86 5. TOPIC AK-5
Determine equation φ
2
= f(t) of the rotational motion of the link 2 of the mechanism, the circular force S at the
contact point between the links 1 and 2, and the tension T of the rope at the time instant t
1
= 1 s (Figure 5.31).
1
M
A
B
2
3
Mc
Figure 5.31.
5.3 SOLUTION
e forces applied to the link 1 of the mechanism are the gravity force G
⃗
1
, the driving moment M, the components
Y
⃗
A
and Z
⃗
A
of the reaction force at the pin support A, the circular force S
⃗
1
, and the normal reaction force N
⃗
1
of the
link 2 (Figure 5.32).
e forces applied to the link 2 of the mechanism are the gravity force G
⃗
2
, the moment of the resistance forces M
C
,
the components Y
⃗
B
and Z
⃗
B
of the reaction force at the pin support B, the tension T
⃗
of the rope holding the load 3,
the circular force S
⃗
2
, and the normal reaction force N
⃗
2
of the link 1 (Figure 5.32).
e forces applied to the load 3 are the gravity force G
⃗
3
and the tension T
⃗
'.
It is obvious that:
S
⃗
2
= –S
⃗
1
, N
⃗
1
= –N
⃗
2
, and T
⃗
' = –T
⃗
.
87
e dierential equation of the rotational motion of the link 1 about xed axis x
1
:
x
1
φ
̈
1
= M
x
1
.
e total moment of external forces M
x
1
applied to the link 1 of the given mechanical system (Figure 5.32) about
axis x
1
:
M
x
1
= M – S
1
R
1
.
e moment M brings to the motion the whole system and therefore it is considered with a positive sign, but the
moment created by the force S
⃗
1
resists to the rotation of the link 1 and therefore it has a negative sign.
en a dierential equation of the rotational motion of the link 1:
x
1
φ
̈
1
= M – S
1
R
1
. (5.1)
e angular velocity φ
̈
1
of the link 1 can be expressed through the angular velocity φ
̈
2
of the link 2.
As:
φ
̈
1
=
R
2
, then φ
̈
1
= φ
̈
2
∙
R
2
.
φ
̈
2
R
1
R
1
en, Equation (5.1) can be written as:
x
1
φ
̈
1
R
2
= M – S
1
R
1
. (5.2)
R
1
An angular momentum principle for the mechanical system can be applied to write a dierential equation of the
rotational motion of the link 2 (which holds the load 3) about axis x
2
:
dL
x
2
= M
x
2
. (5.3)
R
1
e angular momentum of the system 2 – 3 about axis x
2
:
L
x
2
=
x
2
ω
2
+ m
3
vr
2
,
where
x
2
ω
2
– is the angular momentum of the link 2 rotating with the angular velocity ω
2
around xed axis x
2
;
m
3
vr
2
– is the angular momentum of the load 3 in translational motion with the velocity v.
Since = ω
2
r
2
:
L
x
2
= (
x
2
+ m
3
r
2
2
) ω
2
= '
x
2
φ
̇
2
,
where '
x
2
=
x
2
+ m
3
r
2
2
– the moment of inertia of the system 2 – 3 about axis x
2
.
e total moment M
x
2
of the external forces applied to the system 2 – 3 (Figure 5.32) about axis x
2
:
M
x
2
= S
2
R
2
– G
3
r
2
– M
C
.
e moment of the force S
⃗
2
brings to the motion the system 2 – 3 and therefore it is considered with a positive
sign, but the moment created by the load G
⃗
3
and the moment of the resistance forces M
⃗
C
resist to the motion of the
system, and therefore they have negative signs.
