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```Physics 231 Lecture 23
•
•
Main points of today’s lecture:
Gravitation potential energy
PE grav = −
•
GM1M 2
r12
Tensile stress and strain
ΔF
ΔL
=Y
A
L0
•
Bulk stress and strain:
ΔF
ΔV
= ΔP = B
A
V
•
Pressure in fluids:
P=
•
F
; Pbot = Ptop + ρgh
A
Buoyancy
FB = Wdisplaced = ρ f Vdisplacedg
Gravitational Potential Energy
•
•
ΔPE = mgh is valid only near the earth’s surface
F objects
For
bj t high
hi h above
b
th
the earth’s
th’ surface,
f
an
alternate expression is needed (You would need
calculus to derive it.)
M Em
PE2 = −G
r
•
– Zero potential energy is defined to be the value
infinitelyy far from the earth
If the total mechanical energy of an object exceeds
zero, the object can escape the gravitation field of
the earth. The escape velocity is when the Emech=0.
At the Earth's surface:
GM Earth m
1
E mech = mv 2 −
2
R Earth
If E mech ≥ 0 corresponding to v ≥ vescape , the object
can escape the Earth's gravitational field
0=
GM Earth m
1
2
mvescape
−
 vescape =
2
R Earth
2GM Earth
R Earth
you will have loncapa
problems on this.
9
Conceptual question
•
A rock,
k iinitially
iti ll att restt with
ith respectt to
t Earth
E th and
d located
l
t d an
infinite distance away is released and accelerates toward Earth.
An observation tower is built 3 Earth-radii high to observe the
rock as it plummets to Earth.
Earth Neglecting friction,
friction the rock’s
rock s
speed when it hits the ground is
KE f + PE f = KE 0 + PE 0 = PE 0
– a) twice
GM E m
PE
=
−
 PE 0 = 0
– b) three
th
times
ti
r
– c) four times
also: KE 0 = 0  KE f + PE f = 0
– d) six times
GM E m
 KE f = − PE f =
– e) eight times
its speed at the top of the tower. KE groundd
4=
KE ground
KEtower
1 2
mvground v 2
ground
=2
= 2
1 2
vtower
mvtower
2
KE tower
rf
GM E m
1
rgroundd r ground
rtower = 4rE = 4
=
=
GM E m
1
rground rE
rtower
rtower
Stress and strain in solids
•
All materials deform when under stress,
stress ie
ie.
when a force is applied to them. The atoms in a
solid are held in place by forces that act like
springs.
p g
•
If you pull on a bar of cross-sectional area A
with a force F, the bar will stretch a distance
ΔL.. In this
t s case, the
t e force
o ce iss perpendicular
pe pe d cu a to
the surface, and the force F and distance ΔL are
related by Young’s modulus Y:
ΔL F
ΔL
F
=Y
; = Stress;
= Strain
A
L0 A
L0
•
The amount of force you need to apply to achieve
the same strain increases with the cross sectional
area of the bar. The force will be linearly
proportional to the strain up to the elastic limit.
Beyond this point, the bar will be permanently
deformed. Stress units Pa=N/m2. It is the force
divided by the area, and has the units of pressure.
Example
•
Bone h
B
has a Y
Young’s
’ modulus
b t 18 x 109 Pa.
P Under
U d
compression, it can withstand a stress of about 160 x 106 Pa
before breaking. Assume that a femur (thigh bone) is 0.50 m
long and calculate the length by which this bone can be
compressed before breaking.
Y
18 x 109 Pa
– a) 4.4 mm
F/A 160 x 106 Pa
– b) 6.4
6 4 mm
L0 0.50 m
– c) 1.6 cm
ΔL ?
– d) 2.0 cm
ΔL
F
= 160x106 N / m 2 = Y
L0
A
0.5m
F L0 = 160x10
6
2
160
10
N
/
m
= 44.4mm
4
 ΔL =
9
2
18x10 N / m
A Y
If this force is uniformly applied, then F/A
F/A=P,
P, where P is the
pressure or force per unit area applied to the end of the femur.
Units for pressure are N/m2 = Pa. (Pa = Pascals)
Bulk modulus
•
If you drop
d
a metal
t l block
bl k in
i the
th ocean, it will
ill sink
i k to
t bottom
b tt
where
h the
th
pressure is much greater than that of the atmosphere. This will
compress the block; the amount of compression is governed by the
bulk modulus B.
B
ΔV
ΔV
ΔP = B
; ΔP = pressure change ;
= Volume strain
V
V
1. The SI unit of pressure is
A.
B.
C.
D
D.
N
kg/m2
Pa
kg/m3
Slide 13-6
Density
Nuclear matter
2.66x1017
Slide 13-12
Pressure in fluids, Pascal’s principle
•
Consider a vessel filled with liquid and a smaller rectangular
volume
l
V off the
th same liquid
li id contained
t i d within
ithi th
the vessel.
l If
the liquid is at rest, there can be no shear force on V, because
it would cause the liquid to move, but there is a pressure
force F=P⋅A
F=P A perpendicular
to all 6 surfaces.
surfaces

y
x
In equilibrium:  F = 0
z
x : Fleft − Fright = 0; Fleft = Pleft A x ; Fright = Pright A x  Pleft = Pright
z: Ffront − Fback = 0; Ffront = Pfront A z ; Fback = Pback A z  Pfront = Pback
W
yy: Fbot − Ftop − W = 0;; Fbot = Pbot A y ; Ftopp = Ptopp A y  Pbot = Ptopp +
Ay
•
•
More detailed considerations show that pressure always exerts a force
perpendicular to a surface and that pressure is the same everywhere at the same
height i.e.
height,
i e Pfront =Pback, and Pleft =Pright and Pfront= Pright, etc.
etc
Defining density ρ of the liquid by ρ=M/V , and using V=Ayh we can simplify the
dependence of P on height as follows:
W = Mg = ρVg = ρA hg
Pbot = Ptop +
ρA y hg
Ay
y
 Pbot = Ptop + ρgh
Pressure and Depth equation
Pbot = Ptop + ρgh
•
•
•
•
If the top is open, Pbot is normal
atmospheric pressure
– 1.013 x 105 Pa = 14.7 lb/in2
The p
pressure does not depend
p
upon the shape of the container
Gauge pressure is the difference
between the pressure in the
fluid and atmospheric pressure.
If you connected a gauge which
measures the pressure relative
to atmospheric pressure to the
fluid at depth h, then
for this specific case:
Pguage = Pbottom,h _ below _ surface − Patmosphere = ρgh
Pascal’s principal and hydraulic machines
•
•
•
Because F=PA,
B
F PA one can make
k hhydraulic
d li
machines that can generate large forces.
Units: One atmosphere (1 atm) =
– 76.0 cm of mercury
– 1.013 x 105 Pa = 1.013 x 105 N/m2
– 14.7 lb/in
Quiz: A simple hydraulic lift is employed to lift a truck that weighs
30000N. If A1=1 cm2 and A2= 1000 cm2, what force F1 is needed to lift
the truck? (Hint: Are the pressures equal under each piston?)
– a) 3N
P1 = P2 ; this is pascal
pascal'ss principal
– b) 300N
– c) 30N
F2
F1
F2
2 30000N
 F1 = A1

=
1cm
=
= 30N
2
– d) .0.33N
0 33N
A2
A
A
1000cm
1
2
```
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