Solutions to Chapter 5 Exercise Problems
Problem 5.1 For the mechanism shown, do the following: a) Write the vector equation of the above linkage. b) Write the x and y displacement equations. c) Find the velocity component equations. d) Find the acceleration component equations.
B
φ
b
a
C
θ
A
c
Solution Position Analysis a+b=c
In component form a cos cos + bcos = ccos0 asin + bsin
=
csin csin 0
Substituting in the constant numbers a cos cos + bcos = c asin + bsin
=
0
Velocity Analysis a sin bsin = c a cos + b cos = 0 ˙
˙
˙
˙
˙
Acceleration Analysis a( sin + 2 cos ) b( sin + 2 cos ) = c a( cos 2 sin ) + b( cos 2 sin ) = 0 ˙ ˙
˙
˙ ˙
˙
˙ ˙
˙ ˙
˙
˙ ˙
˙
- 178 -
Problem 5.2
In the mechanism in Problem 5.1, determine analytically for the following values: ˙
a
=
b
1 cm,
=
4cm,
=
˙
˚
60 ,
=
10 rad / sec
Solution B
r3
2
3
θ
r2
C
φ
A
4
r1
The analysis can be conducted using the equations in Table 5.4 with M=2, J=3
The known input information is: 1 = 0 ° ; r 2
=
AB
=
1 cm;
2 = = 60°;
˙2
r 3
r 4
=
BC
=
4 cm; cm;
=
=
˙
=
10 rad/sec;
0 cm; cm;
Start with the position analysis, and first compute constants A and B: A
=
2r4(cos 1 cos 4 + sin 1sin 4) 2r2 (cos1cos2
+
sin 1 sin 2 )
= 2*1*(cos0 ° cos60° + sin0°sin60°) = 1
B = r22 + r42 r32 2r2r4(cos 2 cos4 + sin 2 sin 4) =
12
42
=
15
The desired configuration of the linkage corresponds to the position of the slider with the larger x coordinates. Therefore = +1 . Then 2 r1 = A + A 4B 2
Then
3 is
( 1) + (1)2 4(15) 2
=
=
4.405
given by
r sin 1 + r4 sin 4 r2 sin 2 1 sin60° 3 = tan1[ 1 ] = tan1[ ] = 12.504° r1 cos1 + r4 cos 4 r2 cos 2 4.405 1 cos60°
For the velocity , solve the linear set of velocity equations, ˙
- 179 -
Problem 5.2
In the mechanism in Problem 5.1, determine analytically for the following values: ˙
a
=
b
1 cm,
=
4cm,
=
˙
˚
60 ,
=
10 rad / sec
Solution B
r3
2
3
θ
r2
C
φ
A
4
r1
The analysis can be conducted using the equations in Table 5.4 with M=2, J=3
The known input information is: 1 = 0 ° ; r 2
=
AB
=
1 cm;
2 = = 60°;
˙2
r 3
r 4
=
BC
=
4 cm; cm;
=
=
˙
=
10 rad/sec;
0 cm; cm;
Start with the position analysis, and first compute constants A and B: A
=
2r4(cos 1 cos 4 + sin 1sin 4) 2r2 (cos1cos2
+
sin 1 sin 2 )
= 2*1*(cos0 ° cos60° + sin0°sin60°) = 1
B = r22 + r42 r32 2r2r4(cos 2 cos4 + sin 2 sin 4) =
12
42
=
15
The desired configuration of the linkage corresponds to the position of the slider with the larger x coordinates. Therefore = +1 . Then 2 r1 = A + A 4B 2
Then
3 is
( 1) + (1)2 4(15) 2
=
=
4.405
given by
r sin 1 + r4 sin 4 r2 sin 2 1 sin60° 3 = tan1[ 1 ] = tan1[ ] = 12.504° r1 cos1 + r4 cos 4 r2 cos 2 4.405 1 cos60°
For the velocity , solve the linear set of velocity equations, ˙
- 179 -
cos1 r 3 sin3 r1 r 22 sin 2 sin 1 r 3 cos 3 3 r 2 2 cos2 ˙
˙
=
˙
or
cos0° 4sin( 12.504°) r1 = 110sin60° sin0° 4cos(12.504°)3 110cos60° ˙
˙
then
1 0.866 r1 8.660 0 3.905 3 5 ˙
=
˙
or
r˙1 9.768 ˙ 1.280 3 =
Therefore, ˙
=
˙3
=
1.280 rad/sec,
CW.
Problem 5.3 ˚
In the mechanism shown, s˙ 10 in/ s and s˙˙ 0 for the position corresponding to = 60 . Find ˙ and ˙˙ for that position using the loop equation approach. =
=
3
s
4
2
10 inches
φ
Solution The vector equation is r3 = r2 + r1
In component form, this equation becomes: r 3 cos3 = r 2 cos 2 + r 1 cos 1 r 3 sin 3 = r 2 sin 2 + r 1 sin 1
- 180 -
s
r1 3
4
r3
2
r2
φ
Substituting the constant values r 3 cos r 3 sin
=
=
1 = 0 ° and 2 = 90 ° gives
r 1 r 2
The component equations for velocity are: ˙ sin r˙3 cos r 3 ˙ cos r˙3 sin + r 3
= r ˙1 =
0
The component equations for acceleration are: ˙ sin r ˙˙sin r 3 ˙ 2 cos ˙r˙3 cos 2r˙3 3 ˙ cos + r 3 ˙˙ cos r 3 ˙ 2 sin ˙r˙3 sin + 2r˙3
=˙ r˙1 =
0
The known input information is: = 60°
r2 = 10 in
r1 = s = 10 in / s ˙
˙
so r2
r3 =
sin
=
10 sin60°
= 11.547
r1 = r3 cos = 11.547cos60° = 5.774
Solve the velocity equations:
cos r 3 sin r3 r1 sin r 3 cos 0 ˙
˙
=
˙
or
cos60° 11.547sin60 ° r3 10 sin60° 11.547cos60° = 0 ˙
˙
or
0.5 0.866 then
10 r3 10 0 5.774 ˙
=
˙
r˙3 5 ˙ 0.75 =
Therefore ˙
=
0.75 rad/sec,
CCW.
- 181 -
r1 = s = 0
˙ ˙
˙ ˙
Solve the acceleration equations:
cos r 3 sin r3 r1 + 2r3sin + r 32 cos sin r 3 cos = 2r3 cos + r 32 sin ˙
˙ ˙
˙ ˙
˙ ˙
˙
or
˙
˙
˙
˙
cos60° 11.547sin60° r3 2(5)(0.75)sin60° + 11.547 0.752 cos60° sin60° 11.547cos60° = 2(5)(0.75)cos60° + 11.547 0.752 sin60° ˙ ˙
˙ ˙
or
10 r3 3.248 5.774 9.375
0.5 0.866
˙ ˙
=
˙ ˙
then
˙r˙3 6.495 ˙˙ 0.650 =
˙˙ Therefore
=
0.65 .650 rad / sec2 ,
CCW.
Problem 5.4
In the mechanism in Problem 5.3 assume that is 10 rad/s CCW. Use the loop equation equation approach to determine the velocity of point B4 for the position defined by = = 60 . ˙
˚
Solution B 4
r2
r1 3
r3
2 φ
The vector equation is: r3 = r2 + r1
In component form, this equation becomes: r 3 cos3 = r 2 cos 2 + r 1 cos 1 r 3 sin 3 = r 2 sin 2 + r 1 sin 1
Substituting the constant values r 3 cos r 3 sin
=
=
1 =
0 and
2 = 90 ° gives
r 1 r 2
The component equations for velocity are:
- 182 -
˙ sin r˙3 cos r 3 ˙ cos r˙3 sin + r 3
= ˙ r1 =
0
The known input information is: ˙
= 60 ° ;
=
10 ; r 2
=
10 inches;
so r3 =
r2 sin
=
10 sin60°
= 11.547
r1 = r3 cos = 11.547cos60° = 5.774
Solve for the velocities: r˙3 =
˙ cos r3 sin
=
11.547 10cos60° sin60°
=
66.667
˙ sin = 66.667 cos60° 11.547 10sin60° = 133.333 r˙1 = r˙3 cos r3
Therefore
vB4
=
133.333 cm/sec.
Problem 5.5
In the mechanism given, point A is moving to the right with a velocity of 10 cm/s. Use the loop equation approach to determine the angular velocity of link 3. Link 3 is 10 cm long, and is 120 in the position shown. ˚
4 B
3 φ
A 2
Solution The vector loop equation is: r2 = r1 + r3
In component form, this equation becomes: r 2 cos 2 = r 1 cos 1 + r 3 cos 3 r 2 sin 2 = r 1 sin 1 + r 3 sin 3
Substituting the constant values 1 = 0, 3 = , and
2 = 90° gives
- 183 -
4 B r3 3 r2 φ
r1 0
2
A
= r 1 + r 3 cos
r 2
= r 3 sin
The component equations for velocity are: 0 r2 ˙
r1 r 3 sin
= ˙
=
˙
r 3 cos ˙
The known input information is: = 120° ;
r 3
=
10 cm;
r˙1
=
vA
=
10 cm/sec
so r 2
= r 3 sin = 10sin120 = 8.66
r 1 =
r 3 cos = 10cos120° = 5.00
Solve for : ˙
˙ =
Therefore ˙
=
˙r1 10 = = 1.155 r 3 sin 10sin120°
1.155
rad/sec, CCW.
