Step 1

We are given the weight as $W = 150\;{\rm{lb}}$, force on A and B as ${F_A} = 100\;{\rm{lb}}$ and ${F_B} = 200\;{\rm{lb}}$, length of the beam as $L = 12\,{\rm{ft}}$.

We are asked to determine themagnitude of the acceleration of the mass centre and the angular acceleration of the beam.

 
Step 2

The free body diagram of the beam is as follows:

To find the mass moment of inertia, we will use the relation,

\[\begin{array}{c} I = \frac{1}{{12}}m{L^2}\\ I = \frac{1}{{12}}\left( {\frac{W}{g}} \right){L^2} \end{array}\]

On plugging the values in the above relation, we get,

\[\begin{array}{c} I = \frac{1}{{12}}\left( {\frac{{150\;{\rm{lb}}}}{{32.2\;{\rm{ft/}}{{\rm{s}}^2}}}} \right){\left( {12\;{\rm{ft}}} \right)^2}\\ I = 55.9\;{\rm{lb}} \cdot {{\rm{s}}^2} \cdot {\rm{ft}} \times \left( {\frac{{1\;{\rm{slug}} \cdot {\rm{f}}{{\rm{t}}^2}}}{{1\;{\rm{lb}} \cdot {{\rm{s}}^2} \cdot {\rm{ft}}}}} \right)\\ I = 55.9\;{\rm{slug}} \cdot {\rm{f}}{{\rm{t}}^2} \end{array}\]
 
Step 3

To find the acceleration in horizontal direction, we will use the relation,

\[\begin{array}{c} \Sigma {F_x} = ma\\ {F_B}\cos 60^\circ = ma \end{array}\]

On plugging the values in the above relation, we get,

\[\begin{array}{c} \left( {200\;{\rm{lb}}} \right)\cos 60^\circ = \left( {\frac{{150\;{\rm{lb}}}}{{32.2\;{\rm{ft/}}{{\rm{s}}^2}}}} \right)a\\ a = 21.47\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\]

To find the acceleration in vertical direction, we will use the relation,

\[\begin{array}{c} \Sigma {F_y} = ma'\\ {F_A} + {F_B}\cos 60^\circ - W = ma'\\ \left( {100\;{\rm{lb}}} \right) + \left( {200\;{\rm{lb}}} \right)\sin 60^\circ - \left( {150\;{\rm{lb}}} \right) = \left( {\frac{{150\;{\rm{lb}}}}{{32.2\;{\rm{ft/}}{{\rm{s}}^2}}}} \right)a'\\ a' = 26.45\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\]
 
Step 4

To find the magnitude of acceleration, we will use the relation,

\[\begin{array}{c} {a_m} = \sqrt {{a^2} + a{'^2}} \\ {a_m} = \sqrt {{{\left( {21.47\;{\rm{ft/}}{{\rm{s}}^2}} \right)}^2} + {{\left( {26.45\;{\rm{ft/}}{{\rm{s}}^2}} \right)}^2}} \\ {a_m} = 34.067\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\]

To find the angular acceleration, we will use the relation,

\[\begin{array}{c} \Sigma {M_G} = I\alpha \\ - {F_A} \times {d_1} + {F_B}\sin 60^\circ \times {d_2} = \left( {m{k_G}^2} \right)\alpha \end{array}\]

Here, ${d_1}$ and ${d_2}$ are the distance from point A and B to center of mass G respectively.

On plugging the values in the above relation, we get,

\[\begin{array}{c} - \left( {100\;{\rm{lb}}} \right)\left( {6\;{\rm{ft}}} \right) + \left( {200\;{\rm{lb}}} \right)\sin 60^\circ \left( {6\;{\rm{ft}}} \right) = \left( {55.9\;{\rm{slug}} \cdot {\rm{f}}{{\rm{t}}^2}} \right)\alpha \\ \alpha = 7.85\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]