Answer to Question #208487 in Mechanical Engineering for Aranya Dhanure

Question

The dimensions of various links in a mechanism, as shown in Fig. 6.34, are as 

follows : A B = 25 mm ; BC = 175 mm ; CD = 60 mm ; AD = 150 mm ; BE = EC ; and EF = FG 

= 100 mm. The crank A B rotates at 200 r.p.m. When the angle BAD is 135°, determine by 

instantaneous centre method : 1. Velocity of G, 2. Angular velocity of EF, and 3. Velocity of 

sliding of EF in the swivel block S. [Ans. 120 mm/s ; 6.5 rad/s ; 400 mm/s]

Answer 1

AB=25 mm, BC =175 mm, CD =60 mm ; AD =150 mm, BE =EC , EF= FG = 100 mm

$\omega_{AB}=200 rpm = \frac{200*2 \pi}{60}=20.94 rad/sec$

$V_B=AB *20.94 rad/sec=25*20.94 = 523.6 mm/sec$

$\frac{V_B}{4.5}=\frac{V_E}{9.6} \implies \frac{523.6}{4.5}=\frac{V_E}{9.6} \implies V_E=1117.01 mm/sec$

$\frac{V_F}{8.9}=\frac{V_G}{9.6} \implies \frac{V_F}{8.9}=\frac{1117.01}{9.6} \implies V_F=1035.56mm/sec$

$V_G=\frac{V_F*I_1G}{I_2F}=\frac{1035.56*4.2}{8.9}=486.691 mm/sec$

$v=r \omega \implies \omega _{EF}=\frac{V_E}{EF}=\frac{1117.01}{100}=11.170 rad/sec$