Problem 5.6 ˚
Resolve Problem 5.5 if is 150 . Solution
- 184 -
4 B r2
r3
3 φ
A 2
r1 Position Analysis
The basic loop equation is: r1 + r3 = r2
In component form r1 cos 1 + r3 cos = r2 cos 2 r1 sin 1 + r3 sin = r2 sin 2
Substituting in the constant numbers r1 cos(0 ) + r3 cos = r2 cos( 90°) r1 sin(0 ) + r3 sin = r2 sin(90°)
or r1
+ r3 cos
= 0
r3 sin = r2
Then, r1 = r3 cos = 10 cos(150°) = 8 .66 r2
= r3 sin = 10 sin(150°) = 5 .0
Velocity Analysis ˙3 sin ˙3 r˙1 = r3 r˙2
˙3 = r3
r˙1 10 = = 2 .0 rad / sec r3 sin 10 sin(150°)
=
cos = 10( 2.0)cos(150°) = 17.32 in / sec
Acceleration Analysis 2
( )
˙˙3 sin + r3 ˙3 ˙˙ r1 = r3
˙˙3 = cos
( )
˙ ˙˙ r1 r3 3
2
cos
r3 sin
- 185 -
=
0 10(2 )2 cos(150°) 10 sin(150°)
= 6.928
rad / sec2
2
( )
˙˙3 cos r3 ˙3 ˙˙ r2 = r3
2
sin = 10( 6.928)cos(150°) 10( 2) sin(150°) = 80 in / sec 2
Then vA 2
=
17.32 in / sec
aA 2
=
80 in / sec2
Problem 5.7
The mechanism shown is a marine steering gear called Raphson’s slide. AB is the tiller, and CD is the actuating rod. If the velocity of rod CD is a constant 10 inches per minute to the right, use the loop-equation approach to determine the angular acceleration of the tiller. ψ
=
˚
300
A
2
6'
C
3
4
1
D
1
D
B
Solution ψ
=
˚
300
A
r1
r2
2
r3
C
3
4 B
The vector equation is: r2 = r1 + r3
In component form this equation becomes:
- 186 -
r 2 cos 2 = r 1 cos 1 + r 3 cos 3 r 2 sin 2 = r 1 sin 1 + r 3 sin 3
Substituting the constant values q 1 = 0, r 2 cos r 2 sin
=
=
3 = 90° gives
r 1
r 3
The component equations for velocity is: r˙2 cos r ˙ sin 2 r˙2 sin
r1 = ˙
+ r 2˙
cos = 0
The component equations for acceleration is: r2 cos 2˙r2 ˙ sin r 2 ˙˙ sin r ˙ 2 cos ˙˙ 2 r2 sin ˙˙
+
2˙r2 ˙ cos
+ r 2 ˙˙
cos
r˙1 =˙
r ˙ 2 sin = 2
0
The known input information is:
= 300 ° ;
r 3
=
6'
;
r˙1
=
10 in/min;
˙˙ r1
=
0;
so r3 6 = = 6.928 sin sin300° r1 = r2 cos = 6.928cos300 ° = 3.464 r2 =
To solve the velocities:
cos sin or
r 2 sin r2 r1 0 r 2 cos ˙
˙
=
˙
cos300° 6.928sin300 ° r2 10 sin300° 6.928cos300 ° = 0 ˙
˙
or
0.5 0.866
r2 10 0 3.464 6
˙
=
˙
then
r˙2 6 ˙ 1.25 =
To solve the accelerations:
cos sin or
r 2 sin r2 r1 + 2r2 sin + r 2 2 cos = 2r2 cos + r 2 2 sin r 2 cos ˙ ˙
˙ ˙
˙ ˙
˙
˙
˙
˙
˙
˙
cos300° 6.928sin300° r2 2(6)(1.25)sin300° + 9.628 1.252 cos300° = sin300° 6.928cos300° 2(6)(1.25)cos300° + 9.628 1.252 sin300° ˙ ˙
˙ ˙
or
- 187 -
0.5 0.866
r2 7.578 16.875 3.464 6
˙ ˙
=
˙ ˙
then
˙˙r2 10.826 ˙˙ 2.165 =
Therefore ˙˙
=
2.165 rad / min 2 CCW.
Problem 5.8
Use loop equations to determine the velocity and acceleration of point B on link 2 when Make point A the origin of your reference coordinate system. 10 in. 2
y
3
B
1ω 4 = 1 rad
sec
1α4 = 0
θ3 4
x
A
Solution
y 2 B
r3 3 θ3
r1 A
The vector equation is: r3 = r1 + r2
In component form, this equation becomes: r 3 cos3 = r 1 cos 1 + r 2 cos2 r 3 sin 3 = r 1 sin 1 + r 2 sin 2
Substituting the constant values 1 = 0 and
2 = 90 ° gives
- 188 -
r2
x
3
˚
= 30 .
r 3 cos3 r 3 sin 3
=
=
r 1 r 2
The component equations for velocity are: ˙ 3 sin 3 = 0 r˙3 cos3 r 3 ˙3 cos3 = ˙r2 r˙3 sin 3 + r 3
The component equations for acceleration are: ˙ 3 sin 3 r ˙3 sin 3 r ˙ 32 cos 3 = 0 r3 cos3 2˙r3 ˙˙ 3˙ 3 ˙ 3 cos ˙ 3 + r ˙˙ ˙ 2 r3 sin 3 + 2˙r3 r2 ˙˙ 33 cos3 r 33 sin 3 = ˙˙
The known input information is: ˙3
3 = 30 ° ;
=
4
=
1 rad/sec;
˙ ˙3
=
4
=
0;
r 1
=
10 inches;
so r3 =
r1 cos3
=
10 = 11.547 cos30°
r2 = r3 sin 3 = 11.547sin30° = 5.774
To solve the velocities: ˙ 3 sin 3 r3
= 11.547 1 sin30° = 6.667 cos3 cos30° ˙ r˙2 = ˙r3 sin 3 + r33 cos3 = 6.667sin30° + 11.547 1 cos30° = 13.333 r˙3 =
Therefore
vB2
r2
= ˙
=
13.333 in/sec.
Solve for the accelerations: r3 = ˙˙
˙ 3 sin 3 + r3˙˙ ˙2 cos 3 2r˙3 3 sin 3 + r3 3 cos3
2 = 2 6.667 1 sin30° + 11.547 1 cos30° = 19.245 cos30° ˙ ˙3 cos 3 r3 ˙ 32 sin 3 r2 = ˙r˙3 sin 3 + 2˙r33 cos 3 + r3˙ ˙˙
= 19.245 sin30° + 2 6.667 1 cos30° 11.547 12 sin30° = 15.397
Therefore
aB2
r2
= ˙ ˙
=
15.397in / sec2 .
- 189 -
Problem 5.9 ˚
In the mechanism shown, 30 , 2 1 rad / s CCW, and determine the velocity and acceleration of point B on link 4. =
=
2
=
0.
Y
3 in A
2
θ
3 B
X
4
Solution
A
Y 2
r2
r3 30
3
r1 B
4
Position Analysis r1 + r3 = r2
In component form, r1 cos0
+ r3 cos90 = r2 cos
r1sin 0 + r3 sin90
= r2 sin
or r 1 r 3
=
=
r 2 cos r 2 sin
Solving for r2 and r3 , r2 r3
=
=
r1 / cos r2 sin
=
=
3 /cos30
=
3.464sin30
3.464 in.
=
1.732 in
Velocity Analysis r ˙1 + r ˙3 = r ˙2
In component form,
- 190 -
˚
X
Use loop equations to
˙ sin r˙1 = r˙2 cos r 2 ˙ cos 0 = ˙r2 sin + r 2
or r˙2 r˙1
=
˙ cos / sin r 2 ˙ sin ˙r2 cos r 2
=
or r˙2 3.464(1)[cos(30) / sin(30)] 6.00 in / sec r˙1 6.00cos30 3.464(1)sin30 6.928 in/ sec =
=
=
=
Acceleration Analysis r ˙˙1 + r ˙˙3 = r ˙˙2
In component form, ˙ sin r 2˙˙ ˙ 2 cos sin r 2 r1 = ˙r˙2 cos 2˙r2 ˙˙ ˙ cos + r 2 ˙˙ cos r 2 ˙ 2 sin 0 = ˙r˙2 sin + 2r˙2
or r2 ˙˙
=
˙˙ r1 =
˙ cos + r 2˙˙ ˙ 2 sin cos r 2 2˙r2 sin 2 ˙ ˙ + r 2˙ ˙ sin ˙r˙2 r 2 cos 2˙r2
[
]
[
]
or r2 ˙˙
=
[
2 2(6.00)(1)cos30 3.464(0)cos30 3.464(1) sin30
]
=
sin30 2 [ 6.00)(1) + 3.464(0) ]sin30 ˙˙ r1 = [ 24.249 3.464(1) ] cos30 2( 2 = 18.000 + 6.000 = 24.000 in /s
Therefore, vB 4
=
r1
aB4
=
r1 24.000 in / s2
˙
˙˙
=
6.928 in / s
=
- 191 -
24.249 in / sec 2
Problem 5.10
In the mechanism for Problem 5.9, assume that vB4 is a constant 10 in/s to the left and Use loop equations to determine the angular velocity and acceleration of link 3. Solution Y A 2
r2 vB 4
B
3
4
r1
1θ
The basic loop equations is: r2 = r1 + r3
In component form this equation becomes: r 2 cos 2 = r 1 cos 1 + r 3 cos 3 r 2 sin 2 = r 1 sin 1 + r 3 sin 3
Substituting the constant values (r 1 = 3, 1 = 0, 3 = 90) gives r 2 cos 2 r 2 sin 2
=
=
r 1 r 3
When 2 = 30, r2 = 3 / c os 2 = 3 / cos30° = 3.464
and r 3 = r 2 sin 2 = 3.464sin30 ° = 1.732
For the velocities, ˙2 = r ˙1 + r ˙3 r
or in component form, ˙ 2 sin 2 r˙2 cos 2 r 2 ˙ 2 cos2 r˙2 sin 2 + r 2
r1 = ˙ =
0
Substituting the know values gives
0.866 (3.464)(0.5) r2 10 0.5 (3.464)(0.866)2 0 ˙
=
˙
or
- 192 -
3
r3 X
is
˚
45 . .
˙r2 8.661 ˙ 2 1.443 =
Therefore, ˙ 2
=
1.443 rad/sec,
CCW.
