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7
Chapter 2
Position and Displacement
2.1
Describe and sketch the locus of a point A which moves according to the equations
RAx  atcos  2t  , RAy  atsin  2 t  , RAz  0 .
2.2
Find the position difference from point P to point Q on the curve y  x 2  x  16 , where
RPx  2 and RQx  4 .
2
RPy   2   2  16  10 ; R P  2ˆi 10ˆj
RQy   4   4  16  4 ;
2
R Q  4ˆi  4ˆj
RQP  RQ  R P  2ˆi  14ˆj  14.14281.9
Ans.
8
2.3
The path of a moving point is defined by the equation y  2 x 2  28 . Find the position
difference from point P to point Q if RPx  4 and RQx  3 .
RPy  2  4   28  4 ;
2
R P  4ˆi  4ˆj
2
RQy  2  3  28  10 ; RQ  3ˆi  10ˆj
R  R  R  7ˆi  14ˆj  15.652243.4
QP
2.4
Q
P
Ans.
The path of a moving point P is defined by the equation y  60  x3 / 3 . What is the
displacement of the point if its motion begins when RPx  0 and ends when RPx  3 ?
RPy  0   60   0  / 3  60 ; R P  0   60ˆj
3
3
RPy  3  60   3 / 3  51 ; R P  3  3ˆi  51ˆj
ΔR  R  3  R  0   3ˆi  9ˆj  9.487  71.6
P
2.5
P
P
Ans.
If point A moves on the locus of Problem 2.1, find its displacement from t = 2 to t =2.5.
R A  2.0   2.0a cos 4 ˆi  2.0a sin 4 ˆj  2.0aˆi
R  2.5  2.5a cos5 ˆi  2.5a sin 5 ˆj  2.5aˆi
A
ΔR A  R A  2.5  R A  2.0   4.5aˆi
Ans.
9
2.6
The position of a point is given by the equation R  100e j 2t . What is the path of the
point? Determine the displacement of the point from t = 0.10 to t = 0.40.
The point moves in a circle of radius 100 with its center at the origin.
R  0.10   100e j 0.628  80.902ˆi  58.779ˆj
R  0.40   100e j 2.513  80.902ˆi  58.779ˆj
ΔR  R  0.40   R  0.10   161.803ˆi  161.803180
2.7

Ans.
Ans.

The equation R  t 2  4 e j t /10 defines the position of a point. In which direction is
the position vector rotating? Where is the point located when t = 0? What is the next
value t can have if the direction of the position vector is to be the same as it is when t =
0? What is the displacement from the first position of the point to the second?
Since the polar angle for the position vector is
   jt /10 , then d / dt is negative and therefore
the position vector is rotating clockwise.
Ans.
2

j
0
Ans.
R  0  0  4 e
 40


The position vector will next have the same direction
when  t /10  2 , that is, when t=20.
Ans.
2
 j 2
R  20    20  4  e
 4040
R  R  20  R  0  4000
2.8
Ans.
2
The location of a point is defined by the equation R   4t  2 e jt / 30 , where t is time in
seconds. Motion of the point is initiated when t = 0. What is the displacement during the
first 3 s? Find the change in angular orientation of the position vector during the same
time interval.
R  0    0  2  e j 0  20  2ˆi
R  3  12  2  e j9 / 30  1454  8.229ˆi  11.326ˆj
2.9
ΔR  R  3  R  0   6.229ˆi  11.326ˆj  12.92661.2
Ans.
  54  0  54 ccw
Ans.
10
2.9
Link 2 in the figure rotates according to the equation   t / 4 . Block 3 slides outward
on link 2 according to the equation r  t 2  2 . What is the absolute displacement
R P from t = 1 to t = 2? What is the apparent displacement R P ?
3
3/ 2
R P3  re j   t 2  2  e j t / 4
R P3 1  345  2.121ˆi  2.121ˆj
R  2   690  6ˆj
P3
ΔR P3  R P3  2   R P3 1  2.121ˆi  3.879ˆj  4.421118.7
R
 re j 0  t 2  2 ˆi

