A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. The minimum value of F which the cube begins to tip about the edge is ....(Assume that the cube does not slide).

Explanation

To solve the problem, we need to determine the minimum force \( F \) required to tip a uniform cube of side \( a \) and mass \( m \) about one of its edges when a horizontal force is applied at a height of \( \frac{3a}{4} \) above the base.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Cube:**
- The weight of the cube \( mg \) acts downwards at its center of mass, which is located at a height of \( \frac{a}{2} \) from the base.
- A horizontal force \( F \) is applied at a height of \( \frac{3a}{4} \) above the base.
- There is also a normal force \( N \) acting upwards at the base of the cube.

2. **Determine the Point of Rotation:**
- The cube will begin to tip about the edge that is in contact with the table. We will consider the point of rotation to be at the edge of the cube.

3. **Set Up the Torque Equation:**
- The torque due to the force \( F \) about the point of rotation (the edge) is given by:
\[ \tau_F = F \cdot \left(\frac{3a}{4}\right) \]
- The torque due to the weight \( mg \) about the same point is given by:
\[ \tau_{mg} = mg \cdot \left(\frac{a}{2}\right) \]

4. **Condition for Tipping:**
- The cube will start to tip when the torque due to the applied force \( F \) is equal to the torque due to the weight of the cube:
\[ F \cdot \left(\frac{3a}{4}\right) = mg \cdot \left(\frac{a}{2}\right) \]

5. **Solve for \( F \):**
- Rearranging the equation gives:
\[ F = \frac{mg \cdot \left(\frac{a}{2}\right)}{\left(\frac{3a}{4}\right)} \]
- Simplifying this expression:
\[ F = \frac{mg \cdot 2}{3} = \frac{2mg}{3} \]

6. **Conclusion:**
- The minimum value of the force \( F \) required to tip the cube is:
\[ F = \frac{2mg}{3} \]