Question 16.SP.10: The extremities of a 4-ft rod weighing 50 lb can move freely......

Kinematics of Motion. Since the motion is constrained, the acceleration of G must be related to the angular acceleration α. To obtain this relation, we first determine the magnitude of the acceleration \mathbf{a} _A of point A in terms of α. Assuming that α is directed counterclockwise and noting that a_{B / A}=4 \alpha, we write

\begin{aligned}\mathbf{a} _B & = \mathbf{a} _A+ \mathbf{a} _{B / A} \\{\left[a_B  ⦪  45^{\circ}\right] } & =\left[a_A \rightarrow\right]+\left[4 \alpha  ⦫  60^{\circ}\right]\end{aligned}

Noting that \phi=75^{\circ} and using the law of sines, we obtain

a_A=5.46 \alpha \quad a_B=4.90 \alpha

The acceleration of G is now obtained by writing

\begin{aligned}& \overline{ \mathbf{a} }= \mathbf{a} _G= \mathbf{a} _A+ \mathbf{a} _{G / A} \\& \overline{ \mathbf{a} }=[5.46 \alpha \rightarrow]+\left[2 \alpha ⦫ 60^{\circ}\right]\end{aligned}

Resolving \overline{ \mathbf{a} } into x and y components, we obtain

\begin{array}{lll}\bar{a}_x=5.46 \alpha-2 \alpha \cos 60^{\circ}=4.46 \alpha & \overline{ \mathbf{a} }_x=4.46 \alpha \rightarrow \\\bar{a}_y=-2 \alpha \sin 60^{\circ}=-1.732 \alpha & \overline{ \mathbf{a} }_y=1.732 \alpha \downarrow\end{array}

Kinetics of Motion. We draw a free-body-diagram equation expressing that the system of the external forces is equivalent to the system of the effective forces represented by the vector of components m \overline{ \mathbf{a} }_x \text { and } m \overline{ \mathbf{a} }_y attached at G and the couple \bar{I} \alpha. We compute the following magnitudes:

\begin{aligned}& \bar{I}=\frac{1}{12} m l^2=\frac{1}{12} \frac{50\text{ lb}}{32.2\text{ ft} / s ^2}(4 \text{ ft} )^2=2.07 \text{ lb} \cdot \text{ ft} \cdot s ^2 \quad \bar{I} \alpha=2.07 \alpha \\& m \bar{a}_x=\frac{50}{32.2}(4.46 \alpha)=6.93 \alpha \quad m \bar{a}_y=-\frac{50}{32.2}(1.732 \alpha)=-2.69 \alpha\end{aligned}

Equations of Motion
\begin{aligned}+\uparrow \Sigma M_E=&\Sigma\left(M_E\right)_{\text {eff }}:\\&(50)(1.732)=(6.93 \alpha)(4.46)+(2.69 \alpha)(1.732)+2.07 \alpha\end{aligned}

α = 12.30 rad/s²             α = 2.30 rad/s² ↑

R_B=22.5\text{ lb} \quad R _B=22.5 \text{ lb}  ⦨  45^{\circ}