For the system shown below, the moving frame is attached to the slotted arm B wi

Question

For the system shown below, the moving frame is attached to the slotted arm B with the origin at B. The angular velocity of bar OP is o.2 [rad/s] clockwise and is constant. The angular velocity of bar O'B is 0.4 [rad/s] clockwise and is constant. The angular velocity of the moving frame (which is attached to the slotted arm B) is 0.428 [rad/s] clockwise. The relative velocity of point P with respect to the moving frame (slotted arm B) is up and to the left at 0.0334 [m/s] at the instant shown. At the instant shown, Find the angular acceleration of slotted arm B (the moving frame) Find the relative acceleration of point P with respect to the slotted arm B (the moving frame)

Explanation / Answer

Link AB rotates about the fixed point A. Hence For link BD For Link DE A + c B -yE = -4 - 2 sin 30°(0) yE = 4ft>s ; Ans. A :+ B 0 = 2 cos 30° vDE vDE = 0 -yEi = (-4 - 2 sin 30° vDE)i + 2 cos 30°vDEj -yEi = -4i + (vDEk) * (2 cos 30°i + 2 sin 30°j) vE = vD + vDE * rE>D rE>D = {2 cos 30°i + 2 sin 30°j} ft vD = {-4i} ft>s vDE = vDE k vE = -yEi (+ c) 0 = vBD - 8 sin 60° vBD = 6.928 rad>s A :+ B -yD = -8 cos 60° yD = 4 ft>s -yD i = -8 cos 60°i + (vBD - 8 sin 60°)j -yDi = (-8 cos 60°i - 8 sin 60°j) + (vBDk) * (1i) vD = vB + vBD * rD>B rD>B = {1i} ft vB = {-8 cos 60°i - 8 sin 60° j} ft>s vD = -yDi vBD = vBD k yB = vAB rAB = 4(2) = 8 ft>s

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