Step 1

We are given the forces ${F_1} = 900\;{\rm{N,}}\;{F_2} = 750\;{\rm{N}}$ and ${F_3} = 650\;{\rm{N}}$.

We are asked to resolve all the forces into their x and y components and also to express each force in the cartesian vector.

 
Step 2

First, we have to resolve each force acting on the gusset plate into their x and y components as shown in figure below,

From the above figure, the angle with force ${F_3}$ with horizontal axis as follows:

\[\begin{array}{c} \tan \theta = \frac{3}{4}\\ \theta = {\tan ^{ - 1}}\left( {0.75} \right)\\ \theta = 36.87^\circ \end{array}\]
 
Step 3

To find the x and y components of force ${F_1}$, we need to apply the parallelogram law of the forces as follows:

\[\begin{array}{c} {F_{1x}} = {F_1}\cos \left( {0^\circ } \right)\\ {F_{1x}} = 900\;{\rm{N}} \end{array}\]

Similarly,

\[\begin{array}{c} {F_{1y}} = {F_{ 1}}\sin \left( {0^\circ } \right)\\ {F_{1y}} = 0\;{\rm{N}} \end{array}\]

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

\[\begin{array}{c} {F_1} = {F_{1x}}\hat i + {F_{1y}}\hat j\\ {F_1} = \left( {900\hat i + 0\hat j} \right)\;{\rm{N}} \end{array}\]
 
Step 4

To find the x and y components of force ${F_2}$, we need to apply the parallelogram law of the forces as follows:

\[\begin{array}{c} {F_{2x}} = {F_2}\cos \left( {45^\circ } \right)\\ {F_{2x}} = \left( {750\;{\rm{N}}} \right) \times \frac{1}{{\sqrt 2 }}\\ {F_{2x}} = 530.33\;{\rm{N}} \end{array}\]

Similarly,

\[\begin{array}{c} {F_{2y}} = {F_1}\sin \left( {45^\circ } \right)\\ {F_{2y}} = \left( {750\;{\rm{N}}} \right) \times \frac{1}{{\sqrt 2 }}\\ {F_{2y}} = 530.33\;{\rm{N}} \end{array}\]

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

\[\begin{array}{c} {F_2} = {F_{2x}}\hat i + {F_{2y}}\hat j\\ {F_2} = \left( {530.33\hat i + 530.33\hat j} \right)\;{\rm{N}} \end{array}\]
 
Step 5

To find the x and y components of force ${F_3}$, we need to apply the parallelogram law of the forces as follows:

\[\begin{array}{c} {F_{3x}} = {F_3}\cos \theta \\ {F_{3x}} = \left( {650\;{\rm{N}}} \right) \times \cos \left( {36.87^\circ } \right)\\ {F_{3x}} = \left( {650\;{\rm{N}}} \right) \times 0.79\\ {F_{3x}} = 513.5\;{\rm{N}} \end{array}\]

Similarly,

\[\begin{array}{l} {F_{3y}} = - {F_3}\sin \theta \\ {F_{3y}} = - \left( {650\;{\rm{N}}} \right) \times \sin \left( {36.87^\circ } \right)\\ {F_{3y}} = - \left( {650\;{\rm{N}}} \right) \times 0.60\\ {F_{3y}} = - 390\;{\rm{N}} \end{array}\]

In the Cartesian vector form, we can write the force ${F_3}$ as:

\[\begin{array}{c} {F_3} = {F_{3x}}\hat i + {F_{3y}}\hat j\\ {F_3} = \left( {513.5\hat i - 390\hat j} \right)\;{\rm{N}} \end{array}\]