A heavy box is to be dragged along a rough horizontal floor. To do so, the person A pushes it at an angle 30° from the horizontal and requires a minimum force $$F_A$$, while the person B pulls the box at an angle 60° from the horizontal and needs minimum force $$F_B$$. If the coefficient of friction between the box and the floor is $$\frac{\sqrt{3}}{5}$$, the ratio $$\frac{F_A}{F_B}$$ is:

A heavy box is dragged along a rough horizontal floor. The coefficient of friction is given as $$\mu = \frac{\sqrt{3}}{5}$$. Person A pushes the box at an angle of 30° from the horizontal with a minimum force $$F_A$$, and person B pulls it at an angle of 60° from the horizontal with a minimum force $$F_B$$. We need to find the ratio $$\frac{F_A}{F_B}$$.

First, consider the forces acting on the box. The weight of the box is $$mg$$, where $$m$$ is the mass and $$g$$ is the acceleration due to gravity. The normal force $$N$$ is the force exerted by the floor perpendicular to the surface. The friction force opposes motion and is given by $$f = \mu N$$. The minimum force required to move the box occurs when the horizontal component of the applied force equals the maximum static friction force.

Start with person A, who pushes the box at 30° to the horizontal. When pushing, the vertical component of the force acts downward, increasing the normal force. Resolve $$F_A$$ into components:

• Horizontal component: $$F_A \cos 30^\circ$$
• Vertical component: $$F_A \sin 30^\circ$$ (downward)

The normal force $$N_A$$ is:

$$ N_A = mg + F_A \sin 30^\circ $$

The friction force is:

$$ f_A = \mu N_A = \mu (mg + F_A \sin 30^\circ) $$

At the point of motion, the horizontal component equals the friction force:

$$ F_A \cos 30^\circ = \mu (mg + F_A \sin 30^\circ) $$

Substitute the trigonometric values: $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$ and $$\sin 30^\circ = \frac{1}{2}$$:

$$ F_A \cdot \frac{\sqrt{3}}{2} = \mu \left( mg + F_A \cdot \frac{1}{2} \right) $$

Given $$\mu = \frac{\sqrt{3}}{5}$$, substitute:

$$ F_A \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{5} \left( mg + \frac{F_A}{2} \right) $$

Divide both sides by $$\sqrt{3}$$ (since $$\sqrt{3} \neq 0$$):

$$ \frac{F_A}{2} = \frac{1}{5} \left( mg + \frac{F_A}{2} \right) $$

Multiply both sides by 5 to eliminate the denominator:

$$ 5 \cdot \frac{F_A}{2} = mg + \frac{F_A}{2} $$

$$ \frac{5F_A}{2} = mg + \frac{F_A}{2} $$

Subtract $$\frac{F_A}{2}$$ from both sides:

$$ \frac{5F_A}{2} - \frac{F_A}{2} = mg $$

$$ \frac{4F_A}{2} = mg $$

$$ 2F_A = mg $$

$$ F_A = \frac{mg}{2} $$

Now, consider person B, who pulls the box at 60° to the horizontal. When pulling, the vertical component acts upward, reducing the normal force. Resolve $$F_B$$ into components:

• Horizontal component: $$F_B \cos 60^\circ$$
• Vertical component: $$F_B \sin 60^\circ$$ (upward)

The normal force $$N_B$$ is:

$$ N_B = mg - F_B \sin 60^\circ $$

The friction force is:

$$ f_B = \mu N_B = \mu (mg - F_B \sin 60^\circ) $$

At the point of motion, the horizontal component equals the friction force:

$$ F_B \cos 60^\circ = \mu (mg - F_B \sin 60^\circ) $$

Substitute the trigonometric values: $$\cos 60^\circ = \frac{1}{2}$$ and $$\sin 60^\circ = \frac{\sqrt{3}}{2}$$:

$$ F_B \cdot \frac{1}{2} = \mu \left( mg - F_B \cdot \frac{\sqrt{3}}{2} \right) $$

Given $$\mu = \frac{\sqrt{3}}{5}$$, substitute:

$$ \frac{F_B}{2} = \frac{\sqrt{3}}{5} \left( mg - \frac{\sqrt{3} F_B}{2} \right) $$

Multiply both sides by 10 to eliminate denominators:

$$ 10 \cdot \frac{F_B}{2} = 10 \cdot \frac{\sqrt{3}}{5} \left( mg - \frac{\sqrt{3} F_B}{2} \right) $$

$$ 5F_B = 2\sqrt{3} \left( mg - \frac{\sqrt{3} F_B}{2} \right) $$

Expand the right side:

$$ 5F_B = 2\sqrt{3} \cdot mg - 2\sqrt{3} \cdot \frac{\sqrt{3} F_B}{2} $$

Simplify the second term:

$$ 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} F_B = \sqrt{3} \cdot \sqrt{3} F_B = 3F_B $$

So:

$$ 5F_B = 2\sqrt{3} \, mg - 3F_B $$

Add $$3F_B$$ to both sides:

$$ 5F_B + 3F_B = 2\sqrt{3} \, mg $$

$$ 8F_B = 2\sqrt{3} \, mg $$

Divide both sides by 2:

$$ 4F_B = \sqrt{3} \, mg $$

$$ F_B = \frac{\sqrt{3} \, mg}{4} $$

Now, find the ratio $$\frac{F_A}{F_B}$$:

$$ \frac{F_A}{F_B} = \frac{\frac{mg}{2}}{\frac{\sqrt{3} \, mg}{4}} = \frac{mg}{2} \cdot \frac{4}{\sqrt{3} \, mg} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}} $$

The ratio $$\frac{F_A}{F_B} = \frac{2}{\sqrt{3}}$$, which matches option B.

Hence, the correct answer is Option B.