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. AE = 0.7" AB = 2.5" AC = 1.0" BC = 2.0" EF = 2.0" CD = 1.0" DF = 1.5" θ 2 = 135˚
Y C
D
5
6 A 3 ω2
2 E
F
θ2
4
B
0.3"
X
Position Analysis Locate 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: v A3 = v A2 = v A2 / E2 v B4 = v B3 = v A3 + v B3 / A3
(1)
Find vC3 by image. vC5 = vC3 vD5 = vD6 = vD6 / F6 = vC5 + vD5 /C5
(2)
Now, v B3 in horizontal direction v A2 / E2 = ω 2 × rA/ E ⇒ v A2 / E2 = ω 2 ⋅ rA/ E = 10 ⋅ 0.7 = 7 in / s (⊥ to rA/ E ) v B3 / A3 = ω 3 × rB/ A ⇒ v B3 / A3 = ω 3 ⋅ rB/ A (⊥ to rB/ A ) Solve Eq. (1) graphically with a velocity polygon. From the polygon,
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Velocity Polygon 2.5 in/s
D
4
C
3
o
6
b3
A
2 F
E
Slider Line
B
d5 c3 c5 a3
v B3 = 3.29 in / s Using velocity image, vC5 = vC3 = 6.78 in/ s Now, vD6 / F6 = ω 6 × rD/ F ⇒ vD6 / F6 = ω 6 ⋅ rD/ F (⊥ to rD/ F ) vD5 /C5 = ω 5 × rD/C ⇒ vD5 /C5 = ω 5 ⋅ rD/C (⊥ to rD/C ) Solve Eq. (2) graphically with a velocity polygon. From the polygon, vD5 = vD6 = vD6 / F6 = 6.78 in / s or
ω6 =
v D6 / F6 6.78 = = 4.52 rad / s CCW rD6 / F6 1.5
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f6
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, 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. Detail of Link 3
H
J
3
F
8 H
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"
E
6 I
D
K
L
7
C
D
G 3
5
C
F
4 2 1ω
2
110˚ A
B
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 = vC2 / A2 = ω 2 × rC2 / A2 ⇒ vC2 = ω 2 rC2 / A2 = 2 ⋅ (15.37) = 30.74 in / s vC3 = vC2 vD3 = vD4 = vD4 / B4 = vC3 + vD3 /C3
(1)
vG5 = vG6 = vG7 = vF5 + vG5 / F5 = vE6 + vG6 / E6 vF5 = vF3 vE6 = vE2 So,
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vF5 + vG5 / F5 = vE6 + vG6 / E6
(2)
vI7 = vI8 = vG7 + vI7 /G7 = vH8 + vI8 / H8 vH8 = vH3 So, vG7 + vI7 /G7 = vH8 + vI8 / H8
(3)
J
E I
7 G
6
D C
5
8 H
3 F 4
2
A h3 Velocity Polygon 10 in/s
f3
g6 i8
c3 d3
e2
j8
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B
o
Now, vC3 = 30.74 in / s (⊥ to rC / A ) vD3 /C3 = ω 3 × rD/C (⊥ to rD/C ) vD4 / B4 = ω 4 × rD/ B (⊥ to rD/ 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 = 3.80 in / s Now, vG5 / F5 = ω 5 × rG/ F (⊥ to rG/ F ) vG6 / E6 = ω 6 × rG/ E (⊥ to rG/ 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, vI7 /G7 = ω 7 × rI /G (⊥ to rI /G ) vI8 / H8 = ω 8 × rI / H (⊥ to rI / H ) Solve Eq. (3) graphically with a velocity polygon. Using velocity polygon of link 8 v J8 = 73.6 in / s
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Problem 2.23 In the mechanism shown, determine the sliding velocity of link 6 and the angular velocities of links 3 and 5. C
B ω 2 = 3 rad s
AB = 12.5" BC = 22.4" DC = 27.9" CE = 28.0" DF = 21.5"
34˚
2
3 4
A
10.4"
F 2.0"
50˚
D 5
E
6 29.5"
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: v B2 = v B2 / A2 = ω 2 × rB/ A v B3 = v B2 vC3 = v B3 + vC3 / B3
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C
B
A
D
F
E
c3 f3
o
Velocity Polygon
d3
10 in/sec b3
Also,
vC3 = vC4 = vE4 + vC4 / E4 = vC4 / E4 where, v B2 = ω 2 rB/ A = 3⋅12.5 = 37.5 in / sec vC3 / B3 = ω 3 × rC / B (⊥ to CB) vC4 / E4 = ω 4 × rC / E (⊥ to CE ) The velocity of C3 (and C4) can then be found using the velocity polygon. After the velocity of C3 is found, find the velocity of D3 by image. Then,
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vD5 = vD3 vF5 = vD5 + vF5 / D5 and vF5 = vF6 where vF5 / D5 = ω 5 × rF / D ⇒ vC5 = ω 5 rF / 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 vC3 / B3 = 26.6 in / sec vF5 / D5 = 18.54 in / sec
ω3 =
vC3 / B3 26.6 = = 1.187 rad / sec 22.4 rC / B
ω5 =
vF5 / D5 18.54 = = 0.862 rad / sec 21.5 rF / D
To determine the direction for ω 3 , determine the direction that rC / B must be rotated to be in the direction of vC3 / B3 . From the polygon, this direction is CCW. To determine the direction for ω 5 , determine the direction that rF / D must be rotated to be in the direction of vF5 / D5 . From the polygon, this direction is CW.
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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.
Y 2
AC = 2.4" BD = 3.7" BC = 1.2"
A
D
6
3 2.05" 1v = A2
5
B
15 m/s
C
45˚
4
X 2.4"
Velocity Analysis: v A3 = v A2 vC4 = vC3 = v A3 + vC3 / A3
(1)
v B3 = v B5 vD5 = vD6 = v B5 + vD5 / B5
(2)
Now, v A3 =15 m / sec in vertical direction vC3 in horizontal direction vC3 / A3 = ω 3 × rC3 / A3 ⇒ vC3 / A3 = ω 3 ⋅ rC3 / A3 (⊥ to rC3 / A3 ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, using velocity image, v B3 = v B5 = 14.44 m / sec Now,
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2
A D 3
6
5 B
d5 C
10 m/sec
4
c3
o
b3 b5 a3 vD5 along the inclined path vD5 / B5 = ω 5 × rD5 / B5 ⇒ v D5 / B5 = ω 5 ⋅ rD5 / B5 (⊥ to rD5 / B5 ) Solve Eq. (2) graphically with a velocity polygon. From the polygon, vD6 = 12.31m / sec Also, vD5 / B5 = 16.61 m / sec or
ω5 =
Velocity Polygon
v D5 / B5 16.605 = = 4.488 rad / sec CCW 3.7 rD/ B
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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.
