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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 2t , 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 14.14281.9 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 15.652243.4 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 2t . 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 161.803180 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 jt /10 , then d / dt is negative and therefore the position vector is rotating clockwise. Ans. 2 j 0 Ans. R 0 0 4 e 40 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 4040 R R 20 R 0 4000 2.8 Ans. 2 The location of a point is defined by the equation R 4t 2 e jt / 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 20 2ˆi R 3 12 2 e j9 / 30 1454 8.229ˆi 11.326ˆj 2.9 ΔR R 3 R 0 6.229ˆi 11.326ˆj 12.92661.2 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 1 345 2.121ˆi 2.121ˆj R 2 690 6ˆj P3 ΔR P3 R P3 2 R P3 1 2.121ˆi 3.879ˆj 4.421118.7 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 6174.5 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 7.71058.5 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 330 2.598ˆi 1.500ˆj Q3 RQ3 3120 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 2.19630 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 j2 RCB e j3 Taking real and imaginary parts, RC RBA cos 2 RCB cos3 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 [Page intentionally blank.]