For the accelerations, r r1 + ˙˙ r3 ˙˙2 = ˙˙
or in component form, ˙2 sin 2 r 2˙ ˙2 sin 2 r ˙ 2 cos2 ˙˙ r2 cos 2 2˙r2 2 2 ˙ 2 cos2 + r 2˙˙ 2 cos 2 ˙˙ r2 sin 2 + 2˙r2
=˙ r˙1
˙2 r 2 2 sin 2 =
0
Substituting the know values gives
0.866 (3.464)(0.5) r2 2(8.661)(1.443)sin30° + (3.464)(1.4432 )cos30° 0.5 (3.464)(0.866)2 = 2(8.661)(1.443)cos30° + (3.464)(1.4432 )sin30° ˙ ˙
˙ ˙
or
˙˙r2 7.213 ˙˙ 2 7.216 =
Therefore, ˙˙2
=
7.216 rad / sec 2 ,
CCW.
Problem 5.11
For the mechanism in the position shown, link 2 is the driver and rotates with a constant angular velocity of 100 rad/s CCW. Write vector loop equations for position, velocity, and acceleration, and solve for the velocity and acceleration of point C on link 4. AB
=
0.9"
AD
=
1.7", BC 2.6", =
h
=
0.8",
1 C
3 B
4
h 2 φ
θ1
A
Solution
- 193 -
D
=
˚
6,
˚
120 =
r'3
Y 3
C
B
r'4
r2
2
4
r4
r3
θ'4
r1 A
θ4
X
D
The basic loop equation for the mechanism is: r1 + r4 = r2 + r3
In component form of this equation becomes: r 1 cos1 + r 4 cos4
= r 2 cos 2 + r 3 cos3
r 1sin 1 + r 4 sin 4 = r 2 sin 2 + r 3 sin 3
r1, r2 , r4, and 1 are constants, and 2
=
+ 1 and
4 = 90° + 3 .
The component equations for the velocities are: r 44 sin 4 ˙
r 4 4 cos4 ˙
˙3
=
=
r 22 sin 2 ˙
+ r3 cos3 ˙
r 33 sin 3 ˙
= r 2 2 cos 2 + r3 sin 3 + r 33 cos 3
˙ ˙4 , 2
˙
˙
˙
=
˙ .
And the component equations for the accelerations are: r 44 sin 4 r 42 4 cos 4 ˙ ˙
=
˙
2 r 22 sin 2 r 22 33 sin 3 r 33 cos 3 2 cos 2 + r3 cos3 2r33 sin 3 r ˙ ˙
˙
˙
˙ ˙
˙ ˙
˙
˙
r 4 4 cos4 r 4 2 4 sin 4 ˙ ˙
=
˙
r 22 cos2 r 22 sin 2 + r3 sin 3 + 2r33 cos 3 + r 33 cos 3 r 32 sin3 2 3 ˙ ˙
˙
˙
˙ ˙
˙ ˙
˙
˙
The known input information is: r 1
=
1.7 ,
r 2
=
0.9 ,
r 3
=
2.6 ,
h=0.8,
1 = 6 ° ,
2 = 126° ,
= 120°
The position equations cannot be solved directly because there are too many unknowns. However, if we rewrite the equations in terms of r 4' and r 3' , they can be solved. Here, r 4' is perpendicular to r 3 , and r 3' is measured along r 3 from B to the intersection of r 4' and r 3 . Then the magnitude of r 4' is h = 0.8. Also, '4 = 90° + 3 . After substituting '4 = 90° + 3 into the equations for position and rewriting the equations, we get: r 1 cos1 r 2 cos2 r 1sin 1 r 2 sin 2
To eliminate 3 ,
'
'
(1)
= r 4 sin 3 + r 3 cos 3
=
r '4 cos3 + r 3' sin 3
(2)
square Eqs. (1) and (2) and add the results. This gives:
- 194 -
r12 + r22 r' 24 2r1r2 cos(1 2) = r' 23
Substitute 2 r 12
r 3' =
+ 1 into the equation above. Then
=
+
1r 2 cos r 22 r' 2 4 2r
or r3' = 1.72 + 0.92 0.82 2(1.7)(0.9)cos120° = 2.14
To solve for
3 ,
use the trigonometric identities
2tan sin 3
=
1 + tan
3
2 2 3 2
1 tan2 cos 3
=
1 + tan
Let
t
=
tan
3 2
3
2
2 3
2
, and substitute the trigonometric identities above into Eq. (1):
A(1 + t 2 ) r3'(1 t2 ) r4' (2t) = 0
Where
A
=
(3)
r 1 cos 1 r 2 cos 2
Collecting terms in Eq (3) gives: (A + r3' )t 2 2r4' t + (A r3' ) = 0
The roots are: t
=
2 2 r 4' + r' 2 4 A + r' 3
A + r 3'
Where = ±1. Then, t1
=
0.8 +
0.82 2.222 + 2.142 2.22 + 2.14
=
0.307
0.8
0.82 2.222 + 2.142 2.22 + 2.14
=
0.0597
and t1 =
To determine the correct value of for the problem, we must first compute a value of value of t using. 3
=
2tan 1 t
- 195 -
3 for
each
Next substitute both values of 3 into Eq. (3). The correct value of will correspond to the value of 3 satisfying Eq. (2). Then, 3 = 2tan 1 0.307 = 34.13° , or 3 = 2tan 1 0.0597 = 6.83°
We know that
3 = 6.83° .
Before solving for r4 =
Therefore, = -1.
4 , solve for r 4
from geometry. Then,
r' 24 +(r3 r3' )2
=
0.82 + (2.6 2.14)2
=
0.923
and
(
h 4 = tan 1 r + 3 = tan 1 r' 3
3
0.8
2.6 2.14
To solve for the velocities, substitute ˙ 3 equation form:
=
)
+ 6.83° = 66.93°
˙ 4 , into the equation for velocity, and rewrite in matrix
cos3 r 3 sin 3 + r 4 cos 3 r3 r 2 2 sin 2 = sin 3 r 3 cos3 + r 4 sin3 3 r 22 cos2 ˙
˙
˙
or
˙
cos6.83 ° 2.6sin6.83 ° + 0.923cos6.83° r3 (0.9)(100)sin126° = sin6.83° 2.6cos6.83° + 0.923sin6.83° 3 (0.9)(100)cos126 ° ˙
˙
or
˙r3 95.21 ˙ 3 14.08 =
To solve for the accelerations, substitute matrix equation form:
˙ ˙3
=
˙˙ 4
, into the equation for acceleration, and rewrite in
cos3 r 3 sin 3 + r 4 cos 3 r3 sin 3 r 3 cos3 + r 4 sin3 3 r 424 cos 4 + r 22 sin2 + r 2 22 cos2 + 2r33 sin3 + r 332 cos3 = r 424 sin 4 r 22 cos2 + r 2 22 sin 2 2r33 cos3 + r 332 sin 3 ˙ ˙
˙ ˙
˙
˙ ˙
˙
˙ ˙
˙
˙
˙
˙
˙
˙
˙
˙
or
cos6.83° 2.6sin6.83° + 0.923cos6.83° r3 sin6.83° 2.6cos6.83° + 0.923sin6.83° 3 (0.923)(14.08)2 cos66.93° + (0.9)(100)2 cos126° +2(95.21)(14.08)sin6.83° + (2.6)(14.08)2 cos6.83° = (0.923)(14.08)2 sin66.93° + (0.9)(100)2 sin126° 2(95.21)(14.08)cos6.83° + (2.6)(14.08)2 sin6.83° ˙ ˙
˙ ˙
or
˙˙r3 132 ˙˙ 3 4185 =
To solve the velocity and acceleration of C in link 4:
- 196 -
r4c = r4(cos 4i + sin 4 j) v4c = r4c = r44( sin 4i + cos4 j) = (0.923)(14.08)( sin66.93 °i + cos66.93° j) = 1366.93°in /sec ˙
˙
a4c = r4c = r44( sin 4i + cos4 j) + r424 (cos 4i sin 4 j) = (0.923)(4185)( sin66.93 °i + cos66.93 ° j) + (0.923)( 14.08)2( cos66.93 °i sin66.93 ° j) = 3625.5 i + 1345.3j = 3867159.6°in / sec2 ˙ ˙
˙
˙ ˙
Therefore, v4c
= 1366.93°in
/ sec
and a4c = 3867159.6°in
/ sec2
Problem 5.12
For the mechanism in the position shown, link 2 is the driver and rotates with a constant angular velocity of 50 rad/s CCW. Write vector loop equations for position, velocity, and acceleration, and solve for the velocity and acceleration of point C on link 3. B 3 d
C 4
2
˚
60 =
h
A
φ
d
=
0.9"
h
=
0.8"
AB
Solution: The vector equation is: r1 + r4 = r2 + r3
In component form of this equation becomes:
- 197 -
=
1.8"
Y
B 3
r3 2
r 1 cos1 + r 4 cos4
C 4
r2 r4
φ
A
d
r1
X
= r 2 cos 2 + r 3 cos3
r 1sin 1 + r 4 sin 4 = r 2 sin 2 + r 3 sin 3
Substituting the constant values
1
=
0 , 4 = 90° ,
2
=
, and 3 = 90° gives
r 1 = r 2 cos + r 3 sin r 4
= r 2 sin
r 3 cos
The component equations for velocity are: ˙ sin + r ˙ cos r˙1 = ˙r2 cos r 2 3 ˙ cos + r 3 ˙ sin 0 = ˙r2 sin + r 2
The component equations for acceleration are: ˙ sin r 2˙˙ ˙˙ cos r 3 ˙ 2 sin sin r 2˙ 2 cos + r 3 r1 = ˙r˙2 cos 2˙r2 ˙˙ ˙ cos + r ˙ 2 cos ˙ cos r 2˙ 2 sin + r 3˙˙sin + r 3 0 = ˙r˙2 sin + 2˙r2 2˙
The known input information is: r3
=
d
=
0.9
r 4
=
0.8 ,
= 60 ° ,
˙
=
50
Solving for the positions: r4 + r3 cos 0.8 + 0.9cos(60°) = = 1.44 sin(60°) sin r1 = r2 cos + r3 sin = 1.44cos(60°) + 0.9sin(60°) = 1.5 r2 =
