Solutions to Chapter 2 Exercise Problems
Problem 2.1
In the mechanism shown below, link 2 is rotating CCW at the rate of 2 rad/s (constant). In the
position shown, link 2 is horizontal and link 4 is vertical. Write the appropriate vector equations,
solve them using vector polygons, and
a) Determine vC4, ω3, and ω4.
b) Determine aC4, α3, and α4.
Link lengths: AB = 75 mm, CD = 100 mm
ω2
A
3
2
C
B
4
50 mm
D
250 mm
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
v B2 = v A2 + v B2 / A2
- 48 -
v A2 = 0
Therefore,
vC3 + v B3 /C3 = v A2 + v B2 / A2
(1)
Now,
ω 2 = 2 rad / s CCW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (2 rad / s)(75 mm) =150 mm / s
vC3 / B3 = ω 3 × rC / B (⊥ to rC / B )
vC4 / D4 = ω 4 × rC / D (⊥ to rC / D )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vC3 / B3 = 156 mm / s
vC4 / D4 = vC4 = 43 mm / s
Now,
ω3 =
vC3 / B3 156
=
= .86 rad / s
rC / B 182
From the directions given in the position and velocity polygons
ω 3 = .86 rad / s CW
Also,
ω4 =
vC4 / D4 43
=
= .43 rad / s
rC / D 100
From the directions given in the position and velocity polygons
ω 4 = .43 rad / s CW
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 = 2 2 ⋅ 75 = 300 mm / s2
- 49 -
(2)
in the direction of - rB2 / A2
aBt 2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 )
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = .862 ⋅182 =134.6 mm / s2
in the direction of - rC / B
aCt 3 / B3 = α 3 × rC / B ⇒ aCt 3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = .432 ⋅100 = 18.5 mm / s 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 an acceleration polygon. From the acceleration polygon,
aCt 3 / B3 = 19.22 mm / s 2
aCt 4 / D4 = 434.70 mm / s 2
Then,
aCt 3 / B3 67, 600
=
= 27, 900 rad / s2
α3 =
2.42
rC / B
α4 =
aCt 4 / D4 434.70
=
= 4.347 rad / s2
100
rC / D
To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to
aCt 3 / B3 . This direction is clearly counter-clockwise.
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 counter-clockwise.
From the acceleration polygon,
aC4 = 435 mm / s 2
- 50 -
Problem 2.2
In the mechanism shown below, link 2 is rotating CCW at the rate of 500 rad/s (constant). In the
position shown, link 2 is vertical. Write the appropriate vector equations, solve them using vector
polygons, and
a) Determine vC4, ω3, and ω4.
b) Determine aC4, α3, and α4.
Link lengths: AB = 1.2 in, BC = 2.42 in, CD = 2 in
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
v B2 = v A2 + v B2 / A2
v A2 = 0
Therefore,
- 51 -
vC3 + v B3 /C3 = v A2 + v B2 / A2
(1)
Now,
ω 2 = 500 rad / s CCW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (500 rad / s)(1.2 in) = 600 in / s
vC3 / B3 = ω 3 × rC / B (⊥ to rC / B )
vC4 / D4 = ω 4 × rC / D (⊥ to rC / D )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vC3 / B3 = 523.5 in / s
vC4 / D4 = vC4 = 858in / s
Now,
ω3 =
vC3 / B3 523.5
=
= 216.3 rad / s
2.42
rC / B
From the directions given in the position and velocity polygons
ω 3 = 216.3 rad / s CCW
Also,
ω4 =
vC4 / D4 858
=
= 429 rad / s
2
rC / D
From the directions given in the position and velocity polygons
ω 4 = 429 rad / s CCW
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 = 500 2 ⋅1.2 = 300000 in / s 2
in the direction of - rB2 / A2
aBt 2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 )
- 52 -
(2)
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 216.32 ⋅ 2.42 = 113, 000 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 )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 4292 ⋅ 2 = 368, 000 in / s 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 an acceleration polygon. From the acceleration polygon,
aCt 3 / B3 = 67561in / s 2
aCt 4 / D4 = 151437 in / s 2
Then,
aCt 3 / B3 67561
=
= 27, 900 rad / s2
α3 =
2.42
rC / B
aCt 4 / D4 151437
=
= 75, 700 rad / s2
α4 =
2
rC / D
To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to
aCt 3 / B3 . This direction is clearly clockwise.
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 clockwise.
From the acceleration polygon,
aC4 = 398, 000 in / s 2
- 53 -
Problem 2.3
In the mechanism shown below, link 2 is rotating CW at the rate of 10 rad/s (constant). In the
position shown, link 4 is vertical. Write the appropriate vector equations, solve them using vector
polygons, and
a) Determine vC4, ω3, and ω4.
b) Determine aC4, α3, and α4.
Link lengths: AB = 100 mm, BC = 260 mm, CD = 180 mm
C
3
B
4
ω2
2
A
D
250 mm
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
v B2 = v A2 + v B2 / A2
v A2 = 0
Therefore,
vC3 + v B3 /C3 = v A2 + v B2 / A2
(1)
- 54 -
Now,
ω 2 =10 rad / s CW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (10 rad / s)(100 mm) =1000 mm / s
vC3 / B3 = ω 3 × rC / B (⊥ to rC / B )
vC4 / D4 = ω 4 × rC / D (⊥ to rC / D )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vC3 / B3 = 31.3 mm / s
vC4 / D4 = vC4 = 990 mm / s
Now,
ω3 =
vC3 / B3 31.3
=
= .12 rad / s
260
rC / B
From the directions given in the position and velocity polygons
ω 3 =.12 rad / s CCW
Also,
ω4 =
vC4 / D4 990
=
= 5.5 rad / s
180
rC / D
From the directions given in the position and velocity polygons
ω 4 =5.5 rad / s CW
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 = 10 2 ⋅100 =10, 000 mm / s 2
in the direction of - rB2 / A2
aBt 2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 )
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = .12 2 ⋅ 260 = 3.744 mm / s2
- 55 -
(2)
in the direction of - rC / B
aCt 3 / B3 = α 3 × rC / B ⇒ aCt 3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 5.52 ⋅180 = 5, 445 mm / s 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 an acceleration polygon. From the acceleration polygon,
aCt 3 / B3 = 4784 mm / s 2
aCt 4 / D4 = 1778 mm / s 2
Then,
α3 =
aCt 3 / B3 4785
=
= 18.4 rad / s2
260
rC / B
α4 =
aCt 4 / D4 1778
=
= 9.88 rad / s2
180
rC / D
To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to
aCt 3 / B3 . This direction is clearly counter-clockwise.
