A uniform cube of side and mass m rests on a rough horizontal surface. A horizontal force F is applied normal to one face at point that is directly above the centre of the face at a height `(a)/(4)` above the centre. The minimum value of F for which the cube begins to topple above an edge without sliding is

Options

A. `(1)/(4)mg`
B. 2mg
C. `(1)/(2)mg`
D. `(2)/(3)mg`

Answer

To solve the problem of finding the minimum value of the horizontal force \( F \) required for the cube to begin to topple without sliding, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Setup**:
- We have a uniform cube with side length \( A \) and mass \( m \) resting on a rough horizontal surface.
- A horizontal force \( F \) is applied at a height of \( \frac{A}{4} \) above the center of one face of the cube.

2. **Identify Key Points**:
- The center of the cube is at height \( \frac{A}{2} \) from the base.
- The point where the force \( F \) is applied is at height \( \frac{A}{4} \) above the center, which means it is at height \( \frac{3A}{4} \) from the base.

3. **Determine the Point of Rotation**:
- The cube will begin to topple about the edge that is in contact with the ground. Let's denote this edge as point \( O \).

4. **Set Up the Moment Equation**:
- For the cube to begin to topple, the moments about point \( O \) must balance. The moment due to the applied force \( F \) must equal the moment due to the weight of the cube \( mg \).
- The distance from point \( O \) to the line of action of force \( F \) is \( \frac{A}{2} \) (the height of the cube) minus \( \frac{A}{4} \) (the height where the force is applied), which gives us \( \frac{A}{4} \).
- The moment due to the force \( F \) is \( F \times \frac{A}{4} \).
- The moment due to the weight \( mg \) is \( mg \times \frac{A}{2} \).

5. **Write the Moment Balance Equation**:
\[ F \cdot \frac{A}{4} = mg \cdot \frac{A}{2} \]

6. **Solve for \( F \)**:
- Rearranging the equation gives:
\[ F = \frac{mg \cdot \frac{A}{2}}{\frac{A}{4}} = mg \cdot \frac{2}{1} = 2mg \]
- However, we need to consider the height at which the force is applied. The correct distance for the force \( F \) is actually \( \frac{3A}{4} \) from point \( O \).
- Thus, the moment equation should be:
\[ F \cdot \frac{3A}{4} = mg \cdot \frac{A}{2} \]
- Solving this gives:
\[ F = \frac{mg \cdot \frac{A}{2}}{\frac{3A}{4}} = mg \cdot \frac{2}{3} = \frac{2}{3} mg \]

7. **Final Result**:
- The minimum value of the force \( F \) for which the cube begins to topple above an edge without sliding is:
\[ F = \frac{2}{3} mg \]