This classic mechanical linkage results in a straight line motion of point $$C$$ as point $$A$$ is rotated about center $$O$$. Both points $$P$$ and $$O$$ are fixed, as links $$PB, PD$$ and $$OA$$ are free to rotate about them.

The lengths of the links are

$$OA=AB=BC=CD=DA=1$$

while $$P, O, E$$ are co-linear, and $$PO=1$$, $$OE=2$$, and $$PE\bot EC$$

and $$PB=PD$$

Force $$F$$ is applied tangentially at point $$A$$ (perpendicular to $$OA$$), resulting in force $$F’$$ along the line $$EC$$ at point $$C$$.

At angle $$\angle AOE=60°$$, find the ratio of the forces $$\dfrac { F }{ F' }$$ accurate to three decimal places. As this is an exercise in classical mechanics, use the Conservation of Energy Law to determine this ratio.

Note: A property of Peaucellier-Lipkin linkages is that points $$P, A, C$$ are always co-linear.

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