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Physics 172H
Lecture 6 – Ball-Spring Model of Solids, Friction
Read 4.1-4.8
d
radial force (N)
Model of solid: chemical bonds
G
F ≈ linear
0
If atoms don’t move too far
away from equilibrium, force
looks like a spring force!
A ball-spring model of a solid
Ball-spring model of a solid
To model need to know:
- spring length s
- spring stiffness
- mass of an atom
Initial conditions for circular motion
Length of a bond: “diameter” of copper atom
density ρ = 8.94 g/cm3:
molecular weight = 63.55 g/mole
NA molecules
1. Number of atoms in one cm3
8.94 g/cm3
atoms
atoms
N=
⋅ 6.022 × 1023
= 8.47 × 1022
63.55 g/mole
mole
cm3
2. Volume per one atom:
3
1
−23 cm
VCu =
= 1.18 ×10
22
3
8.47 ×10 atoms/cm
atom
3. Interatomic spacing:
dCu = 3 1.18 × 10−23 cm 3 = 2.27 × 10−8 cm=2.27 × 10−10 m=2.27
Ball-Spring Model of a Wire
How is the stiffness of the wire related to the stiffness of one of the short
springs (bonds)?
Two Springs in Series
Spring constant k
Mass M
Each spring must supply an upward force equal to Mg, thus, each stretches by s
giving a total stretch of 2s, or an effective spring constant of k/2.
Two Springs in Parallel
Mass M
Each spring provides an upward force of Mg/2, so each stretches s/2,
giving an effective spring constant of 2k.
Stiffness of a Copper Wire
2-meter long Cu wire
8.8 x 109 bonds
in series
Each side = 1 mm
1.94 x 1013 chains in parallel
The stiffness of the wire is much greater than the effective spring
stiffness between atoms due to the large number of chains in
parallel compared to the number of bonds in series.
Estimating interatomic “spring” stiffness
ΔL
strain =
L
stress =
FT
A
tension
stress = Y ⋅ strain
FT
ΔL
=Y
A
L
Y - Young’s modulus
depends only on material
Compare:
Larger Y means stiffer
wire
G
Fspring = k s s
G
Fspring
s
A = ks L
L
GA
Fspring
L s
= ks
A
A L
ks =
A
Y
L
Effective interatomic spring stiffness
ks =
A
Y
L
d2
ks = Y
d
Interatomic spring stiffness
k s = Yd
Limits of applicability of Young’s modulus
Plastic
deformation
stress = Y ⋅ strain
FT
ΔL
=Y
A
L
When the stretch of the material
exceeds the “elastic limit”, the linear
relation between stress and strain fails,
and the material deforms plastically.
This is an irreversible process.
Elastic region
Aluminum alloy
Brick on a table: compression
G
FN
G
Mg
If the weight of the
brick exceeds a limit,
the table will either
break (if it’s brittle) or
bend, (if it’s ductile).
Friction
Exert a force so that the
brick moves to the right
at a constant speed.
What is the net force on
the brick?
Friction Doesn’t Always Oppose Motion—Just
RELATIVE motion of the “frictioning” surfaces!
QUIZ
Box dropped onto moving
conveyor belt. What happens?
What is the state of
motion of the box
when the belt force
has a horizontal
component?
A Decelerating
B Accelerating
C Steady motion
If the green v vector is the
belt’s velocity, what can
you say about the velocity
of the box at the instant of
the diagram?
How is it that a sprinter can accelerate?
Friction Doesn’t Always Oppose Motion—Just
RELATIVE motion of the “frictioning” surfaces!
Box dropped onto moving
conveyor belt. What happens?
There are no other objects or
agents shown.
We are shown ALL the
forces involved. BUT Fbelt
includes two components, FN
and Fk , each acting on the
box.
The force Fk is unbalanced, and
so at the moment diagrammed,
the box is necessarily
accelerating to the right. It will
eventually match velocity with
the belt, at which point Fk will
VANISH
You can further argue
that the existence of
the Fk component
means that there
MUST be relative
motion between the
belt and the box at the
instant shown.
So the box is moving
slower than the belt at
this instant.
The box’s “inertia” is
NOT sufficient to keep
it moving slower than
the belt (or standing
still) – that would take
an extra agent pulling
to the left.
Sliding Friction
• When one object slides on another, the
component of force exerted by one
object on the other has a component
parallel (or antiparallel) to the motion:
• NOTE – the friction always acts against the relative
motion. It acts on both the object and the surface.
• ffriction ~ μkFN
μk is the coefficient of kinetic friction
FN is the “normal force” – the perpendicular component
of the force that is squeezing the two objects into each
other
Static Friction
• If the applied force Fapplied < μkFN the object will
SLOW DOWN (relative to the surface it’s frictioning
with.) What is the kinetic friction at this point?
• The block will eventually come to rest, and THEN,
any (sideways) force applied < μsFN will not cause
the block to move (relative to the surface it’s
touching)..
• Generally the static friction is a bit greater than the
kinetic friction: μk < μs
• Once the static friction breaks, the block will speed
up (if the applied force does not change) since the
applied force is then somewhat greater than the
retarding force of the kinetic friction.