Question

In the mechanism shown in figure, AB and DC are vertical, BC is horizontal. AB = 30 cm, BC = 40 cm and the velocity ratio is $\frac{1}{2}$. The length of the fixed link, is

• A. 40 cm
• B. 30 cm
• C. 60 cm
• D. 50 cm

Answer 1

The correct option is D 50 cm

AB= 30 cm, BC = 40 cm

Velocity ratio = $\frac{{\omega }_4}{{\omega }_2}=\frac{1}{2}$

Instantaneous centers are located in the figure

From Kennedy theorem

${\omega }_2\left(I_{24}{I}_{12}\right)={\omega }_4\left(I_{24}{I}_{14}\right)$

$\frac{{\omega }_4}{{\omega }_2}=\frac{I_{24}{I}_{12}}{I_{24}{I}_{14}}=\frac{1}{2}$


From geometry BE = 40 cm

$AE=\sqrt{{40}^2+{30}^2}=50cm$

$\therefore AE=AD=50cm$