D 6 EA = 0.55" AB = 2.5" AC = 1.0" CB = 1.75" CD = 2.05"
0.75" 5
C
E ω2 60˚
2
33
4 B
A
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: v A3 = v A2 = v A2 / E2 v B4 = v B3 = v A3 + v B3 / A3
(1)
vC5 = vC3 vD5 = vD6 = vC5 + vD5 /C5
(2)
Now, v B3 in horizontal direction
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D 6 3
5
C
E 2
B 4
3
A Velocity Polygon
a3
10 in/sec c3
b3
d5
o
v A2 / E2 = ω 2 × rA2 / E2 ⇒ v A2 / E2 = ω 2 ⋅ rA/ E = 42 ⋅ 0.55 = 23.1 in / sec (⊥ to rA/ E ) v B3 / A3 = ω 3 × rB3 / A3 ⇒ v B3 / A3 = ω 3 ⋅ rB/ A (⊥ to rB/ A ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B4 = 17.76 in / sec Using velocity image, vC3 = 18.615 in / sec Now, vD5 in vertical direction vD5 /C5 = ω 5 × rD/C ⇒ vD5 /C5 = ω 5 ⋅ rD5 /C5 (⊥ to rD/C )
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Solve Eq. (2) graphically with a velocity polygon. From the polygon, vD5 = 5.63 in / sec
Problem 2.26 Given vA4 = 1.0 ft/s to the left, find vB6. Y D
4 A
2 3
157.5˚
0.5"
X
C DE = 1.9" CD = 1.45" BC = 1.1" AD = 3.5" AC = 2.3"
E
5 1.0" B 6
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: v A4 = v A3 vC5 = vC3 = v A3 + vC3 / A3
(1)
v D3 = vD2 vD2 = vE2 + vD2 / E2
(2) - 118 -
v B5 = vC5 + v B5 /C5 Now, v A4 =1.0 ft / sec in horizontal direction vC3 / A3 = ω 3 × rC / A ⇒ vC3 / A3 = ω 3 ⋅ rC / A (⊥ to rC / A ) vD2 / E2 = ω 2 × rD/ E ⇒ vD2 / E2 = ω 2 ⋅ rD/ E (⊥ to rD/ E ) Solve Eq. (1) graphically with a velocity polygon.
From the polygon, using velocity image, vC3 / A3 = 1.28 ft / sec and,
ω3 =
vC3 / A3 1.28 = = 0.56 rad / s 2.3 rC / A
To determine the direction of ω 3 , determine the direction that rC / A must be rotated to be parallel to vC3 / A3 . This direction is clearly clockwise. Now, v B6 is horizontal direction v B5 /C5 = ω 5 × rB/C ⇒ v B5 /C5 = ω 5 ⋅ rB/C (⊥ to rB/C ) - 119 -
Solve Eq. (2) graphically with a velocity polygon. From the polygon, v B6 = 1.23 ft / 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: v A3 = v A2 v B3 = v A3 + v B3 / A3
(1)
v B4 = v B3 v B4 = vF4 + v B4 / F4 = 0 + v B4 / F4
(2)
vE5 = vE4 vG5 = vE5 + vG5 / E5 - 120 -
vG6 = vG5 Now, v A2 =10 cm / sec (⊥ to rA/C ) v B3 / A3 = ω 3 × rB/ A ⇒ v B3 / A3 = ω 3 ⋅ rB/ A (⊥ to rB/ A ) v B4 / F4 = ω 4 × rB/ F ⇒ v B4 / F4 = ω 4 ⋅ rB/ F (⊥ to rB/ F )
From the polygon, v B4 = 6.6 cm / sec Using velocity image, vE4 = 3.12 cm / sec Now, vG6 is horizontal direction vG5 / E5 = ω 5 × rG/ E ⇒ vG5 / E5 = ω 5 ⋅ rG/ E (⊥ to rG/ E ) For the velocity image draw a line ⊥ to rC / E at e draw a line ⊥ to rC / E at g and find the point “c” From the velocity polygon v B6 = 3.65 cm / sec
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Problem 2.28 If vA2 = 10 in/s as shown, find the angular velocity of link 6. vA
AB = 1.0" AD = 2.0" AC = 0.95" CE = 2.0" EF = 1.25" BF = 3.85"
2
E 5
6
A 2 B
C 3
27˚
D 4
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: v A3 = v A2 vD3 = v A3 + vD3 / A3
(1)
vD4 = vD3 vC5 = vC3 vE5 = vC5 + vE5 /C5
(2)
vE6 = vE5 vE6 = vF6 + vE6 / F6 = 0 + vE6 / F6 Now, v A2 =10 in / sec
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vD3 / A3 = ω 3 × rD/ A ⇒ v D3 / A3 = ω 3 ⋅ rD/ A (⊥ to rD/ A )
From the polygon, vD3 / A3 = 9.1 in / sec Using velocity image, rD/ A: rC / A = v D3 / A3 : vC3 / A3 vC3 / A3 = 4.32 in / sec Now, vE5 /C5 = ω 5 × rE /C ⇒ vE5 /C5 = ω 5 ⋅ rE /C (⊥ to rE /C ) vE6 / F6 = ω 6 × rE / F ⇒ vE6 / F6 = ω 6 ⋅ rE / F (⊥ to rE / F ) from the velocity polygon vE6 = 2.75 in / sec and
ω6 =
vE6 / F6 2.75 = = 1.375 rad / s 2 rE / F
To determine the direction of ω 6 , determine the direction that rE / F must be rotated to be parallel to vE6 / F6 . This direction is clearly counterclockwise.
