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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