1. **Problem statement:** A uniform cube of side $a$ and mass $m$ rests on a rough horizontal surface. A horizontal force $F$ is applied at a height $\frac{a}{4}$ above the center of one face. We need to find the minimum value of $F$ for which the cube begins to topple about an edge without sliding. 2. **Key concepts:** - Torque and rotational equilibrium - Friction and no sliding condition - Forces acting on the cube: weight $mg$ acting at the center of gravity (center of cube), applied force $F$ at height $\frac{a}{4}$ above the center 3. **Setup:** - The cube will topple about the bottom edge of the face where force is applied. - The torque due to $F$ tends to rotate the cube about this edge. - The torque due to weight $mg$ resists this rotation. 4. **Calculate torques about the pivot edge:** - Distance from pivot to line of action of $F$ is $\frac{a}{2} + \frac{a}{4} = \frac{3a}{4}$ (since center is $\frac{a}{2}$ from edge, and force is $\frac{a}{4}$ above center). - Torque by $F$ is $F \times \frac{3a}{4}$ (counterclockwise). - Weight $mg$ acts at the center, which is $\frac{a}{2}$ from pivot horizontally. - Torque by weight is $mg \times \frac{a}{2}$ (clockwise). 5. **Condition for toppling without sliding:** - At the threshold, torques balance: $$F \times \frac{3a}{4} = mg \times \frac{a}{2}$$ 6. **Solve for $F$:** $$F = \frac{mg \times \frac{a}{2}}{\frac{3a}{4}} = mg \times \frac{\frac{1}{2}}{\frac{3}{4}} = mg \times \frac{1}{2} \times \frac{4}{3} = \frac{2}{3} mg$$ 7. **Final answer:** The minimum force $F$ to topple the cube without sliding is $$\boxed{\frac{2}{3} mg}$$ This corresponds to option (d).