A particle is being acted upon by four forces of `30 N` due east , `20 N` due north , `50 N` due west and `40 N` due south. The resultant force will be

Options

A. `20 sqrt(2) N , 60^(@)` south to west
B. `20 sqrt(2) N , 45^(@)` south west
C. `20 sqrt(2) N , 45^(@)` north to east
D. `20 sqrt(2) N , 45^(@)` south to east

Answer

To solve the problem of finding the resultant force acting on a particle due to four forces, we can follow these steps:

### Step 1: Identify the Forces
We have four forces acting on the particle:
1. \( F_1 = 30 \, \text{N} \) due East
2. \( F_2 = 20 \, \text{N} \) due North
3. \( F_3 = 50 \, \text{N} \) due West
4. \( F_4 = 40 \, \text{N} \) due South

### Step 2: Assign Unit Vectors
We can represent the forces using unit vectors:
- East direction is represented by the unit vector \( \hat{i} \)
- North direction is represented by the unit vector \( \hat{j} \)
- West direction is represented by \( -\hat{i} \)
- South direction is represented by \( -\hat{j} \)

Thus, we can write the forces as:
- \( F_1 = 30 \hat{i} \)
- \( F_2 = 20 \hat{j} \)
- \( F_3 = -50 \hat{i} \)
- \( F_4 = -40 \hat{j} \)

### Step 3: Calculate the Resultant Force
The resultant force \( \vec{F}_{\text{R}} \) can be found by summing all the forces:
\[ \vec{F}_{\text{R}} = F_1 + F_2 + F_3 + F_4 \]
Substituting the values:
\[ \vec{F}_{\text{R}} = 30 \hat{i} + 20 \hat{j} - 50 \hat{i} - 40 \hat{j} \]

### Step 4: Combine Like Terms
Now, we combine the \( \hat{i} \) and \( \hat{j} \) components:
\[ \vec{F}_{\text{R}} = (30 - 50) \hat{i} + (20 - 40) \hat{j} \]
\[ \vec{F}_{\text{R}} = -20 \hat{i} - 20 \hat{j} \]

### Step 5: Find the Magnitude of the Resultant Force
The magnitude of the resultant force can be calculated using the Pythagorean theorem:
\[ |\vec{F}_{\text{R}}| = \sqrt{(-20)^2 + (-20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, \text{N} \]

### Step 6: Determine the Direction of the Resultant Force
To find the direction, we can use the tangent function:
\[ \tan(\theta) = \frac{F_y}{F_x} = \frac{-20}{-20} = 1 \]
This gives us:
\[ \theta = 45^\circ \]
Since both components are negative, the direction is \( 45^\circ \) south of west.

### Final Result
The resultant force acting on the particle is:
\[ \vec{F}_{\text{R}} = 20\sqrt{2} \, \text{N} \text{ at } 45^\circ \text{ south of west.} \]

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