P3 / 2

Ans.
2
R P3 / 2 1  3ˆi 2
R  2   6ˆi
P3 / 2
2
ΔR P3 / 2  R P3 / 2  2   R P3 / 2 1  3ˆi 2
2.10
Ans.
A wheel with center at O rolls without slipping so that its center is displaced 10 in to the
right. What is the displacement of point P on the periphery during this interval?
Since the wheel rolls without slipping,
RO   RPO .
  RO / RPO
 10 in / 6 in  1.667 rad  95.5
For R PO ,
       270  95.5  174.5
R   6174.5  5.972ˆi  0.574ˆj in
PO

ΔR P  ΔR O  R PO  R PO

 10ˆi  5.972ˆi  0.574ˆj  6ˆj
ΔR P  4.028ˆi  6.574ˆj  7.71058.5 in
2.11
Ans.
A point Q moves from A to B along link 3 while link 2 rotates from  2  30 to
 2  120 . Find the absolute displacement of Q.
R  330  2.598ˆi  1.500ˆj
Q3
RQ3   3120  1.500ˆi  2.598ˆj
ΔRQ3  RQ3   RQ3  4.098ˆi  1.098ˆj
ΔR
 R  6.000ˆi
Q5 / 3
BA
ΔR Q5  ΔR Q3  ΔR Q5 / 3
ΔRQ5  1.902ˆi  1.098ˆj  2.19630 in
11
2.12
The linkage shown is driven by moving the sliding block 2. Write the loop-closure
equation. Solve analytically for the position of sliding block 4. Check the result
graphically for the position where   45 .
The loop-closure
equation is
R A  R B  R AB .
RAe j /12  RB  RAB e
j   
 RB  RAB e j
Taking the imaginary components of this, we get
RA sin15   RAB sin 
sin 
sin  45
RA   RAB
 200
 546 mm
sin15
sin15
2.13
Ans.
The offset slider-crank mechanism is driven by rotating crank 2. Write the loop-closure
equation. Solve for the position of the slider 4 as a function of  2 .
RC  R A  R BA  RCB
RC  RAe j / 2  RBAe j2  RCB e j3
Taking real and imaginary parts,
RC  RBA cos 2  RCB cos3 and
0  RA  RBA sin  2  RCB sin 3
and, solving simultaneously, we get
  R  R sin  2 
3  sin 1  A BA
 with 90  3  90
RCB


