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 solve the problem, we need to find the ratio \( \frac{F_A}{F_B} \) where \( F_A \) is the force required by person A to push the box at an angle of \( 30^\circ \) and \( F_B \) is the force required by person B to pull the box at an angle of \( 60^\circ \). The coefficient of friction \( \mu \) is given as \( \frac{\sqrt{3}}{5} \).

### Step 1: Analyze the forces for person A (pushing)

When person A pushes the box at an angle of \( 30^\circ \):
- The horizontal component of the force \( F_A \) is \( F_A \cos(30^\circ) \).
- The vertical component of the force \( F_A \) is \( F_A \sin(30^\circ) \).

The normal force \( N_A \) acting on the box can be expressed as:
\[ N_A = mg + F_A \sin(30^\circ) \]
where \( mg \) is the weight of the box.

The limiting friction \( F_L \) is given by:
\[ F_L = \mu N_A = \frac{\sqrt{3}}{5} (mg + F_A \sin(30^\circ)) \]

For the box to move, the horizontal component of the pushing force must overcome the limiting friction:
\[ F_A \cos(30^\circ) = F_L \]
Substituting for \( F_L \):
\[ F_A \cos(30^\circ) = \frac{\sqrt{3}}{5} (mg + F_A \sin(30^\circ)) \]

### Step 2: Substitute values for angles

Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \sin(30^\circ) = \frac{1}{2} \):
\[ F_A \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{5} \left(mg + F_A \cdot \frac{1}{2}\right) \]

### Step 3: Rearranging the equation

Multiply through by 10 to eliminate the fractions:
\[ 5\sqrt{3} F_A = 2\sqrt{3} (mg + \frac{1}{2} F_A) \]
Distributing the right side:
\[ 5\sqrt{3} F_A = 2\sqrt{3} mg + \sqrt{3} F_A \]
Rearranging gives:
\[ 5\sqrt{3} F_A - \sqrt{3} F_A = 2\sqrt{3} mg \]
\[ 4\sqrt{3} F_A = 2\sqrt{3} mg \]
Dividing both sides by \( 4\sqrt{3} \):
\[ F_A = \frac{mg}{2} \]

### Step 4: Analyze the forces for person B (pulling)

When person B pulls the box at an angle of \( 60^\circ \):
- The horizontal component of the force \( F_B \) is \( F_B \cos(60^\circ) \).
- The vertical component of the force \( F_B \) is \( F_B \sin(60^\circ) \).

The normal force \( N_B \) acting on the box can be expressed as:
\[ N_B = mg - F_B \sin(60^\circ) \]

The limiting friction \( F_L \) is given by:
\[ F_L = \mu N_B = \frac{\sqrt{3}}{5} (mg - F_B \sin(60^\circ)) \]

For the box to move, the horizontal component of the pulling force must overcome the limiting friction:
\[ F_B \cos(60^\circ) = F_L \]
Substituting for \( F_L \):
\[ F_B \cos(60^\circ) = \frac{\sqrt{3}}{5} (mg - F_B \sin(60^\circ)) \]

### Step 5: Substitute values for angles

Using \( \cos(60^\circ) = \frac{1}{2} \) and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ F_B \cdot \frac{1}{2} = \frac{\sqrt{3}}{5} \left(mg - F_B \cdot \frac{\sqrt{3}}{2}\right) \]

### Step 6: Rearranging the equation

Multiply through by 10 to eliminate the fractions:
\[ 5 F_B = 2\sqrt{3} (mg - \frac{\sqrt{3}}{2} F_B) \]
Distributing the right side:
\[ 5 F_B = 2\sqrt{3} mg - 3 F_B \]
Rearranging gives:
\[ 5 F_B + 3 F_B = 2\sqrt{3} mg \]
\[ 8 F_B = 2\sqrt{3} mg \]
Dividing both sides by 8:
\[ F_B = \frac{\sqrt{3}}{4} mg \]

### Step 7: Find the ratio \( \frac{F_A}{F_B} \)

Now we can find the ratio:
\[ \frac{F_A}{F_B} = \frac{\frac{mg}{2}}{\frac{\sqrt{3}}{4} mg} \]
Cancelling \( mg \) from the numerator and denominator:
\[ \frac{F_A}{F_B} = \frac{1/2}{\sqrt{3}/4} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}} \]

### Final Answer
The ratio \( \frac{F_A}{F_B} \) is \( \frac{2}{\sqrt{3}} \).