# p150c09

```Chapter 9: Mechanical Properties of Matter
Properties of matter and matter in motion
Density
m

V
Mass
Density 
Volume
Kg
SI units : 3
m
g
Kg
1 3  1000 3
cm
m
weight density : D = w/V
specific gravity = density relative to water
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Substance
Hydrogen
Helium
Air
water
alchohol
gasoline
balsa wood
pine
aluminum
iron
gold
Mass density 
(kg/m3)
0.09
0.18
1.30
1000
790
680
130
370
2700
7800
19000
11000
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Example: A 50 g bracelet is suspected of being gold-plated lead instead of pure gold.
When it is dropped into a full bucket of water, 4.0 cm3 overflows. What is the density
of the bracelet?
Example: A well pump 20.0 m deep is to supply .50 m3 of water per minute. How
much mechanical power must be provided?
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Elasticity
“stretchiness/springiness”
-how materials respond to stress
compression
tension
shear
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“Stretchability” = amount of stress (applied force) produces a
strain (elongation/compression/shear)
Hooke’s Law: the amount of stretching is proportional to the
applied force.
F=kx
The details of such springiness
depends upon the size and shape of
the material as well as how the
forces are applied
x
x
1
Ton
2
Tons
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Elastic Limit: the maximum stress (force) which can be applied
to an object without resulting in permanent deformation.
Plastic Deformation: the permanent deformation which results
when a materials elastic limit has been exceeded.
Ultimate strength: greatest tension (or compression or shear) the
material can withstand. *snap*
A malleable or ductile material has a large range of plastic
deformation.
Fatigue: small defects reduce materials strength well below
original strength.
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Young’s Modulus: how things stretch (elastically)
stress: force per area = F/A
A
L0
compression
L
L0
A
tension
A
L
A
strain: fractional change in length
= change in length per original length = DL/Lo
Elastic modulus = stress/strain
Young’s modulus (for stretching in one direction)
Y
F A
DL L0
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From Lab
Force: suspended weight
Area: cross section of wire
elongation causes cylinder to rotate
d
2q
D
q
DL
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Example 9.4: Stress as force per area
A 70 kg swami lies on a bed of nails 1.0 cm apart whose points have an area of 1.0
mm2. If the area of the swami’s body in contact with the bed is 0.50 m2 and the
threshold for pain is 1MPa (1,000,000 Pa, 1Pa = 1 Pascal = 1N/m2 ), is the swami
actually subjected to a painful experience? (i.e. F/A =?)
- # nails
- total area
- force (=weight)
-F/A
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Example 9.4’ (+): A copper wire1.0 mm in diameter and 2.0 m long is used to support a
mass of 5.0 kg. By how much does this wire stretch under this load? What is the
maximum mass which can be supported without exceeding copper’s elastic limit?
Y = 1.1x1011 Pa
elastic limit = 1.5x108 Pa
Example 9.6: A sagging floor is jacked up and a steel girder 3 m long whose cross
sectional area is 30 cm2 is placed underneath. When the jack is removed, a sensitive
strain gage shows that the girder has been compressed by .0020 cm. Find the load
supported by the girder.
Y = 20x1011 Pa
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Shear Modulus:
stress: force per area = F/A
s
F
A
d
F
shear strain cannot be supported in fluids (gas, liquid)
F A
S
sd
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Bulk Modulus:
stress: force per area = F/A
s
compression of solids, liquids and gases
F A
p
B

DV V0
DV V0
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Structure of solids
crystalline solids (most solids are crystalline)
atoms arranged in regular patterns
bonds hold atoms together like microscopic springs
amorphous solid
atoms are not in a well defined ordering
defects in crystalline structure allow permanent
deformation
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```