7
Chapter 2
Position and Displacement 2.1
Describe and sketch the locus of a point A which moves according to the equations y
x = atcos 2πt , R = atsin 2π t , R z = 0 . R A ( ) A ( ) A
2.2
Find the position difference from point P to point Q on the curve y = x 2 + x − 16 , where x R x P = 2 and RQ = 4 .
2 y RP = ( 2 ) + 2 − 16 16 = −10 ; R P = 2ˆi − 10ˆj 2
RQ y = ( 4 ) + 4 − 16 = 4 ;
R Q = 4ˆi + 4ˆj
R QP = R Q − R P = 2ˆi + 14ˆj = 14.142∠81.9° 2.3
Ans.
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 R x P = 4 and R x = −3 . Q
8
2
RP = 2 ( 4 ) − 28 = 4 ; y
R P = 4ˆi + 4ˆj
2 y RQ = 2 ( −3) − 28 = −10 ; R Q = −3ˆi − 10ˆj
R QP = RQ − R P = −7ˆi − 14ˆj = 15.652∠ 243.4°
2.4
Ans.
The path of a moving point P is defined by the equation y = 60 − x3 / 3 . What is the x displacement of the point if its motion begins when R x P = 0 and ends when RP = 3 ?
3 R yP ( 0 ) = 60 − ( 0 ) / 3 = 60 ; R P ( 0 ) = 60ˆj 3 y RP ( 3) = 60 − ( 3) / 3 = 51 ; R P ( 3) = 3ˆi + 51ˆj
ΔR P
2.5
= R P ( 3) − R P ( 0 ) = 3ˆi − 9ˆj = 9.487∠ − 71.6°
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 4π ˆi + 2.0a sin 4π ˆj = 2.0aˆi R A ( 2.5) = 2.5a cos 5π ˆi + 2.5a sin 5π ˆj = − 2.5aˆi ΔR A
2.6
= R A ( 2.5) − R A ( 2.0 ) = −4.5aˆi
Ans.
The position of a point is given by the equation R = 100e j 2 πt . 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.
Ans.
9
R ( 0.10 ) = 100e j
0.628
R ( 0.40 ) = 100e
j 2.513
ΔR
2.7
= 80.902ˆi + 58.779ˆj = − 80.902ˆi + 58.779ˆj
= R ( 0.40) − R ( 0.10) = − 161.803ˆi = 161.803∠180°
Ans.
The equation R = t 2 + 4 e − j πt /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 θ = − j πt /10 , then dθ / dt is negative and therefore the position vector is rotating clockwise. Ans. R ( 0 ) = 02 + 4 e − j 0 = 4∠0°
(
)
Ans.
The position vector will next have the same direction when π t /10 = 2π , that is, when t =20. Ans. R ( 20 ) = ( 20 + 4 ) e 2
− j 2π
= 404∠ 0°
ΔR = 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 πt / 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 π9 / 3 0 = 14∠54° = 8.229ˆi + 11.326ˆj ΔR
= R ( 3) − R ( 0 ) = 6.229ˆi + 11.326ˆj = 12.926∠61.2°
Δθ = 54° − 0° = 54° ccw
Ans. Ans.
10
2.9
Link 2 in the figure rotates according to the equation θ = π 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
= ( t 2 + 2 ) e jπ t / 4
R P3 = re
jθ
R P3 (1) = 3∠ 45° = 2.121ˆi + 2.121ˆj R P3 ( 2 ) = 6∠90° = 6ˆj ΔR P
3
= R P ( 2 ) − R P (1) = −2.121ˆi + 3.879ˆj = 4.421∠118.7° 3
R P3 / 2 = re
j 0
Ans.
3
= ( t 2 + 2 ) ˆi 2
R P3 / 2 (1) = 3ˆi 2 R P3 / 2 ( 2 ) = 6ˆi 2 ΔR P
3
2.10
/2
= R P / 2 ( 2 ) − R P / 2 (1) = 3ˆi 2 3
Ans.
3
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, Δ RO = −Δθ RPO .
Δθ = −Δ RO / RPO = −10 in / 6 in = −1.667 rad = − 95.5° For R PO , θ ′ = θ + Δθ = 270° − 95.5° = 174.5° R PO′ = 6∠174.5° = − 5.972ˆi + 0.574ˆj in ΔR P
(
= ΔR O + R PO′ − R PO
)
= 10ˆi − 5.972ˆi + 0.574ˆj + 6ˆj ΔR P
2.11
= 4.028ˆi + 6.574ˆj = 7.710∠ 58.5° in
Ans.
