cos1 r3 sin3 ˙r1 = r2˙ 2 sin 2
sin1 r3 cos3 ˙ 3 r2 2 cos2 or
cos0° 4sin(12.504°) ˙r1 = 110sin60°
sin0° 4 cos(12.504°)
˙3 110cos60° then
1 0.866 ˙r1 = 8.660 0 3.905 ˙ 3 5 or
˙r1 = 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.
s
3
4
2
10 inches
φ
Solution
The vector equation is
r3 = r2 + r1
In component form, this equation becomes:
r3 cos3 = r2 cos2 + r1 cos1
r3 sin3 = r2 sin2 + r1 sin1
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s
r1 3
4
r3
2
r2
φ
Substituting the constant values 1 = 0° and 2 = 90° gives
r3 cos = r1
r3 sin = r2
The component equations for velocity are:
r˙3 cos r3˙ sin = ˙r1
r˙3 sin + r3˙ cos = 0
The component equations for acceleration are:
˙r˙3 cos 2˙r3˙ sin r3˙˙sin r3˙ 2 cos = ˙r˙1
˙˙ cos r3˙ 2 sin = 0
˙r˙3 sin + 2˙r3˙ cos + r3
The known input information is:
= 60°
so
r2 = 10 in
˙r1 = ˙s = 10 in / s
r3 = r2 = 10 = 11.547
sin sin60°
r1 = r3 cos = 11.547cos60° = 5.774
Solve the velocity equations:
cos r3 sinr˙3 ˙r1
= sin r3 cos ˙ 0
or
or
then
cos60° 11.547sin60° ˙r3 10 sin60° 11.547cos60° ˙ = 0 10 ˙r3 10
0.5
=
0.866 5.774˙ 0 ˙r3 5 ˙ = 0.75
Therefore ˙ = 0.75 rad/sec, CCW.
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˙r˙1 = ˙s˙ = 0
Solve the acceleration equations:
cos r3 sin˙r˙3 ˙r˙1 + 2˙r3˙ sin + r3˙ 2 cos
sin r3 cos ˙˙ = 2˙r3˙ cos + r3˙ 2 sin or
or
then
cos60° 11.547sin60° ˙r˙3 2(5)(0.75)sin60° +11.5470.752 cos60° =
sin60° 11.547cos60° ˙˙ 2(5)(0.75)cos60° +11.547 0.752 sin60°
10 ˙r˙3 3.248
0.5
0.866 5.774
˙˙ = 9.375 ˙r˙3 6.495
˙˙ = 0.650
Therefore ˙˙ = 0.650 rad / sec2 , CCW.
Problem 5.4
In the mechanism in Problem 5.3 assume that ˙ is 10 rad/s CCW. Use the loop equation 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:
r3 cos3 = r2 cos2 + r1 cos1
r3 sin3 = r2 sin2 + r1 sin1
Substituting the constant values 1 = 0 and 2 = 90° gives
r3 cos = r1
r3 sin = r2
The component equations for velocity are:
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6 ˙r˙2 7.578 0.5
=
0.866 3.464 ˙˙ 16.875
then
˙r˙2 10.826
=
˙˙ 2.165
Therefore ˙˙ = 2.165 rad / min2 CCW.
Problem 5.8
Use loop equations to determine the velocity and acceleration of point B on link 2 when 3 = 30˚.
