This document discusses force analysis of mechanisms. It covers key concepts like kinematics, kinetics, determining reaction forces, power requirements, and vibrations from running mechanisms. It also provides equations for calculating angular momentum of rigid bodies about a point, including moments of inertia. It derives the equations relating angular acceleration, applied moments and reaction forces for any point on a rigid body undergoing plane motion.
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Kinematics:
• Dimensions of various links of the mechanism
• Velocity and acceleration of various links
(linear or angular as the case may be)
Kinetics (Force Analysis)
• Determines the forces and moments to obtain the
desired motion
• Power required for running the mechanisms
• Reaction forces in the links and those transferred
to the foundation
• Vibration resulting from running mechanisms
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î
ĵ
k̂
Q’(x,y,z)
dm
R
r
(x,y)
Three dimensional object
performing plane motion
Plane of
motion -xy
R: position vector of a point of the object
r: projection of R on to the plane of motion
P(0,0)
Q
B
ω
Angular momentum about a point
Elemental mass dm at Q’(x,y,z)
Q(x,y) is a project ion of Q’
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î
ĵ
k̂
(x,y,z)
dm
R
r
Q(x,y)
P(0,0)
Angular momentum of a body about a point
ˆ
Q P PQ
v v k r
B
ω
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î
ĵ
k̂
Q’(x,y,z)
dm
R
r
Q(x,y)
P(0,0)
Angular momentum of a body about a point
ˆ
Q P PQ
v v k r
B
P
B
dm
H R v
:angular momentum about the point P
P
H
ω Volume integral
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î
ĵ
k̂
Q’(x,y,z)
dm
R
r
Q(x,y)
P(0,0)
Angular momentum of a body about a point
ˆ
Q P PQ
v v k r
B
P
B
dm
H R v
due to plane
motion
Q
v v
ˆ ˆ ˆ ˆ
P P PQ
B
x y z dm
H i j k v k r
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ˆ ˆ ˆ ˆ
ˆ ˆ
P P
B
x y z x y dm
H i j k v k i j
R PQ
r
Angular momentum of a body about a point
P P
B B
dm dm
R v R v
PC
B
m dm
r R
: position vector of the mass center
PC
r
P PC P
B
dm m
R v r v
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Angular momentum of a body about a point
ˆ ˆ 0
ˆ ˆ ˆ
ˆ ˆ 0
ˆ ˆ ˆ
ˆ ˆ
ˆ
ˆ ˆ
ˆ
i i
i j k
j j
j i k
k i j
k j i
ˆ ˆ ˆ ˆ
ˆ ˆ
B
x y z x y dm
i j k k i j
ˆ ˆ ˆ ˆ
ˆ
B
x y z y x dm
i j k i j
2 2 ˆ ˆ
ˆ
B
x y xz yz dm
k i j
ˆ ˆ ˆ ˆ
ˆ ˆ
P P
B
x y z x y dm
H i j k v k i j
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2 2 ˆ ˆ
ˆ
B
x y xz yz dm
k i j
Angular momentum of a body about a point
2 2
ˆ
B
x y dm
k ˆ
B
xzdm
i ˆ
B
yzdm
j
2 2
P
zz
B
I x y dm
P
XZ
B
I xzdm
P
YZ
B
I yzdm
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Angular momentum of a body about a point
ˆ ˆ ˆ
P P P
P PC P XZ YZ ZZ
m I I I
H r v i j k
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Angular momentum about the mass center
ˆ ˆ ˆ
P P P
P PC P XZ YZ ZZ
m I I I
H r v i j k
ˆ ˆ ˆ
C C C
C XZ YZ ZZ
I I I
H i j k
ˆ ˆ ˆ
C C C
C
XZ YZ ZZ
C
d d
I I I
dt dt
H
M i j k
Moments about the mass center=rate of change of
angular momentum
about the mass center
P≡C
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ˆ ˆ ˆ
C C C
C
XZ YZ ZZ
C
d d
I I I
dt dt
H
M i j k
ˆ ˆ ˆ
ˆ ˆ
ˆ ˆ
C C C
XZ YZ ZZ
C
C C C
XZ YZ ZZ
I I I
I I I
M i j k
k i j k
2 2
2 2
ˆ ˆ ˆ ˆ
ˆ
ˆ ˆ ˆ
C C C C C
XZ YZ ZZ XZ YZ
C
C C C C C
XZ YZ YZ XZ ZZ
I I I I I
I I I I I
M i j k j i
i j k
Moments about the mass center
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2 2
ˆ ˆ ˆ
C C C C C
XZ YZ YZ XZ ZZ
C
I I I I I
M i j k
If the cross moments of inertia are zero
ˆ
C
ZZ
C
I
M k
Moments about the mass center
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Moments about any other point
P C PC C
m
M H r a
ˆ
C
C ZZ
C
I
M H k
ˆ
C
P ZZ PC C
I m
M k r a
ˆ ˆ ˆ
P P P
P PC P XZ YZ ZZ
m I I I
H r v i j k
P C PC C
m
H H r v
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ˆ ˆ ˆ
P P P
P PC P XZ YZ ZZ
m I I I
H r v i j k
P C PC C
m
H H r v
ˆ ˆ ˆ
P P P
C PC C PC P XZ YZ ZZ
m m I I I
H r v r v i j k
Equating the above
Moments about any other point
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ˆ ˆ ˆ
P P P
C PC C PC P XZ YZ ZZ
m m I I I
H r v r v i j k
ˆ ˆ ˆ
C PC C PC C
P P P
XZ YZ ZZ
PC P PC P
m m
d I I I
m m
dt
H r a r v
i j k
r a r v
P
M
Taking time-derivative on both sides
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ˆ ˆ ˆ
P PC P C
P P P
XZ YZ ZZ
PC P
m
d I I I
m
dt
M r v v
i j k
r a
=0
ˆ ˆ ˆ
P P P
XZ YZ ZZ
P PC P
d I I I
m
dt
i j k
M r a
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ˆ
P
P PC P ZZ
m I
M r a k
If the cross products of inertia are zero
If the point P is a fixed point on the link
ˆ
P
P ZZ
I
M k
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ˆ
C
P PC C ZZ
m I
M r a k
ˆ
P
P PC P ZZ
m I
M r a k
ˆ
C
ZZ
C
I
M k
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C
α
F
C
a
ˆ
C
M k
C
C
ZZ
M
I
m
C
F
a
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A 250 mm long link has a mass
of 2 kg. Its center of gravity is
located at its mid-point and it has
a mass moment of inertia of 0.1
kg-m2. It is inclined at an angle
of 60 degrees. At this instant,
the angular velocity is 30 rad/s
(counter clockwise) and angular
acceleration 15 rad/sec2
(clockwise) A force of 200 N is
acting at its free end as shown .
