Question

In the four-bar mechanism shown in figure ABC is the coupler link and all the dimensions are in cms. For the configuration shown, AC is aligned with PA while B is aligned with QB. Then the ratio of instantaneous angular velocity of the link QB to that of the link PA is

• A. 1.5
• B. 0.75
• C. 1
• D. 1.33

Answer 1

The correct option is C 1

${\omega }_{PA}=\frac{V_A}{PA}$

${\omega }_{QB}=\frac{V_B}{QB}$

$\Delta ACB=\text{right angle triangle}$

$\frac{{\omega }_{QB}}{{\omega }_{PA}}=\frac{V_B}{QB}\times \frac{PA}{V_A}=\frac{V_B}{V_A}\times \left(\frac{PA}{QB}\right)$


The component of velocities of A and B along AB should be same

$V_A\cos \theta =V_B\sin \theta$

$\frac{V_B}{V_A}=\frac{1}{\tan \theta }$

$\tan \theta =\frac{6}{8}=\frac{3}{4}(from\Delta ABC\tan \theta =\frac{AC}{BC})$

$\frac{V_B}{V_A}=\frac{4}{3}$

$\frac{{\omega }_{QB}}{{\omega }_{PA}}=\frac{V_B}{V_A}\times \frac{PA}{QB}$

$=\frac{4}{3}\times \frac{24}{32}=1$