AME 352, Fall 12, HW-3 Answers
Problem 1:
a.1)
2
3
A
R2
4 = 0 o
R4
1 = 270 o
R3
R1
O4
C
P
a.2)
R 2 + R 3 R 4 R1 = 0
b.1)
R2 cos 2 + R3 cos 3 R4 cos 4 R1 cos 1 = 0
R2 sin 2 + R3 sin 3 R4 sin 4 R1 sin 1 = 0
Plug in the constants:
1cos 2 + R3 cos 3 2 = 0
1

AME 352, Fall 12, HW-8
Problem 1: One link of a mechanism is shown where AB is 5.0 units and AC is 8.0 units. The
velocity and acceleration of A, and the axes of the velocity and acceleration of B are known.
a) Graphically determine the angular velocity o

AME 352, Fall 12, HW-3
Problem 1: For an inverted slider-crank mechanism the following data are given:
AO2 = 1.0, AC = 5.0, a = 2.0, b = 1.0, AB = 1.2, BP = 2.2
y
a
A
x
(2)
B
O2
(3)
b
(4)
O4
C
P
(Figure is not to scale)
a.1) Define vectors and show the co

AME 352, Fall 2012, HW-2
1. For this six-bar mechanism we have defined several vectors that can be used in formulating the
kinematic equations.
(a) Define angles 1 . 7 for the vectors according to our convention. Show these angles on the
diagram.
(b) Meas

AME 352, Fall 12, HW-4 Answers
Problem 1:
a.1)
L2 cos 2 a cos( 4 + 90 o ) RBO4 cos 4 L1 = 0
L2 sin 2 a sin( 4 + 90 o ) RBO4 sin 4 = 0
Plug in the constants and 2 = 45 :
2.5 cos 45 1.0 cos( 4 + 90 o ) RBO4 cos 4 3.0 = 0
2.5 sin 45 1.0 sin( 4 + 90 o ) RBO4

AME 352, Fall 2012, HW-1 Answers
1.
P
P2
1
P
P
3
P
P
2
1
P
1
S
3
S
2
S
P
3
P
(a)
(b)
1
P
P
2
S
1
(c)
P
P
1
S
2
P
P
2
S
P
3
P
P
3
3
5
P
4P
4
5
P
S
P
(d)
(e)
(f)
P
5
P
S
P
1
P
P
3
2xP
P
2
4
P
S
4
3
2
1
P
1P
P
(g)
3
2
P
4
S
(h)
(i)
P
4
P
P
P
1
3
P
2
3
P-S
P

AME 352, Fall 20112, HW-9
Apply the Acceleration Polygon method to all three problems.
Problem 1: For an inverted slider-crank mechanism the following data are given:
AO2 = 1.0, AC = 5.0, AB = 1.2, BP = 2.2
In the shown configuration 2 = 2.0 rad/sec CCW a

AME 352, Fall 20112, HW-6
Apply the Instant Center method to all three problems.
Problem 1: For an inverted slider-crank mechanism the following data are given:
AO2 = 1.0, AC = 5.0, AB = 1.2, BP = 2.2
In the shown configuration 2 = 2.0 rad/sec CCW.
a) Loc

AME 352, Fall 12, Answers HW-5
Problem 1:
a) The mechanism is drawn for the given crank angle.
b) The polygon is shown.
t
c) The angular velocity of link 4 is 4 = VAO4 / RAO4 = 0.5 rad/sec CCW.
d) The velocity of point P is shown; its magnitude is VP = 3

AME 352, Fall 12, Answers HW-9
Problem 1:
a)
O
O
c
AAO4
AC
s
AAO4
n
AAO2
t
ACA
n
AO2
A
AP
t
AAO4
n
ACA
n
AAO4
t
AAO2
b) 3 =
c)
d)
t
AAO4
RAO4
n
APA
t
AAO2
A
t
APA
= 0.65 CCW
AP = 4.6 as shown
AC = 2.1 as shown
Problem 2:
a)
O
n
AAO4
c
AAB
s
AAB
n
AO2
A
t

AME 352
GRAPHICAL VELOCITY ANALYSIS
5. GRAPHICAL VELOCITY ANALYSIS
Velocity analysis forms the heart of kinematics and dynamics of mechanical systems.
Velocity analysis is usually performed following a position analysis; i.e., the position and
orientation

