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

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