Problem 2.21
Link 2 of the linkage shown in the figure has an angular velocity of 10 rad/s CCW. Find the angular velocity of link 6 and the velocities of points B, C, and D.
A D C B 2 4 6 5 ω2 E F 3
X
Y
0.3" AE = 0.7" AB = 2.5" AC = 1.0" BC = 2.0" EF = 2.0" CD = 1.0" DF = 1.5" θ2 = 135˚ θ2 Position AnalysisLocate points E and F and the slider line for B. Draw link 2 and locate A. Then locate B. Next locate C and then D.
Velocity Analysis: vA3 =vA2 =vA E2/ 2 vB4 =vB3=vA3+vB A3/ 3 (1) Find vC3 by image. vC5=vC3 vD5=vD6 =vD F6/ 6 =vC5+vD C5/ 5 (2) Now, vB3 in horizontal direction vA E2/ 2 =ω2×rA E/ ⇒vA E2/ 2 =ω2⋅rA E/ =10 0 7 7⋅ . = in / s(⊥torA E/ ) vB A3/ 3=ω3×rB A/ ⇒vB A3/ 3 =ω3⋅rB A/ (⊥torB A/ )
E A B C D F 2 3 4 6 Slider Line Velocity Polygon 2.5 in/s a3 o f6 b3 c3 c5 d5 vB3 = .3 29in / s Using velocity image,
vC5=vC3 = .6 78in/ s Now,
vD F6/ 6 =ω6×rD F/ ⇒vD F6/ 6 =ω6⋅rD F/ (⊥torD F/ )
vD C5/ 5=ω5×rD C/ ⇒vD C5/ 5 =ω5⋅rD C/ (⊥torD C/ )
Solve Eq. (2) graphically with a velocity polygon. From the polygon,
vD5=vD6 =vD F6/ 6 =6 78. in / s or ω6 6 6 6 6 6 78 1 5 4 52 =v = = r D F D F / / . . . rad / s CCW
Problem 2.22
The linkage shown is used to raise the fabric roof on convertible automobiles. The dimensions at given as shown. Link 2 is driven by a DC motor through a gear reduction. If the angular velocity, ωωω
ω2 = 2 rad/s, CCW, determine the linear velocity of point J, which is the point where the linkage connects to the automobile near the windshield.
AB = 3.5" AC = 15.37" BD = 16" CD = 3" CE = 3.62" EG = 13.94" GF = 3.62" HF = 3" FC = 13.62" HI = 3.12" GI = 3.62" HL = 0.75" KC = 0.19" JH = 17" B A E C D F G H I J H F C D K L 2 3 4 6 7 8 5 3 1 2 ω Detail of Link 3 110˚ Position Analysis:
Draw linkage to scale. Start with link 2 and locate points C and E. Then locate point D. Then locate points F and H. Next locate point G. Then locate point I and finally locate J.
Velocity Analysis:
The equations required for the analysis are:
vC2 =vC A2/ 2 =ω2×rC A2/ 2 ⇒vC2 =ω2rC A2/ 2 = ⋅2 15 37( . )=30 74. in / s vC3=vC2 vD3=vD4=vD B4/ 4 =vC3+vD C3/ 3 (1) vG5 =vG6=vG7 =vF5+vG F5/ 5 =vE6 +vG E6/ 6 vF5 =vF3 vE6 =vE2 So,
vF5+vG F5/ 5 =vE6 +vG E6/ 6 (2) vI7=vI8 =vG7+vI G7/ 7=vH8+vI H8/ 8 vH8 =vH3 So, vG7 +vI G7/ 7=vH8+vI H8/ 8 (3) c3 d3 f3 h3 e2 g6 i8 j8 A B C D E G F H I J 2 4 3 6 5 7 8 10 in/s Velocity Polygon o
Now,
vC3=30 74. in / s(⊥torC A/ )
vD C3/ 3 =ω3×rD C/ (⊥torD C/ )
vD B4/ 4 =ω4×rD B/ (⊥torD B/ )
Solve Eq. (1) graphically with a velocity polygon. The velocity directions can be gotten directly from the polygon. The magnitudes are given by:
vD4 =30 9. in / s
Using velocity image of link 3, find the velocity of points F and H and of link 2, find the velocity of point E. vF5 =30 5. in / s vH3 =30 3. in / s and vE6 = .3 80in / s Now, vG F5/ 5=ω5×rG F/ (⊥torG F/ ) vG E6/ 6 =ω6×rG E/ (⊥torG E/ )
Solve Eq. (2) graphically with a velocity polygon. The velocity directions can be gotten directly from the polygon. The magnitudes are given by:
vG6 =37 8. in / s Now,
vI G7/ 7 =ω7×rI G/ (⊥torI G/ )
vI H8/ 8 =ω8×rI H/ (⊥torI H/ )
Solve Eq. (3) graphically with a velocity polygon. Using velocity polygon of link 8
Problem 2.23
In the mechanism shown, determine the sliding velocity of link 6 and the angular velocities of links 3 and 5. 2 4 5 6 B C D E F = 3 ω2 rads 50˚ 10.4" 2.0" 29.5" A AB = 12.5" BC = 22.4" DC = 27.9" CE = 28.0" DF = 21.5" 3 34˚ Position Analysis
First locate Points A and E. Next draw link 2 and locate B. Then locate point C by drawing a circle centered at B and 22.4 inches in radius, and finding the intersection with a circle centered at E and of 28 inches in radius. Find D by drawing a line 27.9 inches long at an angle of 34˚ relative to line BC. Locate the slider line 2 inches above point E. Draw a circle centered at D and 21.5 inches in radius and find the intersections of the circle with the slider line. Choose the proper intersection corresponding to the position in the sketch.
Velocity Analysis
Compute the velocity of the points in the same order that they were drawn. The equations for the four bar linkage are:
vB2 =vB A2/ 2 =ω2×rB A/
vB3 =vB2
A B C D 10 in/sec Velocity Polygon o E c3 F b3 d3 f 3 Also, vC3=vC4 =vE4 +vC E4/ 4 =vC E4/ 4 where, vB2 = ω2rB A/ = ⋅3 12 5 37 5. = . in/ sec vC B3/ 3=ω3×rC B/ (⊥to CB) vC E4/ 4 =ω4×rC E/ (⊥to CE)
The velocity of C3 (and C4) can then be found using the velocity polygon. After the velocity of C3
vD5=vD3 vF5 =vD5+vF D5/ 5 and vF5 =vF6 where vF D5/ 5 =ω5×rF D/ ⇒vC5 = ω5rF D/ (⊥to FD)
and vF6 is along the slide direction.. Then the velocity of F5 (and F6) can be found using the
velocity polygon. From the polygon,
vF6 = 43.33 in/sec vC B3/ 3 =26 6. in/ sec vF D5/ 5 =18 54. in/ sec ω3=vrC B3 3 =26 622 4=1 187 C B rad / / . . . / sec ω5 =vrF D5 5 =18 5421 5 =0 862 F D rad / / . . . / sec
To determine the direction for ω3, determine the direction that rC B/ must be rotated to be in the
direction of vC B3/ 3. From the polygon, this direction is CCW.
