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ME 352 - Machine Design I
Name of Student
___________________________
Fall Semester 2011
Lab Section Number
________________________
Homework No. 3 (30 points). Due at the beginning of lecture on Wednesday, September 14th.
Consider the variation of the Scotch-yoke mechanism shown in Problem 3.26 on pages 161
and 162. For the given position of the input angle, that is,
o
2
45 ,
θ
=
perform a position analysis
of the mechanism using trigonometry (that is, the law of sines and the law of cosines).
If the input link 2 of the mechanism is rotating clockwise with a constant angular velocity of
36 rad/s, that is,
2
36 k
ω
= −
rad/s then for the given position of the mechanism, perform a
velocity analysis using:
(i) The method of kinematic coefficients.
(ii) The method of instantaneous centers of velocity. Draw the Kennedy circle and list the
primary instant centers and the secondary instant centers. Clearly write the scale that you used
for your scaled drawing of the mechanism.
Compare the answers that you obtained from Part (i) with the answers that you obtained from
Part (ii).

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Solution to Homework 3 (30 Points).
Position analysis of the mechanism.
The vectors for the variation of the Scotch-yoke mechanism are as shown in Figure 1. Note
that
4
R
is the vector from the ground pivot
2
O
to point B, fixed in link 4, that is, the time rate
of change of this vector is the velocity of link 4. Also, note that
34
R
is the vector from point B
(fixed in link 4) to point A (which is fixed in link 3). The time rate of change of this vector, that
is,
34
R
,
±
is the relative velocity between these two points (which are fixed on different links).
Figure 1. The vectors for a kinematic analysis of the mechanism.
The known constants are:
2
R
250 mm,
=
o
4
0 ,
θ
=
and
o
34
105 .
θ
=
For the given input
position
o
2
45
θ
=
, the unknown position variables
4
R
and
34
R
can be obtained from
trigonometry.
For example, using the law of sines gives
34
2
4
2
2
2
R
R
R
sin
ABO
sin
AO B
sin
O AB
=
=
∠
∠
∠
(1)
Therefore, the distance between point B and point A can be written as
2
34
2
2
R
250
R
sin
AO B
sin 45
sin
ABO
sin 75
=
∠
=
°
∠
°
(2a)
Therefore, the distance between point B (fixed in link 4) and point A (fixed in link 3) is
34
R
183.01mm
=
(2b)
Similarly, the distance between the ground pivot
2
O
and point B can be written as
2
4
2
2
R
250
R
sin
O AB
sin 60
sin
ABO
sin 75
=
∠
=
°
∠
°
(3a)
Therefore, the distance between the ground pivot
2
O
and point B is
4
R
224.14 mm
=
(3b)