1. **State the problem:** Calculate the total torque about the base of a 4-meter ladder due to a 600 N painter standing 3 meters from the bottom and the ladder's weight of 200 N. 2. **Identify forces and distances:** - Painter's weight (force): $F_p = 600$ N - Painter's distance from base: $d_p = 3$ m - Ladder's weight (force): $F_l = 200$ N - Ladder's length: $L = 4$ m 3. **Find the torque caused by each force:** Torque $\tau = F \times r \times \sin(\theta)$, where $r$ is the perpendicular distance to the pivot. Assuming the ladder is vertical against the wall, force acts vertically down, and distance is horizontal from the base, so $\theta = 90^\circ$ and $\sin 90^\circ = 1$. 4. **Calculate torque by painter:** $$\tau_p = F_p \times d_p = 600 \times 3 = 1800$$ N·m 5. **Calculate torque by ladder weight:** The ladder’s weight acts at its center (midpoint), i.e., at 2 m from the base. $$\tau_l = F_l \times \frac{L}{2} = 200 \times 2 = 400$$ N·m 6. **Calculate total torque:** $$\tau_{total} = \tau_p + \tau_l = 1800 + 400 = 2200$$ N·m **Final answer:** The total torque about the base of the ladder is $\boxed{2200}$ N·m.