To determine the direction for ω5, determine the direction that rF D/ must be rotated to be in the
Problem 2.24
In the mechanism shown, vA2 = 15 m/s. Draw the velocity polygon, and determine the velocity of point D on link 6 and the angular velocity of link 5.
1vA2= 15 m/s 2 3 4 5 A C B D 45˚ 2.05" AC = 2.4" BD = 3.7" BC = 1.2" X Y 2.4" 6 Velocity Analysis: vA3 =vA2 vC4 =vC3=vA3+vC A3/ 3 (1) vB3 =vB5 vD5=vD6 =vB5+vD B5/ 5 (2) Now,
vA3 =15m/ sec in vertical direction
vC3 in horizontal direction
vC A3/ 3 =ω3×rC A3/ 3 ⇒vC A3/ 3 =ω3⋅rC A3/ 3 (⊥torC A3/ 3)
Solve Eq. (1) graphically with a velocity polygon. From the polygon, using velocity image,
vB3 =vB5 =14 44. m/ sec Now,
3 4 5 C B D 2 A 6 a3 c3 o 10 m/sec Velocity Polygon b3 bb5 d5
vD5 along the inclined path
vD B5/ 5 =ω5×rD B5/ 5 ⇒vD B5/ 5 =ω5⋅rD B5/ 5 (⊥torD B5/ 5) Solve Eq. (2) graphically with a velocity polygon. From the polygon,
vD6 =12 31. m/ sec Also, vD B5/ 5 =16 61. m/ sec or ω5 =vrD B5 5 =16 6053 7 =4 488 D B rad CCW / / . . . / sec
Problem 2.25
In the mechanism shown below, points E and B have the same vertical coordinate. Find the velocities of points B, C, and D of the double-slider mechanism shown in the figure if Crank 2 rotates at 42 rad/s CCW. A D C 2 6 5 B 4 33 E ω2 0.75" 60˚ EA = 0.55" AB = 2.5" AC = 1.0" CB = 1.75" CD = 2.05" Position Analysis
Locate point E and draw the slider line for B. Also draw the slider line for D relative to E. Draw link 2 and locate A. Then locate B. Next locate C and then D.
Velocity Analysis: vA3 =vA2 =vA E2/ 2 vB4 =vB3=vA3+vB A3/ 3 (1) vC5=vC3 vD5=vD6 =vC5+vD C5/ 5 (2) Now, vB3 in horizontal direction
A D C 2 6 5 B 4 3 3 E 10 in/sec Velocity Polygon b3 a3 c3 d5 o vA E2/ 2 =ω2×rA E2/ 2 ⇒vA E2/ 2 =ω2⋅rA E/ =42 0 55 23 1⋅ . = . in/ sec (⊥torA E/ ) vB A3/ 3=ω3×rB A3/ 3 ⇒vB A3/ 3 =ω3⋅rB A/ (⊥torB A/ )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vB4 =17 76. in/ sec Using velocity image,
vC3=18 615. in/ sec Now,
Solve Eq. (2) graphically with a velocity polygon. From the polygon,
vD5= .5 63in/ sec
Problem 2.26
Given vA4 = 1.0 ft/s to the left, find vB6.
A B C D 2 4 5 E 6 3 X Y 157.5˚ DE = 1.9" CD = 1.45" BC = 1.1" AD = 3.5" AC = 2.3" 1.0" 0.5" Position Analysis
Draw the linkage to scale. Start by locating the relative positions of A, B and E. Next locate C and D. Velocity Analysis: vA4 =vA3 vC5=vC3 =vA3+vC A3/ 3 (1) vD3=vD2 vD2 =vE2+vD E2/ 2 (2)
vB5 =vC5+vB C5/ 5 Now,
vA4 = .1 0 ft/ sec in horizontal direction
vC A3/ 3 =ω3×rC A/ ⇒vC A3/ 3 =ω3⋅rC A/ (⊥torC A/ )
vD E2/ 2 =ω2×rD E/ ⇒vD E2/ 2 =ω2⋅rD E/ (⊥torD E/ ) Solve Eq. (1) graphically with a velocity polygon.
From the polygon, using velocity image,
vC A3/ 3 =1 28. ft/ sec and, ω3=vrC A3 3 =1 282 3 =0 56 C A / / . . . rad / s
To determine the direction of ω3, determine the direction that rC A/ must be rotated to be parallel to
vC A3/ 3. This direction is clearly clockwise. Now,
Solve Eq. (2) graphically with a velocity polygon. From the polygon,
vB6 = .1 23ft/ sec
Problem 2.27
If vA2 = 10 cm/s as shown, find vC5.
Position Analysis
Draw the linkage to scale. Start by locating the relative positions of D, F and G. Next locate A and B. Then locate E and C.
Velocity Analysis: vA3 =vA2 vB3 =vA3+vB A3/ 3 (1) vB4 =vB3 vB4 =vF4+vB F4/ 4 = +0 vB F4/ 4 (2) vE5=vE4 vG5 =vE5+vG E5/ 5
vG6 =vG5 Now,
vA2 =10cm/ sec (⊥ torA C/ )
vB A3/ 3=ω3×rB A/ ⇒vB A3/ 3 =ω3⋅rB A/ (⊥torB A/ )
vB F4/ 4 =ω4×rB F/ ⇒vB F4/ 4 =ω4⋅rB F/ (⊥torB F/ )
From the polygon,
vB4 = .6 6cm/ sec Using velocity image,
vE4 = .3 12cm/ sec Now,
vG6 is horizontal direction
vG E5/ 5 =ω5×rG E/ ⇒vG E5/ 5 =ω5⋅rG E/ (⊥torG E/ ) For the velocity image
draw a line ⊥ torC E/ at e draw a line ⊥ torC E/ at g and find the point “c” From the velocity polygon
Problem 2.28
If vA2 = 10 in/s as shown, find the angular velocity of link 6.
