Mechanics of Machines
Dr. Mohammad Kilani
Class 9
DYNAMIC FORCE ANALYSIS
Suggested Problems
11-(8,12)
INTRODUCTION
 Kinematic analysis of mechanisms
has been used to define the
position, velocity and acceleration
of the elements of a particular
mechanism, Dynamic analysis, is
concerned with the forces and
torques in the system.
 We will take that approach in this
chapter and concentrate on solving
for the forces and torques that
result from, and are required to
drive, our kinematic system in such
a way as to provide the required
accelerations.
INTRODUCTION
 Determine the force
acting on the table
by:
1. Taking a free body
diagram around A
and B together.
2. Taking a free body
diagram around A
and another one
around B.
W = 50 N
W = 100 N
Single Link in Pure Rotation
 Consider the single link in pure
rotation shown. To carry out a
dynamic analysis on the link, the
kinematics of the problem must first
be fully defined. That is, the angular
speed and angular accelerations,
and the linear accelerations of the
CG must be found for the position
of interest.
 The mass and the mass moment of
inertia with respect to the CG must
also be known.
 In addition there may be external
forces or torques applied to any
member of the system. These
must be known as well.
y
P
CG
ω2 , α2
O2
θ2
x
Newtonian Solution Method
 Dynamic force analysis can be
done by any of several
methods. The one which gives
the most information about
forces internal to the
mechanism requires only the
use of Newton's law. These
can be written as a summation
of all forces and torques in the
system.
 In two-dimensional problems,
the above vector equations
reduce to three independent
scalar equations


 F  ma

 M  IG
 F  ma
 F  ma
M  I 
x
x
y
y
G
Newtonian Solution Method
 These three equations must be written for each
moving body in the system which will lead to a
set of linear simultaneous equations for any
system.
 If the kinematic accelerations are large
compared to gravity, which is often the case,
then the weight forces can be ignored in the
dynamic analysis.
 If the machine members are very massive or
moving slowly with small kinematic
accelerations, or both, the weight of the
members may need to be included in the
analysis. The weight can be treated as an
external force acting on the CG of the member
at a constant angle.
 F  ma
 F  ma
M  I 
x
x
y
y
G
G
D’Alembert Principle
 Consider a moving rigid body of mass
m acted upon by any system of
forces, say F1, F2. and F3. Designate
the center of mass of the body as
point G, and find the resultant of the
force system from the equation.
 D’Alemebr principle states that the
inertia terms resulting from the linear
acceleration of the center of mass,
and the angular acceleration of the
body can be treated as forces in the
opposite direction to the inertia
terms.


 F  maG

 M G  I G


F

m
a
  G 0

 M G  I G  0
D’Alembert Principle
 D’Alembert principle is summarized as follows. The vector sum of all the
external forces and the inertia forces acting upon a rigid body is zero.
The vector sum of all the external moments and the inertia torques
acting upon a rigid body is also separately zero.


F

m
a


M

I


G


 F  ma  0

M

I

0

G
D’Alembert Principle
 D’Alembert principle is summarized as follows. The vector sum of all the
external forces and the inertia forces acting upon a rigid body is zero.
The vector sum of all the external moments and the inertia torques
acting upon a rigid body is also separately zero.


F

m
a


M

I


G


 F  ma  0

M

I

0

G
Example 1
 Determine the force F required to
produce a velocity v = 1 m/s for the
mechanism shown when θ3 = 120°.
Assume that the linkage moves in a
frictionless horizontal plane. Link 3 is
100 mm long, has a mass of 1 kg, and
I3 = 2×10-3 kg.m.s2.
Example 1
 Determine the force F required to produce a
velocity v = 1 m/s for the mechanism shown
when θ3 = 120°. Assume that the linkage
moves in a frictionless horizontal plane. Link 3
is 100 mm long, has a mass of 1 kg, and I3 =
2×10-3 kg.m.s2.
Position Loop Closure Equation :
  
r2  r3  r4
r2 cos  2  r3 cos  3  r4 cos  4
r2 sin  2  r3 sin  3  r4 sin  4
DesignParameters : r3  0.1 m,  2  90 ,  4  0
0.1 cos  3  r4  0
0.1sin  3  r2
Position analysisparameter :  3  120
r4  0.1 cos 120  0.05 m
r2  0.1sin 120  0.0866 m
Example 1
 Determine the force F required to produce a
velocity v = 1 m/s for the mechanism shown
when θ3 = 120°. Assume that the linkage
moves in a frictionless horizontal plane. Link 3
is 100 mm long, has a mass of 1 kg, and I3 =
2×10-3 kg.m.s2.
0.1cos 3  r4  0
0.1sin 3  r2
Velocity Loop Closure Equation :
 0.13 sin  3  v4  0
0.13 cos  3  v2
Velocity analysisparameter : v2  1 m/s
3 
v2
1

 20 rad/s
0.1 cos  3 0.1 cos 120
Example 1
 Determine the force F required to produce a
velocity v = 1 m/s for the mechanism shown
when θ3 = 120°. Assume that the linkage
moves in a frictionless horizontal plane. Link 3
is 100 mm long, has a mass of 1 kg, and I3 =
2×10-3 kg.m.s2.
Acceleration Loop Closure Equation :
 0.132 cos  3  0.1 3 sin  3  a4  0
 0.132 sin  3  0.1 3 cos  3  a2
Acceleration analysisparameter : a2  0 m/s 2
0.1 3 cos  3  0.132 sin  3
 3  32 tan  3   20 2 tan 120  692.8 rad/s 2
a4  0.132 cos  3  0.1 3 sin  3
a4  0.1 20  cos 120  0.1  692.8  sin 120
2
a4  80 m/s 2
Acceleration of the center of mass (G) :



aG  aB  aG B



aB  a4  80i m/s 2



aG B  32 rG B u3   3rG B u3 2

0.1  
0.1  


aG B   202
u3    692.8
u3 2
2
2









aG B  20 cos 120 i  sin 120 j



 34.64 cos 210 i  sin 210 j  40i




aG  80i  40i  40i




Example 1
 Determine the force F required to produce a
velocity v = 1 m/s for the mechanism shown
when θ3 = 120°. Assume that the linkage
moves in a frictionless horizontal plane. Link 3
is 100 mm long, has a mass of 1 kg, and I3 =
2×10-3 kg.m.s2.


 ma  40i N

 I G  1.386 N.m
M


F
   ma

M

I


G
Iα = 13.86 N.m
maG= 40 N
C
0
r2
 I G  FAr4
2

1  0.086

FA 
 1.386   62.12 j N
 40
0.05 
2

FG
C