E
E
E
E
E
E
5.3 SOLUTION
88 5. TOPIC AK-5
Hence, from Equation (5.3):
d
'
x
2
φ
̇
2
= S
2
R
2
– G
3
r
2
– M
C
.
dt
We obtain the following dierential equation of the rotation of the link 2:
'
x
2
φ
̈
2
= S
2
R
2
– G
3
r
2
– M
C
. (5.4)
In the system of Equations (5.2) and (5.4):
x
1
φ
̈
2
R
2
=M – S
1
R
1
,
R
1
'
x
2
φ
̈
2
= S
2
R
2
– G
3
r
2
– M
C
the forces S
1
= S
2
= S and the angular acceleration φ
̈
2
are unknown. To exclude S, the rst equation of this system
we multiply to R
2
and the second equation to R
1
, then add the corresponding sides of these equations:
(
x
1
R
2
2
+ '
x
2
∙ R
1
) φ
̈
2
= MR
2
– (G
3
r
2
+ M
C
) R
1
,
R
1
and from here:
φ
̈
2
=
MR
1
R
2
– (G
3
r
2
+ M
C
) R
1
2
.
(5.5)
x
1
R
2
2
+ '
x
2
R
1
2
.
Equation (5.5) denes the angular acceleration of the link 2 of the mechanism.
Using the given values:
x
1
= m
1
i
x
1
= 100(0.2√2)
2
= 8 kg ∙ m
2
,
'
x
2
=
x
2
+ m
3
r
2
2
= m
2
i
x
2
+ m
3
r
2
2
= 150 ∙ 0.3
2
+ 400 ∙ 0.2
2
= 29.5 kg ∙ m
2
.
From Equation (5.5):
φ
̈
2
=
(4,200 + 200t)0.6 ∙ 0.4 – (400 ∙ 9.81 ∙ 0.2 + 2,000) 0.6
2
= 4.034t + 0.4597 (s
–2
).
8 ∙ 0.4
2
+ 29.5 ∙0.6
2
Twice integrating this equation:
φ
̇
2
= 2.017t
2
+ 0.4597t + C
1
,
φ
̇
2
= 0.672t
3
+ 0.23t
2
+ C
1
t + C
2
.
To determine the integral constants we will use the initial conditions: at
t = 0, φ
20
= 0, φ
̇
20
= ω
20
= ω
10
∙
R
1
= 2 ∙
60
= 3 s
–1
.
R
2
40
Hence: φ
̇
20
= C
1
, φ
20
= C
2
, or C
1
= 3 s
–1
, C
2
= 0.
2
2
89
e equation of the angular velocity of the link 2:
φ
̇
2
= 2.017t
2
+ 0.4597t + 3 (s
–1
).
en the equation of the rotational motion of the link 2:
φ
2
= 0.672t
3
+ 0.23t
2
+ 3t (rad).
e circular force S can be determined from Equation (5.4):
S = S
2
=
'
x
2
φ
̈
2
+ G
3
r
2
+ M
C
,
R
2
at t
1
= 1 s
S =
29.5(4.034 ∙ 1 + 0.4597) + 400 ∙ 9.81 ∙ 0.2 + 2,000
= 7295 N.
0.4
M
A
B
Mc
ω
2
x
2
x
1
s
2
s
1
N
2
N
1
r
2
R
2
R
1
y
1
y
2
G
2
G
3
v
T’
2
T
1
ω
1
Y
B
Z
B
G
1
Y
A
Z
A
1
2
3
Figure 5.32.
To determine the tension T of the rope, we will write a dierential equation of the rotation of the link 2 (Figure
5.33):
x
2
φ̈
2
= S
2
R
2
– Tr
2
– M
C
,
5.3 SOLUTION
90 5. TOPIC AK-5
from where:
T =
S
2
R2 – M
C
–
x
2
φ
̈
2,
r
2
at t
1
= 1 s T =
7,295 ∙ 0.4 – 2000 – 13.5(4.0334 ∙ 1 + 0.4597) = 4,285 N
.
0.2
Mc
ω
2
2
G
2
Z
B
Y
B
x
2
y
2
T
z
N
2
s
2
Figure 5.33.
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