1. **State the problem:** We have a beam with a distributed load of 30 N/m over 40 cm on the left side and a point force of 10 N acting to the left at the left end. The beam length is 80 cm. We want to find the equivalent force system at point A, located at the right end of the beam. 2. **Convert units:** Lengths are given in cm, convert to meters for consistency: - Distributed load length $L = 40\text{ cm} = 0.4\text{ m}$ - Total beam length $= 80\text{ cm} = 0.8\text{ m}$ 3. **Calculate the resultant force of the distributed load:** The distributed load intensity is $w = 30\text{ N/m}$ over $0.4\text{ m}$. $$F_d = w \times L = 30 \times 0.4 = 12\text{ N}$$ 4. **Locate the point of action of the distributed load:** The resultant force of a uniformly distributed load acts at the centroid of the load, which is at the midpoint of the 0.4 m segment from the left end. $$x_d = \frac{0.4}{2} = 0.2\text{ m}$$ 5. **Calculate moments about point A:** Point A is at the right end, 0.8 m from the left end. - Moment due to the 10 N force at the left end (acting left): $$M_{10} = 10 \times (0.8) = 8\text{ Nm}$$ (direction depends on force direction; assuming counterclockwise positive) - Moment due to the distributed load resultant force: Distance from point A to the distributed load resultant: $$d = 0.8 - 0.2 = 0.6\text{ m}$$ Moment: $$M_d = 12 \times 0.6 = 7.2\text{ Nm}$$ 6. **Equivalent force system at point A:** - Sum of forces: $$F_{eq} = 12\text{ N (right)} - 10\text{ N (left)} = 2\text{ N (right)}$$ - Sum of moments about A: $$M_{eq} = M_{10} - M_d = 8 - 7.2 = 0.8\text{ Nm}$$ 7. **Final answer:** The equivalent force system at point A is a single force of 2 N to the right and a moment of 0.8 Nm (counterclockwise).