Step 1

We are given the mass of the cart as $m = 30\,{\rm{kg}}$, the horizontal force as $P = 90\,{\rm{N}}$, the mass of the uniform rod $BC$ as ${m_{BC}} = 15\,{\rm{kg}}$, the length of rod $BC$ as $l = 1\,{\rm{m}}$, the angle of $AB$ with vertical as $\theta = 30^\circ $, the angle of $BC$ as $\phi = 30^\circ $ with horizontal.

We are asked to determine the tension in cord $AB$ and the horizontal and vertical components of reaction on end $C$ of the uniform $15\,{\rm{kg}}$ rod $BC$.

 
Step 2

The free body diagram of the given system can be drawn as,

 
Step 3

Find the acceleration of the system by balancing force along $x$-axis using the following relation.

\[P - ma - {m_{BC}}a = 0\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} \left( {90\,{\rm{N}}} \right) - \left( {30\,{\rm{kg}}} \right)a - \left( {15\,\,{\rm{kg}}} \right)a = 0\\ a = \frac{{\left( {90\,{\rm{N}}} \right) \times \left( {\frac{{1\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{1\,{\rm{N}}}}} \right)}}{{45\,{\rm{kg}}}}\\ a = 2\,{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 4

The free body diagram of rod $BC$ can be drawn as shown below.

 
Step 5

Find the angle of the cord $AB$ with the axis of $BC$ using the following relation.

\[\alpha = 90^\circ - \theta - \phi \]

Substitute the known values in the above equation.

\[\begin{array}{c} \alpha = 90^\circ - 30^\circ - 30^\circ \\ = 30^\circ \end{array}\]
 
Step 6

Find the weight of rod $BC$ using the following relation.

\[{W_{BC}} = {m_{BC}}g\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {W_{BC}} = \left( {15\,{\rm{kg}}} \right)\left( {9.81\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\ = \left( {147.15\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {\frac{{1\,{\rm{N}}}}{{1\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ = 147.15\,{\rm{N}} \end{array}\]
 
Step 7

Find the tension in $AB$ by taking moment about $C$ using the following relation.

\[\left[ {\left( {{T_{AB}}\sin \alpha } \right) \times l} \right] - \left[ {\left( {{W_{BC}}\cos \phi } \right) \times \left( {\frac{l}{2}} \right)} \right] + \left[ {\left( {{m_{BC}}a\sin \phi } \right) \times \left( {\frac{l}{2}} \right)} \right] = 0\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} \left[ \begin{array}{l} \left[ {\left( {{T_{AB}}\sin 30^\circ } \right) \times \left( {1\,{\rm{m}}} \right)} \right]\\ - \left[ {\left( {\left( {147.15\,{\rm{N}}} \right)\cos 30^\circ } \right) \times \left( {\frac{{1\,{\rm{m}}}}{2}} \right)} \right]\\ + \left[ {\left( {\left( {15\,{\rm{kg}}} \right)\sin 30^\circ \left( {2\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)} \right) \times \left( {\frac{{1\,{\rm{m}}}}{2}} \right)} \right] \end{array} \right] = 0\\ \left[ {{T_{AB}}\left( {0.5\,m} \right)} \right] - \left[ {63.72\,{\rm{N}} \cdot {\rm{m}}} \right] + \left( {7.5\,{\rm{kg}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {\frac{{1\,{\rm{N}} \cdot {\rm{m}}}}{{1\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right) = 0\\ {T_{AB}} = 112.44\,{\rm{N}} \end{array}\]
 
Step 8

Find the reaction force at $C$ along $x$ by balancing forces along $x$ direction using the following relation.

\[{T_{AB}}\sin \theta - {C_x} = {m_{BC}}a\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} 112.44\,{\rm{N}}\sin 30^\circ - {C_x} = \left( {15\,{\rm{kg}}} \right)\left( {2\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\ 112.44\,{\rm{N}}\sin 30^\circ - {C_x} = \left( {30\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {\frac{{1\,{\rm{N}}}}{{1\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ {C_x} = 26.22\,{\rm{N}} \end{array}\]
 
Step 9

Find the reaction force at $C$ along $y$ by balancing forces along $y$ direction using the following relation.

\[{C_y} + {T_{AB}}\cos \theta - {W_{BC}} = 0\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {C_y} + \left( {112.44\,{\rm{N}}} \right)\cos 30^\circ - 147.15\,{\rm{N}} = 0\\ {C_y} = 49.77\,{\rm{N}} \end{array}\]