Answer
150 N. 0.83 m
Work Step by Step
The normal force $F_N$ just balances the weight of the box: $F_N=250\;N$.
The frictional force, just when the box is about to slide, equals the coefficient of static friction multiplied by the normal force: $F_{fr}=\mu_s F_N=(0.60)(250\;N)=150N$
The minimum pushing force F just balances the frictional force in the horizontal direction: F=150 N.
Assume that the box is just about to tip and to slide, so that all of the force on the box due to contact with the ground is at the bottom right, as shown in the book’s figure. Calculate torques about that corner. At the critical moment, the net torque equals zero. Let clockwise torques be positive.
$$\Sigma \tau =F(h)-mg(0.5\;m)= (150\;N)(h)-(250N)(0.5\;m)=0$$
$$h = 0.83 \;m$$