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 counter-clockwise.
From the acceleration polygon,
aC4 = 5, 700 mm / s 2
- 56 -
Problem 2.4
In the mechanism shown below, link 2 is rotating CW at the rate of 4 rad/s (constant). In the
position shown, θ is 53˚. Write the appropriate vector equations, solve them using vector polygons,
and
a) Determine vC4, ω3, and ω4.
b) Determine aC4, α3, and α4.
Link lengths: AB = 100 mm, BC = 160 mm, CD = 200 mm
A
ω2
2
B
C
3
160 mm
220 mm
4
D
- 57 -
θ
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
v B2 = v A2 + v B2 / A2
v A2 = 0
Therefore,
vC3 + v B3 /C3 = v A2 + v B2 / A2
(1)
Now,
ω 2 = 4 rad / s CW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (4 rad / s)(100 mm) = 400 mm / s
vC3 / B3 = ω 3 × rC / B (⊥ to rC / B )
vC4 / D4 = ω 4 × rC / D (⊥ to rC / D )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vC3 / B3 = 500 mm / s
vC4 / D4 = vC4 = 300 mm / s
Now,
ω3 =
vC3 / B3 500
=
= 3.125 rad / s
rC / B 160
From the directions given in the position and velocity polygons
ω 3 = 3.125 rad / s CCW
Also,
ω4 =
vC4 / D4 300
=
= 1.5 rad / s
200
rC / D
From the directions given in the position and velocity polygons
ω 4 =1.5 rad / s CCW
Acceleration Analysis:
aB3 = aB2 = aB2 / A2
aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
- 58 -
a Cr 4 / D4 + aCt 4 / D4 = aBr 2 / A2 + aBt 2 / A2 + aCr 3 / B3 + aCt 3 / B3
(2)
Now,
a rB2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB/ A = 4 2 ⋅100 =1600 mm / s 2
in the direction of - rB2 / A2
aBt 2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 )
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 3.1252 ⋅160 = 1560 mm / s2
in the direction of - rC / B
aCt 3 / B3 = α 3 × rC / B ⇒ aCt 3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 1.52 ⋅ 200 = 450 mm / s 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 an acceleration polygon. From the acceleration polygon,
aCt 3 / B3 = 618.5 mm / s 2
aCt 4 / D4 = 3, 220 mm / s 2
Then,
α3 =
aCt 3 / B3 618.5
=
= 3.87 rad / s2
160
rC / B
α4 =
aCt 4 / D4 3220
=
= 16.1rad / s2
200
rC / D
To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to
aCt 3 / B3 . This direction is clearly counter-clockwise.
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 counter-clockwise.
From the acceleration polygon,
aC4 = 3250 mm / s 2
- 59 -
Problem 2.5
In the mechanism shown below, link 2 is rotating CCW at the rate of 4 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, ω3, and ω4.
b) Determine aC4, α3, and α4.
Link lengths: AB = 1.25 in, BC = 2.5 in, CD = 2.5 in
D
1.0 in
A
4
0.75 in
3
C
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
- 60 -
2
B
ω2
v B2 = v A2 + v B2 / A2
v A2 = 0
Therefore,
vC3 + v B3 /C3 = v A2 + v B2 / A2
(1)
Now,
ω 2 = 4 rad / s CCW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (4 rad / s)(1.25in) = 5in / s
vC3 / B3 = ω 3 × rC / B (⊥ to rC / B )
vC4 / D4 = ω 4 × rC / D (⊥ to rC / D )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vC3 / B3 = 6.25in / s
vC4 / D4 = vC4 = 3.75in / s
Now,
ω3 =
vC3 / B3 6.25
=
= 2.5 rad / s
2.5
rC / B
From the directions given in the position and velocity polygons
ω 3 = 2.5 rad / s CCW
Also,
ω4 =
vC4 / D4 3.75
=
=1.5 rad / s
2.5
rC / D
From the directions given in the position and velocity polygons
ω 4 =1.5 rad / s CW
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,
- 61 -
(2)
a rB2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB/ A = 4 2 ⋅1.25 = 20 in / s 2
in the direction of - rB2 / A2
aBt 2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 )
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 2.52 ⋅ 2.5 = 15.6 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 )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 1.52 ⋅ 2.5 = 5.6 in / s 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 an acceleration polygon. From the acceleration polygon,
aCt 3 / B3 = 4.69 in / s 2
aCt 4 / D4 = 4.69 in / s 2
Then,
α3 =
aCt 3 / B3 4.69
=
= 1.87 rad / s2
2.5
rC / B
α4 =
aCt 4 / D4 4.69
=
= 1.87 rad / s2
2.5
rC / D
To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to
aCt 3 / B3 . This direction is clearly counter-clockwise.
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 clockwise.
From the acceleration polygon,
aC4 = 7.32 in / s 2
- 62 -
Problem 2.6
In the mechanism shown below, link 2 is rotating CW at the rate of 100 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 and ω3
b) Determine aC4 and α3
Link lengths: AB = 60 mm, BC = 200 mm
B
A
2
120 mm
ω2
3
C
4
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
v B2 = v A2 + v B2 / A2
v A2 = 0
Therefore,
vC3 + v B3 /C3 = v A2 + v B2 / A2
(1)
- 63 -
Now,
ω 2 =100 rad / s CW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (100 rad / s)(60 mm) = 6000 mm / s
vC3 / B3 = ω 3 × rC / B (⊥ to rC / B )
vC4 / D4 → parallel to the ground.