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:
r3 cos3 = r1 cos1 + r2 cos2
r3 sin3 = r1 sin1 + r2 sin2
Substituting the constant values 1 = 0 and 2 = 90° gives
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r2
x
r3 cos3 = r1
r3 sin3 = r2
The component equations for velocity are:
r˙3 cos3 r3˙ 3 sin3 = 0
r˙3 sin3 + r3˙3 cos3 = ˙r2
The component equations for acceleration are:
˙˙3 sin3 r3˙ 32 cos3 = 0
˙r˙3 cos3 2˙r3˙ 3 sin3 r3
˙˙3 cos3 r3˙ 32 sin3 = ˙r˙2
˙r˙3 sin3 + 2˙r3˙ 3 cos˙ 3 + r3
The known input information is:
so
3 = 30°;
˙ 3 = 4 = 1 rad/sec;
˙˙3 = 4 = 0;
r1 = 10 inches;
r1 = 10 = 11.547
cos3 cos30°
r2 = r3 sin3 = 11.547sin30° = 5.774
r3 =
To solve the velocities:
˙
r˙3 = r33 sin3 = 11.5471sin30° = 6.667
cos3
cos30°
˙
r˙2 = ˙r3 sin3 + r33 cos3 = 6.667sin30° + 11.5471 cos30° = 13.333
Therefore vB2 = ˙r2 = 13.333 in/sec.
Solve for the accelerations:
2r˙3˙ 3 sin3 + r3˙˙3 sin3 + r3˙23 cos3
˙r˙3 =
cos3
2
6.6671sin30°
+ 11.54712 cos30° = 19.245
=
cos30°
˙˙3 cos3 r3˙ 32 sin3
r˙˙2 = ˙r˙3 sin3 + 2˙r3˙ 3 cos3 + r3
= 19.245sin30° + 2 6.6671cos30° 11.54712 sin30° = 15.397
Therefore aB2 = ˙r˙2 = 15.397 in / sec2 .
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175 1.750
0 2.500
r˙3 = 0.819 1.750 = 437.5 = 143.4 in / sec
3.051
0.573 2.500
and
0.819 175
0.573 0
˙ 3 = 0.819 1.750 = 100.28 = +32.87 rad / sec (CCW)
3.051
0.573 2.500
Now considering Point D,
r˙D = (r2˙ 2 sin2 r4˙ 3 sin3)i + (r2˙ 2 cos2 + r3˙3 cos3 )j
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,
and
rD3 = 4.096 i-1.117 j = 4.246 -15.25˚
1v
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 2 = 60˚. Also, length
O2 A = 20 inches. Determine 1 vA4 and 1 aA4 using loop equations.
3
A
2
ω2
θ2
O2
1
4
B
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Solution
3
r2
2
A
r3
θ2
O2
1
r1
4
B
For the vector loop given,
r2 = r1 + r3
and in component form
r2 cos2 = r1 cos1 + r3 cos3
r2 sin2 = r1 sin1 + r3 sin3
We know that
1 = 0, 2 = 60˚, and 3 = 90˚
The variables are r1 , r3 , and 2 . Substituting in the know constants gives the following for the
position equations.
r2 cos2 = r1
r2 sin2 = r3
For the given input values (r2 = 20 and 2 = 60˚) it is clear that r1 = 10 and r3 = 17.32.
The velocity equations are:
r˙2 = ˙r1 + ˙r3
and
r˙2 cos2 r2 ˙ 2 sin2 = ˙r1
r˙2 sin2 + r2˙ 2 cos2 = ˙r3
Simplifying:
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r˙3 = r2˙ 2 cos2 = 20(10)cos60 = 100 in / sec
r˙1 = r2˙ 2 sin2 = 20(10)sin60 = 173.2 in / sec
The acceleration equations are:
r˙˙2 = ˙r˙1 + ˙r˙3
and
˙r˙1 = r2˙˙2 sin2 r2˙ 22 cos2
˙r˙3 = r2˙˙2 cos2 r2˙22 sin2
These equations simplify to:
˙˙2 sin2 r2˙ 22 cos2 = 20(100)sin60 20(10)2 cos60 = 2732 in / sec2
˙r˙1 = r2
˙r˙3 = r2˙˙2 cos2 r2˙22 sin2 = 20(100)cos60 20(10)2 sin60 = 732 in / sec2
1v
= -173.2 in/s
1a
-2732 in/s2
A4
A4 =
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
r1 = 10 cm
φ = 30˚
θ 2 = 60˚
B
r2
φ
θ3
θ2
1ω2
r3
O2
O3
r1
Solution
Position Analysis
The basic vector loop equation is
r1 + r3 = r2
and
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