Solving for the velocities: ˙ cos + r3sin ˙ r2 sin (1.44)(50)cos60 ° + (0.9)(50)sin60 ° = = 86.57 sin60° ˙ sin + r3 ˙ cos r˙1 = ˙r2 cos r2 = (86.57)cos60° (1.44)(50)sin60 ° + (0.9)(50)cos60° = 83.14 r˙2 =
- 198 -
Solving for the accelerations: ˙ cos + r2˙cos ˙ ˙ ˙ 2 cos r2˙ 2 sin + r3˙sin 2˙r2 + r3 sin 2(86.57)(50)cos60° (1.44)(50) 2 sin60° + (0.9)(50)2 cos60°
r2 = ˙˙ =
= 7299 sin60° ˙ sin r2˙˙ sin r2˙2 cos + r3˙˙ cos r3˙ 2 sin ˙˙ r1 = ˙˙r2 cos 2˙r2 = 7299cos60° 2(86.57)(50)sin60° (1.44)(50)2 cos60° (0.9)(50)2 sin60° = 7398
Therefore, ˙
r1
=
r1
=
v3c
=
a3c
= ˙˙
and
83.14 in/sec 7398in / sec2
Problem 5.13
In the mechanism in shown, link 3 slides on link 2, and link 4 is pinned to link 3 and slides on the frame. If 1 2 = 10 rad/s CCW (constant), use loop equations to find the acceleration of Link 4 for the position defined by = 90 . ˚
B 4 3 2 1 cm
φ
A
Solution The basic loop equations is: r2 = r1 + r3
In component form this equation becomes: r 2 cos 2 = r 1 cos 1 + r 3 cos 3 r 2 sin 2 = r 1 sin 1 + r 3 sin 3 ˚
Substituting the constant values (r3 = 1, 1 = 0, and 3 = 90 ) gives r 2 cos 2 r 2 sin 2
=
=
r 1 r 3
=
1
- 199 -
Y B 4 3
r2
2
r3 φ
A
X
r1 ˚
When 2 = 90 , r 2
=
1/ sin 2
=
1/1
=
1
and r 1 = r 2 cos2
=
1• 0 = 0
The equation for the velocities is, r ˙2 = r ˙1 + r ˙3
or in component form, ˙ 2 sin 2 r˙2 cos 2 r 2 ˙ 2 cos2 r˙2 sin 2 + r 2
r1 = ˙ =
0
Substituting the know values gives
(1)(10)(1) r1 10 = ˙
r2 ˙
=
=
0
The equation for the accelerations is, r r1 + ˙˙ r3 ˙˙2 = ˙˙
or in component form, ˙2 sin 2 r 2˙ ˙2 sin 2 r ˙ 2 cos2 ˙˙ r2 cos 2 2˙r2 2 2 ˙ 2 cos2 + r 2˙˙ 2 cos 2 ˙˙ r2 sin 2 + 2˙r2
Substituting the know values gives r1 0 ˙˙ ˙˙ r2 (1)(10)2 =
=
Therefore, 1 aB4 = ˙˙ r1
=
r˙1 =˙
˙2 r 2 2 sin 2 =
0 0 cm / sec2 .
- 200 -
0
Problem 5.14
For the mechanism in the position shown, the cam (link 2) rotates with an angular velocity of 200 rad/s. Write the vector loop equations for position, velocity, and acceleration and determine the angular velocity and acceleration of the follower (link 3). Use 60 and neglect the follower thickness (i.e., assume that it is zero). =
˚
AD = 6.5 in AB = 1.0 in r = 2.0 in
3 C r
2
B
φ
D
A
ω2
Solution
Y 3 C 2
r4
B r3 A
r2
D
r1
X
The vector equation is: r1 + r4 = r2 + r3
In component form, this equation becomes: r 1 cos1 + r 4 cos4
= r 2 cos 2 + r 3 cos3
r 1sin 1 + r 4 sin 4 = r 2 sin 2 + r 3 sin 3
Now,
r 1 , r 2 , r 3 ,
and
1
=
0 are
constants, and 2
=
, 4
The component equations for velocity are: ˙ 4 sin 4 r˙4 cos 4 r 4 ˙ 4 cos 4 r˙4 sin 4 + r 4 ˙3
=
˙2 ˙4 ,
=
=
˙ 2 sin 2 r 3 ˙ 3 sin 3 r 2
˙ 2 cos 2 + r ˙ 3 cos3 = r 2 3
˙ .
The component equations for acceleration are:
- 201 -
= 90° + 3 .
˙ 4 sin 4 r 4 ˙˙4 sin 4 r 4 ˙ 2 cos 4 = r 2˙˙ ˙ 2 cos2 r 3 ˙˙3 sin 3 r ˙2 2 sin 2 r 2 ˙˙ r4 cos 4 2˙r4 33 cos 3 4 2 ˙ 4 cos 4 + r 4˙ ˙4 cos 4 r 4 ˙ 2 sin 4 = r 2˙ ˙2 cos2 r 2 ˙2 sin 2 + r 3˙ ˙3 cos 3 r 3 ˙ 2 sin 3 ˙˙ r4 sin 4 + 2˙r4 4 2 3
The known input information is: r 1
6.5
=
r 2
=
r 3
1.0
=
2.0
To solve for the positions, substitute equation. This gives: r 1 r 2 cos2 r 2 sin2
=
˙
= 60 °
=
4 = 90° + 3 into
200
the equation for position and rewrite the (1)
= r 4 sin3 + r 3 cos 3
(2)
r 4 cos3 + r 3 sin 3
To eliminate 3 , square Eqs. (1) and (2) and add. The result is: r 12
+
r 22 r 32 2r 1r 2 cos2
Substitute 2
=
To solve for
, into the equation above to get,
3 , use the
2tan sin 3
=
1 + tan
cos 3
2 2 3
=
1 + tan
=
tan
3 2
trigonometric identities
3
2
1 tan2
t
r 42
r12 + r22 r32 2r1r2 cos = 6.52 +1.0 2 2.02 2(6.5)(1.0)cos60 ° = 5.72
r4 =
Let
=
3
2 2 3 2
, and substitute the trigonometric identities above into Eq. (1). This gives:
A(1 + t 2 ) r3(1 t2 ) r4 (2t) = 0
where
A
=
(3)
r 1 r 2 cos 2
Collecting terms in Eq (3) gives: (A + r3 )t 2 2r4 t + (A r3 ) = 0
The roots are: t
=
r 4 + r 42 A 2 + r 32 A + r 3
=
5.72 + 5.72
2
62
+2
6 +2
or
- 202 -
2
t1
=
0.82 and t 2
=
0.61.
Where = ±1. To determine the correct value of for the problem, we must first compute a value of 3 for each value of t using. 3
=
2tan 1 t
or 3 = 2tan 1 0.82 = 78.7° and 3 = 2tan 1 0.61 = 62.77°
Next substitute both values of 3 into Eq. (3). The correct value of will correspond to the value of 3 satisfying Eq. (2). In this problem, 3 = 62.77° . To solve the velocities, substitute equation form:
cos4 sin 4 or
˙3
=
˙4
, into the equation for velocity, and rewrite in the matrix
r 4 cos 3 + r 3 sin 3 r4 r 22 sin 2 = r 4 cos 3 r 3 cos3 4 r 2 2 cos2 ˙
˙
˙
˙
cos152.77° 5.72cos62.77° + 2sin62.77° r4 (1)(200)sin60° = sin152.77° 5.72sin62.77° 2cos62.77°4 (1)(200)cos60° ˙
˙
or
r˙4 196.39 ˙ 4 1.69 =
To solve the accelerations, substitute ˙˙3 matrix equation form:
=
˙˙ 4 , into the equation for
acceleration, and rewrite in the
cos4 r 3 sin 3 r 4cos 3 r4 sin 4 r 3 cos3 r 4 sin 3 4 r 323 cos 3 r 22 sin 2 r 222 cos 2 + 2r4 4 cos3 r 424 sin3 = r 323 sin 3 + r 2 2 cos 2 r 222 sin 2 + 2r4 4 sin 3 + r 424 cos 3 ˙ ˙
˙ ˙
˙
˙ ˙
˙
˙˙
˙
˙
˙
˙
˙
˙
˙
˙
or
cos152.77° 5.72cos62.77° + 2sin62.77° r4 sin152.77° 5.72sin62.77° 2 cos62.77° 4 2(1.69)2 cos62.77° (1)(200)2 cos60° +2(196.39)(1.69)cos62.77° 5.72( 1.69)2 sin62.77° = 2(1.69)2 sin62.77° (1)(200)2sin60° +2(196.39)(1.69)sin62.77° + 5.72( 1.69)2cos62.77° ˙ ˙
˙ ˙
or
˙˙r4 1.615 10 4 ˙˙ 4 7.10 10 3 =
Therefore, ˙ 4
=
1.69 rad/sec
CW, and
˙ ˙4
=
7.10 103 rad /sec2 CCW.
- 203 -
Problem 5.15
In the mechanism shown, link 3 is perpendicular to link 2. Write the vector loop equations for position and velocity. If the angular velocity of link 2 is 100 rad/s CCW, use the vector loop equations to solve for the velocity of point C 4 for the position corresponding to = 60 . ˚
10"
B 3
2
C
φ
A
4
Solution
Y
B 2
r3 3
r2 A
φ
C
r1
4
X
The basic loop equations is: r2 = r1 + r3
In component form this equation becomes: r 2 cos 2 = r 1 cos 1 + r 3 cos 3 r 2 sin 2 = r 1 sin 1 + r 3 sin 3
Substituting the constant values (r3 = 10, 1 = 0), and r 2 cos 2 r 2 sin 2
=
=
3 = 2 + 90° gives
r 1 r 3 sin 2 r 3 cos 2
When 2 = 60, r2 = 10 cos60° / sin60° = 5.774
and r1 = r2 cos2 + r3 sin 2 = 5.774 cos60° + 10sin60° = 11.547
For the velocities,
- 204 -
r ˙2 = r ˙1 + r ˙3
or in component form, ˙ 2 sin 2 r˙2 cos 2 r 2 ˙ 2 cos2 r˙2 sin 2 + r 2
= r ˙1 =
˙ 2 cos 2 r 3
˙ 2 sin 2 r 3
Substituting the know values gives r˙2 sin60° + (5.774)(100)cos60° = (10)(100)sin60°
or r˙2
=
1333.40
(1333.40)cos60° (5.774)(100)sin60 ° = r1 (10)(100)cos60° ˙
or r˙1
=
666.70
Therefore, 1 vC4 =
r˙1
=
666.70 in/sec.
