Step 1

We are given the different forces acting over the each component.

We are asked the resultant force and its coordinates x and y acts on the x-y plane.

 
Step 2

(a)

The following is the diagram of the system:

We have the force 1 acting over the component is ${F_1} = 200\;{\rm{N}}$.

We have the distance of force 1 from y-axis is ${d_1}' = 1\;{\rm{m}}$.

We have the force 2 acting over the component is ${F_2} = 100\;{\rm{N}}$.

We have the distance of force 2 from x-axis is ${d_2} = 2\;{\rm{m}}$.

We have the distance of force 2 from y-axis is ${d_2}' = 2\;{\rm{m}}$.

We have the force 3 acting over the component is ${F_3} = 200\;{\rm{N}}$.

We have the distance of force 3 from x-axis is ${d_3} = 2\;{\rm{m}}$.

 
Step 3

The magnitude of the resultant force acting in plane along z-axis is,

\[\begin{array}{l} F = \sum {{F_z}} \\ F = {F_1} + {F_2} + {F_3} \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{l} F = 200\;{\rm{N}} + {\rm{10}}0\;{\rm{N}} + 200\;{\rm{N}}\\ F = 500\;{\rm{N}} \end{array}\]
 
Step 5

The resultant moment couple about the x-axis is,

\[\begin{array}{c} \sum {M_x} = 0\\ Fy = {F_2}{d_2} + {F_3}{d_3} \end{array}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} \left( {500\;{\rm{N}}} \right)y = \left( {100\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right) + \left( {200\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right)\\ y = 1.2\;{\rm{m}} \end{array}\]
 
Step 7

The resultant moment couple about the y-axis is,

\[\begin{array}{c} \sum {M_y} = 0\\ Fx = {F_1}{d_1}' + {F_2}{d_2}' \end{array}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} \left( {500\;{\rm{N}}} \right)x = \left( {200\;{\rm{N}}} \right)\left( {1\;{\rm{m}}} \right) + \left( {100\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right)\\ x = 0.8\;{\rm{m}} \end{array}\]
 
Step 9

(b)

The following is the diagram of the system:

We have the force 1 acting over the component is ${F_1} = 100\;{\rm{N}}$.

We have the distance of force 1 from y-axis is ${d_1}' = 2\;{\rm{m}}$.

We have the distance of force 1 from x-axis is ${d_1} = - 1\;{\rm{m}}$.

We have the force 2 acting over the component is ${F_2} = - 100\;{\rm{N}}$.

We have the distance of force 2 from x-axis is ${d_2} = 2\;{\rm{m}}$.

We have the distance of force 2 from y-axis is ${d_2}' = 2\;{\rm{m}}$.

We have the force 3 acting over the component is ${F_3} = 200\;{\rm{N}}$.

We have the distance of force 3 from x-axis is ${d_3} = 2\;{\rm{m}}$.

 
Step 10

The magnitude of the resultant force acting in plane along z-axis is,

\[\begin{array}{l} F = \sum {{F_z}} \\ F = {F_1} + {F_2} + {F_3} \end{array}\]
 
Step 11

Substitute the values in the above expression.

\[\begin{array}{l} F = 100\;{\rm{N}} - {\rm{10}}0\;{\rm{N}} + 200\;{\rm{N}}\\ F = 200\;{\rm{N}} \end{array}\]
 
Step 12

The resultant moment couple about the x-axis is,

\[\begin{array}{c} \sum {M_x} = 0\\ Fy = {F_1}{d_1} + {F_2}{d_2} + {F_3}{d_3} \end{array}\]
 
Step 13

Substitute the values in the above expression.

\[\begin{array}{c} \left( {200\;{\rm{N}}} \right)y = \left( {100\;{\rm{N}}} \right)\left( { - 1\;{\rm{m}}} \right) + \left( { - 100\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right) + \left( {200\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right)\\ y = 0.5\;{\rm{m}} \end{array}\]
 
Step 14

The resultant moment couple about the y-axis is,

\[\begin{array}{c} \sum {M_y} = 0\\ Fx = {F_1}{d_1}' + {F_2}{d_2}' \end{array}\]
 
Step 15

Substitute the values in the above expression.

\[\begin{array}{c} \left( {200\;{\rm{N}}} \right)x = \left( {100\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right) + \left( { - 100\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right)\\ x = 0\;{\rm{m}} \end{array}\]
 
Step 16

(c)

The following is the diagram of the system:

We have the force 1 acting over the component is ${F_1} = 400\;{\rm{N}}$.

We have the distance of force 1 from y-axis is ${d_1}' = 2\;{\rm{m}}$.

We have the force 2 acting over the component is ${F_2} = 200\;{\rm{N}}$.

We have the distance of force 2 from y-axis is ${d_2}' = - 2\;{\rm{m}}$.

We have the force 3 acting over the component is ${F_3} = 300\;{\rm{N}}$.

We have the distance of force 3 from x-axis is ${d_3} = 4\;{\rm{m}}$.

We have the distance of force 3 from y-axis is ${d_3}' = 2\;{\rm{m}}$.

We have the force 4 acting over the component is ${F_4} = 100\;{\rm{N}}$.

We have the distance of force 4 from x-axis is ${d_4} = 4\;{\rm{m}}$.

We have the distance of force 4 from y-axis is ${d_4}' = - 2\;{\rm{m}}$.

 
Step 17

The magnitude of the resultant force acting in plane along z-axis is,

\[\begin{array}{l} F = \sum {{F_z}} \\ F = {F_1} + {F_2} + {F_3} + {F_4} \end{array}\]
 
Step 18

Substitute the values in the above expression.

\[\begin{array}{l} F = 400\;{\rm{N}} + 200\;{\rm{N}} + 300\;{\rm{N}} + 100\;{\rm{N}}\\ F = 1000\;{\rm{N}} \end{array}\]
 
Step 19

The resultant moment couple about the x-axis is,

\[\begin{array}{c} \sum {M_x} = 0\\ Fy = {F_3}{d_3} + {F_4}{d_4} \end{array}\]
 
Step 20

Substitute the values in the above expression.

\[\begin{array}{c} \left( {1000\;{\rm{N}}} \right)y = \left( {300\;{\rm{N}}} \right)\left( {4\;{\rm{m}}} \right) + \left( {100\;{\rm{N}}} \right)\left( {4\;{\rm{m}}} \right)\\ y = 1.6\;{\rm{m}} \end{array}\]
 
Step 21

The resultant moment couple about the y-axis is,

\[\begin{array}{c} \sum {M_y} = 0\\ Fx = {F_1}{d_1}' + {F_2}{d_2}' + {F_3}{d_3}' + {F_4}{d_4}' \end{array}\]
 
Step 22

Substitute the values in the above expression.

\[\begin{array}{c} \left( {1000\;{\rm{N}}} \right)x = \left( {400\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right) + \left( {200\;{\rm{N}}} \right)\left( { - 2\;{\rm{m}}} \right) + \left( {300\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right) + \left( {100\;{\rm{N}}} \right)\left( { - 2\;{\rm{m}}} \right)\\ x = 0.8\;{\rm{m}} \end{array}\]