A point Q moves from A to B along link 3 while link 2 rotates from θ 2 = 30° to θ 2′ = 120° . Find the absolute displacement of Q. R Q3 = 75 ∠30° = 64.95ˆi + 37.5ˆj R Q3 ′ = 75∠120° = − 37.5ˆi + 64.95ˆj ΔR Q
3
ΔR Q
5
= R Q ′ − R Q = −102.45ˆi + 27.45ˆj 3
/3
3
= R BA = 150ˆi
ΔR Q
= ΔR Q + ΔR Q / 3
ΔR Q
= 47.5ˆi + 27.5ˆj = 55∠ 30° mm
5
5
3
5
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 φ = − 45° . The loop-closure equation is R A = R B + R AB . jπ /12
R Ae
= RB + RAB e j (
π +φ )
= R B − RAB e jφ
Taking the imaginary components of this, we get R A sin15° = − RAB sin φ R A = − RAB
2.13
sin φ sin15°
= −200
sin − 45° sin15°
= 546 mm
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 θ 2 . R C = R A + R BA + RCB RC = RAe
jπ / 2
Taking real and imaginary parts, RC = RBA cosθ 2 + RCB cosθ 3 and
+ RBA e jθ + RCB e jθ 2
3
0 = R A + RBA sin θ 2 + RCB sin θ 3
and, solving simultaneously, we get
⎛ − RA − RBA sin θ 2 ⎞ ⎟ with −90° < θ 3 < 90° R CB ⎝ ⎠
θ 3 = sin −1 ⎜
RC = RBA cos θ 2 +
2
RCB − ( RA + RBA sin θ 2 ) = 63cos θ2 + 1043.75 − 125sin θ 2 + 156.25cos θ2 2
2
Ans.
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
A FORTRAN I V PROGRAM TO PLOT THE COUPLER CURVE OF ANY CRANK- ROCKER OR DOUBLE- CRANK FOUR- BAR LI NKAGE, GI VEN I TS DI MESNI ONS. WRI TTEN FOR A DEC 11/ 70 COMPUTER SYSTEM, USI NG SUBROUTI NES FROM TEKTRONI X PLOT- 10 FOR DI SPLAY ON 4010 SERI ES DI SPLAYS. REF: J . J . UI CKER, J R, G. R. PENNOCK, & J . E. SHI GLEY, ‘ THEORY OF MACHI NES AND MECHANI SMS, ’ THI RD EDI TI ON, OXFORD UNI VERSI TY PRESS, 2003. EXAMPLE 2. 6 WRI TTEN BY: J OHN J . UI CKER, J R. ON: 01 J ANUARY 1980
READ I N THE DI MENSI ONS OF THE LI NKAGE. READ( 5, 1000) R1, R2, R3, R4, X5, Y5, I FORM 1000 FORMAT( 6F10. 0, I 2) C C FI ND R5 AND ALPHA. I F( I FORM. LE. 0) THEN R5=SQRT( X5*X5+Y5* Y5) ALPHA=ATAN2( Y5, X5) ELSE R5=X5 ALPHA=Y5/ 57. 29578 END I F Y5=AMAX1( 0. 0, R5*SI N( ALPHA) ) C C I NI TI ALI ZE FOR PLOTTI NG AT 120 CHARACTERS PER SECOND. CALL I NI TT( 1200) C C SET THE WI NDOW FOR THE PLOTTI NG AREA. CALL DWI NDO( - R2, R1+R2+R4, - R4, R4+R4+Y5) C C CYCLE THROUGH ONE CRANK ROTATI ON I N FI VE DEGREE I NCREMENTS. TH2=0. 0 DTH2=5. 0/ 57. 29578 I PEN=- 1 DO 2 I =1, 73 CTH2=COS( TH2) STH2=SI N( TH2) C
13
C
CALCULATE THE TRANSMI SSI ON ANGLE. CGAM=( R3*R3+R4*R4- R1*R1- R2*R2+2. 0*R1*R2*CTH2) / ( 2. 0*R3*R4) I F( ABS( CGAM) . GT. 0. 99) THEN CALL MOVABS( 100, 100) CALL ANMODE WRI TE( 7, 1001) 1001 FORMAT( / / ’ * ** THE TRANSMI SSI ON ANGLE I S TOO SMALL. ** *’ ) GO TO 1 END I F SGAM=SQRT( 1. 0- CGAM*CGAM) GAM=ATAN2( SGAM, CGAM) C C CALCULATE THETA 3. STH3=- R2*STH2+R4* SI N( GAM) CTH3=R3+R1- R2*CTH2- R4*COS( GAM) TH3=2. 0*ATAN2( STH3, CTH3) C C CALCULATE THE COUPLER POI NT POSI TI ON. TH6=TH3+ALPHA XP=R2*CTH2+R5*COS( TH6) YP=R2*STH2+R5*SI N( TH6) C C PLOT THI S SEGMENT OF THE COUPLER CURVE. I F( I PEN. LT. 0) THEN I PEN=1 CALL MOVEA( XP, YP) ELSE I PEN=- 1 CALL DRAWA( XP, YP) END I F TH2=TH2+DTH2 2 CONTI NUE C C DRAW THE LI NKAGE. CALL MOVEA( 0. 0, 0. 0) CALL DRAWA( R2, 0. 0) XC=R2+R3*COS( TH3) YC=R3*SI N( 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 FI NI TT( 0, 0) CALL EXI T 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, γ = 90° − θ 3 . Also from the figure e + r2 sin θ 2 = r 3 cos γ . Differentiating with respect to θ 2 , r2 cosθ 2 = − r 3 sin γ d γ dθ 2
=−
r 2 cos θ 2 r 3 sin γ
d γ d θ 2
;
.