2
RC  RBA cos 2  RCB
  RA  RBA sin  2   2.5cos 2  41.75  5sin  2  6.25cos2  2 Ans.
2
2.14
Write a calculator program to find the sum of any number of two-dimensional vectors
expressed in mixed rectangular or polar forms. The result should be obtainable in either
form with the magnitude and angle of the polar form having only positive values.
Because the variety of makes and models of calculators is vast and no standards exist for
programming them, no standard solution is shown here.
12
2.15
Write a computer program to plot the coupler curve of any crank-rocker or double-crank
form of the four-bar linkage. The program should accept four link lengths and either
rectangular or polar coordinates of the coupler point relative to the coupler.
Again the variety of programming languages makes it difficult to provide a standard
solution. However, one version, written in FORTRAN IV, is supplied here as an
example. There are no accepted standards for programming graphics. Therefore the
Tektronix PLOT-10 subroutine library, for display on Tektronix 4010 series displays, is
chosen as an older but somewhat recognized alternative. The symbols in the program
correspond to the notation shown in Figure 2.19 of the text. The required input data are:
 X5, Y5, -1
R1, R2, R3, R4, 
R5, θ5, 1
The program can be verified using the data of Example 2.7 and checking the results
against those of Table 2.3.
PROGRAM CCURVE
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
A FORTRAN IV PROGRAM TO PLOT THE COUPLER CURVE OF ANY CRANK-ROCKER
OR DOUBLE-CRANK FOUR-BAR LINKAGE, GIVEN ITS DIMESNIONS.
WRITTEN FOR A DEC 11/70 COMPUTER SYSTEM, USING SUBROUTINES FROM
TEKTRONIX PLOT-10 FOR DISPLAY ON 4010 SERIES DISPLAYS.
REF:J.J.UICKER,JR, G.R.PENNOCK, & J.E.SHIGLEY, ‘THEORY OF MACHINES
AND MECHANISMS,’ THIRD EDITION, OXFORD UNIVERSITY PRESS, 2003.
EXAMPLE 2.6
WRITTEN BY: JOHN J. UICKER, JR.
ON:
01 JANUARY 1980
READ IN THE DIMENSIONS OF THE LINKAGE.
READ(5,1000)R1,R2,R3,R4,X5,Y5,IFORM
1000 FORMAT(6F10.0,I2)
FIND R5 AND ALPHA.
IF(IFORM.LE.0)THEN
R5=SQRT(X5*X5+Y5*Y5)
ALPHA=ATAN2(Y5,X5)
ELSE
R5=X5
ALPHA=Y5/57.29578
END IF
Y5=AMAX1(0.0,R5*SIN(ALPHA))
INITIALIZE FOR PLOTTING AT 120 CHARACTERS PER SECOND.
CALL INITT(1200)
SET THE WINDOW FOR THE PLOTTING AREA.
CALL DWINDO(-R2,R1+R2+R4,-R4,R4+R4+Y5)
CYCLE THROUGH ONE CRANK ROTATION IN FIVE DEGREE INCREMENTS.
TH2=0.0
DTH2=5.0/57.29578
IPEN=-1
DO 2 I=1,73
CTH2=COS(TH2)
STH2=SIN(TH2)
13
C
C
C
C
C
C
C
C
C
C
CALCULATE THE TRANSMISSION ANGLE.
CGAM=(R3*R3+R4*R4-R1*R1-R2*R2+2.0*R1*R2*CTH2)/(2.0*R3*R4)
IF(ABS(CGAM).GT.0.99)THEN
CALL MOVABS(100,100)
CALL ANMODE
WRITE(7,1001)
1001
FORMAT(//’ *** THE TRANSMISSION ANGLE IS TOO SMALL. ***’)
GO TO 1
END IF
SGAM=SQRT(1.0-CGAM*CGAM)
GAM=ATAN2(SGAM,CGAM)
CALCULATE THETA 3.
STH3=-R2*STH2+R4*SIN(GAM)
CTH3=R3+R1-R2*CTH2-R4*COS(GAM)
TH3=2.0*ATAN2(STH3,CTH3)
CALCULATE THE COUPLER POINT POSITION.
TH6=TH3+ALPHA
XP=R2*CTH2+R5*COS(TH6)
YP=R2*STH2+R5*SIN(TH6)
PLOT THIS SEGMENT OF THE COUPLER CURVE.
IF(IPEN.LT.0)THEN
IPEN=1
CALL MOVEA(XP,YP)
ELSE
IPEN=-1
CALL DRAWA(XP,YP)
END IF
TH2=TH2+DTH2
2 CONTINUE
DRAW THE LINKAGE.
CALL MOVEA(0.0,0.0)
CALL DRAWA(R2,0.0)
XC=R2+R3*COS(TH3)
YC=R3*SIN(TH3)
CALL DRAWA(XC,YC)
CALL DRAWA(XP,YP)
CALL DRAWA(R2,0.0)
CALL MOVEA(XC,YC)
CALL DRAWA(R1,0.0)
1 CALL FINITT(0,0)
CALL EXIT
STOP
END
14
2.16
For each linkage shown in the figure, find the path of point P: (a) inverted slider-crank
mechanism; (b) second inversion of the slider-crank mechanism; (c) straight-line
mechanism; (d) drag-link mechanism.
(a)
(c)
(b)
(d)
15
2.17
Using the offset slider-crank mechanism of Fig. 2.15, find the crank angles corresponding
to the extreme values of the transmission angle.
As shown,   90  3 .
Also from the figure
e  r2 sin  2  r3 cos  .
