1. **Problem Statement:** Determine the reactions acting on a smooth uniform bar AB of length 4 m and mass 20 kg. The bar is inclined at 30° to the horizontal at point A, resting on a smooth surface at point B which forms a 60° angle with the horizontal. 2. **Known Data:** - Mass of bar, $m = 20$ kg - Length of bar, $L = 4$ m - Angle at A, $\theta_A = 30^\circ$ - Angle at B, $\theta_B = 60^\circ$ 3. **Forces acting on the bar:** - Weight $W = mg = 20 \times 9.8 = 196$ N acting at the midpoint of the bar (2 m from A) - Reaction forces at A and B, denoted as $R_A$ and $R_B$ 4. **Assumptions:** - The bar is in static equilibrium. - The surface at B is smooth, so the reaction force at B is perpendicular to the surface. 5. **Equilibrium conditions:** - Sum of forces in horizontal direction = 0 - Sum of forces in vertical direction = 0 - Sum of moments about any point = 0 6. **Set up coordinate system:** Let horizontal be x-axis and vertical be y-axis. 7. **Resolve reaction at B:** Since surface at B is at 60°, reaction $R_B$ acts perpendicular to it, making 60° with horizontal. 8. **Write equations:** - Horizontal forces: $$ R_{Ax} + R_B \cos 60^\circ = 0 $$ - Vertical forces: $$ R_{Ay} + R_B \sin 60^\circ - 196 = 0 $$ 9. **Moments about A:** Taking moments about A (counterclockwise positive): - Weight acts at 2 m from A vertically downward - Reaction at B acts at 4 m from A at 60° angle Moment due to weight: $$ 196 \times 2 \times \cos 30^\circ $$ Moment due to $R_B$: $$ R_B \times 4 \times \sin 60^\circ $$ Equilibrium of moments: $$ 196 \times 2 \times \cos 30^\circ = R_B \times 4 \times \sin 60^\circ $$ 10. **Calculate $R_B$:** $$ R_B = \frac{196 \times 2 \times \cos 30^\circ}{4 \times \sin 60^\circ} $$ Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$: $$ R_B = \frac{196 \times 2 \times \frac{\sqrt{3}}{2}}{4 \times \frac{\sqrt{3}}{2}} = \frac{196 \times 2 \times \frac{\sqrt{3}}{2}}{4 \times \frac{\sqrt{3}}{2}} = \frac{196 \times 2}{4} = 98 \text{ N} $$ 11. **Calculate $R_{Ax}$:** From horizontal forces: $$ R_{Ax} + 98 \times \cos 60^\circ = 0 $$ $$ R_{Ax} = -98 \times 0.5 = -49 \text{ N} $$ (Negative sign indicates direction opposite assumed) 12. **Calculate $R_{Ay}$:** From vertical forces: $$ R_{Ay} + 98 \times \sin 60^\circ - 196 = 0 $$ $$ R_{Ay} = 196 - 98 \times \frac{\sqrt{3}}{2} = 196 - 98 \times 0.866 = 196 - 84.87 = 111.13 \text{ N} $$ 13. **Final reactions:** - Reaction at A: $R_A = \sqrt{(-49)^2 + (111.13)^2} = \sqrt{2401 + 12348} = \sqrt{14749} \approx 121.45$ N - Reaction at B: $R_B = 98$ N **Answer:** The reactions acting on the bar are approximately: - At A: 121.45 N - At B: 98 N