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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 F 6. 5 F
E
6
EF = 2.5" CD = 0.95" AB = 0.5" BC = 2.0" CE = 2.4" BE = 1.8"
3 C
2"
B
4 0.1" D
ω 2 ,α 2
2
115˚ A
2.44" 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: v B3 = v B2 = v B2 / A2 vC3 = vC4 = vC4 / D4 = v B3 + vC3 / B3
(1)
vE5 = vE3 vF5 = vF6 = vE5 + vF5 / E5
(2)
Now, v B2 / A2 = ω 2 × rB2 / A2 ⇒ v B2 / A2 = ω 2 ⋅ rB2 / A2 = 20 ⋅ 0.5 = 10 in / s (⊥ to rB2 / A2 ) vC4 / D4 = ω 4 × rC4 / D4 ⇒ vC4 / D4 = ω 4 ⋅ rC4 / D4 (⊥ to rC4 / D4 ) vC3 / B3 = ω 3 × rC3 / B3 ⇒ vC3 / B3 = ω 3 ⋅ rC3 / B3 (⊥ to rC3 / B3 ) Solve Eq. (1) graphically with a velocity polygon. From the polygon and using velocity image,
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vC3 / B3 = 6.59 in / s or
ω3 =
vC3 / B3 6.59 = = 3.29 rad / s CCW 2 rC3 / B3
Also, vC4 / D4 = 8.19 in / s or
ω4 =
vC4 / D4 8.19 = = 8.62 rad / s CW rC4 / D4 0.95
And, vE5 = 5.09 in / s Now, vF5 in horizontal direction vF5 / E5 = ω 5 × rF5 / E5 ⇒ vF5 / E5 = ω 5 ⋅ rF5 / E5 (⊥ to rF5 / E5 ) Solve Eq. (2) graphically with a velocity polygon. From the polygon, vF5 / E5 = 3.97 in / s or
ω5 =
vF5 / E5 3.97 = = 1.59 rad/ s CCW 2.5 rF5 / E5
Also, vF6 = 3.79 in / s Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3 a Cr 4 / D4 + aCt 4 / D4 = aBr 2 / A2 + aBt 2 / A2 + aCr 3 / B3 + aCt 3 / B3
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(3)
5
E
F
Velocity Polygon 5 in/sec e5
3
b3
C B
4
2
D
o
f5
A 1 aF
o'
c3
5
f'5
Acceleration Polygon 50 in/s 2
1a r C4 /D4
1a r B2 /A2 1a E
1 at F5 /E 5
5
1 at C4 /D 4
c'3 1a t C3 /B3
1a r F5 / E 5
b'3 1a t B2 /A2
1a r C3 /B3
e'5
aE5 = aE3 aF5 = aF6 = aE5 + aF5 / E5 aF5 = aE5 + aFr 5 / E5 + aFt 5 / E5
(4)
Now, a rB2 / A2 = ω 2 × (ω 2 × rB2 / A2 ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB2 / A2 = 20 2 ⋅ 0.5 = 200 in / s2
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in the direction of - rB2 / A2 aBt 2 / A2 = α 2 × rB2 / A2 ⇒ aBt 2 / A2 = α 2 ⋅ rB2 / A2 = 100 ⋅ 0.5 = 50 in / s2 (⊥ to rB2 / A2 ) a Cr 3 / B3 = ω 3 × (ω 3 × rC3 / B3 ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC3 / B3 = 3.292 ⋅ 2 = 21.6 in / s2 in the direction of - rC3 / B3 aCt 3 / B3 = α 3 × rC3 / B3 ⇒ aCt 3 / B3 = α 3 ⋅ rC3 / B3 (⊥ to rC3 / B3 ) a Cr 4 / D4 = ω 4 × (ω 4 × rC4 / D4 ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC4 / D4 = 8.62 2 ⋅ 0.95 = 70.6 in / s2 in the direction of - rC4 / D4 aCt 4 / D4 = α 4 × rC4 / D4 ⇒ aCt 4 / D4 = α 4 ⋅ rC4 / D4 (⊥ to rC4 / D4 ) Solve Eq. (3) graphically with an acceleration polygon. From the polygon, using acceleration image, aE5 = 252.0 in / s2 Now, aF5 in horizontal direction a rF5 / E5 = ω 5 × (ω 5 × rF5 / E5 ) ⇒ a rF5 / E5 = ω 5 2 ⋅ rF5 / E5 = 1.592 ⋅ 2.5 = 6.32 in / s2 in the direction of - rF5 / E5 aFt 5 / E5 = α 5 × rF5 / E5 ⇒ aFt 5 / E5 = α 5 ⋅ rF5 / E5 (⊥ to rF5 / E5 ) Solve Eq. (4) graphically with an acceleration polygon. From the polygon, aF6 = 152.7 in / s2
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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.
E 5 4 D B
3
AB = 1.8' BC = 3.75' CD = 3.75' AD = 4.5' AE = 4.35' DE = 6.0' EF = 11.1' F
A
6
2 C
60˚
Velocity Analysis:
ω 2 = 130 rpm = 130 2π = 13.614 rad / s 60 vC3 = vC2 = vC2 / B2 vD3 = vD4 = vD4 / A4 = vC3 + vD3 /C3
(1)
vE5 = vE4 vF5 = vF6 = vE5 + vF5 / E5
(2)
Now, vC2 / B2 = ω 2 × rC / B ⇒ vC2 / B2 = ω 2 ⋅ rC / B = 13.614 ⋅ 3.75 = 51.053 ft / s (⊥ to rC / B ) vD3 /C3 = ω 3 × rD/C ⇒ vD3 /C3 = ω 3 ⋅ rD/C (⊥ to rD/C ) vD4 / A4 = ω 4 × rD/ A ⇒ v D4 / A4 = ω 4 ⋅ rD/ A (⊥ to rD/ A ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, using velocity image, vD3 /C3 = 41.3 ft / s or
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E 5 4 D 3
F
A
B
6
2 C
c3
60˚
d3 Velocity Scale 20 ft/s
f5
o Acceleration Scale 200 ft/s 2
c '3
1 ar C2 / B2
o' a'4
f5' d3' 1 a rD /A 4 4
1 atF /E 5 5
1 ar D3 /C3
t 1a D 3 /C 3 t 1 a D /A 4 4
1 arF /E 5 5
e'4
- 129 -
e4
ω3 =
vD3 /C3 41.3 = = 11.0 rad / s CW rD3 /C3 3.75
Also, vD4 / A4 = 37.19 ft / s or
ω4 =
vD4 / A4 37.19 = = 8.264 rad / s CW 4.5 rD4 / A4
And, vE5 = 35.95 ft / s Now, vF5 in horizontal direction vF5 / E5 = ω 5 × rF5 / E5 ⇒ vF5 / E5 = ω 5 ⋅ rF / E (⊥ to rF / E ) Solve Eq. (2) graphically with a velocity polygon. From the polygon, vF5 / E5 = 12.21 ft / s or
ω5 =
vF5 / E5 12.21 = = 1.1 rad / s CCW rF5 / E5 11.1
Acceleration Analysis: aC3 = aC2 = aC2 / B2 t t r r aD + aD = aCr 2 / B2 + aCt 2 / B2 + aD + aD 4 / A4 4 / A4 3 /C3 3 /C3
(3)
aE5 = aE4 aF5 = aF6 = aE5 + aF5 / E5 aF5 = aE5 + aFr 5 / E5 + aFt 5 / E5
(4)
Now, a Cr 2 / B2 = ω 2 × (ω 2 × rC / B ) ⇒ a Cr 2 / B2 = ω 2 2 ⋅ rC / B = 13.614 2 ⋅ 3.75 = 695.0 ft / s2 in the direction of - rC / B
- 130 -
aCt 2 / B2 = α 2 × rC / B ⇒ aCt 2 / B2 = α 2 ⋅ rC / B = 0 ⋅ 3.75 = 0 ft / s2 a rD3 /C3 = ω 3 × (ω 3 × rD/C ) ⇒ a rD3 /C3 = ω 3 2 ⋅ rD/C = 11.0 2 ⋅ 3.75 = 453.8 ft / s2 in the direction of - rD / C t t aD = α 3 × rD/C ⇒ aD = α 3 ⋅ rD/C (⊥ to rD/C ) 3 /C3 3 /C3
a rD4 / A4 = ω 4 × (ω 4 × rD/ A ) ⇒ a rD4 / A4 = ω 4 2 ⋅ rD/ A = 8.264 2 ⋅ 4.5 = 307.3 ft / s2 in the direction of - rD / A t t aD = α 4 × rD4 / A4 ⇒ aD = α 4 ⋅ rD/ A (⊥ to rD/ A ) 4 / A4 4 / A4
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 a rF5 / E5 = ω 5 × (ω 5 × rF / E ) ⇒ a rF5 / E5 = ω 5 2 ⋅ rF / E = 1.12 ⋅11.1 = 13.4 ft / s2 in the direction of - rF / E aFt 5 / E5 = α 5 × rF5 / E5 ⇒ aFt 5 / E5 = α 5 ⋅ rF5 / E5 (⊥ to rF5 / E5 ) Solve Eq. (4) graphically with an acceleration polygon. From the polygon, aF6 = 83.4 ft / s2 Also, aFt 5 / E5 = 325.2 ft / s2 or aFt 5 / E5 325.2 α5 = = = 29.3 rad / s2 CCW 11.1 rF5 / E5
- 131 -
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.
