Friction : A Simple Case Study
Consider an object moving in one dimension projected with initial velocity v0 (in the
positive direction) at time t= 0. If no friction is present then v = v0 at all times t.
If friction is present the object will slow down. Three types of frictional forces will be
considered.
I] A constant frictional force (b) of the sort encountered when the object is sliding over a
surface. The constant, b, has the units of N. F = - b
II] A frictional drag force which varies linearly with the velocity. The constant, b1, has
the units of Ns/m or Kg/s. F = - b v
1
III] A frictional drag force which varies quadratically with the velocity (as is often the
case for higher velocity objects traveling in air). The constant, b2, has the units of Kg/m.
As can be seen from the
calculations that follow (and
the graph) the function al form
of the velocity decay in these 3
cases is very different.
On the next 3 pages the
calculus based solutions
for these cases are shown.
F = - b2 v2
F=-b
Case I
I] A constant frictional force (b) of
the sort encountered when the object
is sliding over a surface. The
constant, b, has the units of NNewton's.
Reformulating the constant b has the
units of acceleration (m/s2).
This is the familiar case of sliding
friction on a surface where the
friction is equal to coefficient of
kinetic friction (µk) times the normal
force (N). In the case of an object on
a flat horizontal plane N = mg so:
b =µkN
b =µkg
The general solution of this constant
frictional force problem goes as
follows.
ma = - b
b
a = - = -b
m
dv
b
= - = -b
dt
m
dv = - bdt
v
t
v0
0
∫ dv = - ∫ bdt
v - v 0 = -bt
v = v 0 - bt
dx
= v 0 - bt
dt
b 2
x = v0 t - t
2
v
b
t
= 1- t = 1v0
v0
τ
v0
τ =
b
Case II
II] A frictional drag force which varies
linearly with the velocity.
The constant, b1, has the units of Ns/m
or Kg/s.
Using Newton’s Law this can be
reformulated with the constant b1 has
the units of 1/s.
It is sometimes useful to parameterize
the friction with in a way that takes
the units into account by b1= 1/τ .
Here τ is a decay time constant and
has the units of sec.
The general solution to this
problem goes as follows.
F = - b1 v
b1
ma = - b 1 v
a = - v = - b1 v
m
b1
dv
v
1
b1 =
= - v = -b1 v = dt
m
τ
τ
dv
= - b 1 dt
v
t
dx
= v0 e τ
v=
v
t
dt
dv
t
∫v v =∫0 b 1 dt
t
x - x 0 = ∫ v 0 e τ dt
0
ln(v) - ln(v o ) = - b 1 t
v
ln( ) = - b 1 t
vo
v
= e - b1 t
vo
0
x - x0 = − v 0 τ e
t
τ t
0
|
t
x - x 0 = v 0 τ (1 - e τ )
Case III
F = - b2v2
III] A frictional drag force which
varies quadratically with the
velocity (as is often the case for
higher velocity objects traveling in
air). Here the constant, b2, has the
units of Kg/m.
Reformulating one has b2, with
the units of 1/m and the decay
length L has the units of m.
For an object moving through a
medium ( air) ρ is the medium
density (in Kg/.m3), C is the drag
coefficient and A is the crosssection of the object perpendicular
to the motion.
ma = - b 2 v
The general solution to this
problem is as follows.
b2 =
2
L=
2m
1
=
CAρ b 2
b2 2
v2
2
a = - v = -b 2 v = m
L
b
dv
= - 2 v 2 = -b 2 v 2
dt
m
dv
= - b 2 dt
2
v
v
t
dv
∫v v 2 = − ∫0 b 2 dt
0
1 1
= - b2t
v vo
v
1
=
vo 1 + vob2 t
v
=
vo
CAρ
2
1
1
=
t
v
1+ o t 1+
τ
L
1 dx
1
=
v o dt 1 + v o b 2 t
dx
vo
=
dt 1 + v o b 2 t
x
t
vo
∫0 dx = ∫0 1 + vo b2 t dt
1
x = ln([1 + v o b 2 t ])
b2
xb 2 = ln([1 + v o b 2 t ])
e xb 2 = [1 + v o b 2 t ]
v
= e- xb 2 = e
vo
-
x
L