Step 1

We are given the weight $W = 700\;{\rm{lb}}$, $W = 600\;{\rm{lb}}$ and $W = 200\;{\rm{lb}}$ in part (a), (b) and (c) respectively.

We are asked to draw the free body diagram and identify the type of force.

 
Step 2

(a)

The free body diagram is shown in the diagram below.

In the above free body diagram, all the forces acting on the rope are tension forces because ropes are not rigid to carry compression forces.

To find the angle between force we will use the relation,

\[\begin{array}{l} {\theta _1} = 90^\circ + {\sin ^{ - 1}}\left( {\frac{4}{5}} \right)\\ {\theta _1} = 143.13^\circ \end{array}\]

To find the angle among AB and AC we will use the relation,

\[\begin{array}{l} {\theta _2} = 90^\circ - {\sin ^{ - 1}}\left( {\frac{4}{5}} \right) + 30^\circ \\ {\theta _2} = 66.87^\circ \end{array}\]

To find the angle among weight and force AC we will use the relation,

\[\begin{array}{c} {\theta _1} + {\theta _2} + {\theta _3} = 360^\circ \\ {\theta _3} = \left( {360^\circ } \right) - \left( {143.13^\circ + 66.87^\circ } \right)\\ {\theta _3} = 150^\circ \end{array}\]
 
Step 3

To find the force we will use sine law.

\[\begin{array}{c} \frac{{200\,{\rm{N}}}}{{\sin {\theta _2}}} = \frac{{{F_B}}}{{\sin {\theta _3}}}\\ \frac{{200\,{\rm{N}}}}{{\sin 66.87^\circ }} = \frac{{{F_B}}}{{\sin 150^\circ }}\\ {F_B} = 108.74\;{\rm{N}} \end{array}\]

To find the forces we will use sine law.

\[\begin{array}{c} \frac{{200\,{\rm{N}}}}{{\sin {\theta _2}}} = \frac{{{F_C}}}{{\sin {\theta _1}}}\\ \frac{{200\,{\rm{N}}}}{{\sin 66.87^\circ }} = \frac{{{F_C}}}{{\sin 143.13^\circ }}\\ {F_C} = 130.48\;{\rm{N}} \end{array}\]
 
Step 4

(b)

The free body diagram is shown in the diagram below.

In the above free body diagram, all the forces acting on the rope are tension forces because ropes are not rigid to carry compression forces.

To find the angle between force we will use the relation,

\[\begin{array}{l} {\theta _1} = 90^\circ + {\sin ^{ - 1}}\left( {\frac{4}{5}} \right)\\ {\theta _1} = 143.13^\circ \end{array}\]

To find the angle among AB and AC we will use the relation,

\[\begin{array}{l} {\theta _2} = 90^\circ + 30^\circ \\ {\theta _2} = 120^\circ \end{array}\]

To find the angle among weight and force AC we will use the relation,

\[\begin{array}{c} {\theta _1} + {\theta _2} + {\theta _3} = 360^\circ \\ {\theta _3} = \left( {360^\circ } \right) - \left( {143.13^\circ + 120^\circ } \right)\\ {\theta _3} = 96.87^\circ \end{array}\]
 
Step 5

To find the force we will use sine law.

\[\begin{array}{c} \frac{{600\,{\rm{N}}}}{{\sin {\theta _2}}} = \frac{{{F_B}}}{{\sin {\theta _1}}}\\ \frac{{600\,{\rm{N}}}}{{\sin 120^\circ }} = \frac{{{F_B}}}{{\sin 143.13^\circ }}\\ {F_B} = 415.69\;{\rm{N}} \end{array}\]

To find the forces we will use sine law.

\[\begin{array}{c} \frac{{600\,{\rm{N}}}}{{\sin {\theta _2}}} = \frac{{{F_C}}}{{\sin {\theta _3}}}\\ \frac{{600\,{\rm{N}}}}{{\sin 120^\circ }} = \frac{{{F_C}}}{{\sin 96.87^\circ }}\\ {F_C} = 687.84\;{\rm{N}} \end{array}\]
 
Step 6

(c)

The free body diagram is shown in the diagram below.

In the above free body diagram, all the forces acting on the rope are tension forces because ropes are not rigid to carry compression forces.

To find the angle between force we will use the relation,

\[\begin{array}{l} {\theta _1} = 90^\circ + 30^\circ \\ {\theta _1} = 120^\circ \end{array}\]

To find the angle among weight and force AC we will use the relation,

\[\begin{array}{l} {\theta _2} = 90^\circ + 45^\circ \\ {\theta _2} = 135^\circ \end{array}\]

To find the angle among AB and AC we will use the relation,

\[\begin{array}{c} {\theta _1} + {\theta _2} + {\theta _3} = 360^\circ \\ {\theta _3} = \left( {360^\circ } \right) - \left( {135^\circ + 120^\circ } \right)\\ {\theta _3} = 105^\circ \end{array}\]
 
Step 7

To find the force we will use sine law.

\[\begin{array}{c} \frac{{500\,{\rm{N}}}}{{\sin {\theta _3}}} = \frac{{{F_B}}}{{\sin {\theta _2}}}\\ \frac{{500\,{\rm{N}}}}{{\sin 105^\circ }} = \frac{{{F_B}}}{{\sin 135^\circ }}\\ {F_B} = 366.02\;{\rm{N}} \end{array}\]

To find the forces we will use sine law.

\[\begin{array}{c} \frac{{500\,{\rm{N}}}}{{\sin {\theta _3}}} = \frac{{{F_C}}}{{\sin {\theta _1}}}\\ \frac{{500\,{\rm{N}}}}{{\sin 105^\circ }} = \frac{{{F_C}}}{{\sin 120^\circ }}\\ {F_C} = 448.28\;{\rm{N}} \end{array}\]