6 D
Y
o
30 5
EA = 1.0" AB = 3.0" BC = 3.0" AC = 1.0" CD = 2.55"
E
X C
90˚
2 3
3
o
30
A B 4 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,
ω 2 = π ⋅ n = π ⋅ 2000 = 209.44 rad / s 30 30 v A2 = v A2 / E2 = v A3 = ω 2 × rA/ E v B3 = v B4 = v A3 + v B3 / A3
(1)
vC3 = vC5
- 132 -
6 D 5 E C 2 3 A B 4 b3 , b4
Velocity Polygon 100 in/s
1a r B3 /A3
a’ 2 o c3 , c5
r 1 aD 5 /C 5
a2 , a3
c’3 1a t B3 /A3
b3 , b’4
d5 , d6 t 1 a D /C 5 5
1 aC
Acceleration Polygon 10000 in/s 2
3
1 arA / E 2 2 1 aD
1 aB 3
o' - 133 -
d5 , d6 5
vD5 = vD6 = vC5 + vD5 /C5
(2)
Now, v A2 = ω 2 rA/ E = 209.44 ⋅1 = 209.44 in / s (⊥ to rA/ E ) v B3 in the direction of rB/ E v B3 / A3 = ω 3 × rB/ A ⇒ v B3 / A3 = ω 3 rB/ A (⊥ to rB/ A ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B3 = v B4 = 212.7 in / s Also, v B3 / A3 = 109.3 in / s or
ω3 =
v B3 / A3 109.3 = = 36.43 rad / s 3 rB/ A
To determine the direction of ω 3 , determine the direction that rB/ A must be rotated to be parallel to v B3 / A3 . This direction is clearly clockwise. Also, vC3 = vC5 = 243.3 in / s Now, vD5 = vD6 in the direction of rD/ E vD5 /C5 = ω 5 × rD/C ⇒ vD5 /C5 = ω 5 rD/C (⊥ to rD/C ) Solve Eq. (2) graphically with a velocity polygon. From the polygon, vD5 = vD6 = 189.2 in / sec Also, vD5 /C5 = 135 in / s or
ω5 =
vD5 /C5 135 = = 52.9 rad / s 2.55 rD/C
- 134 -
To determine the direction of ω 5 , determine the direction that rD/ C must be rotated to be parallel to v D5 /C5 . This direction is clearly clockwise. Acceleration Analysis: a A2 = a A3 = a rA2 / E2 + a tA2 / E2 aB3 = aB4 = a A3 + aB3 / A3 aB3 = a rA2 / E2 + a tA2 / E2 + a rB3 / A3 + a tB3 / A3
(3)
aC3 = aC5 aD5 = aD6 = aC5 + aD5 /C5 aD5 = aD6 = aC5 + a rD5 /C5 + a tD5 /C5
(4)
Now, aB3 in the direction of ± rB/ E a rA2 / E2 = ω 2 × (ω 2 × rA/ E ) ⇒ a rA2 / E2 = ω 2 2 ⋅ rA/ E = 209.44 2 ⋅1 = 43860 in / s2 in the direction of - rA / E a tA2 / E2 = α 2 × rA/ E ⇒ a tA2 / E2 = α 2 ⋅ rA/ E = 0 ⋅1 = 0 in / sec 2 a rB3 / A3 = ω 3 × (ω 3 × rB/ A ) ⇒ a rB3 / A3 = ω 3 2 ⋅ rB/ A = 36.432 ⋅ 3 = 3981 in / s2 in the direction of - rB / A aBt 3 / A3 = α 3 × rB/ A ⇒ aBt 3 / A3 = α 3 ⋅ rB/ A (⊥ to rB/ A ) Solve Eq. (3) graphically with an acceleration polygon. From the polygon, aB3 = 14710 in / s2 Also, aBt 3 / A3 = 38460 in / s2 or
α3 =
aBt 3 / A3 38460 = = 12820 rad / s2 3 rB/ A
- 135 -
To determine the direction of α 3 , determine the direction that rB/ A must be rotated to be parallel to aBt 3 / A3 . This direction is clearly counterclockwise. Also, aC3 = aC5 = 39,300 in / s 2 Now, aD5 in the direction of - rD/ E a rD5 /C5 = ω 5 × (ω 5 × rD/C ) ⇒ a rD5 /C5 = ω 5 2 ⋅ rD/C = 52.92 ⋅ 2.55 = 7136 in / s2 in the direction of - rD/ C t t aD = α 5 × rD/C ⇒ aD = α 5 ⋅ rD/C (⊥ to rD/C ) 5 /C5 5 /C5
Solve Eq. (4) graphically with an acceleration polygon. From the polygon, aD5 = aD6 = 24, 000 in / s2 Also, t = 26,300 in / s2 aD 5 /C5
or
α5 =
t aD 5 /C5 = 26300 = 10,300 rad / s2 2.55 rD/C
To determine the direction of α 5 , determine the direction that rD / C must be rotated to be parallel to t aD 5 /C5 . This direction is clearly clockwise.
Problem 2.32 In the mechanism shown, ω2 = 4 rad/s CCW (constant). Write the appropriate vector equations, solve them using vector polygons, and a) Determine vE3, vE4, and ω3. b) Determine aE3, aE4, and α3. Also find the point in link 3 that has zero acceleration for the position given.