Now, setting d γ / d θ 2 = 0 , we get cosθ 2 = 0 . Therefore, we conclude that θ 2 = ± ( 2k + 1) π / 2 = ±90°, ± 270° ,… Ans. 2.18
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 γ reaches a maximum or minimum when crank 2 is coincident with the line O2 O4 . Show, analytically, that this statement is true. From ΔO4O2 A : s = r1 + r2 − 2 r1r2 cos θ 2 . 2
2
2
Also, from Δ ABO4 : s 2 = r32 + r42 − 2 r3 r4 cos γ .
Equating these we differentiate with respect to θ 2 to obtain γ 2r1r2 sin θ 2 = 2r3 r4 sin d γ dθ 2
Now, for
d γ d θ 2
=
r1r 2 sin θ 2 r3 r 4 sin γ
d γ d θ 2
or
.
= 0 , we have sin θ 2 = 0 . Thus, θ 2 = 0, ± 180°, ± 360° ,…
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 r 1 = 400 mm , r 2 = 200 mm , r 3 = 500 mm , and r 4 = 400 mm .
(a) (b) (c)
Find θ 2 and θ 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
can calculate or measure θ 2 = 29° , θ 4 = 58° and θ 2′ = 248° , θ 4′ = 136° . Then Δθ 4 = θ 4′ − θ 4 = 78° .
Ans.
(c) Finally, from isosceles O4O2 B , γ = 29° and γ ′ = 68° .
triangle Ans.
(b)
2.20
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 r 1 = 14 in , r 2 = 5.5 in , r 3 = 5 in , and r 4 = 12 in . Find θ 2 and θ 4 corresponding to the dead-center position. Is there a limit position? 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 θ 2 = 114.0° , θ 4 = 162.8° and,
symmetrically,
θ 2′′′ = −114.0° , θ 4′′′ = −162.8° . There are also two limit positions, these occur at θ 2′ = 56.5° , θ 4′ = 133.1° and, symmetrically, at θ 2′′ = −56.5° , θ 4′′ = −133.1° .
Ans.
17
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.
From the figure we can see that e = ( r3 + r2 ) sin θ 2 = ( r3 − r2 ) sin (θ 2′ − 180° ) or
⎛
⎞ e ⎞ −1 ⎛ ′ , θ 180 sin = ° + ⎟ 2 ⎜ ⎟ ⎝ r3 + r 2 ⎠ ⎝ r3 − r 2 ⎠ ⎛ e ⎞ e ⎞ −1 ⎛ = θ 2′ − θ 2 = 180° + sin −1 ⎜ ⎟ − sin ⎜ ⎟ r r r r − + 3 2 3 2 ⎝ ⎠ ⎝ ⎠
θ 2 = sin −1 ⎜
Δθ drive
e
Ans.
⎛ e ⎞ e ⎞ −1 ⎛ Δθ return = θ 2 + 360° − θ 2′ = 180° + sin −1 ⎜ ⎟ − sin ⎜ ⎟ ⎝ r3 + r2 ⎠ ⎝ r3 − r2 ⎠
Ans.
Assuming driving when sliding to the right, the crank should rotate clockwise.
Ans.