Problem 5.16
In the simple, two-link mechanism given, approach to determine
1 v and 1 A2
1v B2
is 10 in/s to the right. Use the loop equation
2.
A AB = 10 inches 2 ˚
30
Solution r3 = r1 + r2 r3 = r1 + r2 ˙
˙
˙
In component form r 3 cos3 = r 1 cos 1 + r 2 cos2 r 3 sin 3 = r 1 sin 1 + r 2 sin 2
- 205 -
B
1v B 2
A 2 r3
AB = 10 inches r2
θ2
30
r1
1vB
B
o
2
Substituting the constant angles, 0
= r 1 + r 2 cos 2
r 3 = r 2 sin 2
Substituting values r1 r3
=
=
r2 cos 2
r2 sin 2
=
=
10cos150
10sin150
=
=
8.66 in
5 in
The velocity components are: 0 r3 ˙
r1 2 r 2 sin 2
= ˙
=
˙
2 r 2 cos2 ˙
Solving for the unknowns, ˙2
= ˙r1 / r2 sin 2 = 10 / [10sin150] = 2 rad CCW =1 2
˙ 2 r2 cos2 r˙3 =
sec = 2(10)cos150 = 17.32 in =1vA 2 sec
Problem 5.17
In the mechanism below, the angular velocity of link 2 is 100 rad/s CCW and the dimensions of various links are given. Use loop equations to find the position and velocity of point D on link 3 when 2 is 90 . ˚
B 3
AB = 1.75 in AC = 2.5 in BD = 5 in
θ2
2
C A
4
- 206 -
D
Solution
θ3
B 3
2
θ2
r2
r3 C
A
r1
4
D
Loop equation: r1 = r2 + r3
In component form, r 1 cos1 = r 2 cos2 + r 3 cos 3 r 1sin 1 = r 2 sin 2 + r 3 sin 3
Noting that 1 = 0, these equations can be simplified to: r 1 = r 2 cos2 + r 3 cos 3 0
= r 2 sin 2 + r 3 sin 3
To solve for r3 , eliminate Then, r 1 r 2 cos2 r 2 sin2
=
=
3 by
isolating
3 in
each equation, squaring both equations and adding.
r 3 cos 3
r 3 sin 3
and r 12 2r 1r 2 cos 2 + r 22 cos2 2 = r 32 cos2 3 r 22 sin2 2
=
r 32 sin2 3
and r 12 2r 1r 2 cos 2 + r 22(cos2 2 + sin2 2 ) = r 32(cos2 3 + sin2 3)
Because
sin2 + cos2
r 12 2r 1r 2 cos 2
+
=1
r 22
=
, this equation becomes,
r 32
or r 3 =
2 1
r
2r 1r 2 cos 2 + r 22
For the values given in the problem,
- 207 -
r3
2.52 2(2.5)(1.75)cos90 + 1.752
=
=
3.052in.
and tan 3
=
r 2 sin 2
r 1 r 2 cos2
or tan 3
=
r2 sin 2
r1 r2 cos2
=
(1.75)sin90
2.5 1.75cos90
=
0.70
Therefore, 3
=
˚
34.99
To find the position of Point D let r 4 = BD. Then, rD = r2 + r4
or rD
=
(r2 cos 2 + r4 cos3)i + (r2 sin 2 + r4 sin 3 ) j
Substituting numbers, rD
[1.75cos90 + 5cos(34.99)]i + [1.75 sin90 + 5sin(34.99)] j = 4.096 i 1.117 j = 4.246 15.25 =
˚
The velocity equation is given by: r ˙1 = r ˙2 + r ˙3
And in component form, 0
=
r 22 sin2
0
= r 22 cos 2 + r3 sin 3 + r 33 cos 3
˙
+ r3 cos3 ˙
r 33 sin 3 ˙
˙
˙
˙
or r 2 2 sin 2 ˙
= r3 cos 3 ˙
r 33 sin 3 ˙
r 22 cos 2 = r3 sin3 + r 33 cos 3 ˙
˙
˙
Substituting numbers for this problem, 1.75(100)sin90 = r3 cos(34.99) 3.0523 sin(34.99) 1.75(100)cos90 = r3 sin(34.99) + 3.0523 cos(34.99) ˙
˙
˙
˙
or 175 = 0.819r3 +1.750 3 ˙
˙
0
=
0.573r3 + 2.5003 ˙
˙
175 0.819 0 0.573
or
=
1.750 r3 ˙
2.500 3
Using Cramer's Rule,
- 208 -
˙
r˙3
=
175
1.750
0
2.500
0.819
1.750
0.573
2.500
=
437.5 3.051
=
143.4 in / sec
and ˙3 =
0.819
175
0.573
0 1.750 2.500
0.819 0.573
=
100.28 = +32.87 rad / sec (CCW) 3.051
Now considering Point D, r˙D
=
˙ 3 sin 3)i + (r2 ˙ 2 cos 2 + r3˙3 cos3 ) j (r2˙ 2 sin 2 r4
or rD
[1.75(100)sin90 5(32.87)sin(34.99)]i + [1.75(100)cos90 + 5(32.87)cos(34.99)] j = 80.78i +134.6 j = 156.98120.97 =
˚
Therefore, rD3 = 4.096 i-1.117 j = 4.246
-15.25˚
and 1 v = D3
-80.78 i + 134.6 j = 156.98
120.97˚
Problem 5.18
In the Scotch Yoke mechanism shown, 1 2 = 10 rad/s, 1 2 = 100 rad/s2 , and O2 A = 20 inches. Determine 1 vA and 1 aA using loop equations. 4 4
3
A
2 ω2 θ2
O2 1
4 B
- 209 -
2 =
˚
60 . Also, length
Solution
3
r2
2
A
r3
θ2
O2 1
r1
4 B
For the vector loop given, r2 = r1 + r3
and in component form r 2 cos 2 = r 1 cos 1 + r 3 cos 3 r 2 sin 2 = r 1 sin 1 + r 3 sin 3
We know that 1 =
0,
2 =
˚
60 , and
3 =
The variables are r1 , r3 , and position equations. r 2 cos 2 r 2 sin 2
=
=
˚
90
2 .
Substituting in the know constants gives the following for the
r 1 r 3 ˚
For the given input values (r2 = 20 and 2 = 60 ) it is clear that r 1 = 10 and r3 = 17.32. The velocity equations are: r˙2
= r ˙1 + r˙3
and ˙ 2 sin 2 r˙2 cos 2 r 2 ˙ 2 cos2 r˙2 sin 2 + r 2
= r ˙1
r3 = ˙
Simplifying:
- 210 -
r˙3 r˙1
=
=
˙ 2 cos2 20(10)cos60 100 in / sec r2 ˙ 2 sin 2 20(10)sin60 173.2 in/ sec r2 =
=
=
=
The acceleration equations are: r r1 + ˙˙ r3 ˙˙2 = ˙˙
and ˙˙ r1
˙ 2 cos 2 r 2˙˙ 2 sin 2 r 2 2
=
˙˙ r3
=
˙2 cos2 r 2 ˙2 sin 2 r 2˙ 2
These equations simplify to: ˙˙ r1
=
˙˙ r3
=
1 v = A4 1a = A4
˙ 22 cos 2 r2˙˙ 2 sin 2 r2 ˙2 sin 2 2 cos2 r2 r2˙˙ 2
=
=
20(100)sin60 20(10)2 cos60
20(100)cos60 20(10) 2 sin60
=
=
2732 in
732 in
/ sec2
/ sec2
-173.2 in/s
-2732 in/s2
Problem 5.19 Use loop equations to determine the velocity and acceleration of point B on link 4. The angular velocity of link 2 is constant at 10 rad/s counterclockwise. 4 = 10 cm φ = 30 θ 2 = 60 r1
˚
˚
B
r2
φ
r3
θ3
θ2 1ω 2
O2
O3 r1
Solution Position Analysis The basic vector loop equation is r1 + r3 = r2
and
- 211 -
4
B
r2
r3
φ θ3
θ2 1ω
2
O2
r1
O3
3 = 2 +
In component form, r 1 cos0 r 1sin0
+ r 3 cos3 = r 2 cos 2 + r 3 sin 3 =
r 2 sin 2
or r 1 + r 3 cos 3
= r 2 cos2
r 3 sin 3 = r 2 sin 2
In the position shown, 2 60 . Therefore, by geometry, the triangle is a 30-60 right triangle. Therefore, =
r2
=
r1 / cos2
=
˚
10 / cos(60 )
=
2
=
60 .
Therefore,
20
and r3 r1 / sin 2 =
=
˚
10 / sin(60 )
=
17.321
Velocity Analysis The velocity loop equation is r ˙1 + r ˙3 = r ˙2
In component form, ˙ 3 sin 3 = ˙r2 cos 2 r ˙ 2 sin 2 r˙1 + ˙r3 cos 3 r 3 2 ˙3 cos3 = r˙2 sin 2 + r 2 ˙2 cos 2 r˙3 sin 3 + r 3
and ˙3
=
˙2
Therefore,
= r 2˙ 2sin 2 + r 3˙ 3 sin 3 = ˙ 2[r 3 sin 3 r 2 sin 2] = r 2˙ 2 cos 2 r 3˙ 3 cos3 = ˙2[r 2 cos 2 r 3 cos3 ]
r˙3 cos3 ˙r2 cos 2 r˙3 sin 3 r˙2 sin 2
- 212 -
3
=
90 .