AME 352, Fall 12, HW-4
Problem 1: For this offset inverted slider-crank mechanism, the following data are given:
RAO2 = L2 =2.5, RO4O2 = L1 =3.0, RAB = a = 1.0 , 2 = 1.0 rad/sec CW, 2 = 0
B
R AB
RAO 2
O2
RBO 4
4
A
RO4 O2
O4
(Figure is not to scale)
a) Ana

AME 352, Fall 32, Answers to HW-11
Problem 1:
Spring force: Fs = 25(4.5 2.0) = 62.5
Weights: W2 = 245, W3 = 29.4, W4 = 39.2
(a) Power Formula:
We assume 2 = 1 rad/sec CCW and construct the velocity
polygon. We determine the velocities of the mass centers.

AME 352, Fall 12, Answers HW-6
Problem 1:
a) The ICs are shown.
R
18 mm
b) 3 = 2 1,22,3 = (2.0)
= 0.57 CCW
63mm
R1,32,3
c) VP = 3 RP1,3 = (0.57)(32 mm /18 mm ) = 1.0 as shown
d) VC = 3 RC 1,3 = (0.57)(48 mm /18 mm ) = 1.52 as shown
I 2,3
(2)
I 2,4
B
I 3,4

AME 352, Fall 13, HW-12
Problem 1. For a four-bar mechanism, the following data are given: (SI units)
1 = 3.0, (radius)
2 = 1.0,
3 = 2.5,
4 = 4.0,
5 = 3.0
Link 2 is in the form of a disc with its mass center at O2. The mass centers of links 3 and 4 are at

AME 352, Fall 13, HW-8
Problem 1: Consider the four-bar mechanism shown where
= 4.0 , 2 = 1.0 , 3 = 3.4 , 4 = 2.5 ,
1
Link 2 rotates with an angular velocity of 1.0 rad/sec
CCW and an angular acceleration of 2 = 1.0 CW. In
the configuration shown a veloci

AME 352, Fall 13, HW-13
1. For the four-bar mechanism shown, the following data are given: (SI units)
1 = 1.2,
2 = 1.0,
3 = 2.0,
4 = 2.5,
Link 2 has its mass center at O2 . The mass center of link 3 is shown. The mass center of link
4 is at its geometric

AME 352, Fall 13, Answers HW-6
Problem 1:
a) The ICs are shown.
b)
R
3 = 2 1,22,3
R1,32,3
= (2.0)
c)
I 2,3
(3)
15mm
= 0.36 CW
83mm
B
I 3,4
I 1,2
I 1,4
VP = 3 RP1,3
P
= (0.36)(45mm /15mm) = 1.0
as shown
d)
(2)
I 2,4
VP
(4)
VC = 3 RC1,3
= (0.36)(82mm /15mm

AME 352, Fall 13, HW-11
1. For a four-bar mechanism, the following data are given: (SI units)
1 = 3.0, (radius)
2 = 1.0,
3 = 2.5,
4 = 4.0,
5 = 3.0
Link 2 is in the form of a disc with its mass center at O2. The mass centers of links 3 and 4
are at their c

AME 352, Fall 13, HW-7
Problem 1: Consider the four-bar mechanism shown where
= 4.0 , 2 = 1.0 , 3 = 3.4 , 4 = 2.5 , BC = 1.0 , CP = 1.7
1
Link 2 rotates with an angular velocity of 1.0 rad/sec CW. For the configuration shown:
(a) Construct the velocity po

AME352 Dynamics of Machines
Fall 2016 Project 1
Recliner chairs come in different shapes and sizes. Some
employ simple mechanisms to perform the reclining
operation, and some use more complex mechanisms. Some
recliners have an attached leg-recliner (footr

AME 352
In-Class Exercise 7
Name: _
1. A six-bar mechanism
For dimensions take direct measurements from the figure either in inches or in centimeters.
B
(3)
(4)
C
(5)
A
(2)
O2
(6)
O4
D

AME 352
In-Class Exercise 6
Name: _
1. For a four-bar mechanism the following data are given:
L1 = 4.0, L2 = 2.0, L3 = 2.5, L4 = 3.0
2. For a slider-crank mechanism (inversion 1) the following data are given:
L2 = 2.0, L3 = 3.0
3. For an offset slider-cra