2 3 4 5 6 A v 2 A 27˚ AB = 1.0" AD = 2.0" AC = 0.95" CE = 2.0" EF = 1.25" BF = 3.85" B C D E F Position Analysis
Draw the linkage to scale. Start by locating the relative positions of B, D and F. Next locate A and C. Then locate E. Velocity Analysis: vA3 =vA2 vD3=vA3+vD A3/ 3 (1) vD4 =vD3 vC5=vC3 vE5=vC5+vE C5/ 5 (2) vE6 =vE5 vE6 =vF6+vE F6/ 6 = +0 vE F6/ 6 Now, vA2 =10in/ sec
vD A3/ 3 =ω3×rD A/ ⇒vD A3/ 3 =ω3⋅rD A/ (⊥torD A/ )
From the polygon,
vD A3/ 3 =9 1. in/ sec Using velocity image,
rD A/ : rC A/ =vD A3/ 3:vC A3/ 3
vC A3/ 3 =4 32. in/ sec Now,
vE C5/ 5=ω5×rE C/ ⇒vE C5/ 5 =ω5⋅rE C/ (⊥torE C/ )
vE F6/ 6 =ω6×rE F/ ⇒vE F6/ 6 =ω6⋅rE F/ (⊥torE F/ ) from the velocity polygon
vE6 = .2 75in/ sec and ω6 =vrE F6 6 =2 752 =1 375 E F / / . . rad / s
To determine the direction of ω6, determine the direction that rE F/ must be rotated to be parallel to
Problem 2.29
The angular velocity of link 2 of the mechanism shown is 20 rad/s, and the angular acceleration is 100 rad/s2 at the instant being considered. Determine the linear velocity and acceleration of point
F6. D 2 3 4 5 6 C B E F ω2,α2 A 115˚ 2" 0.1" 2.44" EF = 2.5" CD = 0.95" AB = 0.5" BC = 2.0" CE = 2.4" BE = 1.8" Position Analysis
Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B and then C. Then locate E and finally F.
Velocity Analysis:
The required equations for the velocity analysis are:
vB3 =vB2=vB A2/ 2 vC3=vC4 =vC D4/ 4 =vB3+vC B3/ 3 (1) vE5=vE3 vF5 =vF6 =vE5+vF E5/ 5 (2) Now, vB A2/ 2 =ω2×rB A2/ 2 ⇒vB A2/ 2 =ω2⋅rB A2/ 2 =20 0 5 10⋅ . = in / s (⊥torB A2/ 2) vC D4/ 4 =ω4×rC D4/ 4 ⇒vC D4/ 4 =ω4⋅rC D4/ 4 (⊥torC D4/ 4) vC B3/ 3=ω3×rC B3/ 3 ⇒vC B3/ 3 =ω3⋅rC B3/ 3 (⊥torC B3/ 3)
vC B3/ 3 =6 59. in / s or ω3 3 3 3 3 6 59 2 3 29 =vrC B = = C B / / . . rad / s CCW Also, vC D4/ 4 =8 19. in / s or ω4 4 4 4 4 8 19 0 95 8 62 =vrC D = = C D / / . . . rad / s CW And, vE5= .5 09in / s Now, vF5 in horizontal direction vF E5/ 5 =ω5×rF E5/ 5 ⇒vF E5/ 5 =ω5⋅rF E5/ 5 (⊥torF E5/ 5) Solve Eq. (2) graphically with a velocity polygon. From the polygon,
vF E5/ 5 =3 97. in / s or ω5 5 5 5 5 3 97 2 5 1 59 =v = = r F E F E / / . . . rad/ s CCW Also, vF6 = .3 79in / s Acceleration Analysis: aB3=aB2=aB A2/ 2 aC3 =aC4 =aC D4/ 4 =aB3+aC B3/ 3 aC Dr4/ 4+aC Dt4/ 4 =arB A2/ 2+aB At2/ 2 +aC Br3/ 3+aC Bt3/ 3 (3)
b3 e5 o 5 in/sec Velocity Polygon f5 b'3 c'3 e'5 o' 50 in/s Acceleration Polygon 2 f'5 1a F5 /E5t 1aF5 1aE 5 D 2 3 4 5 C B E F A 1a F5 /E5 r 1a C3 /B3 t 1a C3 /B3 r 1a C4/D4 t 1a C4 /D4 r 1a B2 /A2 r 1a B2 /A2 t c3 aE5 =aE3 aF5=aF6 =aE5+aF E5/ 5 aF5=aE5+arF E5/ 5+aF Et5/ 5 (4) Now, aB Ar rB A a r B A r B A 2 2 2 2 2 2 2 2 2 2 2 2 20 0 5 2002 / =ω ×(ω × / )⇒ / =ω ⋅ / = ⋅ . = in / s2
in the direction of - rB A2/ 2 aB At rB A a r r B At B A B A 2/ 2 =α2× 2/ 2⇒ 2/ 2 =α2⋅ 2/ 2 =100 0 5 50⋅ . = in / s2(⊥to 2/ 2) aC Br3/ 3=ω3×(ω3×rC B3/ 3)⇒arC B3/ 3 =ω32⋅rC B3/ 3 =3 29 2 21 6. 2⋅ = . in / s2 in the direction of - rC B3/ 3 aC Bt rC B a r r C Bt C B C B 3/ 3 =α3× 3/ 3⇒ 3/ 3 =α3⋅ 3/ 3 (⊥to 3/ 3) aC Dr r a r C D C Dr C D 4 4 4 4 4 4 4 4 4 4 4 2 8 62 0 95 70 62 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ . = . in / s2 in the direction of - rC D4/ 4 aC Dt rC D a r r C Dt C D C D 4/ 4 =α4× 4/ 4⇒ 4/ 4 =α4⋅ 4/ 4 (⊥to 4/ 4)
Solve Eq. (3) graphically with an acceleration polygon. From the polygon, using acceleration image, aE5 =252 0. in / s2 Now, aF5 in horizontal direction aF Er5/ 5=ω5×(ω5×rF E5/ 5)⇒arF E5/ 5 =ω52⋅rF E5/ 5 =1 59 2 5 6 32. 2⋅ . = . in / s2 in the direction of - rF E5/ 5 aF Et rF E a r r F E t F E F E 5/ 5=α5× 5/ 5⇒ 5/ 5 =α5⋅ 5/ 5 (⊥to 5/ 5)
