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ME 352  Machine Design I
Name of Student__
_________________________
Spring Semester 2011
Lab Section Number
________________________
Homework No. 5 (30 points).
Due at the beginning of lecture on Monday, February 21st.
Consider the slidingblock linkage shown in Example 2.5, Figure 2.24, page 77, and repeated
here as Figure 1. The length of the ground link is
14
2
RO
O9
i
n
c
h
e
s
=
=
and the length of link 2
is
22
A
4
.
5
i
n
c
h
e
s
.
==
The position solution for the input angle
o
2
135
θ=
can be obtained
from trigonometry and the answers are
34
4
R
O A
12.59 inches
=
=
and
.
o
3
434
104.64
The angular velocity and the angular acceleration of the input link 2 are
2
5krad s
ω=−
/
and
2
2
10 k rad s ,
α=
/
respectively. The length of link 4 is
4
OC 2
0
i
n
c
h
e
s
=
and note that the origin
of the XY reference frame is now chosen to be coincident with the ground pin O
2
, see Figure 1.
Part A.
Use the vector loop approach to determine the firstorder and the secondorder kinematic
coefficients of this linkage in the given position.
Part B. Using the method of kinematic coefficients determine:
(i) the relative velocity and acceleration between links 3 and 4 (denote as
34
R
±
and
34
R
±±
).
(ii) the angular velocity and acceleration of link 4. Give the magnitudes and the directions.
(iii) the velocity and acceleration of point C. Give the magnitudes and the directions.
(iv) the unit tangent and normal vectors to the path of point C. Show the directions of these
vectors on a figure.
(v) the radius of curvature of the path of point C.
(vi) the X and Y coordinates of the center of curvature of the path of point C.
Figure 1. The SlidingBlock Linkage.
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Solution to Homework Set 5.
Part A. A suitable set of vectors for the slidingblock linkage are shown in Figure 2.
Figure 2. The vectors for the slidingblock linkage.
The vector loop equation (VLE) for the linkage is
??
0
123
4
I
RRR
√√
√
+
−=
(1)
The X and Y components of Equation (1) are
11
2 2
3
4 3
4
cos
cos
cos
0
RR
R
θ
θθ
+
(2a)
and
3
4
sin
sin
sin
0
R
+
(2b)
Differentiating Equations (2) with respect to the input angle
θ
2
gives
2
2
34
34 34
34
34
sin
sin
cos
0
R
′
′
−+
−
=
(3a)
and
2
2
34
34 34
34
34
cos
cos
sin
0
R
′
′
−
(3b)
3
Equations (3) can be written in matrix form as
34
34
34
34
22
34
34
34
34
sin
θ
cos
θθ
sin
θ
cos
θ
sin
θ
R
cos
θ
R
R
R
R
′
−
⎡⎤
⎡
⎤
⎡
⎤
=
⎢⎥
⎢
⎥
⎢
⎥
′
−−
−
⎣
⎦
⎣⎦
⎣
⎦
(4)
The determinant of the coefficient matrix is
34
34
34
34
34
34
34
34
34
34
34
sin
θ
cos
θ
sin
θ
cos
θ
cos
θ
sin
θ
R
DET
R
R
R
R
−
==
−
−
=
−
(5)
Using Cramer’s rule, the firstorder kinematic coefficient for link 3 can be written as
3
4
3
4
2
2
34
2
2
34
34
34
sin
cos
cos
sin
sin
sin
cos
cos
R
R
RR
DET
R
θ
−
′
−
(6a)
which can be written as
3
4
34
34
cos (
)
R
R
−
′ =
(6b)
The firstorder kinematic coefficient for links 3 and 4 can be written as
34
34
2
2
34
34
2
2
2
34
2
34
2
34
2
34
34
34
sin
sin
cos
cos
cos
sin
sin
cos
R
DET
R
−+
′
−
(7a)
which can be written as
23
4
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This note was uploaded on 04/23/2011 for the course ME 352 taught by Professor Staff during the Spring '08 term at Purdue University.
 Spring '08
 Staff
 Machine Design

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