Therefore,
or r˙3 cos90 r˙2 cos60 = 10[17.321sin90 20sin60] r˙3 sin90 ˙r2 sin60 = 10[20cos60 17.321cos90]
or
0.5r2 0.0 r3 0.866r2 ˙
=
˙
and
˙
r˙2 r˙3
100
0.0
=
=
=
100
Then, vB4 = r1 + r3 = ˙
˙
(r3 cos3 r33 sin 3, r3 sin 3 + r33 cos 3) ˙
˙
˙
˙
Substituting in numbers,
= ( 173.21,100 ) = 200 in / sec149.98
˚
vB4
Acceleration Analysis The basic acceleration loop equation is r ˙˙1 + r ˙˙3 = r ˙˙2
In component form, ˙˙ ˙ 3 sin 3 r ˙2 ˙˙ ˙ 2 sin2 r 2 ˙ 2 cos 2 ˙˙ r3 cos3 r r3 r2 33 sin 3 2˙ 33 cos 3 = ˙r˙2 cos 2 r 2 2 sin 2 2˙ 2 ˙˙ ˙ 3 cos3 r ˙2 ˙˙2 cos2 + 2˙r2 ˙ 2 cos 2 r ˙2 ˙˙ r3 sin 3 + r r3 r2 sin 2 + r 2 33 cos3 + 2˙ 33 sin 3 = ˙˙ 2 2 sin 2
and ˙˙3
=
˙˙ 2
=
0
Then, ˙˙ r3 cos3 ˙r˙2 cos 2 ˙˙ r3 sin 3 ˙r˙2 sin 2
=
=
˙ 2 sin 2 r 2 ˙ 2 cos 2 + 2˙r3 ˙ 3 sin 3 2˙ r2 2
˙2
+ r 33 cos 3
˙ 2 cos2 r 2 ˙2 sin 2 2˙r3 ˙ 3 cos3 + r ˙2 2˙r2 33 sin 3 2
Substituting in numbers, ˙ 2 sin60 r2 ˙ 2 cos60 + 2˙r3 ˙3 sin90 + r3 ˙ 2 cos90 ˙˙ r3 cos90 ˙r˙2 cos60 = 2˙r2 2 3 ˙ 2 sin60 2˙r3 ˙ 3 cos90 + r3 ˙ 2 sin90 ˙˙ r3 sin90 ˙˙ r2 sin60 = 2˙r2˙2 cos60 r2 2 3
or 0.5r2 = 0.5(20)(100) + 2(100)(10) = 1000 r2 = 2000 in / sec2 r3 0.866r2 = 0.886(20)(100) + 17.321(100) = 0 r3 = (0.866)2000
˙˙
˙ ˙
˙ ˙
˙ ˙
˙ ˙
1732 in / sec2
=
aB4 = r1 + r3 ˙˙
=
˙ ˙
(r3 cos3 r33 sin3 2r33 sin 3 r323 cos 3, ˙ ˙
˙ ˙
˙
˙
˙
- 213 -
r3 sin 3 + r33 cos 3 + 2r33 cos 3 r323 sin3 ) ˙ ˙
˙ ˙
˙
˙
˙
Substituting in numbers, aB4
=
(2r33, r3 r323 ) (2(100)10, 1732 17.321(100)) ˙
˙
vB 4
=
200 in / s
aB4
=
4000 in / s2
˙
˙ ˙
=
=
(2000, 3464)
Problem 5.20
The oscillating fan shown below is to be analyzed as a double rocker. The fan is link 2, the motor shaft is connected to link 3, and link 4 is connected from the coupler to the frame. The actual input of the mechanism is the coupler, and 2 3 that is a constant 956 (rad/s) in the counterclockwise direction. Compute the angular velocity and angular acceleration of link 2 if = 120 , AD = 0.75 in, AB = DC = 3.0 in, BC = 0.50 in. ˚
2ω
3
B
3
C
2
4
θ
A
Solution The basic loop equations is: r1 + r4 = r2 + r3
In component form, this equation becomes: r 1 cos1 + r 4 cos4
= r 2 cos 2 + r 3 cos3
r 1sin 1 + r 4 sin 4 = r 2 sin 2 + r 3 sin 3
- 214 -
D
3
B
Y
C
r3
r4 4 2
r2
θ
D
A
r1 Substituting the constant values ( 1 r 1 + r 4 cos4 r 4 sin 4
To solve for
=
X
0 ) gives
= r 2 cos 2 + r 3 cos3
= r 2 sin 2 + r 3 sin 3
3 and 4 ,
rewrite the equations above
r 1 + r 4 cos4 r 2 cos 2 = r 3 cos3 r 4 sin 4 r 2 sin 2
= r 3 sin 3
Now square both equations and add. The resulting equation is: 2r4 (r1 r2 cos2 )cos 4 2r4r2 sin 2 sin 4
+
(r1 r2 cos2 )2 + r22 sin2 2 + r42 r32
=
0
Let A = 2r4(r1 r2 cos 2) = 2 3 (0.75 3cos120°) = 13.5 B = 2r4 r2 sin 2 = 2 3 3sin120° = 15.588 C = (r1 r2 cos 2 )2 + r22 sin2 2 + r42 r32 = (0.75 3cos120°)2 + 32 sin2 120° + 32
0.52 = 20.563
Then rewrite the equation above Acos 4 + Bsin 4 + C
=
(1)
0
Simplify using the following trigonometric itentities 2tan
sin 4
=
4
2 1 + tan2 4 2
- 215 -
cos 4
1 tan2 =
1 + tan2
Let
t
=
tan
4 2
4 2 4 2
, and substitute the trigonometric identities into Eq. (1). Then:
A(1 t 2) + B(2t) + C(1 + t 2 ) = 0
Collecting terms gives: (C A)t 2 + 2Bt + (C + A) = 0
The roots are: t
B + B2 C2 + A2 CA
=
=
15.588 + (15.588)2 20.5632 + 13.52 20.563 13.5
=
15.588 ±1.549 7.063
Determine the sign for the square root that corresponds to this problem. t1 t2
=
2.426 ;
4 = 2tan 1 2.426 = 135.197°
1.988 ;
4 = 2tan 11.988 = 126.586°
=
According to the problem figure, 3 = sin 1
r 4 sin 4 r 2 sin 2 r 3
4 = 126.586° is the = sin 1
correct root. Then
3sin126.586° 3sin120° = 22.233° 0.5
The component form for the velocities are: r 44 sin 4 ˙
r 4 4 cos4 ˙
Substituting
=
r 22 sin 2 r 33 sin 3 ˙
˙
= r 2 2 cos 2 + r 33 cos 3 ˙
˙3 = ˙2
+
˙
2 3
, into the equations above and simplifying, we get
r44 sin 4 + (r2 sin 2 + r3 sin 3)2
=
r3 (2 3) sin 3
r44 cos4
=
r3 (2 3 ) cos3
˙
˙
˙
(r2 cos2 + r3 cos 3)2 ˙
Substituting the pertinent values into the equations gives
3sin126.586° (3sin120° + 0.5sin( 22.233°)) 4 0.5 956sin(22.233°) 3cos126.586° (3cos120° + 0.5cos(22.233°))2 = 0.5 956cos(22.233°) ˙
˙
or
˙ 4 486.365 ˙ 2 411.362 =
The component equations for acceleration are:
- 216 -
r 44 sin 4 r 42 4 cos 4 ˙ ˙
˙
r 4 4 cos4 r 4 2 4 sin 4 ˙ ˙
˙
Substituting
˙ ˙3
=
=
2 2 r 2 2 sin 2 r 22 cos 2 r 33 sin3 r 33 cos3 ˙ ˙
˙
= r 2 2 cos 2 ˙˙
r 22 2 sin 2 ˙
˙ ˙
˙
+ r 33 cos3 ˙ ˙
2 r 33 sin 3 ˙
˙˙ 2 , into the equations above and rewriting, we get
r44 sin 4 + (r2 sin 2 + r3 sin 3)2
=
r4 24 cos4 r222 cos 2 r323 cos 3
r44 cos4
=
r4 24 sin 4 r2 22 sin 2 r323 sin 3
˙ ˙
˙ ˙
˙ ˙
(r2 cos2 + r3 cos 3)2 ˙ ˙
˙
˙
˙
˙
˙
˙
Substituting the pertinent values into the equations gives
3sin126.586° (3sin120° + 0.5sin( 22.233°)) 4 3cos126.586° (3cos120° + 0.5cos(22.233°))2 3(486.365)2 cos126.586° 3(411.362)2 cos120° 0.5(411.362 + 956)2 cos(22.233°) = 3(486.365)2 sin126.586° 3(411.362)2 sin120° 0.5(411.362 + 956)2sin(22.233°) ˙ ˙
˙ ˙
or
˙˙ 4 4.244 105 ˙˙ 2 5.516 105 =
Therefore
˙2
=
411.362 rad/sec
CW; and
˙ ˙2
=
5.516 10 5 rad / sec2 CW.
