Answer to Question #200285 in Mechanical Engineering for Muhammad Idrees

Question #200285

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.



1
Expert's answer
2021-06-07T05:45:52-0400


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"


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