Question 16.SP.8: A 6 × 8 in. rectangular plate weighing 60 lb is suspended fr...
A 6 \times 8 in. rectangular plate weighing 60 lb is suspended from two pins A and B. If pin B is suddenly removed, determine (a) the angular acceleration of the plate, (b) the components of the reaction at pin A immediately after pin B has been removed.
STRATEGY: You are asked to determine forces and the angular acceleration of the plate, so use Newton’s second law.
MODELING: Choose the plate to be your system and model it as a rigid body. Observe that as the plate rotates about point A, its mass center G describes a circle with a radius \bar{r} and its center at A (Fig. 1). The freebody diagram and kinetic diagram for this system are shown in Fig. 2. The plate is released from rest (\omega=0), so the normal component of the acceleration of \bar{a} is zero. The magnitude of the acceleration a of the mass center G is thus \bar{a}=\bar{r} \alpha.
ANALYSIS:
a. Angular Acceleration. Using your free-body diagram and kinetic diagram, you can take moments about A to find
+\circlearrowright \Sigma M_A=\bar{I} \alpha+m \overline{ a} d_{\perp}: \quad W \bar{x}=\bar{I} \alpha+(m \bar{a}) \bar{r}
Since \bar{a}=\bar{r} \alpha, you have
W \bar{x}=\bar{I} \alpha+(m \bar{r} \alpha) \bar{r} \alpha=\frac{W \bar{x}}{\frac{W_{-2}}{g} r^2+\bar{I}} (1)
The centroidal moment of inertia of the plate is
\begin{aligned}\bar{I}=\frac{m}{12}\left(a^2 + b^2\right) &=\frac{60 \mathrm{lb}}{12\left(32.2 \mathrm{ft} / \mathrm{s}^2\right)}\left[\left(\frac{8}{12} \mathrm{ft}\right)^2+\left(\frac{6}{12} \mathrm{ft}\right)^2\right] \\&=0.1078 \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2\end{aligned}
Substituting this value of \bar{I} together with W=60 \mathrm{lb}, \bar{r}=\frac{5}{12} \mathrm{ft}, and \bar{x}=\frac{4}{12} \mathrm{ft} ft into Eq. (1), you obtain
\alpha=+46.4 \mathrm{rad} / \mathrm{s}^2 \quad \alpha=46.4 \mathrm{rad} / \mathrm{s}^2 \circlearrowright
b. Reaction at A. Using the computed value of α, determine the magnitude of the vector m \overline{\mathbf{a}} attached at G as
m \bar{a}=m \bar{r} \alpha=\frac{60 \mathrm{lb}}{32.2 \mathrm{ft} / \mathrm{s}^2}\left(\frac{5}{12} \mathrm{ft}\right)\left(46.4 \mathrm{rad} / \mathrm{s}^2\right)=36.0 \mathrm{lb}
Applying Newon’s second law in the x and y directions gives
\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: \quad A_x=-\frac{3}{5}(36 \mathrm{lb})
=-21.6 \mathrm{lb} \quad \mathbf{A}_x=21.6 \mathrm{lb} \leftarrow
+\uparrow \Sigma F_y=m \bar{a}_y: \quad A_y – 60 \mathrm{lb}=-\frac{4}{5}(36 \mathrm{lb})
A_y=+31.2 \mathrm{lb} \quad \mathbf{A}_y=31.2 \mathrm{lb} \uparrow
REFLECT and THINK: If you had chosen to take moments about the center of gravity rather than point A, the two reaction forces A_x and A_y would have been in the resulting equation; that is, you would have had one equation and three unknowns, and you could not solve for α directly. Therefore, you would also need to use the equations from the x and y directions to solve for the three unknowns. Note that for convenience, we used a non-right handed coordinate system.
