3Forces

IA Dynamics and Relativity 3.6 Friction
At an atomic level, energy is always conserved. However, in many everyday
processes, this does not appear to be the case. This is because friction tends to
take kinetic energy away from objects.
In general, we can divide friction into dry friction and fluid friction.
Dry friction
When solid objects are in contact, a normal reaction force
N
(perpendicular
to the contact surface) prevents them from interpenetrating, while a frictional
force
F
(tangential to the surface) resists relative tangential motion (sliding or
slipping).
N
F
mg
If the tangential force is small, it is insufficient to overcome friction and no
sliding occurs. We have static friction of
|F| µ
s
|N|,
where µ
s
is the coefficient of static friction.
When the external force on the object exceeds
µ
s
|N|
, sliding starts, and we
have a kinetic friction of
|F| = µ
k
|N|,
where µ
k
is the coefficient of kinetic friction.
These coefficients are measures of roughness and depend on the two surfaces
involved. For example, Teflon on Teflon has coefficient of around 0.04, while
rubber on asphalt has about 0.8, while a hypothetical perfectly smooth surface
has coefficient 0. Usually, µ
s
> µ
k
> 0.
Fluid drag
When a solid object moves through a fluid (i.e. liquid or gas), it experiences a
drag force.
There are two important regimes.
(i)
Linear drag: for small things in viscous fluids moving slowly, e.g. a single
cell organism in water, the friction is proportional to the velocity, i.e.
F = k
1
v.
where
v
is the velocity of the object relative to the fluid, and
k
1
>
0 is a
constant. This
k
1
depends on the shape of the object. For example, for a
sphere of radius R, Stoke’s Law gives
k
1
= 6πµR,
where µ is the viscosity of the fluid.
(ii)
Quadratic drag: for large objects moving rapidly in less viscous fluid, e.g.
cars or tennis balls in air, the friction is proportional to the square of the
velocity, i.e.
F = k
2
|v|
2
ˆ
v.
In either case, the object loses energy. The power exerted by the drag force is
F · v =
(
k
1
|v|
2
k
2
|v|
3
Example.
Consider a projectile moving in a uniform gravitational field and
experiencing a linear drag force.
At t = 0, we throw the projectile with velocity u from x = 0.
The equation of motion is
m
dv
dt
= mg kv.
We first solve for v, and then deduce x.
We use an integrating factor exp(
k
m
t) to obtain
d
dt
e
kt/m
v
= e
kt/m
g
e
kt/m
v =
m
k
e
kt/m
g + c
v =
m
k
g + ce
kt/m
Since v = u at t = 0, we get c = u
m
k
g. So
v =
˙
x =
m
k
g +
u
m
k
g
e
kt/m
.
Integrating once gives
x =
m
k
gt
m
k
u
m
k
g
e
kt/m
+ d.
Since x = 0 at t = 0. So
d =
m
k
u
m
k
g
.
So
x =
m
k
gt +
m
k
u
m
k
g
(1 e
kt/m
).
In component form, let x = (x, y), u = (u cos θ, u sin θ), g = (0, g). So
x =
mu
k
cos θ(1 e
kt/m
)
y =
mgt
k
+
m
k
u sin θ +
mg
k
(1 e
kt/m
).
We can characterize the strength of the drag force by the dimensionless constant
ku/(mg), with a larger constant corresponding to a larger drag force.
Effect of damping on small oscillations
We’ve previously seen that particles near a potential minimum oscillate indefi-
nitely. However, if there is friction in the system, the oscillation will damp out
and energy is continually lost. Eventually, the system comes to rest at the stable
equilibrium.
Example.
If a linear drag force is added to a harmonic oscillator, then the
equation of motion becomes
m
¨
x =
2
x k
˙
x,
where
ω
is the angular frequency of the oscillator in the absence of damping.
Rewrite as
¨
x + 2γ
˙
x + ω
2
x = 0,
where γ = k/2m > 0. Solutions are x = e
λt
, where
λ
2
+ 2γλ + ω
2
= 0,
or
λ = γ ±
p
γ
2
ω
2
.
If
γ > ω
, then the roots are real and negative. So we have exponential decay.
We call this an overdamped oscillator.
If 0
< γ < ω
, then the roots are complex with
Re
(
λ
) =
γ
. So we have
decaying oscillations. We call this an underdamped oscillator.
For details, refer to Differential Equations.