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
vC3 / B3 = 7, 500 mm / s
vC4 / D4 = vC4 = 4500 mm / s
Now,
ω3 =
vC3 / B3 7500
=
= 37.5 rad / s
200
rC / B
From the directions given in the position and velocity polygons
ω 3 =.12 rad / s CW
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 = 100 2 ⋅ 60 = 600, 000 mm / s 2
in the direction of - rB2 / A2
aBt 2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 )
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 37.52 ⋅ 200 = 281, 000 mm / s2
in the direction of - rC / B
aCt 3 / B3 = α 3 × rC / B ⇒ aCt 3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B )
a C4 / D4 = a C4 → parallel to ground
Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
- 64 -
(2)
aCt 3 / B3 = 211, 000 mm / s 2
a C4 / D4 = a C4 = 248, 000 mm / s 2
Then,
α3 =
aCt 3 / B3 211, 000
=
= 1060 rad / s2
200
rC / B
To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to
aCt 3 / B3 . This direction is clearly counter-clockwise.
From the acceleration polygon,
aC4 = 248, 000 mm / s 2
Problem 2.7
In the mechanism shown below, link 4 is moving to the left at the rate of 4 ft/s (constant). Write the
appropriate vector equations, solve them using vector polygons, and
a) Determine ω3 and ω4.
b) Determine α3 and α4.
Link lengths: AB = 10 ft, BC = 20 ft.
B
3
8.5 ft
vC 4
C
4
- 65 -
2
120˚
A
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
v B2 = v A2 + v B2 / A2
v A2 = 0
Therefore,
vC3 + v B3 /C3 = v A2 + v B2 / A2
(1)
Now,
vC4 = 4 ft / s parallel to the ground
v B3 /C3 = ω 3 × rB/C (⊥ to rB/C )
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
v B3 /C3 = 2.3 ft / s
v B2 / A2 = 2.3 ft / s
or
ω3 =
v B3 /C3 2.3
=
= .115 rad / s
20
rB/C
From the directions given in the position and velocity polygons
- 66 -
ω 3 =.115 rad / s CW
Also,
ω2 =
v B2 / A2 2.3
=
= .23 rad / s
10
rB/ A
From the directions given in the position and velocity polygons
ω 2 =.23 rad / s CCW
ω 4 = 0 rad / s since it does not rotate
Acceleration Analysis:
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 = .232 ⋅10 = .529 ft / s 2
in the direction of - rB2 / A2
aBt 2 / A2 = α 2 × rB/ A ⇒ aBt 2 / A2 = α 2 ⋅ rB/ A (⊥ to rB/ A )
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = .1152 ⋅ 20 = .264 ft / s2
in the direction of - rC / B
aCt 3 / B3 = α 3 × rC / B ⇒ aCt 3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B )
a C4 / D4 = 0 link 4 is moving at a constant velocity
Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
aCt 3 / B3 = 0.045 ft / s 2
a tB2 / A2 = 0.017 ft / s 2
Then,
aCt 3 / B3 0.45
=
= .023 rad / s2
α3 =
20
rC / B
aBt 2 / A2 0.017
=
= .0017 rad / s2
α2 =
10
rB/ A
- 67 -
(2)
To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to
aCt 3 / B3 . This direction is clearly clockwise.
To determine the direction of α2, determine the direction that rB/ A must be rotated to be parallel to
aBt 2 / A2 . This direction is clearly counter-clockwise.
Problem 2.8
In the mechanism shown below, link 4 is moving to the right at the rate of 20 in/s (constant). Write
the appropriate vector equations, solve them using vector polygons, and
a) Determine ω3 and ω4.
b) Determine α3 and α4.
Link lengths: AB = 5 in, BC = 5 in.
A
45˚
2
7 in
B
3
C
4
- 68 -
vC 4
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
v B2 = v A2 + v B2 / A2
v A2 = 0
Therefore,
vC3 + v B3 /C3 = v A2 + v B2 / A2
(1)
Now,
vC4 = 20 in / s parallel to the ground
v B3 /C3 = ω 3 × rB/C (⊥ to rB/C )
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
v B3 /C3 = 14.1in / s
v B2 / A2 = 14.1in / s
or
ω3 =
v B3 /C3 14.1
=
= 2.82 rad / s
5
rB/C
From the directions given in the position and velocity polygons
ω 3 = 2.82 rad / s CCW
Also,
ω2 =
v B2 / A2 14.1
=
= 2.82 rad / s
5
rB/ A
From the directions given in the position and velocity polygons
ω 2 = 2.82 rad / s CCW
ω 4 = 0 rad / s since it doesn’t rotate
Acceleration Analysis:
aB3 = aB2 = aB2 / A2
aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
- 69 -
a Cr 4 / D4 + aCt 4 / D4 = aBr 2 / A2 + aBt 2 / A2 + aCr 3 / B3 + aCt 3 / B3
(2)
Now,
a rB2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB/ A = 2.82 2 ⋅ 5 = 39.8in / s 2
in the direction of - rB2 / A2
aBt 2 / A2 = α 2 × rB/ A ⇒ aBt 2 / A2 = α 2 ⋅ rB/ A (⊥ to rB/ A )
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 2.82 2 ⋅ 5 = 39.8 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 )
a C4 / D4 = 0 link 4 is moving at a constant velocity
Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
aCt 3 / B3 = 38.8 in / s 2
a tB2 / A2 = 38.8 in / s 2
Then,
α3 =
aCt 3 / B3 38.8
=
= 7.76 rad / s2
5
rC / B
α2 =
aBt 2 / A2 38.8
=
= 7.76 rad / s2
5
rB/ A
α 4 = 0 (link 4 isnot rotating)
To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to
aCt 3 / B3 . This direction is clearly counter-clockwise.
