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We next focus on the dynamics. This starts with the free body diagram for member AB:Adesh said:We can move on now. Everything is cleared till that far.
View attachment 1588073847854.png
The above FBD shows the forces acting on strut AB. The tension F/4 at pin A is the portion of the imposted load F/2 that acts on strut AB. The tension R at pin A represents the contact force that strut DA imposes at A to keep vertex A from moving in the y direction. The tension T at pin B represents the contact force that strut CB imposes at B to pull vertex B back.
Next, we write the Newton's 2nd law force- and moment balances on strut AB. The force balances on AB in the x- and y-directions, respectively (based on the free body diagram) are:
$$\frac{F}{4}-T=ma_{cmx}=m\frac{a_{Ax}}{2}\tag{1}$$
$$-R=mA_{cmy}=-m\frac{a_{Ax}}{2}\tag{2}$$where use has been made here of some of the kinematic relations in post #29 to express the acceleration components of the center of mass of strut AB in terms of the x-direction acceleration of vertex A relative to the center of mass of the overall structure.
A clockwise moment balance on strut AB about its center of mass gives:
$$R\frac{b}{2\sqrt{2}}-\frac{F}{4}\frac{b}{2\sqrt{2}}-T\frac{b}{2\sqrt{2}}=I\frac{d^2 \theta}{dt^2}\tag{3}$$where I is the moment of inertial of the strut about its center of mass: $$I=m\frac{b^2}{12}\tag{4}$$If we combine Eqns. 3 and 4, we obtain:
$$R-\frac{F}{4}-T=\frac{m}{3}\frac{b}{\sqrt{2}}\frac{d^2 \theta}{dt^2}\tag{5}$$But, from post #29, $$\frac{b}{\sqrt{2}}\frac{d^2 \theta}{dt^2}=-a_{Ax}$$Therefore, $$\frac{F}{4}+T-R=\frac{m}{3}a_{Ax}\tag{6}$$
If we now combine Eqns. 1,2, and 6 to eliminate T and R, we obtain: $$a_{Ax}=\frac{3}{8}\frac{F}{m}$$Based on this and the results of Part 1 loading, we find that the absolute acceleration of vertex A is ##\frac{F}{4m}+\frac{3}{8}\frac{F}{m}=\frac{5}{8}\frac{F}{m}## and that the absolute acceleration of vertex C is ##\frac{F}{4m}-\frac{3}{8}\frac{F}{m}=-\frac{1}{8}\frac{F}{m}##