Solve Eq. (4) graphically with an acceleration polygon. From the polygon,
Problem 2.30
In the drag-link mechanism shown, link 2 is turning CW at the rate of 130 rpm. Construct the velocity and acceleration polygons and compute the following: aE5, aF6, and the angular acceleration of link 5. AB = 1.8' BC = 3.75' CD = 3.75' AD = 4.5' AE = 4.35' DE = 6.0' EF = 11.1' 60˚ A C B D E F 3 2 4 5 6 Velocity Analysis: ω2=130rpm=130 260π=13 614. rad / s vC3=vC2=vC B2/ 2 vD3=vD4 =vD A4/ 4 =vC3+vD C3/ 3 (1) vE5=vE4 vF5 =vF6 =vE5+vF E5/ 5 (2) Now, vC B2/ 2 =ω2×rC B/ ⇒vC B2/ 2 =ω2⋅rC B/ =13 614 3 75 51 053. ⋅ . = . ft / s(⊥torC B/ ) vD C3/ 3 =ω3×rD C/ ⇒vD C3/ 3 =ω3⋅rD C/ (⊥torD C/ ) vD A4/ 4 =ω4×rD A/ ⇒vD A4/ 4 =ω4⋅rD A/ (⊥torD A/ )
Solve Eq. (1) graphically with a velocity polygon. From the polygon, using velocity image,
vD C3/ 3 =41 3. ft / s or
20 ft/s Velocity Scale c3 d3 o e4 f5 1a C2/ B2 r 1a D3 /C3 r 1a D3 /C3t 1aD4 /A 4r 1a D4 /A4 t d3' c3' o' 200 ft/s Acceleration Scale 2 a4' 60˚ A C B D E F 3 2 4 5 6 e'4 f5' 1a F5 /E5 t 1a F5 /E5 r
ω3 3 3 3 3 41 3 3 75 11 0 =vrD C = = D C / / . . . rad / s CW Also, vD A4/ 4 =37 19. ft / s or ω4 4 4 4 4 37 19 4 5 8 264 =vrD A = = D A CW / / . . . rad / s And, vE5=35 95. ft / s Now, vF5 in horizontal direction vF E5/ 5 =ω5×rF E5/ 5 ⇒vF E5/ 5 =ω5⋅rF E/ (⊥torF E/ )
Solve Eq. (2) graphically with a velocity polygon. From the polygon,
vF E5/ 5 =12 21. ft / s or ω5 5 5 5 5 12 21 11 1 1 1 =vrF E = = F E / / . . . rad / s CCW Acceleration Analysis: aC3 =aC2=aC B2/ 2 aD Ar a a a a a D A t C Br C Bt D Cr D Ct 4/ 4+ 4/ 4 = 2/ 2+ 2/ 2 + 3/ 3+ 3/ 3 (3) aE5 =aE4 aF5=aF6 =aE5+aF E5/ 5 aF5=aE5+arF E5/ 5+aF Et5/ 5 (4) Now, aC Br rC B a r C B r C B 2 2 2 2 2 2 2 2 13 614 3 75 695 02 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ . = . ft / s2 in the direction of - rC B/
aC Bt2/ 2 =α2×rC B/ ⇒aC Bt2/ 2 =α2⋅rC B/ = ⋅0 3 75 0. = ft / s2 aD Cr rD C a r D C r D C 3 3 3 3 3 3 3 2 11 0 3 75 453 82 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ . = . ft / s2 in the direction of - rD C/ aD Ct rD C a r r D C t D C D C 3/ 3 =α3× / ⇒ 3/ 3 =α3⋅ / (⊥to / ) aD Ar4/ 4=ω4×(ω4×rD A/ )⇒aD Ar4/ 4 =ω42⋅rD A/ =8 264 4 5 307 3. 2⋅ . = . ft / s2 in the direction of - rD A/ aD At rD A a r r D At D A D A 4/ 4 =α4× 4/ 4⇒ 4/ 4 =α4⋅ / (⊥to / )
Solve Eq. (3) graphically with an acceleration polygon. From the polygon, using acceleration image,
aE5 =308 0. ft / s2 Now,
aF5 is in the horizontal direction
aF Er5/ 5=ω5×(ω5×rF E/ )⇒aF Er5/ 5 =ω52⋅rF E/ =1 1 11 1 13 4. 2⋅ . = . ft / s2 in the direction of - rF E/
aF Et5/ 5=α5×rF E5/ 5⇒aF Et5/ 5 =α5⋅rF E5/ 5 (⊥torF E5/ 5)
Solve Eq. (4) graphically with an acceleration polygon. From the polygon,
aF6 =83 4. ft / s2 Also, aF Et5/ 5=325 2. ft / s2 or α5 5 5 5 5 325 2 11 1 29 3 =a = = r F E t F E / / . . . rad / s CCW2
Problem 2.31
The figure shows the mechanism used in two-cylinder 60-degree V-engine consisting, in part, of an articulated connecting rod. Crank 2 rotates at 2000 rpm CW. Find the velocities and acceleration of points B, C, and D and the angular acceleration of links 3 and 5.
A C 2 3 D 5 B 4 E 3 X EA = 1.0" AB = 3.0" BC = 3.0" AC = 1.0" CD = 2.55" 30o 30o Y 6 90˚ Position Analysis
Draw the linkage to scale. First locate the two slider lines relative to point E. Then draw link 2 and locate point A. Next locate points B and C. Next locate point D.
