1. **Problem statement:** Find the algebraic measure of the moment of the force $20\sqrt{2}$ newton with respect to point A. 2. **Given:** Force magnitude $F = 20\sqrt{2}$ N, angle $\theta = 45^\circ$ from vertical, vertical segment length $1.2$ m, horizontal segment length $0.2$ m from A to vertical segment. 3. **Formula for moment:** $$ M = F \times d \times \sin(\alpha) $$ where $d$ is the perpendicular distance from point A to the line of action of the force, and $\alpha$ is the angle between the force and the lever arm. 4. **Calculate components:** - Horizontal distance from A to force line: $0.2$ m - Vertical distance: $1.2$ m - The force acts at $45^\circ$ from vertical, so the force components are: $$ F_x = F \sin 45^\circ = 20\sqrt{2} \times \frac{\sqrt{2}}{2} = 20 $$ $$ F_y = F \cos 45^\circ = 20\sqrt{2} \times \frac{\sqrt{2}}{2} = 20 $$ 5. **Calculate moment about A:** Moment due to horizontal component $F_x$ acting at vertical distance $1.2$ m: $$ M_x = F_x \times 1.2 = 20 \times 1.2 = 24 $$ Moment due to vertical component $F_y$ acting at horizontal distance $0.2$ m: $$ M_y = F_y \times 0.2 = 20 \times 0.2 = 4 $$ 6. **Determine sign of moments:** - $F_x$ tends to rotate counterclockwise (positive moment) - $F_y$ tends to rotate clockwise (negative moment) 7. **Algebraic sum of moments:** $$ M = M_x - M_y = 24 - 4 = 20 $$ 8. **Final answer:** The algebraic measure of the moment of the force about point A is $20$ newton.meter. **Answer: a) 20**