- 136 -
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: v B2 = v B3 = v B2 / A2 vC3 = v B3 + vC3 / B3 = vC4 = vC4 / D4 Then, vC4 / D4 = vC3 / B3 + v B2 / A2 and the vectors are: v B2 / A2 = ω 2 × rB/ A ⇒ v B2 / A2 = ω 2 rB/ A = 4 ⋅ 0.5 = 2 m / s (⊥ to rB/ A ) vC3 / B3 = ω 3 × rC / B (⊥ to rC / B ) vC4 / D4 = ω 4 × rC / D (⊥ to rC / 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: vC3 / B3 = 2.400∠ − 46.6° m / s vC4 / D4 = 0.650∠ − 1˚ m / s vE3 = 2.522∠93.9° m / s (by image) - 137 -
Now,
ω3 =
vC3 / B3 2.400 = = 3.0 rad / s 0.8 rC / B
ω4 =
vC4 / D4 0.65 = = 0.81 rad / s 0.8 rC / D
To determine the direction of ω 3 , determine the direction that rC / B must be rotated to be parallel to vC3 / B3 . This direction is clearly clockwise. To determine the direction of ω 4 , determine the direction that rC / D must be rotated to be parallel to vC4 / D4 . 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 = 2.522 m / s vE4 = 0.797 m / s Acceleration Analysis The graphical acceleration analysis follows the same points as in the velocity analysis. Start at link 2. aB2 / A2 = aBt 2 / A2 + aBr 2 / A2 aBt 2 / A2 = α 2 × rB/ A = 0 since α 2 = 0 aBr 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ aBr 2 / A2 = ω 2
2 rB/ A
or aBr 2 / A2 = (4.0)2 (0.5) = 8.0 m / s 2 from B to A
- 138 -
C
e3
b3
E
3 4
e4
2
Velocity Scale
B D
A
1 m/s o
c3
1aC
o' d'4 1a r C
3
1 at C 4 /D 4
c'3
1 ar B 2 / A2
b'3 1 r aC / B 3 3 1 t aC / B 3
1a
E3
3
Acceleration Scale 4 m/s 2
e'3 1a
e'4
- 139 -
E4
4
/D
4
Now go to Point C and follow the same path as was used with velocities. aC4 / D4 = aCt 4 / D4 + aCr 4 / D4 = α 4 × rC / D + ω 4 × vC4 / D4 Also aC4 / D4 = aC3 / B3 + aB2 / A2 = aCt 3 / B3 + aCr 3 / B3 + aB2 / A2 = α 3 × rC / B + ω 3 × vC3 / B3 + aB2 / A2 Therefore, aCt 4 / D4 + aCr 4 / D4 = aCt 3 / B3 + aCr 3 / B3 + aB2 / A2 and aCr 4 / D4 = ω 4 × vC4 / D4 = [(0.812)(0.650) = 0.528] from C to D aCt 4 / D4 = α 4 × rC / D = ? ⊥ rC / D aCr 3 / B3 = ω 3 × vC3 / B3 = [(3.0)(2.4) = 7.2] from C to B aCt 3 / B3 = α 3 × 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, aCt 3 / B3 = 16.13 m / s2 Then,
α3 =
aCt 3 / B3 16.13 = = 20.16 rad / s2 0.8 rC / B
To determine the direction of α 3 , determine the direction that rC / B must be rotated to be parallel to aCt 3 / B3 . This direction is clearly counterclockwise. and aE3 = 15.84 m / s2 aE4 = 32.19 m / s2
- 140 -
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. Y FE = 1.35" ED = 1.5" BD = CD = 1.0" AB = 3.0"
vA2= 10 in/s (constant)
4 B
Square
D 90˚
3
A
84˚ C
2
X
5 E F (-1.0", -0.75")
6
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: v A3 = v A2 v B3 = v B4 v B3 = v A3 + v B3 / A3 = v B4 = v B4 /C4
(1)
Now, v A2 = 10 in / s in the horizontal direction v B3 = v B4 = ω 4 × rB4 /C4 ⇒ v B4 = ω 4 ⋅ rB/C (⊥ to rB/C ) v B3 / A3 = ω 3 × rB3 / A3 ⇒ v B3 / A3 = ω 3 rB/ A (⊥ to rB/ A )
- 141 -
B D
A C b3
b4 Velocity Scale 5 in/s
c4 o
d4
Acceleration Scale
a3 r 1 a B /A 3 3
30 in/s 2
a'2 a'3 c'4 o'
r 1 a B /C 4 4
d4'
1a t B 3 /A 3
t 1a B /C 4 4
b3' ' b4 Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B3 / A3 = 6.64 in / s or
- 142 -
ω3 =
v B3 / A3 6.64 = = 2.21 rad / s 3 r B/ A
To determine the direction of ω 3 , determine the direction that rB/ A must be rotated to be parallel to v B3 / A3 . This direction is clearly counterclockwise. Also, v B4 = 9.70 in / s and
ω4 =
v B4 /C4 9.70 = = 6.86 rad / s 1.414 rB/C
To determine the direction of ω 4 , determine the direction that rB/ C must be rotated to be parallel to v B4 /C4 . This direction is clearly clockwise. Also, vD4 = 6.77 in / s (⊥ to rD/C ) Draw the image of link 4 on the velocity polygon. The image is a square. Acceleration Analysis: a A2 = a A3 aB3 = aB4 = a A3 + aB3 / A3 = aB4 /C4 aBr 4 /C4 + aBt 4 /C4 = a A3 + a rB3 / A3 + a tB3 / A3
