A uniform rectangular bookcase of height \(H\) and width \(W=H / 2\) is to be pushed at a constant velocity across a level floor. The bookcase is pushed horizontally at its top edge, at the distance \(H\) above the floor. What is the maximum value the coefficient of kinetic friction between the bookcase and the floor can have if the bookcase is not to tip over while being pushed?

The forces acting on the bookcase are the gravitational force (weight) acting on the center of mass, the normal force acting on the bottom edge, the friction force between the bookcase and the floor, and the applied horizontal force pushing the bookcase at the top edge.

Choose a coordinate system with its origin at the bottom right edge of the bookcase (i.e., the point where the bookcase would tip over if the force is too strong). Positive torque would be counterclockwise and negative torque would be clockwise.

Let's denote the applied force as \(F_A\). The torque \(\tau_A\) due to the applied force is given by $$\tau_A = F_A \times H,$$ where \(H\) is the distance between the tipping point and the point where the force is applied.

Let's denote the weight of the bookcase as \(W\), the width of the bookcase as \(W_\text{bookcase}\), and the height of the bookcase as \(H_\text{bookcase}\). Since the weight acts on the center of mass, which is at \((W_\text{bookcase}/2, H_\text{bookcase}/2)\), the torque \(\tau_W\) is given by $$\tau_W = - W \times \frac{W_\text{bookcase}}{2},$$ where the negative sign indicates that the torque is in the clockwise direction.

Since the bookcase is pushed horizontally, there is no vertical acceleration, so the normal force \(N\) is equal to the weight \(W\). The friction force \(F_f\) is given by $$F_f = \mu N = \mu W,$$ where \(\mu\) is the coefficient of kinetic friction.

To find the threshold at which the bookcase will start to tip, we can set the net torque on the bookcase to zero, so the torque due to the applied force is equal to the torque due to the friction force: $$\tau_A = \tau_W + \tau_f \Rightarrow F_A \times H = -W \times \frac{W_\text{bookcase}}{2} + F_f \times H_\text{bookcase}.$$ Since \(F_f = \mu W\) and \(W_\text{bookcase} = H_\text{bookcase} / 2\), we can rewrite the equation as $$F_A \times H = -W \times \frac{H}{2} + \mu W \times H_\text{bookcase}.$$ To solve for the maximum coefficient of kinetic friction, we can assume that the bookcase is just about to tip, so the horizontal force is very small and can be set to zero: $$0 = -W \times \frac{H}{2} + \mu W \times H_\text{bookcase}.$$

Now we can solve for the maximum \(\mu\): $$\mu_\text{max} = \frac{H}{2 H_\text{bookcase}} = \frac{H}{2 H} = \frac{1}{2}.$$ Thus, the maximum value the coefficient of kinetic friction can have, to ensure that the bookcase does not tip over while being pushed, is 0.5.