ex2010.2

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EG3170 Modelling and simulation of engineering systems

Simple translational problems

1. Consider the system depicted below in which the displacement of the mass, x(t) is taken to be the dis-

placement corresponding to unstretched/uncompressed spring conditions. Draw a Free Body Diagram of

the system and show that the equation of motion can be expressed as

Mx + Bx + Kx = Fa(t)

x

B K

M

Fa(t)

Simple mass-spring-damper system

2. Consider the system depicted below in which, again, the displacements of the masses M1 and M2 are

measured with respect to unstretch/uncompressed springs.

x1(t) x2(t)

B1

K1

M1

K2

M2Fa(t)

Simple mass-spring-damper-pulley system

Stating appropriate assumptions on x1(t), x2(t) and Fa(t), draw Free Body Diagrams of the system and

prove that the equations of motion can be written as

M1x1 + B1x1 + K1x1 = Fa(t) + B1x2 + K1x2

M2x2 + B1x2 + (K1 + K2)x2 = B1x1 + K1x1

3. Consider the system depicted below in which at rest (Fa(t) 0) the displacements x = 0 and z = 0

correspond to unstretch spring positions. The displacement x(t) is measured with respect to the fixed

reference frame of the ground, whereas the displacement z(t) is taken with respect to the movement of the

mass M1.

Draw Free Body Diagrams of the system and show that the equations of motion are given by:

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B2

x1(t)

z t

B

B1

K1

M1

K2

M2Fa(t)

Mass-spring-damper system with relative motion

4. Consider the system depicted below.

d0 + x(t)

B

K

M Fa(t)


When the system is at rest and Fa(t) = 0, the mass is a constant distance, d0, from the wall. When Fa(t) is

applied, the mass moves a distance x(t) further to the right. Draw a Free Body Diagram for the system

and prove that the equations of motion can be written as:


5. Consider the system depicted below where F1(t) is a constant force and Fa(t) is a time-varying applied

force.

B

K

MFa(t)

F1


Find an expression for the constant (static) extension fo the spring due to F1(t).

Show that the differential equation describing the system motion has the form

Mx + B x + Kx = Fa(t)

and state what x represents.

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Solutions

1. Damper extended, spring compressed. Hence Free Body Diagram looks like:

Bx

KxMx

M

Fa(t)

Applying DAlemberts Law:

i fi = 0

Mx + Bx + Kx Fa(t) = 0


2. Asssume that

F(t) is sufficiently large such that all pulley cables remain in tension at all times

x1(t) > x2(t), which implies x(t) = x1(t) x2(t) > 0

Free Body Diagrams are

B1(x1 x2)

B1(x1 x2)

K1(x1 x2)

K1(x1 x2)

M1

M1x

K2zM2

M2x

Fa(t)

Hence Applying DAlemberts Law (

i fi = 0) to both masses gives:

M1 : M1x1 + B1(x1 x2) + K1(x1 x2) Fa(t) = 0

M1x1 + B1x1 + K1xs = Fa(t) + B1x2 + K1x2

M2 : M2x2 + K2x2 B1(x1 x2)K1(x1 x2) = 0

M2x2 + B2x2 + (K1 + K2)x2 = B1x1 + K1x1

3. Damper B1 and Spring K1 extended; Damper B extended; Damper B2 and Spring K2 compressed.

Free Body Diagrams are:

Bz

BzB1x

K1xM1

M1x

K2(x + z)

B2(x + z)M2M2(x + z)

Fa(t)

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Hence Applying DAlemberts Law (

i fi = 0) to both masses gives:

M1 : M1x + B1x + K1x Bz = 0

M1x + B1x + K1x = Bz

M2 : M2(x2 + z) + B2(x + z) + K2(x + z) + Bz = Fa(t)

4. When system is at rest and Fa(t) = 0, system is at distance d0m from the wall. Hence restraining forces

are purely due to extra displacement x (and subsequent derivatives) which Fa(t) causes. Hence

FK = KxFB = Bx

FI = Mx

A simple Free Body Diagram and application of DAlemberts Law then yields the differential equations

as given in the question.

5. First assume Fa(t) 0 and system is at rest. As F1 is a constant force to the rigth, mass will move a

(constant) distance x0 to right, assuming this measurement is taken from the unstretched/uncompressed

spring length. Hence at static equilibrium we have the Free Body Diagram:

Kx0 M F1

Note that as x0 = 0 and x0 = 0, both the inertial and damper forces are zero. Therefore

F1 = Kx0 x0 = F1/K

Now assume Fa(t) = 0. Now, with x(t) assumed to be the displacement of the mass to the right (a

measurement corresponding to unstretched/uncompressed springs) the Free Body Diagram becomes:

Kx

MMxBx

F1

Fa(t)

Applying DAlemberts Law (

i fi = 0) gives

Mx + Bx + Kx = Fa(t) + F1

Now let x = x + x0, then as x0 is a constant, it follows that x = x and x = x. Substituting for x we thus

have

Mx + B x + K(x + x0) = Fa(t) + F1

But as F1 = Kx0, this simplifies to

Mx + B x + Kx = Fa(t)

as suspected. Note that x represents the distance from static equilibrium not that distance from the point

corresponding to no spring extension/compression (which is represented by x).

ex2010.2

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