Determine the reaction force at
the hinge and the torque that
has to be applied.
600
250
F
O
P
C
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0 0
ˆ ˆ
0.250 cos60 sin 60
ˆ ˆ
0.125 0.2165 m
OP
r i j
i j
0 0
ˆ ˆ
200 cos25 sin 25
ˆ ˆ
181.26 84.52 N
F i j
i j
ˆ ˆ
ˆ ˆ
30 0.125 0.2165 m
ˆ ˆ
6.48 3.75 m/s
P O OP
v v k r k i j
i j
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2
2
2
ˆ
ˆ ˆ
ˆ
15 0.125 0.2165
ˆ ˆ
30 0.125 0.2165
ˆ ˆ
110.625 196.275 m/s
P O OP OP
a a k r r
k i j
i j
i j
2
1 ˆ ˆ
110.625 196.275 m/s
2
C
a i j
Since the acceleration of P is a factor of the length OP,
the acceleration of C is obtained by dividing the acceleration
of P by 2
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ˆ ˆ ˆ ˆ ˆ
181.26 84.52 2 9.81
1 ˆ ˆ
2 110.625 196.275
2
ˆ ˆ
181.26 84.52 2 9.81
ˆ ˆ
110.625 196.275
C
x y
x y
m
F F
F F
F a
i j i j j
i j
i j
i j
F
O
P
C
T
Fy
Fx
mg
291.885
x
F N
300.415
y
F N
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ˆ
C
P PC C ZZ
m I
M r a k
ˆ
P
P PC P ZZ
m I
M r a k
ˆ
C
ZZ
C
I
M k
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ˆ
C
O OC C ZZ
m I
M r a k F
O
P
C
T
Fy
Fx
mg
ˆ
ˆ
1 ˆ ˆ ˆ
ˆ 0.125 0.2165 2 9.81
2
ˆ ˆ ˆ ˆ
0.125 0.2165 181.26 84.52
O OC OP
T mg
T
M k r j r F
k i j j
i j i j
ˆ
(T 29.9040)
O
M k
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ˆ
1
ˆ ˆ ˆ ˆ
0.125 0.2165 2 110.625 196.275
2
ˆ ˆ
0.1 15 2.0841
C
OC C ZZ
m I
r a k
i j i j
k k
ˆ
(T 29.9040)
O
M k
ˆ ˆ
2.0841
C
OC C ZZ
m I
r a k k
T=27.82 N-m
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O P
C
ℓ
The uniform rod of mass m is released from
rest in the horizontal position indicated in the
figure. Determine the reaction at the pin as a
function of the angle between the horizontal
and the rod measured in the clockwise
direction
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Fx
Fy
mg
ĵ
θ
-θ
O
C
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0
2
2
^
2
^
j
12 2
j
3
o zz
OC
OC
M I
m
mg m
m
mg
r
r
(1)
cos -sin
2
OC
^ ^
r i j
(2)
2
^
cos -sin j
2 3
m
mg
^ ^
i j
(3)
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2
^
cos -sin j
2 3
m
mg
^ ^
i j
3
cos (clockwise)
2
g
(4)
3
cos
2
g
(5)
2
1
3
sin
2 2
g
C
(6)
2
3
cos
2
3
sin
g
g
(7)
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2
2 2
2
3 3
cos cos -sin sin cos -sin
2 2 2
3 3
sin cos cos sin cos sin
4 2
3 sin cos 3 (1 sin )
4 4
c o OC OC
g g
g g
g g
^
^ ^ ^ ^ ^
^ ^ ^ ^
^ ^
a a k r r
k i j i j
i j i j
i j
(8)
2
3 sin cos
4
3 (1 sin )
4
x
y
mg
F
mg
F mg
(9)
