Question 16.SP.3: The thin plate ABCD has a mass of 8 kg and is held in the po...
The thin plate ABCD has a mass of 8 kg and is held in the position shown by the wire BH and two links AE and DF. Neglecting the mass of the links, determine immediately after wire BH has been cut (a) the acceleration of the plate, (b) the force in each link.
STRATEGY: Since you are asked to determine the acceleration and forces, use Newton’s second law. After wire BH has been cut, corners A and D move along parallel circles, each with a radius of 150 mm centered, respectively, at E and F. The motion of the plate is thus a curvilinear translation (Fig. 1); the particles forming the plate move along parallel circles each with a radius of 150 mm.
MODELING: Choose the plate to be your system and model it as a rigid body. To draw the kinetic diagram, you need to consider the kinematics of the motion. At the instant wire BH is cut, the velocity of the plate is zero. Thus, the acceleration of the mass center G of the plate is tangent to the circular path described by G (Fig. 1). The free-body diagram and kinetic diagram for this system are shown in Fig. 2. The external forces consist of the weight W and the forces \mathbf{F}_{A E} and \mathbf{F}_{D F} exerted by the links. Since the plate is in translation, the kinetic diagram is the vector m \overline{\mathbf{a}} attached at G and directed along the t axis.
ANALYSIS:
a. Acceleration of the Plate.
\begin{aligned}W \cos 30^{\circ} &=m \bar{a} \\m g \cos 30^{\circ} &=m \bar{a} \\\bar{a}=g \cos 30^{\circ} &=\left(9.81 \mathrm{~m} / \mathrm{s}^2\right) \cos 30^{\circ}\end{aligned} (1)
\overline{\mathbf{a}}=8.50 \mathrm{~m} / \mathrm{s}^2 \text { ⦫ } 60^{\circ}
b. Forces in Links AE and DF.
+\nwarrow \Sigma F_n=m \bar{a}_n: \quad F_{A E} + F_{D F} – W \sin 30^{\circ}=0 (2)
+\circlearrowright \Sigma M_G=\bar{I} \alpha: \begin{aligned}\left(F_{A E} \sin 30^{\circ}\right)(250 \mathrm{~mm}) &-\left(F_{A E} \cos 30^{\circ}\right)(100 \mathrm{~mm}) \\+&\left(F_{D F} \sin 30^{\circ}\right)(250 \mathrm{~mm}) + \left(F_{D F} \cos 30^{\circ}\right)(100 \mathrm{~mm})=0\end{aligned}\begin{gathered}38.4 F_{A E} + 211.6 F_{D F}=0 \\F_{D F}=-0.1815 F_{A E}\end{gathered} (3)
Substituting F_{DF} from Eq. (3) into Eq. (2), you have
\begin{gathered}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 \mathrm{~W})=-0.1109 \mathrm{~W}\end{gathered}
Noting that W = mg = (8 kg)(9.81 m/s²) = 78.48 N, you have
\begin{aligned}&F_{A E}=0.6109(78.48 \mathrm{~N}) \quad F_{A E}=47.9 \mathrm{~N} \mathrm{~T}\\&F_{D F}=-0.1109(78.48 \mathrm{~N}) \quad F_{D F}=8.70 \mathrm{~N} C\end{aligned}
where bar AE is in tension and bar DF is in compression.
REFLECT and THINK: If AE and DF had been cables rather than links, the answers you just determined indicate that DF would have gone slack (i.e., you can’t push on a rope), since the analysis showed that it would be in compression. Therefore, the plate would not be undergoing curvilinear translation, but it would have been undergoing general plane motion. It is important to note that that there is always more than one way to solve problems like this, since you can choose to take moments about any point you wish. In this case, you took them about G, but you could have also chosen to take them about A or D.