To determine the direction of α 2 , determine the direction that rB/ A must be rotated to be parallel to
aBt 2 / A2 . This direction is clearly clockwise.
- 70 -
Problem 2.9
In the mechanism shown below, link 4 is moving to the left at the rate of 0.6 ft/s (constant). Write
the appropriate vector equations, solve them using vector polygons, and determine the velocity and
acceleration of point A3.
Link lengths: AB = 5 in, BC = 5 in.
A
3
B
135˚
2
4
C
vC 4
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
v B3 = v A3 + v B3 / A3
Therefore,
vC3 + v B3 /C3 = v A3 + v B3 / A3
(1)
- 71 -
Now,
vC4 = .6 ft / s parallel to the ground
v B3 /C3 = ω 3 × rB/C (⊥ to rB/C )
v B3 / A3 = ω 3 × rB/ A (⊥ to rB/ A )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
v B3 /C3 = .85 ft / s
or
ω3 =
v B3 /C3
= .85 = 2.04 rad / s
(5 /12)
rB/C
From the directions given in the position and velocity polygons
ω 3 = 2.04 rad / s CW
Now,
v B3 / A3 = ω 3 × rB/ A (⊥ to rB/ A ) = (2.04)(5 /12) = .85 ft / s
Using velocity image,
v A3 =1.34 ft / s
Acceleration Analysis:
aC4 = aC3 = 0
aB3 = aB2 = aB3 /C3 = a r B3 /C3 + a t B3 /C3
(2)
Now,
a rB3 /C3 = ω 3 × (ω 3 × rB/C ) ⇒ a rB3 /C3 = ω 3 2 ⋅ rB/C = 2.04 2 ⋅ (5 /12) = 1.73 ft / s 2
in the direction of - rB3 /C3
aBt 3 /C3 = α 3 × rB/C ⇒ aBt 3 /C3 = α 3 ⋅ rB/C (⊥ to rB/C )
a C4 / D4 = 0 link 4 is moving at a constant velocity
Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
aBt 3 /C3 = 1.73 ft / s 2
Then,
- 72 -
α3 =
aBt 3 /C3
= 1.73 = 4.15 rad / s2
(5 /12)
rB/C
To determine the direction of α3, determine the direction that rB/C must be rotated to be parallel to
aBt 3 /C3 . This direction is clearly clockwise.
Using acceleration image,
a A3 = 4.93 ft / s 2
Problem 2.10
In the mechanism shown below, link 4 moves to the right with a constant velocity of 75 ft/s. Write
the appropriate vector equations, solve them using vector polygons, and
a) Determine vB2, vG3, ω2, and ω3.
b) Determine aB2, aG3, α2, and α3.
Link lengths: AB = 4.8 in , BC = 16.0 in , BG = 6.0 in
B
G
3
2
A
42˚
C
4
Position Analysis: Draw the linkage to scale.
- 73 -
B
AB = 4.8"
BC = 16.0"
BG = 6.0"
AC = 19.33"
G
3
2
42˚
C
A
ov
c 3, c4
a 1, a 2
25 ft/sec
Velocity Polygon
g3
b2, b3
Velocity Analysis:
v B3 = vC3 + v B3 /C3
v B3 = v B2
- 74 -
v B2 = v A2 + v B2 / A2
v A2 = 0
Therefore,
vC3 + v B3 /C3 = v A2 + v B2 / A2
(1)
Now,
vC3 = 75 ft / s in the direction of rC / A
v B3 /C3 = ω 3 × rB/C (⊥ to rB/C )
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A )
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
v B3 /C3 = 69.4 ft / s
or
ω3 =
v B3 /C3
= 69.4 = 52 rad / s
rB/C 16(1 /12)
From the directions given in the position and velocity polygons
ω 3 = 52 rad / s CCW
Also,
ω2 =
v B2 / A2
= 91.5 = 228 rad / s
4.8(1 /12)
rB/ A
From the directions given in the position and velocity polygons
ω 2 = 228 rad / s CW
To compute the velocity of G3,
vG3 = v B3 + vG3 / B3 = v B3 + ω 3 × rG3 / B3
Using the values computed previously
ω 3 × rG3 / B3 = 52(6.0) = 312 in / s
and from the directions given in the velocity and position diagrams
ω 3 × rG3 / B3 = 312 in / s ⊥ rG3 / B3
Now draw vG3 on the velocity diagram
vG3 = 79.0 ft / s in the direction shown.
- 75 -
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
(2)
Now,
a rB2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB/ A = 2282 ⋅ (4.8 /12) = 20, 900 ft / s 2
in the direction of - rB2 / A2
aBt 2 / A2 = α 2 × rB/ A ⇒ aBt 2 / A2 = α 2 ⋅ rB/ A (⊥ to rB/ A )
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 52 2 ⋅ (16 /12) = 3605 ft / s2
in the direction of - rC / B
aCt 3 / B3 = α 3 × rC / B ⇒ aCt 3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B )
a C4 / D4 = 0 link 4 is moving at a constant velocity
Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
aCt 3 / B3 = 28, 700 ft / s 2
a tB2 / A2 = 20, 000 ft / s 2
Then,
α3 =
aCt 3 / B3 28, 700
=
= 21, 500 rad / s2
(16 /12)
rC / B
α2 =
aBt 2 / A2
= 20, 000 = 50, 000 rad / s2
(4.8 /12)
rB/ A
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 clockwise.
To determine the direction of α 2 , determine the direction that rB/ A must be rotated to be parallel to
aBt 2 /C2 . This direction is clearly counter-clockwise.
From the acceleration polygon,
aB2 = 28, 900 ft / s 2
- 76 -
To compute the acceleration of G3, use acceleration image. From the acceleration polygon,
aG3 = 18, 000 ft / s 2
Problem 2.11
For the four-bar linkage, assume that ω2 = 50 rad/s CW and α2 = 1600 rad/s2 CW. Write the
appropriate vector equations, solve them using vector polygons, and
a) Determine vB2, vC3, vE3, ω3, and ω4.
b) Determine aB2, aC3, aE3,α3, and α4.
