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Example: Derive EOM for simple 1DOF mechanical system
The figure shows a mechanical system comprised of two
blocks connected by a pulley and (inextensible) cable
system. The cable does NOT slip on the pulley. The Fixed
wall
block of mass M1 slides on a smooth surface and is
connected to a fixed rigid wall through an elastomeric
cable represented by stiffness K and viscous damping
coefficient C.
The system is initially at rest at its static equilibrium
position.
The external force F(t), applied to block of mass M2 for time
t≥0 s , drives the system into motion.
a) Select suitable coordinates for motion of the two blocks,
show them on the Figure and explain rationale for your
choice.
b)
Identify the kinematical constraint relating motions of
the two blocks.
c)
Find the static deflection (s) of the elastomeric cable
attached to fixed wall
d)
Draw free body diagrams for motion of blocks and
applicable for time t>0 s. Label all forces and define
their constitutive relation in terms of the motion
coordinates.
e) Using Netwon’s Laws, state a EOM for each block (t≥0
s ), combine them to obtain a single EOM.
K
elastomeric
cable
g
C
M1
Smooth
surface
Non extensible
cable
=20
M2
F(t),
applied at
t>=0
2012 Luis San Andres©
Definition of SEP (Static equilibrium
position): System is NOT in MOTION + there
are NO external forces applied on the system.
Free Body Diagram – STATIC EQUILIBRIUM
Parameters:
M: mass
K : stiffness coefficient
Forces:
W: weight
N: normal to wall
Fs: elastic force from top cable
T: cable tension (inextensible)
Spring force (static):
Fse = Ks
Wall
reaction
force
M1
W1 sin
Te

W1
N=W1 cos
Te (cable tension)
Te
2Te
BALANCE OF STATIC FORCES
2  Te
cable tension
M2
W2
Static spring force
Fse
hence
Fse
W2
must hold weight 2
Te  W1  sin  
W2
2
 W1  sin  
must hold a fraction of
weight 1+ cable tension
K s
(2)
s is the static deflection of spring needed to hold the system together at the SEP
s 
1
2
W2  W1 sin 
K
(3)
Free Body Diagram Assumed state of motion to draw FBD :
System moving:
X>0, Y>0
Parameters:
M: mass
K : stiffness coefficient
C: viscous damping coefficient
Spring force
Fs = K (s +Y)
Wall
reaction
force
DEFINITIONS:
SEP
Y
W1 sin
M1
Dashpot force:
T
FD =CdY/dt

W1
N=W1 cos
T (cable tension)
T
2T
SEP
X
Variables:
X, Y: coordinates for motion of
block 2 and block 1, resp. (absolute
M2
W2
Forces:
W: weight
N: normal to wall
Fs: elastic force from top cable
FD: viscous damper force top cable
T: cable tension (inextensible)
F: external force
F(t),
applied at t=0
Static equilibrium position (SEP) defines
origin of coordinates X, Y describing the motion
of blocks 2 and 1, respectively.
coords., with origin at equilibrium
state)
Kinematic constraint – inextensible cable
2 T X = T Y, hence 2 X = Y
2 X = Y
(1)
DERIVE EOMs
M1 Y  T  W1 sin   FS  FD
Spring force:
FD =CdY/dt
(4)
SEP
Fs = K (s +Y)
Dashpot force:
Block 1 (top)
M1 Y  T W1 sin   K Y   s   CY
Y
W1 sin
M1
M 1 Y  K Y  C Y  T  W1 sin   K  s
T

W1
T (cable tension)
T
2 X = Y
2T
Block 2
M 2 X W2  F(t )  2T
(5)
SEP
X
M2
W2
F(t),
applied at t=0
X>0, Y>0
Free Body Diagram System moving
In Eq. (4), isolate the tension (T) and
substitute constraint Y=2 X
T  M 1 Y  K Y  C Y  W1 sin   K  s 
 2M 1 X  2 K X  2C X  W1 sin   K  s  (6)
Substitute Eq. (6) into Eq. (5) to obtain
DERIVE final EOM
Spring force:
FD =CdY/dt
(6)
SEP
Fs = K (s +Y)
Dashpot force:
M 2 X W2  F(t )  2T
T  2M1 X  2K X  2C X  W1 sin   K s 
Y
W1 sin
M1
T
M 2 X W2  F(t )  4M1 X  4K X  4C X  2 W1 sin   K s 