Velocity Analysis:
Find angular velocity of link 2,
ω2 = ⋅ = ⋅π30n π 200030 =209 44. rad / s vA2 =vA E2/ 2 =vA3 =ω2×rA E/
vB3 =vB4 =vA3+vB A3/ 3 (1)
10000 in/s Acceleration Polygon 2 E C A B D 2 4 5 6 3 100 in/s Velocity Polygon a2,a3 b3,bb4 c3, c5 d5,d6 o a’2 b3,bb’4 1a A2 /E2 r 1a B3 /A3r 1a c’3 d5 , d6 1aC 3 1aD5 /C5r 1aD5 /C5t 1aD5 1a B3 /A3t
vD5=vD6 =vC5+vD C5/ 5 (2) Now,
vA2 =ω2rA E/ =209 44 1 209 44. ⋅ = . in / s(⊥torA E/ )
vB3 in the direction of rB E/
vB A3/ 3=ω3×rB A/ ⇒vB A3/ 3 =ω3rB A/ (⊥torB A/ )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vB3 =vB4 =212 7. in / s Also, vB A3/ 3=109 3. in / s or ω3=vrB A3 3 =109 33 =36 43 B A / / . . rad / s
To determine the direction of ω3, determine the direction that rB A/ must be rotated to be parallel to
vB A3/ 3. This direction is clearly clockwise. Also,
vC3=vC5 =243 3. in / s Now,
vD5=vD6 in the direction of rD E/
vD C5/ 5=ω5×rD C/ ⇒vD C5/ 5 =ω5rD C/ (⊥torD C/ )
Solve Eq. (2) graphically with a velocity polygon. From the polygon,
vD5=vD6 =189 2. in/ sec Also, vD C5/ 5=135in / s or ω5 =vrD C5 5 =2 55135 =52 9 D C / / . . rad / s
To determine the direction of ω5, determine the direction that rD C/ must be rotated to be parallel to
vD C5/ 5. This direction is clearly clockwise. Acceleration Analysis: aA2 =aA3 =arA E2/ 2+atA E2/ 2 aB3=aB4 =aA3+aB A3/ 3 aB3=arA E2/ 2+atA E2/ 2+arB A3/ 3+atB A3/ 3 (3) aC3 =aC5 aD5 =aD6 =aC5+aD C5/ 5 aD5 =aD6 =aC5+arD C5/ 5+atD C5/ 5 (4) Now, aB3in the direction of ± rB E/ aA Er2/ 2=ω2×(ω2×rA E/ )⇒arA E2/ 2 =ω22⋅rA E/ =209 44 1 43860. 2⋅ = in / s2 in the direction of - rA E/ aA Et rA E a r A E t A E in 2/ 2 =α2× / ⇒ 2/ 2 =α2⋅ / = ⋅ =0 1 0 / sec2 aB Ar3/ 3=ω3×(ω3×rB A/ )⇒arB A3/ 3 =ω32⋅rB A/ =36 43 3 3981. 2⋅ = in / s2 in the direction of - rB A/ aB At r a r r B A B At B A B A 3/ 3 =α3× / ⇒ 3/ 3 =α3⋅ / (⊥to / )
Solve Eq. (3) graphically with an acceleration polygon. From the polygon,
aB3=14710in / s2 Also, aB At3/ 3 =38460in / s2 or α3= 3 3 =384603 =12820 a r B A t B A / / rad / s 2
To determine the direction of α3, determine the direction that rB A/ must be rotated to be parallel to
aB At3/ 3. This direction is clearly counterclockwise. Also, aC3 =aC5 =39 300, in s/ 2 Now, aD5 in the direction of - rD E/ aD Cr rD C a r D C r D C 5 5 5 5 5 5 5 2 52 9 2 55 71362 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ . = in / s2 in the direction of - rD C/ aD Ct5/ 5 =α5×rD C/ ⇒aD Ct5/ 5 =α5⋅rD C/ (⊥ torD C/ )
Solve Eq. (4) graphically with an acceleration polygon. From the polygon,
aD5 =aD6 =24 000, in / s2 Also, aD Ct5/ 5 =26 300, in / s2 or α5 = 5 5 =263002 55 =10 300 a r D C t D C / / . , rad / s 2
To determine the direction of α5, determine the direction that rD C/ must be rotated to be parallel to aD Ct
5/ 5. This direction is clearly clockwise.
Problem 2.32
In the mechanism shown, ωωωω2 = 4 rad/s CCW (constant). Write the appropriate vector equations, solve them using vector polygons, and
a) Determine vE3, vE4, and ωωωω3. b) Determine aE3, aE4, and αααα3.
Position Analysis
Locate pivots A and D. Draw link 2 and locate B. Then locate point C. Finally locate point E. Velocity Analysis
For the velocity analysis, the basic equation is:
vB2 =vB3 =vB A2/ 2
vC3=vB3+vC B3/ 3 =vC4 =vC D4/ 4 Then,
vC D4/ 4 =vC B3/ 3+vB A2/ 2 and the vectors are:
vB A2/ 2 =ω2×rB A/ ⇒vB A2/ 2 =ω2rB A/ = ⋅4 0 5 2. = m / s(⊥torB A/ )
vC B3/ 3 =ω3×rC B/ (⊥torC B/ )
vC D4/ 4 =ω4×rC D/ (⊥torC D/ )
The basic equation is used as a guide and the vectors are added accordingly. Each side of the equation starts from the velocity pole. The directions are gotten from a scaled drawing of the mechanism.
The graphical solution gives:
vC B3/ 3 =2 400. ∠ −46 6. ° m /s
Now, ω3=vC B3 3 =2 4000 8 =3 0 C B r / / . . . rad / s ω4 =vC D4 4 =0 650 8 =0 81 C D r / / . . . rad / s
To determine the direction of ω3, determine the direction that rC B/ must be rotated to be parallel to vC B3/ 3. This direction is clearly clockwise.
To determine the direction of ω4, determine the direction that rC D/ must be rotated to be parallel to vC D4/ 4. This direction is clearly clockwise.
Find the velocity of E3 and E4 by image. The directions are given on the polygon. The magnitudes
are given by,
vE3 = .2 522m / s
vE4 = .0 797m / s Acceleration Analysis
The graphical acceleration analysis follows the same points as in the velocity analysis. Start at link 2. a a a a r a r a r B A B At B Ar B At B A B Ar B A B Ar B A ce 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 / / / / / / / / / sin = + = × = = = ×
(
×)
⇒ = α α ω ω ω or aB Ar2 2 4 0 2 0 5 8 0 / =( . ) ( . )= . m / s from B to A21 m/s Velocity Scale C B D 2 3 4 E b o o' 1a Br2/ A2 1a Ct3/ B3 c3 3 e3 e4 4 m/s Acceleration Scale d'4 1a E4 2 A e'3 c'3 1a Cr3/ B3 1 aE3 b'3 1aCr4/D4 1a C4 /D4 t 1a C3
Now go to Point C and follow the same path as was used with velocities. aC D4/ 4 =aC Dt4/ 4+aC Dr4/ 4 =α4×rC D/ +ω4×vC D4/ 4 Also a a a a a a r v a C D C B B A C Bt C Br B A C B C B B A 4 4 3 3 2 2 3 3 3 3 2 2 3 3 2 2 3 3 / / / / / / / / / = + = + + =α × +ω × + Therefore, aC Dt4/ 4 +aC Dr4/ 4 =aC Bt3/ 3+aC Br3/ 3+aB A2/ 2 and aC Dr vC D 4/ 4 =ω4× 4/ 4 =
[
( .0 812 0 650)( . )=0 528.]
from C to D aC Dt4/ 4 =α4×rC D/ = ⊥? rC D/ aC Br vC B 3/ 3 =ω3× 3/ 3 =[
( . )( . )3 0 2 4 =7 2.]
from C to B aC Bt3/ 3 =α3×rC B/ = ⊥? rC B/These values permit us to solve for the unknown vectors. We can then find “e” by acceleration image. From the acceleration polygon,
aC Bt 3/ 3 =16 13. m / s2 Then, α3= 3 3 =16 130 8 =20 16 a r C Bt C B / / . . . rad / s2
To determine the direction of α3, determine the direction that rC B/ must be rotated to be parallel to
aC Bt
3/ 3. This direction is clearly counterclockwise. and
aE3 =15 84. m / s2 aE4 =32 19. m / s2
Problem 2.33
In the mechanism shown, point A lies on the X axis. Draw the basic velocity and acceleration polygons and use the image technique to determine the velocity and acceleration of point D4. Then
determine the velocity and acceleration images of link 4. Draw the images on the velocity and acceleration polygons. A B D E F (-1.0", -0.75") 2 3 4 5 6 v = 10 in/sA2 (constant) C X Y FE = 1.35" ED = 1.5" BD = CD = 1.0" AB = 3.0" 84˚ 90˚ Square Position Analysis:
Plot the linkage to scale. Start by drawing point D and the rest of link 4. Next draw link B and finally draw link 3. Links 5 and 6 do not need to be drawn because they do not affect the information that is requested.