E
3
C
B
2
120˚
AB = 1.75"
AD = 3.55"
CD = 2.75"
BC = 5.15"
BE = 2.5"
EC = 4.0"
4
D
A
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.
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 = 50 ⋅1.75 = 87.5 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 )
- 77 -
E
C
3
B
Velocity Scale
4
2
50 in/s
b3
A
D
e3
o' d'4
o
c3
aCr 3 / B3
aCr
4 /D 4
aBr 2 / A2
atC
3 / B3
b'3
t
aB 2 / A 2
e'3
atC /D
4 4
Acceleration Scale
c'3
2000 in/s 2
Solve Eq. (1) graphically with a velocity polygon and locate the velocity of point E3 by image.
From the polygon,
vC3 / B3 = 65.2 in / s
vC4 / D4 = 92.6 in / s
and
vE3 = 107.8 in / s
in the direction shown.
Now
ω3 =
vC3 / B3 65.2
=
= 12.7 rad / s
5.15
rC / B
- 78 -
and
ω4 =
vC4 / D4 92.6
=
= 33.7 rad / s
2.75
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.
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
(2)
Now,
a rB2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB/ A = 50 2 ⋅1.75 = 4375 in / s2
in the direction of - rB2 / A2
aBt 2 / A2 = α 2 × rB/ A ⇒ aBt 2 / A2 = α 2 rB/ A = 1600 ⋅1.75 = 2800 in / s2 (⊥ to rB/ A )
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 12.72 ⋅ 5.15 = 830.6 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 )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 33.72 ⋅ 2.75 = 3123 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 an acceleration polygon and determine the acceleration of point E3 by
image. From the acceleration polygon,
aCt 3 / B3 = 1563 in / s2
aCt 4 / D4 = 4881 in / s2
Then,
- 79 -
α3 =
aCt 3 / B3 1563
=
= 303 rad / s2
5.15
rC / B
α4 =
aCt 4 / D4 4881
=
= 1775 rad / s2
2.75
rC / D
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 clockwise.
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 clockwise.
Also
aE3 = 5958 in / s2
Problem 2.12
Resolve Problem 2.11 if ω2 = 50 rad/s CCW and α 2 = 0 .
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.
Velocity Analysis:
The velocity analysis is similar to that in Problem 2.18.
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 = 50 ⋅1.75 = 87.5 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 and locate the velocity of point E3 by image.
From the polygon,
vC4 / D4 = 103.1 in / s
and
vE3 = 116 in / s
- 80 -
in the direction shown.
Now
ω3 =
vC3 / B3 88.8
=
= 17.2 rad / s
5.15
rC / B
ω4 =
vC4 / D4 103.1
=
= 37.5 rad / s
2.75
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 counterclockwise.
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.
- 81 -
E
C
3
B
Velocity Scale
4
2
50 in/s
A
c3
D
o' d'4
aBr 2/ A 2
o
g3
b'3
b3
aCr 3 / B3
Acceleration Scale
e'3
1000 in/s 2
aCt
r
aC
4 / D4
atC
4 /D 4
- 82 -
c'3
3 / B3
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
(2)
Now,
a rB2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB/ A = 50 2 ⋅1.75 = 4375 in / s2
in the direction of - rB2 / A2
aBt 2 / A2 = α 2 × rB/ A ⇒ aBt 2 / A2 = α 2 rB/ A = 0 ⋅1.75 = 0
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 17.24 2 ⋅ 5.15 = 1530 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 )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 37.492 ⋅ 2.75 = 3865 in / s2
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 an acceleration polygon and determine the acceleration of point E3 by
image. From the acceleration polygon,
aCt 3 / B3 = 2751 in / s2
aCt 4 / D4 = 1405 in / s2
Then,
aCt 3 / B3 2751
=
= 534 rad / s 2
5.15
rC / B
at
= 511 rad / s2
α 4 = C4 / D4 = 1405
2.75
rC / D
α3 =
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 clockwise.
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 clockwise.
- 83 -
Also
aE3 = 2784 in / s2
Problem 2.13
In the mechanism shown below, link 2 is rotating CW at the rate of 180 rad/s. Write the
appropriate vector equations, solve them using vector polygons, and
a) Determine vB2, vC3, vE3, ω3, and ω4.
b) Determine aB2, aC3, aE3, α3, and α4.
Link lengths: AB = 4.6 in, BC = 12.0 in, AD = 15.2 in, CD = 9.2 in, EB = 8.0 in, CE = 5.48 in.
Y
C
3
B
4
E
120˚
2
D
A
Position Analysis: Draw the linkage to scale.
