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. `25 kgf`
B. `2.5 kgf`
C. `12.5 kgf`
D. `6.25 kgf`

Explanation

To find the least force required to drag a heavy body of mass 25 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 body experiences:
- Weight \( W = mg \) acting downwards.
- Normal force \( N \) acting upwards.
- Applied force \( F \) at an angle \( \alpha \) to the horizontal.
- Frictional force \( f \) acting opposite to the direction of motion.

### Step 2: Calculate the Weight of the Body
Given:
- Mass \( m = 25 \, \text{kg} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

The weight \( W \) is calculated as:
\[ W = mg = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 245 \, \text{N} \]

### Step 3: Set Up the Equations
The normal force \( N \) can be expressed as:
\[ N = W - F \sin \alpha \]
The frictional force \( f \) is given by:
\[ f = \mu N = \mu (W - F \sin \alpha) \]

### Step 4: Apply Newton's Second Law
In the horizontal direction, the applied force \( F \cos \alpha \) must overcome the frictional force:
\[ F \cos \alpha = f \]
Substituting for \( f \):
\[ F \cos \alpha = \mu (W - F \sin \alpha) \]

### Step 5: Rearrange the Equation
Rearranging gives:
\[ F \cos \alpha + \mu F \sin \alpha = \mu W \]
Factoring out \( F \):
\[ F (\cos \alpha + \mu \sin \alpha) = \mu W \]
Thus,
\[ F = \frac{\mu W}{\cos \alpha + \mu \sin \alpha} \]

### Step 6: Maximize the Denominator
To minimize \( F \), we need to maximize the denominator \( \cos \alpha + \mu \sin \alpha \). We can use calculus for this:
1. Differentiate \( \cos \alpha + \mu \sin \alpha \) with respect to \( \alpha \).
2. Set the derivative to zero to find the angle \( \alpha \) that maximizes the expression.

### Step 7: Differentiate and Solve
The derivative is:
\[ -\sin \alpha + \mu \cos \alpha = 0 \]
This leads to:
\[ \mu \cos \alpha = \sin \alpha \quad \Rightarrow \quad \tan \alpha = \mu \]
Given \( \mu = \frac{1}{\sqrt{3}} \):
\[ \tan \alpha = \frac{1}{\sqrt{3}} \quad \Rightarrow \quad \alpha = 30^\circ \]

### Step 8: Substitute Back to Find Minimum Force
Substituting \( \alpha = 30^\circ \) into the equation for \( F \):
\[ F = \frac{\frac{1}{\sqrt{3}} \times 245}{\cos 30^\circ + \frac{1}{\sqrt{3}} \sin 30^\circ} \]
Calculating \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) and \( \sin 30^\circ = \frac{1}{2} \):
\[ F = \frac{\frac{245}{\sqrt{3}}}{\frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}} \]
Simplifying the denominator:
\[ \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}} = \frac{3 + 1}{2\sqrt{3}} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}} \]
Thus,
\[ F = \frac{\frac{245}{\sqrt{3}}}{\frac{2}{\sqrt{3}}} = \frac{245}{2} = 122.5 \, \text{N} \]

### Final Step: Convert to kgf
To convert to kgf:
\[ F = \frac{122.5}{9.8} \approx 12.5 \, \text{kgf} \]

### Final Answer
The least force required to drag the body is approximately **12.5 kgf**.