1. **Problem statement:** A rectangular block of mass $m$, width $w$, and height $h$ is hinged at its lower right corner on a tabletop. A force $F$ is applied perpendicular to the left side of the block at a distance $y$ above the bottom. The block is in rotational equilibrium. We need to find the expression for $y$. 2. **Relevant formula:** For rotational equilibrium, the sum of torques about the hinge must be zero: $$\sum \tau = 0$$ Torque $\tau$ is given by: $$\tau = \text{force} \times \text{lever arm}$$ 3. **Identify torques:** - The force $F$ creates a counterclockwise torque about the hinge: $$\tau_F = F \times y$$ - The weight of the block $mg$ acts at the center of mass, which is at the midpoint of the block. The horizontal distance from the hinge to the center of mass is $\frac{w}{2}$, so the weight creates a clockwise torque: $$\tau_{mg} = mg \times \frac{w}{2}$$ 4. **Set torques equal for equilibrium:** $$F y = mg \frac{w}{2}$$ 5. **Solve for $y$:** $$y = \frac{mg \frac{w}{2}}{F} = \frac{w mg}{2F}$$ 6. **Interpretation:** The distance $y$ above the bottom where the force $F$ must be applied to hold the block in equilibrium is: $$\boxed{\frac{w mg}{2F}}$$ This corresponds to option A.