Question 16.SP.2: The thin plate ABCD of mass 8 kg is held in the position sho......

Kinematics of Motion. After wire BH has been cut, we observe that corners A and D move along parallel circles of radius 150 mm centered, respectively, at E and F. The motion of the plate is thus a curvilinear translation; the particles forming the plate move along parallel circles of radius 150 mm.
At the instant wire BH is cut, the velocity of the plate is zero. Thus the acceleration \overline{\mathbf{a}} of the mass center G of the plate is tangent to the circular path which will be described by G.
Equations of Motion. The external forces consist of the weight W and the forces F _{A E} \text { and } F _{D F} exerted by the links. Since the plate is in translation, the effective forces reduce to the vector m\overline{\mathbf{a}} attached at G and directed along the t axis. A free-body-diagram equation is drawn to show that the system of the external forces is equivalent to the system of the effective forces.
a. Acceleration of the Plate.

\begin{aligned}+\swarrow \Sigma F_t=\Sigma\left(F_t\right)_{\text{eff}}: & \\W \cos 30^{\circ} & =m \bar{a} \\m \mathrm{g} \cos 30^{\circ} & =m \bar{a} \\\bar{a}=\mathrm{g} \cos 30^{\circ} & =\left(9.81  m / s ^2\right) \cos 30^{\circ}&(1)\end{aligned}

\overline{\mathbf{a}}=8.50 m / s ^2  ⦫  60^{\circ}

b. Forces in Links AE and DF.

\begin{aligned}& +\nwarrow \Sigma F_n=\Sigma\left(F_n\right)_{\text{eff}}: \quad F_{A E}+F_{D F}-W \sin 30^{\circ}=0&(2) \\& +\downarrow \Sigma M_G=\Sigma\left(M_G\right)_{\text{eff}}:\end{aligned}
\begin{aligned}\left(F_{A E} \sin 30^{\circ}\right)(250  mm )&-\left(F_{A E} \cos 30^{\circ}\right)(100  mm ) \\&+\left(F_{D F} \sin 30^{\circ}\right)(250  mm )+\left(F_{D F} \cos 30^{\circ}\right)(100  mm )=0 \\&38.4 F_{A E}+211.6 F_{D F}=0 \\&F_{D F}=-0.1815 F_{A E}&(3)\end{aligned}

Substituting for F_{D F} from (3) into (2), we write

\begin{aligned}& F_{A E}-0.1815 F_{A E}-W \sin 30^{\circ}=0 \\& F_{A E}=0.6109 W \\& F_{D F}=-0.1815(0.6109 W)=-0.1109 W\end{aligned}

Noting that W = mg = (8 kg)(9.81 m/s²) = 78.48 N, we have

\begin{array}{lll}F_{A E}=0.6109(78.48  N ) & F_{A E}=47.9  N  T  \\F_{D F}=-0.1109(78.48  N ) & F_{D F}=8.70  N  C\end{array}