Velocity Analysis:
v B2 = v B2 / A2 = ω 2 × rB2 / A2 ⇒ v B2 = ω 2 rB2 / A2 = 180(4.6) = 828 in / s
v B3 = v B2
vC3 = v B3 + vC3 / B3
vC3 = vC4 = vD4 + vC4 / D4
and
vD4 = 0
- 84 -
X
C
AD = 15.2"
DC = 9.2"
BC = 12.0"
AB = 4.6"
EC = 5.48"
EB = 8.0"
3
4
B
E
120˚
2
D
A
b 2, b 3
Velocity Polygon
400 in/sec
ov
e3
a1 , a2
c 3, c 4
Therefore,
vC4 / D4 = v B3 + vC3 / B3
(1)
Now,
v B3 = 828 ft / s (⊥ to rB/ A)
vC3 / B3 = ω 3 × rC / B (⊥ to rC / B )
vC4 / D4 = ω 4 × rC / D (⊥ to rC / D )
Solve Eq. (1) graphically with a velocity polygon. The velocity directions can be gotten directly
from the polygon. The magnitudes are given by:
vC3 / B3 = 583 in / s ⇒ ω 3 =
vC3 / B3 583
=
= 48.6 rad / s
12
rC / B
From the directions given in the position and velocity polygons
ω 3 = 48.6 rad / s CCW
Also,
- 85 -
vC4 / D4 = 475 rad / s ⇒ ω 4 =
vC4 / D4 475
=
= 51.6 rad / s
9.2
rC / D
From the directions given in the position and velocity polygons
ω 4 = 51.6 rad / s CW
To compute the velocity of E3,
vE3 = v B3 + vE3 / B3 = vC3 + vE3 /C3
(1)
Because two points in the same link are involved in the relative velocity terms
vE3 / B3 = ω 3 × rE / B (⊥ to rE / B )
and
vE3 /C3 = ω 3 × rE /C (⊥ to rE /C )
Equation (2) can now be solved to give
vE3 = 695 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 = 180 2 ⋅ 4.6 = 149, 000 in / s2
in the direction of - rB2 / A2
aBt 2 / A2 = α 2 × rB/ A ⇒ aBt 2 / A2 = α 2 rB/ A = 0 ⋅ 4.6 = 0
a Cr 3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a Cr 3 / B3 = ω 3 2 ⋅ rC / B = 48.992 ⋅12 = 28,800 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 )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 50.4 2 ⋅ 9.2 = 23,370 in / s2
in the direction of - rC / D
aCt 4 / D4 = α 4 × rC / D ⇒ aCt 4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D )
- 86 -
(2)
Solve Eq. (2) graphically with an acceleration polygon and determine the acceleration of point E3 by
image.
From the acceleration polygon,
aCt 3 / B3 = 96,880 in / s2
aCt 4 / D4 = 9785 in / s2
Then,
α3 =
aCt 3 / B3 96876
=
= 8073 rad / s 2
12
rC / B
α4 =
aCt 4 / D4 9785.5
=
= 1063.6 rad / s2
9.2
rC / D
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 clockwise.
- 87 -
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 clockwise.
Also
aE3 = 123, 700 in / s2
and
aC3 = 149, 780 in / s2
Problem 2.14
The accelerations of points A and B in the coupler below are as given. Determine the acceleration of
the center of mass G and the angular acceleration of the body. Draw the vector representing aG
from G.
B
- 63˚
G
aA = 7000 in/s 2
a B = 7000 in/s 2
AG = 1.5"
BG = 1.5"
AB = 2.8"
aA
50˚
aB
22˚
A
Acceleration Analysis:
Draw the accelerations of points A and B on an acceleration polygon. Then locate the accleration of
point G by image.
For the angular acceleration of the body, resolve the acceleration aBt / A in terms of components
along and perpendicular to rB/ A . The tangential component is perpendicular to rB/ A .
aBt / A = α × rB/ A ⇒ aBt / A = α rB/ A
and
at
= 1122 rad / s2
α = r B/ A = 3141
2.8
B/ A
- 88 -
B
a'
G
aG
o'
A
g'
Acceleration Scale
2000 in/s 2
atB/ A
b'
To determine the direction of α , determine the direction that rB/ A must be rotated to be parallel to
aBt / A . This direction is clearly clockwise.
Also
aG = 6980 in / s2
in the direction shown.
- 89 -
Problem 2.15
Crank 2 of the push-link mechanism shown in the figure is driven at a constant angular velocity ω2
= 60 rad/s (CW). Find the velocity and acceleration of point F and the angular velocity and
acceleration of links 3 and 4.
Y
C
4
3
E
AB = 15 cm
BC = 29.5 cm
CD = 30.1 cm
AD = 7.5 cm
BE = 14.75 cm
EF = 7.5 cm
F
D
A
X
30˚
2
B
Position Analysis:
Draw the linkage to scale. First located the pivots A and D. Next locate B, then C, then E, then F.
Velocity Analysis:
vA2 = vB3 = vB2 / A2 =ω 2·rB2 / A2
vB2 = ω 2 rB2 / A2 = 60 (0.15) = 9 m / s
vC3 = vB3 + vC3 / B3
(1)
vC3 = vC4 = vC4 / D4 =ω 4·rC / D
Now,
vB3 = 9 m / s
(^ to rB / A)
vC3 / B3 =ω 3·rC / B (^ to rC / B)
vC4 =ω 4·rC / D (^ to rC / D)
Solve Eq. (1) graphically with a velocity polygon. The velocity directions can be gotten directly
from the polygon. The magnitudes are given by:
vC3 / B3 = 12.82 m / s ω 3 =
vC3 / B3 12.82
=
= 43.45 rad / s
rC / B
0.295
Using velocity image,
vF3 = 4.94 m / s
- 90 -
in the direction shown.
C
5 cm
4
3
E
F
A
Acceleration Polygon
100 m/s 2
b'2 , b'3
2
B
a rB2 /A2
30°
Velocity Polygon
2 m/s
f '3
o'
D
c 3 , c4
b2 , b3
aCr 3 /B 3
arC 4 /D4
c' 3 , c'4
atC3 /B3
t
aC4
/D 4
a2 o
From the directions given in the position and velocity polygons
ω 3 = 43.45 rad / s CW
Also,
vC4 / D4 = 11.39 m / s
ω4 =
vC4 / D4 11.39
=
= 37.84 rad / s
rC / D
0.301
From the directions given in the position and velocity polygons
ω 4 = 37.84 rad / s CW
Acceleration Analysis:
aB2 = aB3 = ar B2 / A2 =ω 2· (ω 2·rB / A)
- 91 -
aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
atC4 / D4 + arC4 / D4 = ar B2 / A2 + at B2 / A2 + arC3 / B3 + atC3 / B3
(2)
Now,
arC4 / D4 = ω 4 2 rC / D = 37.842 0.301 = 430.99 m / s2 in the direction opposite to rC/ D )
arC3 / B3 = ω 3 2 rC / B = 43.452 0.295 = 556.93 m / s2 in the direction opposite to rC/ B )
ar B2 / A2 = ω 2 2 rB / A = 602 0.15 = 540 m / s2 in the direction opposite to rB/ A )
atC3 / B3 = α 3·rC / B α 3 =
atC3 / B3
(^ to rC / B)
rC / B
atC4 / D4 = α 4·rC / D α 4 =
atC4 / D4
(^ to rC / D)
rC / D
Solve Eq. (2) graphically with an acceleration polygon. The acceleration directions can be gotten
directly from the polygon. The magnitudes are given by:
α3 =
atC3 / B3 142.79
=
= 484 rad / s2 CW
0.295
rC / B
α4 =
atC4 / D4 41.01
=
= 136 rad / s2 CCW
0.301
rC / D
Also,
Using acceleration polygon,
aF3 = 256 m / s2
in the direction shown.
