Step 1

We are given the mass of plate $B$ as ${m_B} = 20\;{\rm{kg}}$, the coefficient of friction between the plate $B$ and the ground as ${\mu _B} = 0.3$, the coefficient of friction between the plate $B$ and the block $A$ as ${\mu _A} = 0.2$, and the coefficient of friction between the cable and the peg $C$ as ${\mu _C} = 0.3$.

We are asked to determine the smallest mass of the block $A$.

 
Step 2

We will draw a free body diagram of the block $A$.

Here, ${T_1}$ is the tension in the upper rope, ${W_A}$ is the is the weight of block $A$, ${N_A}$ is the normal force, and ${f_A} = {\mu _A}{N_A}$ is the friction force.

 
Step 3

We will resolve the forces along the plane.

\[\begin{array}{c} {T_1} - {f_A} - {W_A}\sin 30^\circ = 0\\ {T_1} - {\mu _A}{N_A} - {m_A}g\sin 30^\circ = 0\\ {N_A} = \frac{{{T_1} - {m_A}g\sin 30^\circ }}{{{\mu _A}}}\;\;\;...\left( 1 \right) \end{array}\]

Here, ${m_A}$ is the mass of the block $A$ and $g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ is the gravitational acceleration.

 
Step 4

We will resolve the forces in the direction of vertical to the plane.

\[\begin{array}{c} {N_A} - {W_A}\cos 30^\circ = 0\\ {N_A} = {m_A}g\cos 30^\circ \;\;\;\;...\left( 2 \right) \end{array}\]

Substitute the given expression in the above equation.

\[\begin{array}{c} \frac{{{T_1} - {m_A}g\sin 30^\circ }}{{{\mu _A}}} = {m_A}g\cos 30^\circ \\ {T_1} - {m_A}g\sin 30^\circ = {\mu _A}\left( {{m_A}g\cos 30^\circ } \right)\\ {T_1} = {\mu _A}\left( {{m_A}g\cos 30^\circ } \right) + {m_A}g\sin 30^\circ \\ {T_1} = {m_A}g\left( {{\mu _A}\cos 30^\circ + \sin 30^\circ } \right)\,\;\;\;\;\;\;\;\;...\left( 3 \right) \end{array}\]
 
Step 5

Substitute the given value in equation (3).

\[\begin{array}{c} {T_1} = {m_A}\left( {9.81} \right)\left( {0.2\cos 30^\circ + \sin 30^\circ } \right)\\ {T_1} = \left( {6.5727{m_A}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...\left( 4 \right) \end{array}\]
 
Step 6

We will draw a free body diagram of the block $B$.

Here, ${T_2}$ is the tension in the lower rope, ${W_B}$ is the is the weight of block $B$, ${N_B}$ is the normal force, and ${f_B} = {\mu _B}{N_B}$ is the friction force.

 
Step 7

We will resolve the forces along the plane.

\[\begin{array}{c} {T_2} + {f_B} - {W_B}\sin 30^\circ + {f_A} = 0\\ {T_2} + {\mu _B}{N_B} - {m_B}g\sin 30^\circ + {\mu _A}{N_A} = 0\;\;\;...\left( 5 \right) \end{array}\]
 
Step 8

We will resolve the forces in the direction of vertical to the plane.

\[\begin{array}{c} {N_B} - {N_A} - {W_B}\cos 30^\circ = 0\\ {N_B} = {N_A} + {W_B}\cos 30^\circ = 0\\ {N_B} = {N_A} + {m_B}g\cos 30^\circ \;\;\;...\left( 6 \right) \end{array}\]

Substitute the value of equation (2) and (6) in equation (5).

\[\begin{array}{c} \left[ \begin{array}{l} {T_2} + {\mu _B}\left( {{N_A} + {m_B}g\cos 30^\circ \;} \right) - \\ {m_B}g\sin 30^\circ + {\mu _A}{N_A} \end{array} \right] = 0\\ {T_2} = \left[ \begin{array}{l} {m_B}g\sin 30^\circ - \\ {\mu _B}\left( {{m_A}g\cos 30^\circ + {m_B}g\cos 30^\circ \;} \right) - \\ {\mu _A}\left( {{m_A}g\cos 30^\circ } \right) \end{array} \right]\\ {T_2} = g\left[ \begin{array}{l} {m_B}\sin 30^\circ - {\mu _B}{m_A}\cos 30^\circ - {\mu _B}{m_B}\cos 30^\circ \; - \\ {\mu _A}{m_A}\cos 30^\circ \end{array} \right]\\ {T_2} = g\left[ \begin{array}{l} {m_B}\left( {\sin 30^\circ - {\mu _B}\cos 30^\circ } \right)\; - \\ {m_A}\left( {{\mu _B}\cos 30^\circ + {\mu _A}\cos 30^\circ } \right) \end{array} \right]\;\;\;\;\;\;\;\;...\left( 7 \right) \end{array}\]
 
Step 9

Substitute the given value in equation (7).

\[\begin{array}{l} {T_2} = \left( {9.81} \right)\left[ \begin{array}{l} \left( {20\;{\rm{kg}}} \right)\left( {\sin 30^\circ - 0.3\cos 30^\circ } \right)\; - \\ {m_A}\left( {0.3\cos 30^\circ + 0.2\cos 30^\circ } \right) \end{array} \right]\\ {T_2} = \left[ {47.1 - 4.25{m_A}} \right] \end{array}\]
 
Step 10

The formula of the tensions in the upper cable and lower cable is given blow.

\[{T_2} = {T_1}{e^{{\mu _C}\beta }}\]

Here, $\beta = \pi $ is the angle mage by the cable.

Substitute the given value in the above equation.

\[\begin{array}{c} \left[ {47.1 - 4.25{m_A}} \right] = \left[ {\left( {6.5727{m_A}} \right)} \right]{e^{0.3\left( \pi \right)}}\\ 47.1 - 4.25{m_A} = \left( {6.5727{m_A}} \right){e^{0.3\left( \pi \right)}}\\ {m_A}\left[ {\left( {6.5727} \right){e^{0.3\left( \pi \right)}} + 4.25} \right] = 47.1\\ {m_A} \approx 2.23\;{\rm{kg}} \end{array}\]