1. **State the problem:** Calculate the moment $M_A$ about point A caused by a force $F = 12$ lb applied at an angle of $30^\circ$ to the hammer handle. 2. **Formula used:** The moment about a point is given by $$M = F \times d_\perp$$ where $d_\perp$ is the perpendicular distance from the point to the line of action of the force. 3. **Identify distances:** The hammer length is 18 in, and the horizontal offset from the handle to point A is 0.5 in. 4. **Calculate the perpendicular distance:** The force is applied at $30^\circ$ to the handle, so the perpendicular distance from A to the force line is $$d_\perp = 0.5 + 18 \times \sin(30^\circ)$$ 5. **Evaluate $\sin(30^\circ)$:** $$\sin(30^\circ) = 0.5$$ 6. **Calculate $d_\perp$:** $$d_\perp = 0.5 + 18 \times 0.5 = 0.5 + 9 = 9.5 \text{ in}$$ 7. **Calculate the moment $M_A$:** $$M_A = F \times d_\perp = 12 \times 9.5 = 114 \text{ lb} \cdot \text{in}$$ **Final answer:** $$M_A = 114 \text{ lb} \cdot \text{in}$$