Problem 5.21
The rear motorcycle suspension can be analyzed as an inverted slider-crank mechanism. The frame of the motorcycle is link 1, the tire assemble is attached to link 2 at point C . The shock absorber is links 3 and 4. As the bicycle goes over a bump in the position shown, the angular velocity of link 2 relative to the frame is 1 2 is 5 (rad/s), and the angular acceleration is 1 2 is 45 (rad/s2 ), both in the clockwise direction. Compute the angular velocity and angular acceleration of link 3 for the position defined by = 187 . ˚
(-9.262, 10.728) 14.17"
A
Y
3
4
˚
4.7
D
B C
2 19.27"
- 217 -
θ
1ω
X 2
Solution
A
r1
3
r3
r2
4
B C
Y
D
X
2
The basic loop equations is: r2 = r1 + r3
In component form this equation becomes: r 2 cos 2 = r 1 cos 1 + r 3 cos 3 r 2 sin 2 = r 1 sin 1 + r 3 sin 3
The known values are: r 1
r 2
=
=
14.17 ;
1 = 180° + tan 1(10.728 / (9.262)) = 130.806°
19.27 ;
2 = + 4.7° = 187° + 4.7° = 191.7°
To solve for
3 and r 3 ,
rewrite the equations above as
r 2 cos 2 r 1 cos1 r 2 sin 2 r 1 sin 1
=
=
(1)
r 3 cos 3
r 3 sin 3
(2)
Divide Eq. (2) by Eq. (1), we get 3 = 180° + tan1
r2 sin 2 r1 sin 1 r2 cos2 r1 cos1
= 180° + tan1
19.27sin191.7° 14.17sin130.806° 19.27cos191.7° 14.17 cos130.806°
= 236.708°
Then, r3 =
r2 sin 2 r1 sin 1 sin 3
=
19.27 sin191.7° 14.17sin130.806° = 17.507 sin236.708°
The component equations for the velocities are: r 22 sin 2 = r3 cos3 r 33 sin 3 ˙
˙
˙
r 2 2 cos2 ˙
= r3 sin 3 + r 33 cos 3 ˙
˙
- 218 -
Substitute the relevant values
cos236.708° 17.507sin236.708° r3 19.27 (5)sin191.7° sin236.708° 17.507cos236.708° 3 = 19.27 (5)cos191.7° ˙
˙
or
˙r3 68.128 ˙ 3 3.891 =
The component equations for acceleration are: 2 r 22 sin 2 r 2 2 cos2 ˙ ˙
˙
= r3 cos3 ˙ ˙
2 2r33 sin 3 r 33 sin 3 r 33 cos 3 ˙
˙ ˙
˙
˙
2 r 2 2 cos2 r 22 33 cos3 r 33 sin 3 2 sin 2 = r3 sin3 + 2r33 cos3 + r ˙ ˙
˙
˙
˙ ˙
˙ ˙
˙
˙
Rewrite the equations to put the unknowns on the same side of the equations 2 2 r 22 sin 2 r 2 2 cos2 + 2r33 sin 3 + r 33 cos3 = r3 cos 3 r 33 sin 3 ˙ ˙
˙
˙
˙
˙ ˙
˙
˙ ˙
2 r 2 2 cos2 r 22 33 sin 3 2 sin 2 2r33 cos 3 + r ˙ ˙
˙
˙
˙
˙
= r3 sin 3 + r 33 cos3 ˙ ˙
˙ ˙
Substitute the pertinent values
cos236.708° 17.507sin236.708° r3 sin236.708° 17.507cos236.708° 3 19.27(45)sin191.7° 19.27(5)2 cos191.7° +2(68.128)(3.891)sin236.708° + 17.507(3.891)2 cos236.708° = 19.27(45)cos191.7° 19.27(5)2 sin191.7° 2(68.128)(3.891)cos236.708° +17.507(3.891)2 sin236.708° ˙ ˙
˙ ˙
or
˙˙r3 688.691 ˙˙ 3 45.842 =
Therefore
˙3
=
3.891 rad/sec,
CW; and
˙ ˙3
=
45.842 rad / sec2 ,
- 219 -
CW.
Problem 5.22
The door-closing linkage shown is to be analyzed as a slider-crank linkage. Link 2 is the door, and links 3 and 4 are the two links of the door closer. Assume that the angular velocity of the door (link 2) is a constant at 3.71 radians per second clockwise. Compute the angular velocity and angular acceleration of link 4 if the dimensions are as follows: Coordinates of D (-2.5, -3.0) AB = 17.0 inches
1ω
2
y
˚
6
Door
2
C
A
x B
3 4 D
Solution Y
r2 r1
2
X
C
A
B
r3
4
3
D
The basic loop equations is: r
1
= r + r 2
3
In component form, this equation becomes: r 1 cos1 = r 2 cos2 + r 3 cos 3 r 1sin 1 = r 2 sin 2 + r 3 sin 3
The known values are: r1 = r 2
=
2.52 + 3.02
=
3.905 ;
3 and r 3 ,
r 1 cos1 r 2 cos2
2.5) = 50.194°
2 = 174°
17.0 ;
To solve for
1 = tan1(3.0 /
rewrite the equations above =
(1)
r 3 cos 3
- 220 -
r 1sin 1 r 2 sin 2
=
(2)
r 3 sin 3
Dividing Eq. (2) by (1), we get 3 = tan1
r3 =
r1 sin 1 r2 sin 2 r1 cos 1 r2 cos2
r1 sin 1 r2 sin 2 sin 3
=
3.905sin50.194° 17sin174° = 3.606° 3.905cos50.194° 17cos174°
= tan 1
3.905sin50.194° 17 sin174° = 19.445 sin3.606°
The component equations for the velocities are: r 2 2 sin 2 ˙
= r3 cos 3 ˙
r 33 sin 3 ˙
r 22 cos 2 = r3 sin3 + r 33 cos 3 ˙
˙
˙
Substitute the relevant values
cos3.606° 19.445sin3.606° r3 17 (3.71)sin174° = sin3.606° 19.445cos3.606° 3 17 (3.71)cos174° ˙
˙
or
r˙3 10.525 ˙ 3 3.198 =
The component equations for the acceleration are: 2 r 2 2 sin 2 + r 22 33 sin3 r 33 cos 3 2 cos 2 = r3 cos3 2r33 sin 3 r ˙ ˙
˙
˙
˙ ˙
2 r 22 cos 2 + r 2 2 sin 2 ˙ ˙
˙
˙ ˙
˙
˙
= r3 sin 3 + 2r33 cos 3 + r 33 cos3 ˙
˙ ˙
˙ ˙
˙
2 r 33 sin 3 ˙
Rewrite the equations to isolate the unknowns on the same side of the equations 2 r 2 2 sin 2 + r 22 33 cos 3 = r3 cos 3 r 33 sin 3 2 cos 2 + 2r33 sin 3 + r ˙ ˙
˙
˙
˙
˙ ˙
˙
˙ ˙
2 2 r 22 cos 2 + r 2 2 sin 2 2r33 cos 3 + r 33 sin 3 = r3 sin 3 + r 33 cos 3 ˙ ˙
˙
˙
˙
˙
˙ ˙
˙ ˙
Substituting the pertinent values gives
cos3.606° 19.445sin3.606° r3 sin3.606° 19.445cos3.606° 3 17( 3.71)2 cos174° + 2(10.525)(3.198)sin3.606° + (19.445)( 3.198)2 cos3.606° = 17( 3.71)2 sin174° 2(10.525)(3.198)cos3.606° + (19.445)(3.198)2 sin3.606° ˙ ˙
˙ ˙
or
˙˙r3 31.842 ˙˙ 3 1.454 =
Therefore
˙3
=
3.198 rad/sec,
CW; and
˙ ˙3
=
1.454 rad / sec2 ,
- 221 -
CW.
Problem 5.23
The general action of a person who is doing pushups can be modeled as a four-bar linkage as shown below. The floor is the base link, and link 4 is the back and legs. Link 2 is the forearm, and link 3 is the upper arm. For the purposes of analysis, the motion that is controlled is the motion of link 3 relative to link 2 (elbow joint). Assume that 2 3 is a constant 6.0 rad/s in the counterclockwise direction. Compute the angular velocity and angular acceleration of link 4 if link 2 is oriented at 45 to the horizontal. ˚
14.0" C 2
3
ω3
2 A
4
B 45
57.7"
˚
D 12.0" 52.0"
Solution Y C
r3 A
B
r4
r2
D
X
r1
The basic loop equations is: r1 + r4 = r2 + r3
In component form, this equation becomes: r 1 cos1 + r 4 cos4
= r 2 cos 2 + r 3 cos3
r 1sin 1 + r 4 sin 4 = r 2 sin 2 + r 3 sin 3
Substituting the constant values ( 1 = 0) gives r 1 + r 4 cos4 r 4 sin 4
= r 2 cos 2 + r 3 cos3
= r 2 sin 2 + r 3 sin 3
To solve for 3, 4, rewrite the equations above by isolating the terms which contain r 1 + r 4 cos4 r 2 cos 2 = r 3 cos3 r 4 sin 4 r 2 sin 2
= r 3 sin 3
- 222 -
3 .
Then,
Square both sides of both equations and add the results. This gives 2r4 (r1 r2 cos2 )cos 4 2r4r2 sin 2 sin 4
+
(r1 r2 cos2 )2 + r22 sin2 2 + r42 r32
=
0
(1)
Let A = 2r4(r1 r2 cos 2) = 2 57.7 (52.0 12.0cos45°) = 5021.6 B = 2r4 r2 sin 2 = 2 57.7 12.0sin45° = 979.2 C = (r1 r2 cos 2 )2 + r22 sin2 2 + r42 r32 = (52.0 12.0cos45°)2 + 12.02 sin2 45° + 57.72 14.02 = 5098.8
Equation (1) can now be simplified as Acos 4 + Bsin 4 + C
=
0
(2)
Simplify the resulting equation using sin 4
=
where
t
2t 1 + t2
=
tan
and 4 2
cos 4
=
1 t2 1 + t2
.
. Then, Eq. (2) becomes
A(1 t 2) + B(2t) + C(1 + t 2 ) = 0
(3)
Collecting terms in Eq. (3) gives: (C A)t 2 + 2Bt + (C + A) = 0
The roots are: t
=
B + B2 C2 + A 2 CA
979.2 + (979.2)2 5098.82 + 5021.62 5098.8 5021.6
=
where b = = ±1. Then, t1
t2
=
18.142 which gives 4 = 2tan
=
7.226 which
gives 4
= 2tan
1
1
18.142 = 173.69°
7.226 = 164.242°
According to the picture shown, the appropriate value is 3 = sin 1 = sin 1
r4 sin 4 r2 sin 2 r3
57.7sin164.242° 12.0sin45° = 149.12° 14.0
The component form of velocity equations is: r 44 sin 4 ˙
r 4 4 cos4 ˙
4 = 164.242 ° .
