2
44
Problems and Solutions Section 2.7 (2.63 through 2.79)
2.63
Consider a springmass sliding along a surface providing Coulomb friction, with stiffness
1.2
×
10
4
N/m and mass 10 kg, driven harmonically by a force of 50 N at 10 Hz.
Calculate the approximate amplitude of steadystate motion assuming that both the mass
and the surface that it slides on, are made of lubricated steel.
Solution:
Given:
m
= 10 kg,
k
= 1.2x
10
4
N/m,
F
o
= 50 N,
ω
=10(2
!
) = 20
!
rad/s
ω
=
k
m
= 34.64 rad/s
for lubricated steel,
μ
= 0.07
From Equation (2.109)
X
=
F
o
k
1
!
4
μ
mg
"
(
F
o
)
#
$
%
%
&
’
(
(
2
(1
!
r
2
)
X
=
50
1.2
!
10
4
1
"
4(.07)(10)(9.81)
#
(50)
$
%
&
’
(
)
2
(1
"
20
#
34.64
*
+
,

.
/
2
)
X
=1.79
×
10
!
3
m
2.64
A springmass system with Coulomb damping of 10 kg, stiffness of 2000 N/m, and
coefficient of friction of 0.1 is driven harmonically at 10 Hz. The amplitude at steady
state is 5 cm. Calculate the magnitude of the driving force.
Solution:
Given:
m
= 10 kg,
k
= 2000 N/m,
μ
= 0.1,
ω
=10(2
!
) = 10(2
!
) = 20
!
rad/s,
ω
n
=
k
m
= 14.14 rad/s,
X
= 5 cm
Equation (2.108)
X
=
F
0
k
(1
!
r
2
)
2
+
4
μ
mg
"
kX
#
$
%
&
’
(
2
)
F
0
=
Xk
(1
!
r
2
)
2
+
4
μ
mg
"
kX
#
$
%
&
’
(
2
F
0
=
(0.05)(2000)
1
!
20
"
14.14
#
$
%
&
’
(
2
)
*
+
+
,

.
.
2
+
4(0.1)(10)(9.81)
"
(2000)(.05)
)
*
+
,

.
2
=
1874 N
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2
45
2.65
A system of mass 10 kg and stiffness 1.5
×
10
4
N/m is subject to Coulomb damping. If
the mass is driven harmonically by a 90N force at 25 Hz, determine the equivalent
viscous damping coefficient if the coefficient of friction is 0.1.
Solution:
Given:
m
= 10 kg,
k
= 1.5x
10
4
N/m,
F
0
= 90 N,
ω
= 25(2
!
) = 50
!
rad/s,
ω
n
=
k
m
= 38.73 rad/s,
μ
= 0.1
Steadystate Amplitude using Equation (2.109) is
X
=
F
0
k
1
!
4
μ
mg
"
(
F
o
)
#
$
%
&
’
(
2
(1
!
r
2
)
=
90
1.5
)
10
4
1
!
4(0.1)(10)(9.81)
"
(90)
#
$
%
&
’
(
2
1
!
50
"
38.73
*
+
,

.
/
2
=
3.85
)
10
!
4
m
From equation (2.105), the equivalent Viscous Damping Coefficient becomes:
c
eq
=
4
μ
mg
!"
X
=
4(0.1)(10)(9.81)
!
(50
!
)(3.85
#
10
$
4
)
= 206.7 Ns/m