A horizontal force `F = (mg)/(3)` is applied on the upper surface of a uniform cube of mass `m` and slide a which is resting on a rough horizontal surface having `mu = (1)/(2)`. The distance between lines of action of `mg` and normal reation is

Options

A. `(a)/(2)`
B. `(a)/(3)`
C. `(a)/(4)`
D. None of these

Explanation

To solve the problem, we need to determine the distance \( x \) between the line of action of the gravitational force \( mg \) and the normal reaction force \( N \) acting on the cube. Let's break down the solution step by step.

### Step 1: Identify the forces acting on the cube
The forces acting on the cube are:
- The gravitational force \( mg \) acting downward at the center of mass of the cube.
- The normal force \( N \) acting upward.
- The applied horizontal force \( F = \frac{mg}{3} \).
- The frictional force \( f \) acting in the opposite direction of the applied force.

### Step 2: Determine the frictional force
Since the cube is resting and not sliding, the frictional force \( f \) must balance the applied force \( F \). Therefore:
\[ f = F = \frac{mg}{3} \]

### Step 3: Set up the moment balance about point C
Let’s consider point C as the center of the cube. The moment due to the normal force \( N \) and the moment due to the applied force \( F \) must balance each other about point C. The distance from point C to the line of action of the normal force is \( x \), and the distance from point C to the line of action of the applied force \( F \) is \( \frac{a}{2} \) (where \( a \) is the side length of the cube).

The moment balance equation can be expressed as:
\[ N \cdot x = F \cdot \frac{a}{2} \]

### Step 4: Substitute the known values
We know that \( F = \frac{mg}{3} \). Therefore, substituting this into the moment balance equation gives:
\[ N \cdot x = \left(\frac{mg}{3}\right) \cdot \frac{a}{2} \]

### Step 5: Determine the normal force
The normal force \( N \) can be determined from the equilibrium of vertical forces. Since the cube is not accelerating vertically:
\[ N = mg \]

### Step 6: Substitute \( N \) into the moment balance equation
Now substituting \( N = mg \) into the moment balance equation:
\[ mg \cdot x = \left(\frac{mg}{3}\right) \cdot \frac{a}{2} \]

### Step 7: Simplify the equation
We can cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ x = \frac{1}{3} \cdot \frac{a}{2} \]
\[ x = \frac{a}{6} \]

### Step 8: Conclusion
Thus, the distance \( x \) between the line of action of \( mg \) and the normal reaction \( N \) is:
\[ \boxed{\frac{a}{6}} \]