Step 1

We are given the forces ${F_1} = 50\;{\rm{N,}}\;{F_2} = 80\;{\rm{N}}$ and ${F_3} = 30\;{\rm{N}}$.

We are asked to express each force in the cartesian vector form and also to determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis.

 
Step 2

Resolve each force into their x and y components as shown in figure below,

 
Step 3

In the Cartesian vector form, we can write the force ${F_1}$ as:

\[\begin{array}{c} {F_1} = {F_{1x}}\hat i + {F_{1y}}\hat j\\ {F_1} = \left( {{F_1}\cos \theta } \right)\hat i + \left( {{F_1}\sin \theta } \right)\hat j\\ {F_1} = \left( {50\;{\rm{N}}} \right) \times \left( {\frac{3}{5}} \right)\hat i + \left( {50\;{\rm{N}}} \right) \times \left( {\frac{4}{5}} \right)\hat j\\ {F_1} = \left( {30\hat i + 40\hat j} \right)\;{\rm{N}} \end{array}\]

In the Cartesian vector form, we can write the force ${F_2}$ as:

\[\begin{array}{c} {F_2} = - {F_{2x}}\hat i - {F_{2y}}\hat j\\ {F_2} = - \left( {{F_2}\sin 15^\circ } \right)\hat i - \left( {{F_2}\cos 15^\circ } \right)\hat j\\ {F_2} = - \left[ {\left( {80\;{\rm{N}}} \right) \times \sin 15^\circ } \right]\hat i - \left[ {\left( {80\;{\rm{N}}} \right) \times \cos 15^\circ } \right]\hat j\\ {F_2} = \left( { - 20.71\hat i - 77.27\hat j} \right)\;{\rm{N}} \end{array}\]

Similarly, we can write the force ${F_3}$ in Cartesian form as:

\[\begin{array}{c} {F_3} = {F_{3x}}\hat i - {F_{3y}}\hat j\\ {F_3} = \left( {30\hat i + 0\hat j} \right)\;{\rm{N}} \end{array}\]
 
Step 4

To find the resultant vector, add all three forces ${F_1},\;{F_2}$ and ${F_3}$, we get:

\[\begin{array}{c} {F_R} = {F_1} + {F_2} + {F_3}\\ {F_R} = \left( {30\hat i + 40\hat j} \right)\;{\rm{N}} + \left( { - 20.71\hat i - 77.27\hat j} \right)\;{\rm{N}} + \left( {30\hat i + 0\hat j} \right)\;{\rm{N}}\\ {F_R} = \left( {30\hat i - 20.711\hat i + 30\hat i} \right)\;{\rm{N}} + \left( {40\hat j - 77.27\hat j + 0\hat j} \right)\;{\rm{N}}\\ {F_R} = \left( {39.29\hat i} \right)\;{\rm{N}} - \left( {37.27\hat j} \right)\;{\rm{N}} \end{array}\]

To find the magnitude of the resultant force, we have:

\[\begin{array}{c} {F_R} = \sqrt {{{\left( {39.29\;{\rm{N}}} \right)}^2} + {{\left( { - 37.27\,{\rm{N}}} \right)}^2}} \\ {F_R} = \sqrt {2932.757} \;{\rm{N}}\\ {F_R} = 54.2\;{\rm{N}} \end{array}\]

To find the direction measured clockwise from the positive x-axis, take the ratio of the resultant force in y direction to the x direction then, we get:

\[\begin{array}{c} \tan \phi = \frac{{37.27\;{\rm{N}}}}{{39.29\;{\rm{N}}}}\\ \phi = {\tan ^{ - 1}}\left( {0.99} \right)\\ \phi = 44.71^\circ \end{array}\]