Answer to Question #200285 in Mechanical Engineering for Muhammad Idrees

Question

In a mechanism as shown in Fig. 8.39, the link AB rotates with a uniform angular velocity of 30 rad/s. The lengths of various links are : A B = 100 mm ; BC = 300 mm ; BD = 150 mm ; DE = 250 mm ; EF = 200 mm ; DG = 165 mm.

Determine the velocity and acceleration of G for the given configuration.

Answer 1

ad$=3.5 cm = \frac{3.5}{1.667}= 2.09m/sec=V_D$

$ag=1cm = \frac{1}{1.667}=0.59m/sec=V_G \space velocity \space of \space G =0.59m/sec$

$ae=0.5cm=\frac{0.5}{1.667}=0.29m/sec =V_E$

$bc=4.5cm=2.699m/sec=V_{BC}$

$wbc=\frac{V{BC}}{BC}=\frac{2.699}{0.3}=8.99rad/sec$

$de=3.8=2.27m/sec=V_{DE}$

$w_{DE}=\frac{V_{DE}}{DE}=\frac{2.27}{0.25}=9.11rad/sec$

$w_{EF}=\frac{VE}{EF}=\frac{0.29}{0.2}=1.45 rad/sec$

$a_{BN}=AB(wAB)^2$

$=0.1(30)^2=90m/s^2$

$a_{Bcn}=BC(w_{Bc})^2$

$=0.3(8.99)^2=24.24m/s^2$

$a_{DE}=BE(w_{DE})^2$

$=0.25(9.11)^2=20.74m/s^2$

$a_{EF}=EF(w{EF})^2$

$0.2(1.45)^2=0.4205m/s^2$

$(SF)_a=\frac{10}{90}=0.1112cms^2/m$

$ac=18cm=ac=\frac{18}{0.1112}=162.002m/sec^2$

$ad=10cm=ad=\frac{10}{0.112}=90.0009m/s^2$

$ag=19.3cm=ag=\frac{19.3}{0.112}=173.70002m/s^2$

Acceleration of G $=1737m/sec^2$