3

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.

4

(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).