146 7. TOPIC AK-7
7.2 SAMPLE PROBLEM
Given parameters: Q = 100 N; c = 5 N/cm; r
1
= 20 cm; r
2
= 40 cm; r
3
= 10 cm; OA = l = 50 cm; α =30°; β = 90°
(Figure 7.31).
Ignoring the weights of the links OA and AB, dene the spring deformation h.
Q
r
3
O
A
r
1
r
2
G
1
G
3
3
1
2
O
1
1
K
s
s
1
3
s
Q
B
F
s
A
s
B
P
y
x
Figure 7.31.
7.3 SOLUTION
e following balanced forces are acting on the given mechanism:
F
⃗
– spring elasticity force;
G
⃗
1
– gravity force of the disc 1 with the gear 2;
G
⃗
3
– gravity force of the gear 3;
Q
⃗
– gravity force of the load; and reaction forces of supports.
e supports applied to the mechanism allow the following virtual displacements of its links:
– an angle of turn δφ
1
of the disc 1 along with the gear 2;
147
– an angle of turn δφ
3
of the gear 3;
– a translational vertical displacement δs
Q
of the load Q;
– a horizontal displacement δs
B
of the slider B;
– a displacement δs
A
of the particle A perpendicular to OA.
e sum of work performed in the virtual displacements:
Qδs
Q
– Fδs
B
= 0. (7.1)
Next, we will determine a relationship between the virtual displacements of the particles of the system. e dis-
placements of the load Q and the point of the rim of the disc 1 are equal as a thread holding the load Q is inelastic
and the slipping between the thread and the disc 1 is ignored. erefore, the virtual angle of turn of the disc 1 along
with the gear 2:
δφ
1
=
δs
Q
.
r
1
A virtual displacement of the point K of the rim of the gear 2:
δs
1
= r
2
δφ
1
=
r
2
δs
Q
.
r
1
e virtual displacements of the contact points of the gears 2 and 3 are equal as the slipping between them is ignored.
en, the virtual angle of turn of the gear 3:
δφ
3
=
δs
1
=
r
2
δs
Q
.
r
3
r
1
r
3
e virtual displacement of the point A of the crank rigidly attached to the gear 3:
δs
A
= OA δφ
3
=
r
2
l
δs
Q
.
r
1
r
3
To dene a relationship between the virtual displacements δs
A
and δs
B
we nd a position of the instantaneous center
of rotation P of the link AB. en:
δs
B
=
PB
.
δs
A
PA
From here:
δs
B
=
PB
δs
A
.
PA
From the triangle ∆APB:
PB
=
1
.
PA cos 30°
Hence,
δs
B
=
r
2
l
δs
Q
.
r
1
r
3
cos 30°
e force of the elasticity of the spring is proportional to its deformation:
F = ch.
7.3 SOLUTION
148 7. TOPIC AK-7
Substituting the expressions for the force of the elasticity and the virtual displacements of the system particles into
Equation (7.1), we will obtain:
Qδs
Q
– ch
r
2
l
δs
Q
= 0.
r
1
r
3
cos 30°
From here:
h =
Q
∙
r
1
r
3
cos 30°
=
100 ∙ 20 ∙ 10 ∙ 0.87
= 1.74 cm.
c r
2
l 5 ∙ 40 ∙ 50
Hence, the spring is compressed 1.74 cm.
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