Problem 2.16
For the straight-line mechanism shown in the figure, ω2 = 20 rad/s (CW) and α2 = 140 rad/s2
(CW). Determine the velocity and acceleration of point B and the angular acceleration of link 3.
B
D
DA = 2.0"
AC = 2.0"
AB = 2.0"
2
15
o
A
3
C
4
- 92 -
Velocity Analysis:
vA2 = vA2 / D2 = vA3 =ω 2·rA2 / D2
vC3 = vC4 = v A3 + vC3 / A3
(1)
Now,
v A3 = ω 2 rA2 / D2 = 20 ⋅ 2 = 40 in / s (⊥ to rA2 / D2 )
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,
v B3 = 77.3 in / s
Also,
vC3 / A3 = 40 in / s
or
ω3 =
vC3 / A3 40
= = 20 rad / s CCW
2
rC3 / A3
Also,
vC3 = 20.7 in / s
Acceleration Analysis:
aC3 = aC4 = a A3 + aC3 / A3
aC3 = a Ar 2 / D2 + a tA2 / D2 + aCr 3 / A3 + aCt 3 / A3
Now,
aC3 in horizontal direction
a rA2 / D2 = ω 2 × (ω 2 × rA2 / D2 ) ⇒ a rA2 / D2 = ω 2 2 ⋅ rA/ D = 20 2 ⋅ 2 = 800 in / s2
in the direction opposite to rA/ D
a tA2 / D2 = α 2 × rA2 / D2 ⇒ a tA2 / D2 = α 2 ⋅ rA/ D = 140 ⋅ 2 = 280 in / s2 (⊥ to rA/ D )
a Cr 3 / A3 = ω 3 × (ω 3 × rC3 / A3 ) ⇒ a Cr 3 / A3 = ω 3 2 ⋅ rC3 / A3 = 20 2 ⋅ 2 = 800 in / s2
in the direction opposite to rC3 / A3
- 93 -
(2)
aCt 3 / A3 = α 3 × rC3 / A3 ⇒ aCt 3 / A3 = α 3 ⋅ rC3 / A3 (⊥ to rC3 / A3 )
Solve Eq. (2) graphically with a acceleration polygon. From the polygon,
aB3 = 955 in / s2
aC3 / A3 = 280 in / s2
Also,
α3 =
aCt 3 / A3 280
=
= 140 rad / s2 CCW
2
rC3 / A3
B
2
D
A
3
C
4
o
v C3
v A3
c3 , c 4
vC 3 / A 3
Acceleration Polygon
400 in/s 2
c '3
o' a C3
t
aC
3 / A3
v B3
r
a A2 / D 2
t
a2 , a3
a A2/D 2
r
a C 3 / A3
a 2' , a 3'
a B3
v B3 / A3
b 3'
Velocity Polygon
20 in/s
b3
- 94 -
Problem 2.17
For the data given in the figure below, find the velocity and acceleration of points B and C. Assume
vA = 20 ft/s, aA = 400 ft/s2, ω2 = 24 rad/s (CW), and α 2 = 160 rad/s2 (CCW).
C
vA
α2
AB = 4.05"
AC = 2.5"
BC = 2.0"
90˚
A
B
15 o
ω2
aA
Position Analysis
Draw the link to scale
Velocity Polygon
a2
5 ft/sec
Acceleration Polygon
80 ft/sec 2
o'
c2
b'2
b2
a tB2 /A 2
a'2
r
aB
2 /A 2
c'2
o
Velocity Analysis:
(1)
v B2 = v A2 + v B2 / A2
Now,
- 95 -
v A2 = 20 ft / sec in the positive vertical direction
v B2 / A2 = ω 2 × rB/ A ⇒ v B2 / A2 = ω 2 ⋅ rB/ A = 24 ⋅ 4.05 = 97.2in / sec
v B2 / A2 = 8.1 ft / sec(⊥ to rB/ A ) in the positive vertical direction
Solve Eq. (1) graphically with a velocity polygon. From the polygon,
v B2 = 11.9 ft / sec
Also, from the velocity polygon,
vC2 = 15.55 ft / sec in the direction shown
Acceleration Analysis:
aB2 = a A2 + aB2 / A2 = a A2 + aBt 2 / A2 + aBr 2 / A2
Now,
a A2 = 400 ft / sec 2 in the given direction
aBt 2 / A2 = α 2 ⋅ rB/ A = 160 ⋅ 4.05 = 648 in / sec 2 = 54 ft / sec 2
aBr 2 / A2 = ω 2 2 ⋅ rB/ A = 24 2 ⋅ 4.05 = 2332 in / sec 2 = 194 ft / sec 2
Solve Eq. (2) graphically with an acceleration polygon. From the polygon,
aB2 = 198.64 ft / sec 2
in the direction shown. Determine the acceleration of point C by image. From the acceleration
image,
aC2 = 289.4 ft / sec 2
in the direction shown.
- 96 -
(2)
Problem 2.18
In the mechanism shown below, link 2 is turning CCW at the rate of 10 rad/s (constant). Draw the
velocity and acceleration polygons for the mechanism, and determine aG3 and α4.
