the IV
bag be hung in order for fluid to f low into the vein? Assume the fluid in IV has the
same density as blood.
At a minimum, the gauge pressures must be equal. When h2 is large
enough, fluid will flow from high pressure to low pressure.
Pgauge = ρ Hg gh1 = ρ blood gh2
ρ Hg gh1
h2 =
ρ blood g
⎛ ρ Hg ⎞
⎛ 13,600 kg/m 3 ⎞
⎟⎟h1 = ⎜⎜
⎟(12 mm)
= ⎜⎜
3 ⎟
⎝ 1060 kg/m ⎠
⎝ ρ blood ⎠
= 154 mm
Question
Suppose that you placed an extended
object in the water. How does the
pressure at the top of this object
relate to the pressure at the
bottom?
1. It’s the same.
2. The pressure is greater at the top.
3. The pressure is greater at the
bottom.
4. Whatever…
Since pressure is related to force: there is a force present!
Buoyant Force
  This
force is called the buoyant force.
  What
is the magnitude of that force?
ΔF = B = (P2 − P1 )A, but :
P2 = P1 + ρ gh, so :
P 1A
B = (P1 + ρ gh − P1 )A = ρ fluid ghA = ρ fluid gV !
mg
P 2A
Buoyant Force
•  The magnitude of the buoyant force always equals the
weight of the displaced fluid
B = ρ fluidVg = w fluid
•  The buoyant force is the same for a totally submerged
object of any size, shape, or density
•  The buoyant force is exerted by the fluid
•  Whether an object sinks or floats depends on the
relationship between the buoyant force and the weight
Archimedes' Principle
Any object completely or partially submerged
in a fluid is buoyed up by a force whose
magnitude is equal to the weight of the fluid
displaced by the object.
This force is buoyant force.
Physical cause: pressure difference between the top and the bottom of the object
Example: Archimedes’ Principle for
Totally Submerged Object
•  The upward buoyant force is B=ρfluidgVobj
•  The downward gravitational force is w=mg=ρobjgVobj
•  The net force is B-w=(ρfluid-ρobj)gVobj
Depending on the direction
of the net force, the object
will either float up or sink!
The net force is B-w=(ρfluid-ρobj)gVobj
•  The object is less dense
 
than the fluid ρfluid<ρobj
•  The object experiences a
net upward force
The object is more dense
than the fluid ρfluid>ρobj
 
The net force is downward,
so the object accelerates
downward
Archimedes’ Principle:
Floating Object
•  The object is in static equilibrium
•  The upward buoyant force is
balanced by the downward force of
gravity
•  Volume of the fluid displaced
corresponds to the volume of the
object beneath the fluid level
If B = mg : ρ fluid gV fluid = ρ object gVobject , or
ρ obj V fluid
=
ρ fluid Vobj
Example: floating frog
A frog in a hemispherical pod finds that
he just floats without sinking in a fluid of
density 1.35 g/cm3. If the pod has a
radius of 6.00 cm and negligible mass,
what is the mass of the frog?
Question 1
Suppose that you have a steel bar. Will
it float on water? Why?
Question 2
Suppose that you have a steel bar. Will
it float on water? Why?
How come that ships (which are made
of steel) can float?
Question 3
Suppose that your friend gave you a necklace
(crown, piece of yellow metal, …). He claims
that this object is made of pure gold. How can
you check his statement (without going through
his credit history)?
Question 3
Suppose that your friend gave you a necklace
(crown, piece of yellow metal, …). He claims
that this object is made of pure gold. How can
you check his statement (without going through
his credit history)?
Idea: determine density! Let’s weight the
object in and outside the water container:
Out:
mg = ρ gV or
mg
ρ=
gV
In:
" weight" = mg − B
= mg − ρ fluid gV
V=
" weight"−mg
ρ fluid g
If ρ is not the same as ρgold, your friend is lying…
ConcepTest 2
Two identical glasses are filled to the same level with water.
One of the two glasses has ice cubes floating in it.Which
weighs more?
1. The glass without ice cubes.
2. The glass with ice cubes.
3. The two weigh the same.
ConcepTest 2
Two identical glasses are filled to the same level with water.