(3)
Now, a A3 = 0 aBr 4 /C4 = ω 4 × (ω 4 × rB/C ) ⇒ a rB4 /C4 = ω 4 2 ⋅ rB/C = 6.862 ⋅1.414 = 66.54 in / s2 in the direction of - rB/ C aBt 4 /C4 = α 4 × rB/C ⇒ aBt 4 /C4 = α 4 ⋅ rB/C (⊥ to rB/C ) a rB3 / A3 = ω 3 × (ω 3 × rB/ A ) ⇒ a rB3 / A3 = ω 3 2 ⋅ rB/ A = 2.212 ⋅ 3 = 14.6 in / s2 in the direction of - rB/ A aBt 3 / A3 = α 3 × rB3 / A3 ⇒ aBt 3 / A3 = α 3 ⋅ rB/ A (⊥ to rB/ A )
- 143 -
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. CF = 1.95" FE = 1.45" ED = 1.5" CD = 1.0" BC = 1.45" BD = 1.05" AB = 3.0" vA2 = 10 in/s (constant) 2
E B
D
5 6
3
4
A
C
103˚ F
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: v A3 = v A2 v B3 = v B4 v B3 = v A3 + v B3 / A3
(1)
v D5 = vD4 vE5 = vE6 = vD5 + vE5 / D5
(2)
Now, v A2 = 10 in / s in the horizontal direction
- 144 -
v B3 = v B4 = ω 4 × rB/C ⇒ v B4 = ω 4 ⋅ rB/C (⊥ to rB/C ) v B3 / A3 = ω 3 × rB/ A ⇒ v B3 / A3 = ω 3 rB/ A (⊥ to rB/ A ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B3 / A3 = 5.97 in / s
ω3 =
v B3 / A3 5.97 = = 1.99 rad / s 3 rB/ A
Also, v B4 = 9.42 in / s or
ω4 =
v B4 /C4 9.42 = = 6.50 rad / s CW rB4 /C4 1.45
Now, vD4 = 6.50 in / s (⊥ to rD/C ) vE5 / D5 = ω 5 × rE / D ⇒ vE5 / D5 = ω 5 rE / D (⊥ to rE / D ) vE5 = vE6 = vE6 / F6 = ω 6 × rE / F ⇒ vE6 / F6 = ω 6 rE / F (⊥ to rE / F ) Solve Eq. (2) graphically with a velocity polygon. From the polygon, vE6 = 5.76 in / s or
ω6 =
vE6 / F6 5.76 = = 3.97 rad / s 1.45 rE / F
Acceleration Analysis: a A2 = a A3 aB3 = aB4 = a A3 + aB3 / A3 aBr 4 /C4 + aBt 4 /C4 = a rB3 / A3 + a tB3 / A3
(3)
aD5 = aD4 aE5 = aE6 = aE6 / F6 = aD5 + aE5 / D5
- 145 -
aEr 6 / F6 + aEt 6 / F6 = aD5 + a rE5 / D5 + a tE5 / D5
(4)
Now, 1 ar B3 /A 3
Acceleration Polygon
o'
1 at B3 /A 3
1a r E 6 /F6
1 at E 6 /F6
r 1 a B /C 4 4
e'5 O
1 at E5 /D 5
e3
25 in/s 2
d3
Velocity Polygon 2.5 in/s
d4'
b3 , b4
r 1 aE 5/ D 5
t 1aB /C 4 4
b 3’ , b’4 3 e5 a2 , a3
o d4 , d5 E B D 3
2
5
6
D 4
A
C
or aBr 4 /C4 = ω 4 × (ω 4 × rB/C ) ⇒ a rB4 /C4 = ω 4 2 ⋅ rB/C = 6.50 2 ⋅1.45 = 61.3 in / s2 in the direction of - rB/ C aBt 4 /C4 = α 4 × rB/C ⇒ aBt 4 /C4 = α 4 ⋅ rB/C (⊥ to rB/C )
- 146 -
F
a rB3 / A3 = ω 3 × (ω 3 × rB/ A ) ⇒ a rB3 / A3 = ω 3 2 ⋅ rB/ A = 1.992 ⋅ 3 = 11.9 in / s2 in the direction of - rB / A aBt 3 / A3 = α 3 × rB/ A ⇒ aBt 3 / A3 = α 3 ⋅ rB/ A (⊥ to rB/ A ) Solve Eq. (3) graphically with an acceleration polygon. From the polygon, aBt 3 / A3 = 65.3 in / s2 or aBt 3 / A3 65.3 α3 = = = 21.8 rad / s 2 CW 3 rB/ A And, aD3 = 86.6 in / s2 Also, a rE5 / D5 = ω 5 × (ω 5 × rE / D ) ⇒ a rE5 / D5 = ω 5 2 ⋅ rE / D = 1.732 ⋅1.5 = 4.49 in / s2 in the direction of - rE / D aEt 5 / D5 = α 5 × rE / D ⇒ aEt 5 / D5 = α 5 ⋅ rE / D (⊥ to rE / D ) a rE6 / F6 = ω 6 × (ω 6 × rE / F ) ⇒ a rE6 / F6 = ω 6 2 ⋅ rE / F = 5.762 ⋅1.45 = 48.1 in / s2 in the direction of - rE / F aEt 6 / F6 = α 6 × rE / F ⇒ aEt 6 / F6 = α 6 ⋅ rE / F (⊥ to rE / F ) Now solve Eq. (4) using the acceleration polygon. Then, aEt 6 / F6 = 38.2 in / s2 or
α6 =
aEt 6 / F6 38.2 = = 26.4 rad / s2 CCW 1.45 rE / F
Using the image concept, the velocities of points D3 and E3 are vD3 = 10.2 in / s
- 147 -
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 oa3b3 .
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. AB = 3.75" A
aA = 1500 in/s 2
5˚ B
70˚
a B = 1050 in/s
2
145˚
aB aA
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
- 148 -
o'
c'
b' A B 1 in a' Acceleration Scale 500 in/s 2
C
Problem 2.36 The following are given for the mechanism shown in the figure:
α 2 = 40 rad/ s2 (CCW)
ω 2 = 6.5 rad/ s (CCW);
Draw the velocity polygon, and locate the velocity of Point E using the image technique. D (2.2", 1.1") Y
B
2 55˚ A
E
4 3 X
AB = DE = 1.0 in BC = 2.0 in CD = 1.5 in
C Position Analysis Locate the two pivots A and D. Draw link 2 and locate pivot B. Then find point C and finally locate E.