C
3
G
4
AB = 1.0"
BC = 2.0"
BG = 1.0"
CD = 3.0"
AD = 3.0"
B
2
A
ω 2 , α2
90˚
D
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:
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 = 10 ⋅ 2 = 20 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 )
- 97 -
Solve Eq. (1) graphically with a velocity polygon.
From the polygon,
vC3 / B3 = 14.4 in / s
vC4 / D4 = 13.7 in / s
in the direction shown.
Now
ω3 =
vC3 / B3 11.4
=
= 5.7 rad / s
2
rC / B
ω4 =
vC4 / D4 13.7
=
= 4.57 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 G3
vG3 = v B3 + vG3 / B3 = ω 3 × rG/ B = ω 3 ⋅ rG/ B = 5.7 ⋅1 = 5.7 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,
- 98 -
(2)
a rB2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a rB2 / A2 = ω 2 2 ⋅ rB/ A = 10 2 ⋅1 = 100 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 = 5.72 ⋅ 2 = 64.98 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 )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 4.572 ⋅ 3 = 62.66 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 for the accelerations.
- 99 -
From the acceleration polygon,
aCt 3 / B3 = 38 in / s2
aCt 4 / D4 = 128 in / s2
Then,
α3 =
aCt 3 / B3 38
= = 19 rad / s2
2
rC / B
α4 =
aCt 4 / D4 128
=
= 42.67 rad / s2
3
rC / D
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.
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 clockwise.
Determine the acceleration of point G3
r
aG3 = aB3 + aG3 / B3 = aB2 + a G
+ aGt 3 / B3
3 / B3
aGt 3 / B3 = α 3 × rG/ B ⇒ aGt 3 / B3 = α 3 ⋅ rG/ B = 19 ⋅1 = 19 in / sec 2
r
r
aG
= ω 3 × (ω 3 × rG/ B ) ⇒ a G
= ω 3 2 ⋅ rG/ B = 5.72 ⋅1 = 32.49 in / sec 2
3 / B3
3 / B3
From the acceleration polygon,
aG3 = 116 in / s2
Problem 2.19
If ω2 = 100 rad/s CCW (constant) find the velocity and acceleration of point E.
D
A
ω2
70˚
2
E
4
B
3
C
- 100 -
AB = 1.0"
BC = 1.75"
CD = 2.0"
DE = 0.8"
AD = 3.0"
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.
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 = 100 ⋅1 = 100 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 = 77.5 in / s
vC4 / D4 = 71 in / s
in the direction shown.
- 101 -
Now
ω3 =
vC3 / B3 77.5
=
= 44.29 rad / s
rC / B 1.75
ω4 =
vC4 / D4 71
= = 35.5 rad / s
2
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
vE4 = vD4 + vE4 / D4 = ω 4 × rD/ E = ω 4 ⋅ rD/ E = 35.5 ⋅ 0.8 = 28.4 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 = 100 2 ⋅1 = 10, 000 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 = 44.292 ⋅1.75 = 3432.8 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 )
a Cr 4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a Cr 4 / D4 = ω 4 2 ⋅ rC / D = 35.52 ⋅ 2 = 2520.5 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.
- 102 -
(2)
From the acceleration polygon,
aCt 3 / B3 = 3500 in / s2
aCt 4 / D4 = 10, 900 in / s2
Then,
α3 =
aCt 3 / B3 3500
=
= 2000 rad / s2
1.75
rC / B
α4 =
aCt 4 / D4 10, 900
=
= 5450 rad / s2
2
rC / D
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.
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 clockwise.
Determine the acceleration of point E4
aE4 = aD4 + aE4 / D4 = a rE4 / D4 + aEt 4 / D4
- 103 -
aEt 4 / D4 = α 4 × rE / D ⇒ aEt 4 / D4 = α 4 ⋅ rE / D = 5450 ⋅ 0.8 = 4360 in / sec 2
a rE4 / D4 = ω 4 × (ω 4 × rE / D ) ⇒ a rE4 / D4 = ω 4 2 ⋅ rE / D = 35.52 ⋅ 0.8 = 1008.2 in / sec 2
From the acceleration polygon,
aE4 = 4600 in / s2
Problem 2.20
Draw the velocity polygon to determine the velocity of link 6. Points A, C, and E have the same
vertical coordinate.
2
1 ω2 =
B
AB = 1.80"
BC = 1.95"
CD = 0.75"
DE = 2.10"
rad
6 s
3
50˚
A
4
E
C
6
5
D
Velocity Analysis:
v B3 = v B2 = v B2 / A2
vC4 = vC3 = v B3 + vC3 / B3
(1)
v D5 = vD3
vE5 = vE6 = vD5 + vE5 / D5
(2)
Now,
v B2 / A2 = ω 2 × rB2 / A2 ⇒ v B2 / A2 = ω 2 ⋅ rB2 / A2 = 6 ⋅1.8 = 10.8 in / s (⊥ to rB2 / A2 )
vC3 is in the vertical direction. Then,
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,
vD3 = 18.7 in / s
- 104 -
B
2
3
C
A
E
6
4
D
5
o
e5
b3
Velocity Polygon
c3
d3
10 in/sec
d5
Now,
vE5 in horizontal direction
vE5 / D5 = ω 5 × rE / D ⇒ vE5 / D5 = ω 5 ⋅ rE / D (⊥ to rE / D )
Solve Eq. (2) graphically with a velocity polygon. From the polygon, using velocity image,
vE5 = vE6 = 8.0 in / s
- 105 -
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,
- 106 -
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
- 107 -
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,
- 108 -
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
- 109 -
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
- 110 -
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
- 111 -
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,
- 112 -
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.
- 113 -
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,
- 114 -
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
- 115 -
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
- 116 -
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 )
- 117 -
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
- 121 -
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
- 122 -
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.
- 123 -
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,
- 124 -
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
- 125 -
(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
- 126 -
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
- 127 -
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
- 128 -
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
- 158 -
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,
- 163 -
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
- 164 -
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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