1. 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 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.
Select suitable coordinates for motion of the two blocks, show them on the Figure and explain rationale for your choice.
Identify the kinematical constraint relating motions of the two blocks.
Find the static deflection (s) of the elastomeric cable attached to fixed wall
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.
Using Netwon’s Laws, state a EOM for each block (t≥0 s ), combine them to obtain a single EOM.
2012 Luis San Andres©

2. Fse = Kds N=W1 cosq Free Body Diagram – STATIC EQUILIBRIUM W1 sinq W1
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 = Kds
Wall reaction force M1
W1 sinq Te q W1
N=W1 cosq
Te (cable tension) Te
BALANCE OF STATIC FORCES
2Te M2 W2
ds is the static deflection of spring needed to hold the system together at the SEP
(3)

3. M2 Free Body Diagram DEFINITIONS: System moving: Fs = K (ds +Y) Y M1
Assumed state of motion to draw FBD :
X>0, Y>0
DEFINITIONS:
Parameters:
M: mass
K : stiffness coefficient
C: viscous damping coefficient
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
Variables:
X, Y: coordinates for motion of block 2 and block 1, resp. (absolute coords., with origin at equilibrium state)
Spring force
Fs = K (ds +Y)
Wall reaction force
SEP Y M1
W1 sinq
Dashpot force:
FD =CdY/dt T q W1
T (cable tension)
N=W1 cosq T 2T
SEP M2 X W2
F(t),
applied at t=0
Kinematic constraint – inextensible cable
2 T dX = T dY, hence 2 dX = dY
Static equilibrium position (SEP) defines origin of coordinates X, Y describing the motion of blocks 2 and 1, respectively.
2 dX = dY
(1)

4. DERIVE EOMs (4) (5) M2 (6) Fs = K (ds +Y) Y M1 W1 sinq FD =CdY/dt W1 T
Block 1 (top)
(4) Y X
Dashpot force:
FD =CdY/dt
Spring force:
Fs = K (ds +Y) M2 2T W2 T
T (cable tension) M1 W1
W1 sinq
F(t),
applied at t=0
2 dX = dY
SEP q
Block 2
(5)
In Eq. (4), isolate the tension (T) and
substitute constraint dY=2 dX
X>0, Y>0
(6)
Free Body Diagram System moving
Substitute Eq. (6) into Eq. (5) to obtain

5. DERIVE final EOM
(6)
From SEP: balance of forces for static equilibrium
(3)
Cancel forces from SEP to obtain:
Move to LHS terms related to motion:
Final EOM:
If using Y as the independent coordinate:
dY=2 dX

6. ENERGIES for system components
Assume a state of motion:
X > 0, Y >0
Kinetic energy T
(1a)
Potential energy V = strain energy in cables + gravitational potential energy change
(1b)
Includes static deflection from spring. Datum for potential energy is SEP
Viscous dissipated power =
(1c)
External power
(1d)
Substitute above constraint relating motion of blocks: dY=2 dX

7. ENERGIES for system components
Substitute constraint relating motion of blocks: dY=2 dX
Kinetic energy T
(1a)
Potential energy V = strain energy in cables + gravitational potential energy change
(1b)
Datum is SEP
Viscous dissipated power
(1c)
External power
(1d)

8. Derive EOM from PCME Substitute Eqs. (3) into Eq. (2) to obtain (2)
Note how static SEP forces cancel
Substitute Eqs. (3) into Eq. (2) to obtain

9. Derive EOM from PCME Cancel velocity dX/dt to obtain final EOM: (2)
(5)