A heavy box is to be dragged along a rough horizontal floor. To do so, person A pushes it at an angle `30^(@)` from the horizontal and requires a minimum force `F_(A)`, while person B pulls the box at angle `60^(@)` from the horizontal and needs minimum force `F_(B)`. If the coefficient of friction between the box and the floor is `sqrt(3)/(5)` , the ratio is `(F_(A))/(F_(B))`

Options

A. `sqrt(3)`
B. `(5)/sqrt(3)`
C. `sqrt(3)/(2)`
D. `(2)/sqrt(3)`

Explanation

To find the ratio \( \frac{F_A}{F_B} \) where person A pushes the box at an angle of \( 30^\circ \) and person B pulls it at an angle of \( 60^\circ \), we will analyze the forces acting on the box in both scenarios.

### Step 1: Identify the Forces
1. **For Person A (Pushing)**:
- The applied force is \( F_A \) at an angle of \( 30^\circ \).
- The horizontal component of the force is \( F_A \cos(30^\circ) \).
- The vertical component of the force is \( F_A \sin(30^\circ) \).

2. **For Person B (Pulling)**:
- The applied force is \( F_B \) at an angle of \( 60^\circ \).
- The horizontal component of the force is \( F_B \cos(60^\circ) \).
- The vertical component of the force is \( F_B \sin(60^\circ) \).

### Step 2: Write the Equations for Normal Force
- **For Person A**:
- The normal force \( N_A \) can be expressed as:
\[ N_A = mg + F_A \sin(30^\circ) = mg + \frac{F_A}{2} \]

- **For Person B**:
- The normal force \( N_B \) can be expressed as:
\[ N_B = mg - F_B \sin(60^\circ) = mg - \frac{\sqrt{3}}{2} F_B \]

### Step 3: Write the Frictional Force Equations
- The frictional force \( f \) is given by:
\[ f = \mu N \]
where \( \mu = \frac{\sqrt{3}}{5} \).

- **For Person A**:
\[ f_A = \mu N_A = \frac{\sqrt{3}}{5} \left( mg + \frac{F_A}{2} \right) \]

- **For Person B**:
\[ f_B = \mu N_B = \frac{\sqrt{3}}{5} \left( mg - \frac{\sqrt{3}}{2} F_B \right) \]

### Step 4: Set Up the Equations for Motion
- Since the box is moving at constant velocity, the applied horizontal forces must equal the frictional forces.

1. **For Person A**:
\[ F_A \cos(30^\circ) = f_A \]
Substituting for \( f_A \):
\[ F_A \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{5} \left( mg + \frac{F_A}{2} \right) \]

2. **For Person B**:
\[ F_B \cos(60^\circ) = f_B \]
Substituting for \( f_B \):
\[ F_B \cdot \frac{1}{2} = \frac{\sqrt{3}}{5} \left( mg - \frac{\sqrt{3}}{2} F_B \right) \]

### Step 5: Solve the Equations
1. **For Person A**:
Rearranging gives:
\[ F_A \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{5} mg + \frac{\sqrt{3}}{10} F_A \]
Collecting \( F_A \):
\[ F_A \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{10} \right) = \frac{\sqrt{3}}{5} mg \]
Simplifying:
\[ F_A \cdot \frac{4\sqrt{3}}{10} = \frac{\sqrt{3}}{5} mg \implies F_A = \frac{5mg}{4} \]

2. **For Person B**:
Rearranging gives:
\[ F_B \cdot \frac{1}{2} = \frac{\sqrt{3}}{5} mg - \frac{3}{10} F_B \]
Collecting \( F_B \):
\[ F_B \left( \frac{1}{2} + \frac{3}{10} \right) = \frac{\sqrt{3}}{5} mg \]
Simplifying:
\[ F_B \cdot \frac{8}{10} = \frac{\sqrt{3}}{5} mg \implies F_B = \frac{5mg}{8} \]

### Step 6: Find the Ratio \( \frac{F_A}{F_B} \)
\[ \frac{F_A}{F_B} = \frac{\frac{5mg}{4}}{\frac{5mg}{8}} = \frac{8}{4} = 2 \]

### Final Result
The ratio \( \frac{F_A}{F_B} = 2 \).