Let the side length of the cube be $a$. The weight $W$ acts at the center of mass (height $a/2$), and the force $F$ is applied at the top (height $a$).

For tipping, the box pivots about the bottom edge P. At the tipping point, the sum of torques about P is zero. The normal force $N$ acts at P.

$$\sum \tau_P = W \left(\frac{a}{2}\right) - F(a) = 0$$ $$F = \frac{W}{2}$$

For the box not to slide, the static friction force $f_s$ must be less than or equal to the maximum possible friction, $f_s \leq \mu_s N$. From force equilibrium: Horizontal: $F - f_s = 0 \implies f_s = F$ Vertical: $N - W = 0 \implies N = W$

Substituting these into the no-slip condition at the moment of tipping:

$$F \leq \mu_s N$$ $$\frac{W}{2} \leq \mu_s W$$ $$\mu_s \geq \frac{1}{2}$$

The minimum required coefficient of static friction is $\mu_{s,min} = 0.5$.