W1
T (cable tension)
T
2 X = Y
2T
1
2
SEP
X
M2
W2
From SEP: balance of forces for static equilibrium
F(t),
applied at t=0
W2  W1 sin   Fse  K s (3)
Cancel forces from SEP to obtain:
M 2 X  F(t )  4M1 X  4K X  4C X
Move to LHS terms related to motion:
Final EOM:
If using Y as the
independent
coordinate:
Y=2 X
 M 2  4M1  X  4 K X  4 C X  F(t )
 14 M 2  M1  Y  K Y  CY  12 F(t )
Spring force:
ENERGIES for system components
SEP
Fs = K (s +Y)
Dashpot force:
FD =CdY/dt
Assume a state of motion:
Y
W1 sin
M1
T
X > 0, Y >0

W1
T
Kinetic energy
2 X = Y
T  12 M 2 X 2  12 M1 Y 2 (1a)
T (cable tension)
T
2T
SEP
X
M2
W2
V = strain energy in cables + gravitational potential energy change
2
1 K  Y
 s  W2 X  W1 Y sin  (1b)
2
Potential energy
V
Includes static deflection from spring. Datum for potential energy is SEP
Viscous dissipated power =   F Y  CY 2
vis
D
External power
ext  F(t ) X
Substitute above constraint relating motion of blocks:
(1d)
Y=2 X
(1c)
F(t),
applied at t=0
Spring force:
ENERGIES for system components
Dashpot force:
Substitute constraint relating motion of blocks:
SEP
Fs = K (s +Y)
FD =CdY/dt
Y=2 X
Y
W1 sin
M1
T

W1
Kinetic energy
T T  1 2 M 2 X  1 2 4M1 X
2
2 X = Y
2
(1a)
T (cable tension)
T
2T
SEP
X
M2
W2
V = strain energy in cables + gravitational potential energy change
2
1 K   2X
 s
 W2 X  W1  2 X sin   (1b)
2
Potential energy
V
Datum is SEP
Viscous dissipated power
External power
vis  4 C X 2
ext  F(t ) X
(1d)
(1c)
F(t),
applied at t=0
Spring force:
Derive EOM from PCME
Dashpot force:
d
T  V  ext vis (2)
dt
ext  F(t ) X vis  4 C X 2
T  1 2  M 2  4M 1  X
V
1
SEP
Fs = K (s +Y)
Y
W1 sin
M1
FD =CdY/dt
T

W1
T (cable tension)
T
2 X = Y
2T
SEP
X
M2
2
dT
  M 2  4M1  X X
dt
W2
(3a)
2 K  s  2 X  W2 X  W1  2 X sin  
2
dV
 K  s  2 X  2 X  W2 X  2W1 X sin  
dt
  4 K X  2  K s  W1 sin   12 W2   X
4K X  X
Substitute Eqs. (3) into Eq. (2) to obtain
(3b)
Note how static SEP forces cancel
s 
1
2
W2  W1 sin 
K
F(t),
applied at t=0
Derive EOM from PCME
Spring force:
Dashpot force:
d
T  V  ext vis (2)
dt
2
ext  F(t ) X vis  4 C X
SEP
Fs = K (s +Y)
FD =CdY/dt
Y
W1 sin
M1
T

W1
2 X = Y
2T
SEP
X
d T  V 
  M 2  4M1  X X  4K X X  F(t ) X  4CX 2
dt
Cancel velocity dX/dt to obtain final EOM:
 M 2  4M1  X  4KX  4CX  F(t )
T (cable tension)
T
(5)
M2
W2
F(t),
applied at t=0