To solve this problem, we need to break down each force into its components along the North-South and East-West axes.

1. For the first force of 20 N inclined at 30º towards North of East, the components are:
- East: 20 cos(30) = 17.32 N
- North: 20 sin(30) = 10 N

2. The second force of 25 N towards North has no East-West component, so:
- East: 0 N
- North: 25 N

3. For the third force of 30 N towards North West, we can assume it makes a 45º angle with both North and West, so the components are:
- West: 30 cos(45) = 21.21 N
- North: 30 sin(45) = 21.21 N

4. For the fourth force of 35 N inclined at 40º towards South of West, the components are:
- West: 35 cos(40) = 26.79 N
- South: 35 sin(40) = 22.52 N

Now, we add up all the components along each axis:

- Total East: 17.32 N
- Total West: 21.21 N + 26.79 N = 48 N
- Total North: 10 N + 25 N + 21.21 N = 56.21 N
- Total South: 22.52 N

Net force along East-West axis = Total East - Total West = 17.32 N - 48 N = -30.68 N (This is a force towards West because it's negative)

Net force along North-South axis = Total North - Total South = 56.21 N - 22.52 N = 33.69 N (This is a force towards North because it's positive)

Now, we can find the magnitude of the resultant force using Pythagoras' theorem:

Resultant force = sqrt((Net force along East-West axis)^2 + (Net force along North-South axis)^2) = sqrt((-30.68 N)^2 + (33.69 N)^2) = 45.67 N

The direction of the resultant force can be found using the arctan function:

Direction = arctan((Net force along North-South axis) / (Net force along East-West axis)) = arctan(33.69 N / 30.68 N) = 47.27º

This angle is measured from the West towards North because the net force along the East-West axis was negative (West) and the net force along the North-South axis was positive (North).

So, the magnitude of the resultant force is 45.67 N and its direction is 47.27º towards North of West.

Question

To solve this problem, we need to break down each force into its components along the North-South and East-West axes.

1. For the first force of 20 N inclined at 30º towards North of East, the components are:
   - East: 20 cos(30) = 17.32 N
   - North: 20 sin(30) = 10 N

2. The second force of 25 N towards North has no East-West component, so:
   - East: 0 N
   - North: 25 N

3. For the third force of 30 N towards North West, we can assume it makes a 45º angle with both North and West, so the components are:
   - West: 30 cos(45) = 21.21 N
   - North: 30 sin(45) = 21.21 N

4. For the fourth force of 35 N inclined at 40º towards South of West, the components are:
   - West: 35 cos(40) = 26.79 N
   - South: 35 sin(40) = 22.52 N

Now, we add up all the components along each axis:

- Total East: 17.32 N
- Total West: 21.21 N + 26.79 N = 48 N
- Total North: 10 N + 25 N + 21.21 N = 56.21 N
- Total South: 22.52 N

Net force along East-West axis = Total East - Total West = 17.32 N - 48 N = -30.68 N (This is a force towards West because it's negative)

Net force along North-South axis = Total North - Total South = 56.21 N - 22.52 N = 33.69 N (This is a force towards North because it's positive)

Now, we can find the magnitude of the resultant force using Pythagoras' theorem:

Resultant force = sqrt((Net force along East-West axis)^2 + (Net force along North-South axis)^2) = sqrt((-30.68 N)^2 + (33.69 N)^2) = 45.67 N

The direction of the resultant force can be found using the arctan function:

Direction = arctan((Net force along North-South axis) / (Net force along East-West axis)) = arctan(33.69 N / 30.68 N) = 47.27º

This angle is measured from the West towards North because the net force along the East-West axis was negative (West) and the net force along the North-South axis was positive (North).

So, the magnitude of the resultant force is 45.67 N and its direction is 47.27º towards North of West.

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