- 149 -
Velocity Analysis: v B3 = v B2 = v B2 / A2 vC3 = vC4 = vC4 / D4 = v B3 + vC3 / B3
(1)
Now, v B2 / A2 = ω 2 × rB2 / A2 ⇒ v B2 / A2 = ω 2 ⋅ rB2 / A2 = 6.5 ⋅1 = 6.5 in / sec (⊥ to rB2 / A2 ) vC4 / D4 = ω 4 × rC4 / D4 ⇒ vC4 / D4 = ω 4 ⋅ rC / D (⊥ to rC / D ) vC3 / B3 = ω 3 × rC3 / B3 ⇒ vC3 / B3 = ω 3 ⋅ rC3 / B3 (⊥ to rC3 / B3 ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, using velocity image, vE4 = 5.4042 in / sec
b3 Velocity Polygon 2 in/sec
c3
o
E D B A C e4
- 150 -
Problem 2.37 In the mechanism shown, find ω6 and α3. Also, determine the acceleration of D3 by image. Y B CD = 1.0" BD = 1.05" BC = 1.45" ED = 1.5" FE = 1.4" AB = 3.0"
D vA2 = 10 in/s (constant) 2
3
5 4
E
81˚
A
C 6
X
F (-1.0", -0.75")
Velocity Analysis: v A3 = v A2 v B3 = v B4 v B3 = v A3 + v B3 / A3
(1)
v D5 = vD4 vE5 = vE6 = vD5 + vE5 / D5
(2)
Now, v A2 = 10 in / sec in the horizontal direction v B3 = v B4 = ω 4 × rB4 /C4 ⇒ v B4 = ω 4 ⋅ rB4 /C4 (⊥ to rB4 /C4 ) v B3 / A3 = ω 3 × rB3 / A3 ⇒ v B3 / A3 = ω 3 rB3 / A3 (⊥ to rB3 / A3 ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B3 / A3 = 6.282 in / sec or
ω3 =
v B3 / A3 6.282 = = 2.094 rad / sec 3 rB3 / A3 - 151 -
B 3
5
E A
D 4
C
2 6 F
Acceleration Polygon r 1a B /A 3 3
o'
25 in/sec 2
b 3 , b4
1a r B4 /C 4
1 t a B3 / A3
Velocity Polygon 2.5 in/sec
e5
t 1 a B /C 4 4
b3’ , b’4
o d4 , d5
a 2 ,a3 d3'
Also, v B4 = 9.551 in / sec or
ω4 =
v B4 /C4 9.551 = = 6.587 rad / sec CW rB4 /C4 1.45
Now, vD4 = 6.587 in / sec (⊥ to rD4 / C4 )
- 152 -
vE5 / D5 = ω 5 × rE5 / D5 ⇒ vE5 / D5 = ω 5 rE5 / D5 (⊥ to rE5 / D5 ) vE5 = vE6 = vE6 / F6 = ω 6 × rE6 / F6 ⇒ vE6 / F6 = ω 6 rE6 / F6 (⊥ to rE6 / F6 ) Solve Eq. (2) graphically with a velocity polygon. From the polygon, vE6 = 6.132 in / sec or
ω6 =
vE6 / F6 6.132 = = 4.229 rad / sec CW rE6 / F6 1.45
Acceleration Analysis: a A2 = a A3 aB3 = aB4 = a A3 + aB3 / A3 aBr 4 /C4 + aBt 4 /C4 = a rB3 / A3 + a tB3 / A3
(3)
Now, aBr 4 /C4 = ω 4 × (ω 4 × rB4 /C4 ) ⇒ a rB4 /C4 = ω 4 2 ⋅ rB4 /C4 = 6.5872 ⋅1.45 = 62.913 in / sec 2 r in the direction of B4 / C 4 aBt 4 /C4 = α 4 × rB4 /C4 ⇒ aBt 4 /C4 = α 4 ⋅ rB4 /C4 (⊥ to rB4 /C4 ) a rB3 / A3 = ω 3 × (ω 3 × rB3 / A3 ) ⇒ a rB3 / A3 = ω 3 2 ⋅ rB/ A = 2.094 2 ⋅ 3 = 13.155 in / sec 2 in the direction of rB / A aBt 3 / A3 = α 3 × rB3 / A3 ⇒ aBt 3 / A3 = α 3 ⋅ rB3 / A3 (⊥ to rB3 / A3 ) Solve Eq. (3) graphically with a acceleration polygon. From the polygon, aBt 3 / A3 = 68.568 in / sec 2 or
α3 =
aBt 3 / A3 68.568 = = 22.856 rad / sec 2 CW 3 rB3 / A3
Also, aD3 = 91.390 in / sec 2
- 153 -
Problem 2.38 In the mechanism shown, ω2 = 1 rad/s (CCW) and α2 = 0 rad/s2. Find ω5, α5, vE6, aE6 for the position given. Also find the point in link 5 that has zero acceleration for the position given.
AD = 1 m AB = 0.5 m BC = 0.8 m CD = 0.8 m BE = 0.67 m
6 E
C 3
4
5
0.52 m
B
2 A
30˚
D
Velocity Analysis v B2 = v B3 = v B5 = v B2 / A2 vE5 = vE6 = v B5 + vE5 / B5
(1)
Now, v B2 / A2 = ω 2 × rB2 / A2 ⇒ v B2 / A2 = ω 2 ⋅ rB2 / A2 = 1⋅ 0.5 = 0.5 m / sec (⊥ to rB2 / A2 ) 1vE5
in the horizontal direction vE5 / B5 = ω 5 × rE5 / B5 ⇒ vE5 / B5 = ω 5 ⋅ rE5 / B5 (⊥ to rE5 / B5 )
Solve Eq. (1) graphically with a velocity polygon. From the polygon, vE5 / B5 = 0.47313 m / sec or
ω5 =
vE5 / B5 0.47313 = = 0.706 rad / sec CCW 0.67 rE5 / B5
Also,
- 154 -
C
3 O'
6 E
5
4 B
2 D
A o'
b5
r 1 a B /A 2 2
Velocity Polygon 0.1 m/sec
1 aE
5
e'5
27.8° b'3
2.2° 1a r E 5 /B5
o
t 1 aE /B 5 5
Acceleration Polygon 0.1 m/sec2
e5
vE6 = 0.441 m / sec Acceleration Analysis: aB2 = aB3 = aB5 = aB2 / A2 aE5 = aE6 = aB5 + aE5 / B5 aE5 = a rB2 / A2 + a tB2 / A2 + a rE5 / B5 + a tE5 / B5 Now,
- 155 -
(2)
aE5 in horizontal direction a rB2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB/ A = 12 ⋅ 0.5 = 0.5 m / sec 2 in the direction opposite to rB / A aBt 2 / A2 = α 2 × rB/ A ⇒ aBt 2 / A2 = α 2 ⋅ rB/ A = 0 ⋅1 = 0 m / sec 2 aEr 5 / B5 = ω 5 × (ω 5 × rE / B ) ⇒ a rE5 / B5 = ω 5 2 ⋅ rE / B = 0.7062 ⋅ 0.67 = 0.334 m / sec 2 in the direction opposite to rE / B aEt 5 / B5 = α 5 × rE / B ⇒ aEt 5 / B5 = α 5 ⋅ rE / B (⊥ to rE / B ) Solve Eq. (2) graphically with an acceleration polygon. From the polygon, aE6 = 0.042 m / sec 2 Also, aEt 5 / B5 = 0.42 m / sec 2 or
α5 =
aEt 5 / B5 0.42 = = 0.626 rad / sec 2 CW rE5 / B5 0.67
Using acceleration image, point O' is in the location in which the acceleration of link five is zero.
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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 ω7 and α7 for the position given. Also find the velocity of G7 by image.
Y
E 7
D
X
6
8
1.6"
1.25"
G DE = 1.5" DG = 0.7" GE = 1.65" v D 7 = 5.0 320˚ in/sec Velocity Analysis Compute the velocity of Points E7 and E8. vE7 = vE8 vE7 = vD7 + vE7 / D7 and because points on the same link are involved, vD7 + vE7 / D7 = vD7 + ω 7 × rE / D = vE8 From the velocity polygon: vE8 = 2.884 in / sec in the direction shown, and vE7 / D7 = 3.405 in / sec in the direction shown.
