A heavy body of mass `25kg` is to be dragged along a horizontal plane (mu = `(1)/(sqrt(3))`. The least force required is `(1 kgf = 9.8 N)`

Options

A. a) `25 kgf`
B. b)`2.5 kgf`
C. c)`12.5 kgf`
D. d)`6.25 kgf`

Explanation

To solve the problem of dragging a heavy body of mass \( 25 \, \text{kg} \) along a horizontal plane with a coefficient of friction \( \mu = \frac{1}{\sqrt{3}} \), we will follow these steps:

### Step 1: Understand the Forces Acting on the Body
The forces acting on the body include:
- The weight of the body \( W = mg \) acting downwards.
- The normal force \( N \) acting upwards.
- The applied force \( F \) making an angle \( \alpha \) with the horizontal.
- The frictional force \( f \) acting in the opposite direction of the motion.

### Step 2: Write the Equations for Forces
1. **Vertical Forces**:
\[ N + F \sin \alpha = mg \]
(Equation 1)

2. **Horizontal Forces**:
\[ f = \mu N = F \cos \alpha \]
(Equation 2)

### Step 3: Substitute for Normal Force
From Equation 2, we can express \( N \) in terms of \( F \):
\[ N = \frac{F \cos \alpha}{\mu} \]

### Step 4: Substitute \( N \) into the Vertical Forces Equation
Substituting \( N \) into Equation 1:
\[ \frac{F \cos \alpha}{\mu} + F \sin \alpha = mg \]

### Step 5: Solve for Force \( F \)
Rearranging gives:
\[ F \left( \frac{\cos \alpha}{\mu} + \sin \alpha \right) = mg \]
Thus,
\[ F = \frac{mg}{\frac{\cos \alpha}{\mu} + \sin \alpha} \]

### Step 6: Find the Minimum Force
To minimize \( F \), we need to maximize the denominator:
\[ \frac{\cos \alpha}{\mu} + \sin \alpha \]
Taking the derivative with respect to \( \alpha \) and setting it to zero:
\[ -\sin \alpha + \mu \cos \alpha = 0 \]
This leads to:
\[ \tan \alpha = \mu \]
Given \( \mu = \frac{1}{\sqrt{3}} \), we find:
\[ \alpha = 30^\circ \]

### Step 7: Substitute \( \alpha \) Back to Find \( F \)
Now substituting \( \alpha = 30^\circ \) into the equation for \( F \):
\[ F = \frac{mg}{\frac{\cos 30^\circ}{\frac{1}{\sqrt{3}}} + \sin 30^\circ} \]
Calculating the trigonometric values:
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 30^\circ = \frac{1}{2} \)

Substituting these values:
\[ F = \frac{25 \times 9.8}{\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{3}}} + \frac{1}{2}} = \frac{25 \times 9.8}{\frac{3}{2} + \frac{1}{2}} = \frac{25 \times 9.8}{2} = \frac{245}{2} = 122.5 \, \text{N} \]

### Step 8: Convert to kgf
Since \( 1 \, \text{kgf} = 9.8 \, \text{N} \):
\[ F = \frac{122.5}{9.8} \approx 12.5 \, \text{kgf} \]

### Conclusion
The least force required to drag the body is \( 12.5 \, \text{kgf} \).