Show Full work The parallelogram linkage shown moves in the vertical plane with
ID: 2240796 • Letter: S
Question
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The parallelogram linkage shown moves in the vertical plane with the uniform 8-kg bar EF attached to the plate at E which is welded both to the plate and to the bar. The mass moment of inertia of the bar about an axis through its mass center is IG = mL2/12. A torque (not shown) is applied to link AB through its lower pin to drive the links in a clockwise direction. When theta = 60degree, the links have an angular acceleration and an angular velocity of 6 rad/s2 and 4 rad/s, respectively. For this instant, calculate The acceleration aGx and aGy of the mass center of the bar EF. The force (magnitude and direction) exerted on the bar EF by the pin at E. The moment ME (magnitude and direction) exerted on the bar EF by the pin at E.Explanation / Answer
Linear acceleration of plate E is a =epsilon*R = 6*0.8 =4.8 m/s^2
The linear acceleration is tangent to the trajectory, thus perperendicular to AB and/or CD.
When theta = 60 deg the angle between linear acceleration and horizontal is (90-theta)
Thus ax = a*cos(90-theta) = 4.8*cos(30) =4.157 m/s^2
ay = a*sin(90-theta) =4.8*sin(30) =2.4 m/s^2
The total acceleration f the CM of EF bar is a =4.8 m/s^2 at 30 degree above the horizontal (counterclockwise).
F = m*a =8*4.8 =38.4 N.
The angle is opposed acceleration angle with 180 degree, therefore is makin with direction EF an angle of 30+180 =210 degree (above the horizontal) counterclockwise.
Moment of force in center of mass of EF is
M = F x r = 38.4* 0.6*sin(30) =11.52 N*m rotating anticlockwise (direction out of the page).
Moment from pin E at point E is the same, rotating opposite:
M(E) =11.52 N*m rotating clockwise (direction into the page)