Three known forces `overset(rarr)Fm,overset(rarr)F` and `2 overset(rarr)F` acting on a particle along x-axis, `120^(@)` with + ve x-axis and `240^(@)` with the + ve x-axis, respectively. The particle is moving along `60^(@)` with +ve x-axis. Choose correct options.

Options

A. There must be a fourth force acting on the particle
B. There may be a fourth force acting on the particle
C. There must be only given three forces acting on the particle
D. There may be only three given forces acting on the particle

Explanation

To solve the problem, we need to analyze the forces acting on the particle and their resultant. Let's break it down step by step.

### Step 1: Identify the Forces
We have three forces acting on a particle:
1. Force \( \vec{F_1} \) at an angle of \( 120^\circ \) with the positive x-axis.
2. Force \( \vec{F_2} \) at an angle of \( 240^\circ \) with the positive x-axis.
3. Force \( \vec{F_3} = 2\vec{F} \) which is also acting along the x-axis.

### Step 2: Resolve the Forces into Components
We will resolve each force into its x and y components.

- For \( \vec{F_1} \):
\[ F_{1x} = F_1 \cos(120^\circ) = F_1 \left(-\frac{1}{2}\right) = -\frac{F_1}{2} \]
\[ F_{1y} = F_1 \sin(120^\circ) = F_1 \left(\frac{\sqrt{3}}{2}\right) = \frac{F_1 \sqrt{3}}{2} \]

- For \( \vec{F_2} \):
\[ F_{2x} = F_2 \cos(240^\circ) = F_2 \left(-\frac{1}{2}\right) = -\frac{F_2}{2} \]
\[ F_{2y} = F_2 \sin(240^\circ) = F_2 \left(-\frac{\sqrt{3}}{2}\right) = -\frac{F_2 \sqrt{3}}{2} \]

- For \( \vec{F_3} = 2\vec{F} \):
\[ F_{3x} = 2F \quad (acting along the x-axis) \]
\[ F_{3y} = 0 \]

### Step 3: Calculate the Resultant Force Components
Now, we can find the resultant force components \( R_x \) and \( R_y \):

\[ R_x = F_{1x} + F_{2x} + F_{3x} = -\frac{F_1}{2} - \frac{F_2}{2} + 2F \]
\[ R_y = F_{1y} + F_{2y} + F_{3y} = \frac{F_1 \sqrt{3}}{2} - \frac{F_2 \sqrt{3}}{2} + 0 \]

### Step 4: Analyze the Motion of the Particle
The particle is moving at an angle of \( 60^\circ \) with the positive x-axis. This implies that the direction of the resultant force \( R \) must also be at \( 60^\circ \).

The tangent of the angle gives us the relationship between the y and x components of the resultant force:
\[ \tan(60^\circ) = \sqrt{3} = \frac{R_y}{R_x} \]

### Step 5: Set Up the Equation
Substituting the expressions for \( R_x \) and \( R_y \):
\[ \sqrt{3} = \frac{\frac{F_1 \sqrt{3}}{2} - \frac{F_2 \sqrt{3}}{2}}{-\frac{F_1}{2} - \frac{F_2}{2} + 2F} \]

### Step 6: Analyze the Options
1. **There must be a fourth force**: This is incorrect because the motion can be satisfied by the three forces.
2. **There may be a fourth force**: This is correct because a fourth force could exist that aligns with the resultant.
3. **There must be only three forces**: This is incorrect as we have established that a fourth force could exist.
4. **There may be three forces**: This is correct since the motion can be sustained with just the three forces.

### Conclusion
The correct options are:
- **Option B**: There may be a fourth force.
- **Option D**: There may be three forces.