ω7 =
vE7 / D7 3.405 = = 2.27 rad / sec 1.50 rE / D
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a D 7 = 40 260˚ in/s 2
E 7
D 6 G
8
Velocity Scale 2 in/sec
o'
e'7 e 7 , e8
o
1a t E7 / D 7
Acceleration Scale g7
20 in/sec 2
d7
d'7 1a r E7 / D 7
The direction can be found by rotating rE / D 90˚ in the direction of ω 7 to get vE7 / D7 . From the polygon, the direction must be counter clockwise. Therefore,
ω 7 = 2.27 rad / sec CCW The velocity of G7 is found by image. The magnitude of the velocity is: vG7 = 6.016 in / sec in the direction shown. Acceleration Analysis Use the same points as were used in the velocity analysis. aD7 + aE7 / D7 = aF8 = aD7 + aEr 7 / D7 + α 7 × rE7 / D7 = aF8 where aEr 7 / D7 =
vE7 / D7 2 3.4052 = = 7.729 in / sec 2 1.50 rE / D
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in the direction opposite rE / D . From the polygon, aEt 7 / D7 = α 7 × rE7 / D7 = 0 Therefore,
α7 =
aEt 7 / D7 42.59 = = 28.40 rad / sec 2 1.50 rE / D
The direction can be found by rotating rE / D 90˚ in the direction of ω 7 to get aEt 7 / D7 . From the polygon, the direction must be counter clockwise. Therefore, a7 = 28.40 rad / sec 2 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, ω3, and ω4. b) Determine aC4, aE4, α3, and α4. Link lengths: AB = 3 in, BC = BE = CE = 5 in, CD = 3 in E
3 C
4
ω2 A
2
B
D
7" 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.
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Velocity Analysis: v B3 = v B2 = v B2 / A2 vC3 = vC4 = vC4 / D4 = v B3 + vC3 / B3
(1)
Now, v B2 / A2 = ω 2 × rB/ A ⇒ v B/ A = ω 2 ⋅ rB/ A = 3⋅ 3 = 9 in / s (⊥ to rB/ A ) vC4 / D4 = ω 4 × rC / D ⇒ vC4 / D4 = ω 4 ⋅ rC / D (⊥ to rC / D ) vC3 / B3 = ω 3 × rC / B ⇒ vC3 / B3 = ω 3 ⋅ rC / B (⊥ to rC / B ) Solve Eq. (1) graphically with a velocity polygon.
From the polygon, vC3 / B3 = 11.16 in / s vC4 / D4 = 6.57 in / s in the direction shown.
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Now
ω3 =
vC3 / B3 11.16 = = 2.232 rad / s 5 rC / B
ω4 =
vC4 / D4 6.57 = = 2.19 rad / s 3 rC / D
and
To determine the direction of ω 3 , determine the direction that rC / B must be rotated to be parallel to vC3 / B3 . This direction is clearly clockwise. To determine the direction of ω 4 , determine the direction that rC / D must be rotated to be parallel to vC4 / D4 . This direction is clearly counterclockwise. The velocity of point E3 vE3 = v B3 + vE3 / B3 = v B3 + ω 3 × rE / B from the polygon vE3 = 13.5 in / s vC4 = 6.57 in / s Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3 a Cr 4 / D4 + aCt 4 / D4 = aBr 2 / A2 + aBt 2 / A2 + aCr 3 / B3 + aCt 3 / B3 Now, a rB2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB/ A = 32 ⋅ 3 = 27 in / s2 in the direction of - rB2 / A2 aBt 2 / A2 = α 2 × rB/ A ⇒ aBt 2 / A2 = α 2 rB/ A = 0 a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 2.232 2 ⋅ 5 = 24.9 in / s2 in the direction of - rC / B aCt 3 / B3 = α 3 × rC / B ⇒ aCt 3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B )
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(2)
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 2.192 ⋅ 3 = 14.39 in / sec 2 in the direction of - rC / D aCt 4 / D4 = α 4 × rC / D ⇒ aCt 4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically with acceleration.
From the acceleration polygon, aCt 3 / B3 = 0 in / s2 aCt 4 / D4 = 46.71 in / s2 Then,
α3 =
aCt 3 / B3 0 = = 0 rad / s2 2 rC / B
aCt 4 / D4 46.71 = = 15.57 rad / s2 α4 = 3 rC / D To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to aCt 4 / D4 . This direction is clearly counterclockwise. Determine the acceleration of point E3 aE3 = aB3 + aE3 / B3 = aB2 + a rE3 / B3 + aEt 3 / B3
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aEt 3 / B3 = α 3 × rE / B ⇒ aEt 3 / B3 = α 3 ⋅ rE / B = 0 ⋅1 = 0 in / sec 2 a rE3 / B3 = ω 3 × (ω 3 × rE / B ) ⇒ a rE3 / B3 = ω 3 2 ⋅ rE / B = 2.232 2 ⋅ 5 = 24.9 in / sec 2 From the acceleration polygon, aE3 = 34.29 in / s2 aC4 = 48.87 in / s2
Problem 2.41 Part of a 10-link mechanism is shown in the figure. Links 7 and 8 are drawn to scale, and the velocity and acceleration of points D7 and F8 are given. Find ω8 and α7 for the position given. Also find the velocity of G7 by image. Y E DE = 1.5" EF = 1.45" DG = 0.7" EG = 1.65"
7 D
X 8
6 G
F (1.8", -1.05") 9
v D 7 = 6.0 353˚ in/s
aD 7 = 40 235˚ in/s 2
v F 8 = 7.5 54˚ in/s
a F 8 = 30 305˚ in/s 2
Velocity Analysis Compute the velocity of Points E7 and E8. vE7 = vE8 vE7 = vD7 + vE7 / D7 vE8 = vF8 + vE8 / F8 Therefore, vD7 + vE7 / D7 = vF8 + vE8 / F8 and because points on the same link are involved,
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vD7 + vE7 / D7 = vF8 + vE8 / F8 = vD7 + ω 7 × rE7 / D7 = vF8 + ω 8 × rE8 / F8 From the velocity polygon: vE8 / F8 = 0 so
ω8 = 0 The velocity of G7 is found by image. The magnitude of the velocity is vG7 = 9.1 in / s E 7
D
8
6 G
F 9
Velocity Polygon
f 8 , e8 , e7 o'
5 in/s
Velocity Polygon 20 in/s 2
g7
o
f'8
d7 d'7 e'8 , e'7 vG7 = 9.1 in / s Acceleration Analysis
Use the same points as were used in the velocity analysis for the acceleration analysis. aD7 + aE7 / D7 = aF8 + aE8 / F8 = aD7 + aEr 7 / D7 + α 7 × rE7 / D7 = aF8 + aEr 8 / F8 + α 8 × rE8 / F8
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vE7 / D7 2 6.92 2 = = 32.0 in / s2 1.50 rE7 / D7 2 v 02 = E8 / F8 = = 0 in / s2 1.44 rE8 / F8
aEr 7 / D7 = aEr 8 / F8
From the polygon, aEt 7 / D7 = α 7 × rE7 / D7 = 0 Therefore,
α7 = 0
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