Velocity Analysis:
vA3 =vA2
vB3 =vB4
vB3 =vA3+vB A3/ 3 =vB4 =vB C4/ 4 (1)
Now,
vA2 =10in / s in the horizontal direction
vB3 =vB4 =ω4×rB C4/ 4 ⇒vB4 =ω4⋅rB C/ (⊥torB C/ )
30 in/s Acceleration Scale 2 A B C D 5 in/s Velocity Scale a3 o b3 c4 b4 d4 1aB 3 /A3r 1a B 4/C4 r b3' o' a'2 a'3 c'4 1a B 3 /A3 t 1aB4 /C 4t b4' d4'
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vB A3/ 3=6 64. in / s or
ω3=vrB A3 3 =6 643 =2 21 B A
/
/
. . rad / s
To determine the direction of ω3, determine the direction that rB A/ must be rotated to be parallel to
vB A3/ 3. This direction is clearly counterclockwise. Also, vB4 = .9 70in / s and ω4 =vrB C4 4 =1 4149 70 =6 86 B C / / . . . rad / s
To determine the direction of ω4, determine the direction that rB C/ must be rotated to be parallel to
vB C4/ 4. This direction is clearly clockwise. Also,
vD4 =6 77. in / s(⊥torD C/ )
Draw the image of link 4 on the velocity polygon. The image is a square. Acceleration Analysis: aA2 =aA3 aB3=aB4 =aA3+aB A3/ 3=aB C4/ 4 aB Cr a a a a B Ct A rB A tB A 4/ 4+ 4/ 4 = 3+ 3/ 3+ 3/ 3 (3) Now, aA3 =0 aB Cr rB C a r B C r B C 4 4 4 4 4 4 4 2 6 86 1 414 66 542 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ . = . in / s2 in the direction of - rB C/ aB Ct4/ 4 =α4×rB C/ ⇒aB Ct4/ 4 =α4⋅rB C/ (⊥torB C/ ) aB Ar rB A a r B A r B A 3 3 3 3 3 3 3 2 2 21 3 14 62 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ = . in / s2 in the direction of - rB A/ at =α3×rB A/ ⇒at =α3⋅rB A/ (⊥torB A/ )
Solve Eq. (3) graphically with a acceleration polygon. From the polygon,
aD4 =54 0. in / s2
The image of link 4 is a square as shown on the acceleration polygon.
Problem 2.34
In the mechanism shown below, the velocity of A2 is 10 in/s to the right and is constant. Draw the
velocity and acceleration polygons for the mechanism, and record values for angular velocity and acceleration of link 6. Use the image technique to determine the velocity of points D3, and E3, and
locate the point in link 3 that has zero velocity.
A C B D E F 2 3 4 5 6 v = 10 in/sA2 (constant) CF = 1.95" FE = 1.45" ED = 1.5" CD = 1.0" BC = 1.45" BD = 1.05" AB = 3.0" 103˚ Position Analysis:
Locate points C and F and the line of action of A. Draw link 6 and locate pont E. Then locate point D. Next locate point B and finally locate point A.
Velocity Analysis:
The equations required for the velocity analysis are:
vA3 =vA2 vB3 =vB4 vB3 =vA3+vB A3/ 3 (1) vD5=vD4 vE5=vE6 =vD5+vE D5/ 5 (2) Now,
vB3 =vB4 =ω4×rB C/ ⇒vB4 =ω4⋅rB C/ (⊥torB C/ )
vB A3/ 3=ω3×rB A/ ⇒vB A3/ 3 =ω3rB A/ (⊥torB A/ )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vB A3/ 3=5 97. in / s ω3=vrB A3 3 =5 973 =1 99 B A / / . . rad / s Also, vB4 = .9 42in s/ or ω4 4 4 4 4 9 42 1 45 6 50 =v = = r B C B C / / . . . rad / s CW Now, vD4 =6 50. in s/ (⊥torD C/ ) vE D5/ 5=ω5×rE D/ ⇒vE D5/ 5 =ω5rE D/ (⊥torE D/ ) vE5=vE6 =vE F6/ 6 =ω6×rE F/ ⇒vE F6/ 6 =ω6 rE F/ (⊥to rE F/ ) Solve Eq. (2) graphically with a velocity polygon. From the polygon,
vE6 = .5 76in / s or ω6 =vrE F6 6 =5 761 45=3 97 E F / / . . . rad / s Acceleration Analysis: aA2 =aA3 aB3=aB4 =aA3+aB A3/ 3 aB Cr4/ 4+aB Ct4/ 4 =arB A3/ 3+atB A3/ 3 (3) aD5 =aD4 aE5 =aE6 =aE F6/ 6 =aD5+aE D5/ 5
aE Fr a a a a E Ft D E Dr E Dt 6/ 6+ 6/ 6 = 5 + 5/ 5+ 5/ 5 (4) Now, A C B D E F 2 3 4 6 o' 25 in/s Acceleration Polygon 2 1a B3 /A3 r 1a B3/A3 t 1aB4 /C4r b3’ b, b’4 d4' 1aE 5/ D 5r 1a E5/D 5 t 1a E 6 /F6r e'5 1a B4/ C4 t 1a E 6 /F6 t O 2.5 in/s Velocity Polygon a2 a3 , b3,bb4 d4,d5 e5 3 o d3 e3 5 D or aB Cr4/ 4 =ω4×(ω4×rB C/ )⇒arB C4/ 4 =ω42⋅rB C/ =6 50 1 45 61 3. 2⋅ . = . in / s2 in the direction of - rB C/ aB Ct rB C a r r B Ct B C B C 4/ 4 =α4× / ⇒ 4/ 4 =α4⋅ / (⊥to / )
aB Ar rB A a r B A r B A 3 3 3 3 3 3 3 2 1 99 3 11 92 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ = . in / s2 in the direction of - rB A/ aB At rB A a r r B At B A B A 3/ 3 =α3× / ⇒ 3/ 3 =α3⋅ / (⊥to / )
Solve Eq. (3) graphically with an acceleration polygon. From the polygon,
aB At3/ 3 =65 3. in / s2 or α3= 3 3 =65 33 =21 8 2 a r B A t B A rad s CW / / . . / And, aD3 =86 6. in / s2 Also, aE Dr rE D a r E D r E D 5 5 5 5 5 5 5 2 1 73 1 5 4 492 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ . = . in / s2 in the direction of - rE D/ aE Dt rE D a r r E Dt E D E D 5/ 5 =α5× / ⇒ 5/ 5 =α5⋅ / (⊥to / ) aE Fr6/ 6=ω6×(ω6×rE F/ )⇒arE F6/ 6 =ω62⋅rE F/ =5 76 1 45 48 1. 2⋅ . = . in / s2 in the direction of - rE F/ aE Ft6/ 6 =α6×rE F/ ⇒atE F6/ 6 =α6⋅rE F/ (⊥torE F/ ) Now solve Eq. (4) using the acceleration polygon. Then,
aE Ft 6/ 6 =38 2. in / s2 or α6 = 6 6 =1 4538 2=26 4 a r E F t E F / / . . . rad / s CCW2
Using the image concept, the velocities of points D3 and E3 are vD3=10 2. in / s
vE3=12 7. in / s
The point in link 3 with zero velocity is shown on position diagram. The point is found by finding the position image of oa b3 3.