Differentiating with
respect to  2 ,
d
r2 cos  2   r3 sin 
;
d 2
r cos  2
d
 2
.
d 2
r3 sin 
Now, setting d / d 2  0 , we get cos 2  0 .
Therefore, we conclude that 2    2k  1 / 2  90, 270,
2.18
Ans.
In Section 1.10 it is pointed out that the transmission angle reaches an extreme value for
the four-bar linkage when the crank lies on the line between the fixed pivots. Referring to
Fig. 2.19, this means that  reaches a maximum or minimum when crank 2 is coincident
with the line O2 O4 . Show, analytically, that this statement is true.
From O4O2 A :
s 2  r12  r22  2r1r2 cos 2 .
Also, from ABO4 :
s 2  r32  r42  2r3r4 cos  .
Equating these we differentiate
with respect to  2 to obtain
d
2r1r2 sin  2  2r3 r4 sin 
or
d 2
r r sin  2
d
 12
.
d 2 r3r4 sin 
Now, for
d
 0 , we have sin  2  0 . Thus,  2  0,  180,  360,
d 2
Q.E.D.
16
2.19
The figure illustrates a crank-and-rocker linkage in the first of its two limit positions. In a
limit position, points O2 , A, and B lie on a straight line; that is, links 2 and 3 form a
straight line. The two limit positions of a crank-rocker describe the extreme positions of
the rocking angle. Suppose that such a linkage has r1  400 mm , r2  200 mm ,
r3  500 mm , and r4  400 mm .
(a)
(b)
(c)
Find  2 and  4 corresponding to each limit position.
What is the total rocking angle of link 4?
What are the transmission angles at the extremes?
(a)
From isosceles triangle O4O2 B we
calculate or measure  2  29 ,
 4  58 and  2  248 ,  4  136 .
(b)
can
(c)
   68 .
2.20
Then  4   4   4  78 .
Ans.
Finally, from isosceles triangle O4O2 B ,   29
Ans.
A double-rocker mechanism has a dead-center position and may also have a limit position
(see Prob. 2.19). These positions occur when links 3 and 4 in the figure lie along a
straight line. In the dead-center position the transmission angle is 180 and the
mechanism is locked. The designer must either avoid such positions or provide the
external force, such as a spring, to unlock the linkage. Suppose, for the linkage shown in
the figure, that r1  14 in , r2  5.5 in , r3  5 in , and r4  12 in . Find  2 and  4
corresponding to the dead-center position. Is there a limit position?
and
17
For the given dimensions, there are
two dead-center positions, and they
correspond to the two extreme
travel positions of crank O2 A .
From O4 AO2 using the law of
cosines, we can find  2  114.0 ,
 4  162.8 and, symmetrically,
 2  114.0 ,  4  162.8 . There
are also two limit positions, these
occur at  2  56.5 ,  4  133.1
and, symmetrically, at  2  56.5 ,
Ans.
 4  133.1 .
2.21
The figure shows a slider-crank mechanism that has an offset e and that is placed in one
of its limiting positions. By changing the offset e, it is possible to cause the angle that
crank 2 makes in traversing between the two limiting positions to vary in such a manner
that the driving or forward stroke of the slider takes place over a larger angle than the
angle used for the return stroke. Such a linkage is then called a quick-return mechanism.
The problem here is to develop a formula for the crank angle traversed during the forward
stroke and also develop a similar formula for the angle traversed during the return stroke.
The ratio of these two angles would then constitute an time ratio of the drive to return
strokes. Also determine which direction the crank should rotate.
18
From the figure we can see that e   r3  r2  sin 2   r3  r2  sin 2 180 or

e 
e 
1 
 ,  2  180  sin 

 r3  r2 
 r3  r2 
 2  sin 1 
 e 
e 
1 
 drive   2   2  180  sin 1 
  sin 

 r3  r2 
 r3  r2 
 e 
e 
1 
 return   2  360   2  180  sin 1 
  sin 

 r3  r2 
 r3  r2 
Assuming driving when sliding to the right, the crank should rotate clockwise.
Ans.
Ans.
Ans.
19
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