Solution: vOv C vCBDDC 3 vC 3 5 cm CD v BDv 5 60 5 60...
Transcript of Solution: vOv C vCBDDC 3 vC 3 5 cm CD v BDv 5 60 5 60...
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Solution:
s
cm
Bv 60
s
cm
Bv 60
Dv
Dv
BDv
C
5Cv
35CCv
5Cv
CDv
3Cv
CDv
3Cv
35CCv
(a)
60B
cmv
s
vO
5Cv3C Dv
3Cv
BDv
3 5C Cv
Dv
(b)
Fig. 1.
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(a) From the figure, the velocity of point B is the same as that for point A, i.e.,
𝑣𝐵 = 𝑣𝐴 𝑎𝑛𝑑 |𝑣𝐵| = |𝑣𝐴| = 60𝑐𝑚
𝑠
(b) velocity of point D will be vertical with link 4 to the left, as shown in the figure and also in the velocity
polygon. Furthermore, the velocity 𝑣𝐵𝐷 is vertical with link BD, as shown in the below velocity
polygon.
60B
cmv
s
vODv
BDv
Measuring the distance, we can obtain:
𝑣𝐷 = 𝑣𝐵 − 𝑣𝐵𝐷 and |𝑣𝐷| = 17.69𝑐𝑚
𝑠
(c) Assume the contact point of link 3 and link 5 be point C. Since B,C,D are on the same link, the
velocity of C3 can be derived to be
𝑣𝐶3= 𝑣𝐷 + 𝑣𝐶3𝐷 and |𝑣𝐶3
| = 70.85𝑐𝑚
𝑠
3C Dv
3Cv
Dv
vO
(d) The velocity of C5 is vertical with link 5, and the relative velocity of C5 with respect to C3 is along the
link 5 direction. From the below velocity polygon, we can obtain that
𝑣𝐶5= 𝑣𝐶3
− 𝑣𝐶3𝐶5 and |𝑣𝐶5
| = 48.81𝑐𝑚
𝑠
5
vO
5Cv
3Cv
3 5C Cv
(e) From the figure, the distance between the O6 and C5 is 7.6cm, i.e.,
|𝑂6𝐶5| = 7.6𝑐𝑚
The rotation rate of link 5 can be calculated by
𝜔5 =|𝑣𝐶5
|
|𝑂6𝐶5|≈ 6.42
𝑟𝑎𝑑
𝑠
The rotation direction of link5 is clockwise, as shown in the figure (a).