Step 1

We have given a block A of weight ${W_A} = 8\;{\rm{lb}}$ and a block B of weight ${W_B} = 15\;{\rm{lb}}$.

We have given that a force $P = 12\;{\rm{lb}}$ is applied horizontally to block A.

We are asked to calculate the acceleration of block B.

 
Step 2

Draw a free-body diagram of block A.

Here, ${N_A}$ is the normal reaction force applied by the ground surface, ${N_B}$ is the normal force reaction applied by the block B on block A, and ${a_A}$ is the acceleration of block A.

 
Step 3

Apply equilibrium equation of motion in the horizontal direction:

\[\begin{array}{c} \sum {{F_x}} = {m_A}{a_A}\\ P - {N_B}\sin 15^\circ = \left( {\frac{{{W_A}}}{g}} \right) \times {a_A} \end{array}\]

Substitute the value of $P$ and ${W_A}$in the above equation:

\[\begin{array}{c} 12\;{\rm{lb}} - {N_B}\sin 15^\circ = \left( {\frac{{8\;{\rm{lb}} \times \frac{{1\;{\rm{lbm}} \cdot {\rm{ft}}/{{\rm{s}}^2}}}{{1\;{\rm{lb}}}}}}{{32.2\;{\rm{ft}}/{{\rm{s}}^2}}}} \right) \times {a_A}\\ 12\;{\rm{lb}} - {N_B}\sin 15^\circ = \left( {\frac{{8\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_A}\\ {N_B}\sin 15^\circ = 12\;{\rm{lb}} - \left( {\frac{{8\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_A} \end{array}\]……(1)
 
Step 4

Draw a free-body diagram of block B.

Here, ${N_C}$ is the normal reaction force applied by the wall and ${a_B}$ is the acceleration of block B.

 
Step 5

Apply equilibrium equation of motion in the vertical direction:

\[\begin{array}{c} \sum {{F_y}} = {m_B}{a_B}\\ {N_B}\cos 15^\circ - 15\;{\rm{lb}} = \left( {\frac{{{W_B}}}{g}} \right) \times {a_B} \end{array}\]

Substitute the value of ${W_B}$ in the above equation:

\[\begin{array}{c} {N_B}\cos 15^\circ - 15\;{\rm{lb}} = \left( {\frac{{15\;{\rm{lb}} \times \frac{{1\;{\rm{lbm}} \cdot {\rm{ft}}/{{\rm{s}}^2}}}{{1\;{\rm{lb}}}}}}{{32.2\;{\rm{ft}}/{{\rm{s}}^2}}}} \right) \times {a_B}\\ {N_B}\cos 15^\circ - 15\;{\rm{lb}} = \left( {\frac{{15\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_B}\\ {N_B}\cos 15^\circ = \left( {\frac{{15\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_B} + 15\;{\rm{lb}} \end{array}\]……(2)
 
Step 6

Divide equation (1) by equation (2):

\[\begin{array}{c} \frac{{{N_B}\sin 15^\circ }}{{{N_B}\cos 15^\circ }} = \frac{{12\;{\rm{lb}} - \left( {\frac{{8\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_A}}}{{\left( {\frac{{15\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_B} + 15\;{\rm{lb}}}}\\ \tan 15^\circ = \frac{{12\;{\rm{lb}} - \left( {\frac{{8\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_A}}}{{\left( {\frac{{15\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_B} + 15\;{\rm{lb}}}}\\ 12\;{\rm{lb}} - \left( {\frac{{8\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_A} = \tan 15^\circ \times \left( {\frac{{15\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_B} + \tan 15^\circ \times 15\;{\rm{lb}}\\ \tan 15^\circ \times \left( {\frac{{15\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_B} + \left( {\frac{{8\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_A} = 12\;{\rm{lb}} - \tan 15^\circ \times 15\;{\rm{lb}} \end{array}\]

Solving further the above equation:

\[\begin{array}{c} \tan 15^\circ \times \left( {\frac{{15\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_B} + \left( {\frac{{8\;{\rm{lbm}}}}{{32.2}}} \right) \times {a_A} = 7.9807\;{\rm{lb}}\\ \tan 15^\circ \times 15\;{\rm{lbm}} \times {a_B} + 8\;{\rm{lbm}} \times {a_A} = 256.98\;{\rm{lb}} \end{array}\]

Divide the above equation by 1 lbm on both sides:

\[\begin{array}{c} \frac{{\tan 15^\circ \times 15\;{\rm{lbm}} \times {a_B} + 8\;{\rm{lbm}} \times {a_A}}}{{1\;{\rm{lbm}}}} = \frac{{256.98\;{\rm{lb}} \times \frac{{1\;{\rm{lbm}} \cdot {\rm{ft}}/{{\rm{s}}^2}}}{{{\rm{lb}}}}}}{{1\;{\rm{lbm}}}}\\ \tan 15^\circ \times 15 \times {a_B} + 8 \times {a_A} = 256.98\;{\rm{ft}}/{{\rm{s}}^2} \end{array}\]……(3)
 
Step 7

The kinematic of distance ${S_A}$ travelled by block A and the distance ${S_B}$ travelled by block B is shown by the geometry drawn below.

 
Step 8

From the above geometry, we get:

\[\begin{array}{c} \tan 15^\circ = \frac{{{S_B}}}{{{S_A}}}\\ {S_B} = \tan 15^\circ \times {S_A} \end{array}\]

Differentiate the above equation with respect to time t:

\[\begin{array}{c} \frac{d}{{dt}}\left( {{S_B}} \right) = \tan 15^\circ \times \frac{d}{{dt}}\left( {{S_A}} \right)\\ {v_B} = \tan 15^\circ \times {v_A} \end{array}\]

Again differentiate the above equation with respect to time t:

\[\begin{array}{c} \frac{d}{{dt}}\left( {{v_B}} \right) = \tan 15^\circ \times \frac{d}{{dt}}\left( {{v_A}} \right)\\ {a_B} = \tan 15^\circ \times {a_A}\\ {a_A} = \frac{{{a_B}}}{{\tan 15^\circ }} \end{array}\]……(4)
 
Step 9

Substitute the value of ${a_A}$ from equation (4) in equation (3):

\[\begin{array}{c} \tan 15^\circ \times 15 \times {a_B} + 8 \times \left( {\frac{{{a_B}}}{{\tan 15^\circ }}} \right) = 256.98\;{\rm{ft}}/{{\rm{s}}^2}\\ {a_B}\left( {\tan 15^\circ \times 15 + 8 \times \frac{1}{{\tan 15^\circ }}} \right) = 256.98\;{\rm{ft}}/{{\rm{s}}^2}\\ {a_B}\left( {33.87} \right) = 256.98\;{\rm{ft}}/{{\rm{s}}^2}\\ {a_B} = 7.59\;{\rm{ft}}/{{\rm{s}}^2} \end{array}\]