Problem 2.35
The instant center of acceleration of a link can be defined as that point in the link that has zero acceleration. If the accelerations of Points A and B are as given in the rigid body shown below, find the Point C in that link at which the acceleration is zero.
B A A
a
a
B 70˚ 5˚ 145˚ A a = 1500 in/s2 = 1050 in/s aB 2 AB = 3.75" Acceleration Analysis:Draw the accelerations of points A and B on an acceleration polygon. Then o' will correspond to the instant center of acceleration. Find the image of o' on the position diagram, and that will be the location of C
500 in/s Acceleration Scale 2 o' a' b' c' A B C 1 in Problem 2.36
The following are given for the mechanism shown in the figure: ω2 =6 5. rad/ s (CCW); α2 =40rad/ s (CCW)
2
Draw the velocity polygon, and locate the velocity of Point E using the image technique.
A C B D (2.2", 1.1") 2 3 4 E 55˚ X Y AB = DE = 1.0 in BC = 2.0 in CD = 1.5 in Position Analysis
Locate the two pivots A and D. Draw link 2 and locate pivot B. Then find point C and finally locate E.
Velocity Analysis: vB3 =vB2=vB A2/ 2 vC3=vC4 =vC D4/ 4 =vB3+vC B3/ 3 (1) Now, vB A2/ 2 =ω2×rB A2/ 2 ⇒vB A2/ 2 =ω2⋅rB A2/ 2 =6 5 1 6 5. ⋅ = . in/ sec (⊥torB A2/ 2) vC D4/ 4 =ω4×rC D4/ 4 ⇒vC D4/ 4 =ω4⋅rC D/ (⊥torC D/ ) vC B3/ 3 =ω3×rC B3/ 3 ⇒vC B3/ 3 =ω3⋅rC B3/ 3 (⊥torC B3/ 3)
Solve Eq. (1) graphically with a velocity polygon. From the polygon, using velocity image,
vE4 = .5 4042in/ sec c3 b3 o 2 in/sec Velocity Polygon e4 A C B D E
Problem 2.37
In the mechanism shown, find ωωωω6 and αααα3. Also, determine the acceleration of D3 by image.
A C B D E F (-1.0", -0.75") 2 3 5 4 6 v = 10 in/sA2 (constant) X Y 81˚ CD = 1.0" BD = 1.05" BC = 1.45" ED = 1.5" FE = 1.4" AB = 3.0" Velocity Analysis: vA3 =vA2 vB3 =vB4 vB3 =vA3+vB A3/ 3 (1) vD5=vD4 vE5=vE6 =vD5+vE D5/ 5 (2) Now,
vA2 =10in/ sec in the horizontal direction
vB3 =vB4 =ω4×rB C4/ 4 ⇒vB4 =ω4⋅rB C4/ 4 (⊥torB C4/ 4)
vB A3/ 3=ω3×rB A3 3/ ⇒vB A3/ 3 =ω3rB A3 3/ (⊥torB A3 3/ )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vB A3/ 3=6 282. in/ sec or
6 282
A C B D E F 3 4 5 6 2 d4,d5 2.5 in/sec Velocity Polygon o b3,bb4 a2,a3 e5 25 in/sec Acceleration Polygon 2 o' 1a B3 /A3 r 1 aBt3/ A3 1aB4/C4t 1a B4/C4r b3’,bb’4 d3' Also, vB4 = .9 551in/ sec or ω4 4 4 4 4 9 551 1 45 6 587 =vrB C = = B C rad CW / / . . . / sec Now, vD4 = .6 587in/ sec (⊥ torD C4/ 4)
vE D5/ 5=ω5×rE D5/ 5 ⇒vE D5/ 5 =ω5rE D5/ 5 (⊥torE D5/ 5)
vE5=vE6 =vE F6/ 6 =ω6×rE F6/ 6 ⇒vE F6/ 6 =ω6 rE F6/ 6 (⊥torE F6/ 6) Solve Eq. (2) graphically with a velocity polygon. From the polygon,
vE6 = .6 132in/ sec or ω6 6 6 6 6 6 132 1 45 4 229 =vrE F = = E F rad CW / / . . . / sec Acceleration Analysis: aA2 =aA3 aB3=aB4 =aA3+aB A3/ 3 aB Cr a a a B C t B A r B A t 4/ 4+ 4/ 4 = 3/ 3+ 3/ 3 (3) Now, aB Cr rB C a r B C r B C in 4 4 4 4 4 4 4 4 4 4 4 2 6 587 1 45 62 9132 2 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ . = . / sec in the direction of rB4/C4 aB Ct rB C a r B C t B C 4/ 4 =α4× 4/ 4 ⇒ 4/ 4 =α4⋅ 4/ 4 (⊥ torB C4/ 4) aB Ar3/ 3=ω3×(ω3×rB A3/ 3)⇒arB A3/ 3 =ω32⋅rB A/ =2 094 3 13 155. 2⋅ = . in/ sec2 in the direction of rB A/ aB At3/ 3 =α3×rB A3/ 3⇒atB A3/ 3 =α3⋅rB A3/ 3 (⊥torB A3/ 3)
Solve Eq. (3) graphically with a acceleration polygon. From the polygon,
aB At in 3/ 3 =68 568. / sec2 or α3 3 3 2 3 3 68 568 3 22 856 = arB A = = t B A rad CW / / . . / sec Also,
Problem 2.38
In the mechanism shown, ωωωω2 = 1 rad/s (CCW) and αααα2 = 0 rad/s2. Find ωωωω
5, αααα5, vE6, aE6 for the position given. Also find the point in link 5 that has zero acceleration for the position given.
