1. **Problem statement:** A non-homogeneous bar AB with weight 230 N slides freely on a frictionless cylindrical surface. Given dimensions are diameter $D=72$ cm, $x=36$ cm, and $y=27$ cm. We need to find the angle $\beta$, the reaction force at point A, and the reaction force at point B. 2. **Setup and assumptions:** - The bar AB rests on a cylindrical surface of diameter $D$. - The centroid CG is given, and the bar is free to slide without friction. - The weight $W=230$ N acts at the centroid CG. - Reactions at A and B are normal forces from the surface. 3. **Geometry and angle $\beta$:** - The bar contacts the cylinder at points A and B. - The horizontal distance $x=36$ cm and vertical distance $y=27$ cm relate to the geometry of the bar and cylinder. - The angle $\beta$ is the angle between the bar and the horizontal. 4. **Calculate angle $\beta$:** Using the right triangle formed by $x$ and $y$: $$\tan(\beta) = \frac{y}{x} = \frac{27}{36} = 0.75$$ $$\beta = \arctan(0.75) \approx 36.87^\circ$$ 5. **Equilibrium equations:** - Sum of vertical forces: $$R_A \sin(\beta) + R_B \sin(\beta) = W$$ - Sum of horizontal forces: $$R_A \cos(\beta) = R_B \cos(\beta)$$ (since the bar is in equilibrium and frictionless, horizontal components balance) - Sum of moments about A: Let $L$ be the length of the bar, and $d$ the distance from A to CG along the bar. 6. **From horizontal force balance:** $$R_A \cos(\beta) = R_B \cos(\beta) \implies R_A = R_B$$ 7. **From vertical force balance:** $$R_A \sin(\beta) + R_B \sin(\beta) = W \implies 2 R_A \sin(\beta) = W \implies R_A = \frac{W}{2 \sin(\beta)}$$ 8. **Calculate $R_A$ and $R_B$:** $$R_A = R_B = \frac{230}{2 \times \sin(36.87^\circ)} = \frac{230}{2 \times 0.6} = \frac{230}{1.2} \approx 191.67 \text{ N}$$ **Final answers:** - Angle $\beta \approx 36.87^\circ$ - Reaction at A: $R_A \approx 191.67$ N - Reaction at B: $R_B \approx 191.67$ N