=
r 22 sin 2 r 33 sin 3 ˙
˙
= r 2 2 cos 2 + r 33 cos 3 ˙
˙
- 223 -
Then,
When we substitute
r4 4 cos4 ˙
=
˙ 2 + 2 3 ,
into the equations above and rewrite, we get
(r2 sin 2 + r3 sin 3)2 = r32 3 sin 3 (r2 cos2 + r3 cos 3)2 = r3 2 3 cos 3
r44 sin 4 ˙
˙3
+
˙
˙
Substituting the pertinent values into the equations and rewriting them in matrix form gives
57.7sin164.242° 12.0sin45° +14.0sin149.12° 4 14.0 6.0sin149.12° 57.7cos164.242° (12.0cos45° + 14.0cos149.12°) 2 = 14.0 6.0cos149.12° ˙
˙
or
˙ 4 1.2 ˙ 2 1.55 =
The component equations for the accelerations are: r 44 sin 4 r 42 4 cos 4 ˙ ˙
˙
r 4 4 cos4 r 4 2 4 sin 4 ˙ ˙
Substitute
˙˙3
˙
=
˙˙ 2 into
r44 sin 4 ˙ ˙
r4 4 cos4 ˙ ˙
+
=
2 2 r 2 2 sin 2 r 22 cos 2 r 33 sin3 r 33 cos3 ˙ ˙
˙
= r 2 2 cos 2 ˙ ˙
˙ ˙
˙
2 r 22 33 cos3 r 33 sin 3 2 sin 2 + r ˙
˙ ˙
˙
the equations above and rewrite. This gives
(r2 sin 2 + r3 sin 3)2 ˙˙
(r2 cos2
+
r3 cos 3)2 ˙ ˙
=
r4 24 cos4 r222 cos 2 r323 cos 3
=
r424 sin 4 r2 22 sin 2 r323 sin 3
˙
˙
˙
˙
˙
˙
Now substitute the pertinent values into the equations to get
57.7sin164.242° (12.0sin45° +14.0sin149.12°) 4 57.7cos164.242° (12.0cos45° + 14.0cos149.12°) 2 57.7(1.2)2 cos164.242° 12.0(1.55)2 cos45° 14.0(1.55 + 6.0)2 cos149.12° = 57.7(1.2)2 sin164.242° 12.0(1.55)2 sin45° 14.0(1.55 + 6.0)2 sin149.12° ˙ ˙
˙ ˙
or
˙˙ 4 3.29 ˙˙ 2 12.07 =
Therefore
˙4
=
1.2 rad/s,
CCW and
˙ ˙4
=
3.29rad / s2 ,
- 224 -
CCW.
Problem 5.24
A carousel mechanism can be modeled as an inverted slider-crank mechanism as shown. Point D is the location of the saddle on the horse. Assume that the angular velocity of the driver (Link 2) is a constant 2 rad/s counterclockwise. Compute the velocity and acceleration of D3 in the position shown if AB = 8.0 in, BC = 96.0 in, and BD = 54 in. 2 A
45
B
1ω2
3
D
4
C Solution The basic loop equations is: r1 = r2 + r3
In component form, this equation becomes: r 1 cos1 = r 2 cos2 + r 3 cos 3 r 1sin 1 = r 2 sin 2 + r 3 sin 3 r 2
r3
=
=
8.0 ; 2 = 45°
96.0 ;
3 = cos1( r2 cos 2 / r3) = cos1(8cos(45°) / 96) = 93.378°
1 = 90°
r1 =
r2 sin 2 + r3 sin 3 8sin(45°) + 96sin( 93.378°) = = 101.490 sin1 sin(90°)
- 225 -
˚
In component form, the velocities are: r 2 2 sin 2 ˙
= r3 cos 3 ˙
r 33 sin 3 ˙
r 22 cos 2 = r3 sin3 + r 33 cos 3 ˙
˙
˙
Substitute the relevant values
Y A
r2 2
X
B
3
r1
r3
D
4
C
cos(93.378°) 96sin(93.378°) r3 8(2)sin(45°) sin(93.378°) 96 cos( 93.378°) 3 = 8(2)cos(45°) ˙
˙
or
˙r3 11.964 ˙ 3 0.111 =
To solve for rC /D
vD 3
=
,
rC/ D(cos 3i + sin3 j)
vD 3 = rC /D = rC / D(cos 3i + sin 3 j) rCD3( sin 3i + cos3 j) = [(11.964)cos(93.378°) + 42(0.111)sin( 93.378°)]i + [( 11.964)sin(93.378°) 42(0.111)cos(93.378°)] j = 5.359i + 11.669 j = 12.84165.333° ˙
˙
Therefore
vD 3
˙
= 12.84165.333° in/sec.
- 226 -
The component equations for acceleration are: 2 r 2 2 sin 2 + r 22 33 sin3 r 33 cos 3 2 cos 2 = r3 cos3 2r33 sin 3 r ˙ ˙
˙
˙
˙ ˙
2 r 22 cos 2 + r 2 2 sin 2 ˙ ˙
˙
˙ ˙
˙
˙
= r3 sin 3 + 2r33 cos 3 + r 33 cos3 ˙
˙ ˙
˙ ˙
˙
2 r 33 sin 3 ˙
Rewriting the equations 2 r 2 2 sin 2 + r 22 33 cos 3 = r3 cos 3 r 33 sin 3 2 cos 2 + 2r33 sin 3 + r ˙ ˙
˙
˙
˙
˙ ˙
˙
˙ ˙
2 2 r 22 cos 2 + r 2 2 sin 2 2r33 cos 3 + r 33 sin 3 = r3 sin 3 + r 33 cos 3 ˙ ˙
˙
˙
˙
˙
˙ ˙
˙ ˙
Substituting the pertinent values
cos(93.378°) 96sin(93.378°) r3 sin(93.378°) 96 cos( 93.378°) 3 8(2)2 cos( 45°) + 2(11.964)(0.111)sin(93.378°) + 96(0.111)2 cos(93.378°) = 8(2)2 sin(45°) 2(11.964)(0.111)cos( 93.378°) + 96(0.111)2 sin( 93.378°) ˙ ˙
˙ ˙
or
˙˙r3 22.444 ˙˙ 3 0.277 =
To solve for
aD 3
,
aD3 = rCD = rCD (cos3i + sin 3 j) 2rCD3( sin 3i + cos3 j) rCD3( sin 3i + cos 3 j) + rCD2 3 (cos3i + sin 3 j) = (22.444)(cos(93.378 °)i + sin(93.378 °) j) 2(11.964)( 0.111)( sin(93.378 °)i + cos(93.378 °) j) 42(0.277)( sin( 93.378 °)i + cos(93.378 °) j) +42(0.111)2 (cos(93.378 °)i + sin(93.378 °) j) = 7.669i + 22.411 j = 23.687108.891 ° ˙ ˙
˙ ˙
Therefore
˙
˙ ˙
˙
˙
aD 3 = 23.687108.891 ° in / sec2 .
Problem 5.25
The shock absorber mechanism on a mountain bicycle is a four-bar linkage as shown. The frame of the bike is link 1, the fork and tire assembly is link 3, and the connecting linkage are links 2 and 4. As the bicycle goes over a bump in the position shown, the angular velocity of link 2 relative to 2 the frame is 2 is 205 (rad/s) CW, and the angular acceleration is 2 is 60 (rad/s ) CW. Compute the angular velocity and angular acceleration of link 3 for the position shown
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˚
201 A
2
B
˚
1.64"
D
3 4 C
2.22"
Solution
A 2
B
1 ω2
1
r3
r4
3
r1 D
4 C
The basic loop equations is: r + r 1
2
= r + r 3
4
In component form, this equation becomes: r 1 cos1 + r 2 cos2 r 1sin 1 + r 2 sin 2
= r 3 cos 3 + r 4 cos 4
= r 3 sin 3 + r 4 sin 4
The known values are: r 1
=
1.55"
1 = 119°
r 2
=
1.88"
r 3
=
1.64"
r 4
=
2.22"
2 = 201 °
To solve for 3 , 4 , rewrite the equations above to isolate 4 in both equations. Then,
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1.55" 1
1.88"
r2
61
1ω 2
r 1 cos1 + r 2 cos2 r 3 cos 3 = r 4 cos 4
(1)
r 1sin 1 + r 2 sin 2 r 3 sin 3 = r 4 sin 4
Square both sides of both equations and add the results. This gives 2r3(r1 cos 1 + r2 cos 2)cos3 + 2r3 (r1 sin 1 + r2 sin 2)sin 3 +r 2 r 2 (r1 cos 1 + r2 cos 2 )2 (r1 sin1 + r2 sin 2 )2 = 0 4 3
(2)
To simplify the equation, let A = 2r3 (r1 cos1 + r2 cos2 ) = 2(1.64)(1.55cos119° + 1.88cos201°) = 8.222 B = 2r3(r1 sin 1 + r2 sin 2 ) = 2(1.64)(1.55sin119° + 1.88sin201°) = 2.237 C = r42 r32 (r1 cos 1 + r2 cos 2)2 (r1 sin 1 + r2 sin 2 )2 = 2.222 1.642 (1.55cos119° + 1.88cos201°)2 (1.55sin119° + 1.88sin201°)2 = 4.509
Now rewrite Eq. (2) as Acos 3 + Bsin 3 + C
=
0
To solve for 3 , replace the trignometric functions with t
=
tan
3
sin 3 =
2t 1 + t2
and
cos 3
=
1 t2 1 + t2
. Then
2
A(1 t 2) + B(2t) + C(1 + t 2 ) = 0
(3)
Collecting terms in Eq. (3) gives: (C A)t 2 + 2Bt + (C + A) = 0
The roots are: t
=
B + B2 C2 + A2 CA
=
2.237 + 2.2372 (4.509)2 4.509 + 8.222
+
(8.222)2
where = ±1. The two possible solutions are t1 t2
=
where
1.345 which
=
gives
3 = 2tan 11.345 = 106.7°
2.550 which gives 3 = 2tan1(2.550) = 137.2°
According to the picture shown, the correct value is 3 = 106.7 ° . Then from Eq. (1), tan 4
=
r 1 sin 1 + r 2 sin 2 r 3 sin 3 r 1 cos 1 + r 2 cos 2 r 3 cos3
and
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