A C B D 2 3 4 E 5 6 30˚ AD = 1 m AB = 0.5 m BC = 0.8 m CD = 0.8 m BE = 0.67 m 0.52 m Velocity Analysis vB2 =vB3 =vB5 =vB A2/ 2 vE5=vE6 =vB5+vE B5/ 5 (1) Now, vB A2/ 2 =ω2×rB A2/ 2 ⇒vB A2/ 2 =ω2⋅rB A2/ 2 = ⋅1 0 5 0 5. = . m/ sec (⊥torB A2/ 2)
1vE5 in the horizontal direction
vE B5/ 5 =ω5×rE B5/ 5⇒vE B5/ 5 =ω5⋅rE B5/ 5 (⊥torE B5/ 5) Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vE B5/ 5 =0 47313. m/ sec or ω5 5 5 5 5 0 47313 0 67 0 706 =v = = r E B E B rad CCW / / . . . / sec Also,
A C B D 2 3 4 E 5 6 o' 0.1 m/sec Acceleration Polygon 2 1aE 5 1aB2/A2r 1a E 5/B5r 1aE 5 /B5t b3 b' e'5 2.2° 27.8° O' 0.1 m/sec Velocity Polygon b5 o e5 vE6 = .0 441m/ sec Acceleration Analysis: aB2 =aB3=aB5 =aB A2/ 2 aE5 =aE6 =aB5+aE B5/ 5 aE5 =aB Ar2/ 2+atB A2/ 2+arE B5/ 5+atE B5/ 5 (2) Now,
aE5 in horizontal direction
aB Ar2/ 2=ω2×(ω2×rB A/ )⇒arB A2/ 2 =ω22⋅rB A/ =1 0 5 0 52⋅ . = . m/ sec2 in the direction opposite to rB A/
aB At rB A a r B A t B A m 2/ 2 =α2× / ⇒ 2/ 2 =α2⋅ / = ⋅ =0 1 0 / sec2 aE Br r a r E B E Br E B m 5 5 5 5 5 5 5 2 0 706 0 67 0 3342 2 / =ω ×(ω × / )⇒ / =ω ⋅ / = . ⋅ . = . / sec
in the direction opposite to rE B/
aE Bt5/ 5 =α5×rE B/ ⇒aE Bt5/ 5 =α5⋅rE B/ (⊥torE B/ )
Solve Eq. (2) graphically with an acceleration polygon. From the polygon,
aE6 = .0 042m/ sec2 Also, aE Bt5 5 0 42m 2 / = . / sec or α5 5 5 2 5 5 0 42 0 67 0 626 =a = = r E B t E B rad CW / / . . . / sec
Problem 2.39
Part of an eight-link mechanism is shown in the figure. Links 7 and 8 are drawn to scale, and the velocity and acceleration of point D7 are given. Find ωωωω7 and αααα7 for the position given. Also find
the velocity of G7 by image.
D E 7 6 G 8 X Y 1.25" 1.6" DE = 1.5" DG = 0.7" GE = 1.65" = 5.0 320˚ in/sec = 40 260˚ in/s2 vD7 aD7 Velocity Analysis
Compute the velocity of Points E7 and E8.
v v v v v E E E D E D 7 8 7 7 7 7 = = + /
and because points on the same link are involved,
vD7+vE D7/ 7 =vD7 +ω7×rE D/ =vE8 From the velocity polygon:
vE8 = .2 884in/ sec in the direction shown, and
vE D7/ 7 =3 405. in/ sec in the direction shown.
ω7 = vrE D7 7 =3 4051 50 =2 27 E D rad / / . . . / sec
o' D E 7 7 d' 6 G 8 1a E7/ D7 r 1a E7/ D7 t e'7 2 in/sec Velocity Scale 2 20 in/sec Acceleration Scale 7 d 7 e , e8 7 g o
The direction can be found by rotating rE D/ 90˚ in the direction of ω7 to get vE D7/ 7. From the polygon, the direction must be counter clockwise. Therefore,
ω7= .2 27rad/ secCCW
The velocity of G7 is found by image. The magnitude of the velocity is:
vG7 = .6 016in/ sec in the direction shown. Acceleration Analysis
Use the same points as were used in the velocity analysis.
aD7 +aE D7/ 7 =aF8 =aD7+arE D7/ 7+α7×rE D7/ 7 =aF8 where a v r E Dr E D E D in 7 7 7 7 2 3 4052 2 1 50 7 729 / / / . . . / sec = = =
in the direction opposite rE D/ .
From the polygon,
aE Dt7/ 7 =α7×rE D7/ 7 =0 Therefore, α7 = 7 7 =42 591 50 =28 40 2 a r E Dt E D rad / / . . . / sec
The direction can be found by rotating rE D/ 90˚ in the direction of ω7 to get aE Dt7/ 7. From the polygon, the direction must be counter clockwise. Therefore,
a7=28 40. rad/ sec2 CCW
Problem 2.40
In the mechanism shown below, link 2 is rotating CW at the rate of 3 rad/s (constant). In the position shown, link 2 is horizontal. Write the appropriate vector equations, solve them using vector polygons, and
a) Determine vC4, vE4, ωωωω3, and ωωωω4.
b) Determine aC4, aE4, αααα3, and αααα4.
Link lengths: AB = 3 in, BC = BE = CE = 5 in, CD = 3 in
B C 2 3 4 D ω2 A 7" E Position Analysis
Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B and then C. Then locate G.
Velocity Analysis: vB3 =vB2=vB A2/ 2 vC3=vC4 =vC D4/ 4 =vB3+vC B3/ 3 (1) Now, vB A2/ 2 =ω2×rB A/ ⇒vB A/ =ω2⋅rB A/ = ⋅ =3 3 9in / s(⊥torB A/ ) vC D4/ 4 =ω4×rC D/ ⇒vC D4/ 4 =ω4⋅rC D/ (⊥torC D/ ) vC B3/ 3 =ω3×rC B/ ⇒vC B3/ 3 =ω3⋅rC B/ (⊥torC B/ ) Solve Eq. (1) graphically with a velocity polygon.
From the polygon,
vC B3/ 3 =11 16. in / s
vC D4/ 4 =6 57. in / s in the direction shown.