One of the two glasses has ice cubes floating in it.Which
weighs more?
1. The glass without ice cubes.
2. The glass with ice cubes.
3. The two weigh the same.
Ice cubes displace exactly their own weight in water.
ConcepTest 3
Two identical glasses are filled to the same level with water.
One of the two glasses has ice cubes floating in it.When the
ice cubes melt, in which glass is the level of the water
higher?
1. The glass without ice cubes.
2. The glass with ice cubes.
3. It is the same in both.
ConcepTest 3
Two identical glasses are filled to the same level with water.
One of the two glasses has ice cubes floating in it.When the
ice cubes melt, in which glass is the level of the water
higher?
1. The glass without ice cubes.
2. The glass with ice cubes.
3. It is the same in both.
Previous problem: weight was the same. Since
density is also the same, the volumes are going to
be the same as well.
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Fluid Flow
A moving fluid will exert forces parallel to the surface over which it
moves, unlike a static fluid. This gives rise to a viscous force that
impedes the forward motion of the fluid.
A steady flow is one where the velocity at a given point in a fluid is
constant. Steady flow is laminar; the fluid flows in layers.
An ideal fluid is incompressible, undergoes laminar flow, and has no viscosity.
V1 =
constant
V2 =
constant
v1≠v2
Fluids in Motion: Streamline Flow
•  Streamline flow
•  every particle that passes a particular point moves exactly
along the smooth path followed by particles that passed the
point earlier
•  also called laminar flow
•  Streamline is the path
•  different streamlines cannot cross each other
•  the streamline at any point coincides with the direction of fluid
velocity at that point
Fluids in Motion: Turbulent Flow
•  The flow becomes irregular
•  exceeds a certain velocity
•  any condition that causes abrupt changes in velocity
32
Equation of Continuity
The amount of mass that flows though the cross-sectional area A1 is the
same as the mass that flows through cross-sectional area A2.
ΔV
= Av
Δt
is called the volume flow rate (units m3/s)
33
Δm
= ρAv
Δt
is the mass flow rate (units kg/s)
The continuity equation is
ρ1 A1v1 = ρ2 A2v2
If the fluid is incompressible, then ρ1= ρ2.
34
Example: A garden hose of inner radius 1.0 cm carries water at 2.0 m/s. The
nozzle at the end has radius 0.20 cm. How fast does the water move through
the constriction?
A1v1 = A2v2
⎛ A1 ⎞
⎛ πr12 ⎞
v2 = ⎜⎜ ⎟⎟v1 = ⎜⎜ 2 ⎟⎟v1
⎝ A2 ⎠
⎝ πr2 ⎠
2
⎛ 1.0 cm ⎞
= ⎜
⎟ (2.0 m/s) = 50 m/s
⎝ 0.20 cm ⎠
35
Spring Break!
36
Additional Example: A piece of metal is released under water. The volume of the
metal is 50.0 cm3 and its specific gravity is 5.0. What is its initial acceleration?
(Note: when v = 0, there is no drag force.)
y
FBD for
the metal
Apply Newton’s 2nd Law to the
piece of metal:
FB
∑F = F
B
x
w
− w = ma
The magnitude of the buoyant force equals
the weight of the fluid displaced by the
metal.
FB = ρ waterVg
Solve for a:
⎛ ρ waterV
⎞
ρ waterVg
FB
a=
−g =
− g = g ⎜
− 1⎟
⎜ ρ V
⎟
m
ρobjectVobject
⎝ object object ⎠
37
Example continued:
Since the object is completely submerged V=Vobject.
specific gravity =
ρ
ρ water
where ρwater = 1000 kg/m3 is the density of
water at 4 °C.
Given
ρ object
specific gravity =
= 5.0
ρ water
⎛ ρ waterV
⎞
⎛ 1
⎞
⎛ 1
⎞
⎜
⎟
a=g
− 1 = g ⎜
− 1⎟ = g ⎜
− 1⎟ = −7.8 m/s2
⎜ ρ V
⎟
⎝ S .G. ⎠
⎝ 5.0 ⎠
object
object
⎝
⎠