1. **State the problem:** We have a horizontal member pinned at point A and resting on a smooth support at B. A downward force of 60 N acts 0.5 m from A, and a counterclockwise moment of 90 N·m acts at the right end (1 m from A). The support B is 0.75 m above the member and forms a 30° angle with the horizontal. We need to find the horizontal ($A_x$) and vertical ($A_y$) reaction components at pin A. 2. **Identify forces and reactions:** - At A: reactions $A_x$ (horizontal) and $A_y$ (vertical). - At B: a reaction force $B$ acting perpendicular to the smooth support (since it is smooth, no friction), at 30° angle to horizontal. 3. **Set up equilibrium equations:** Since the member is in equilibrium, sum of forces and moments are zero. - Sum of horizontal forces: $$\sum F_x = 0 = A_x + B \cos 30^\circ$$ - Sum of vertical forces: $$\sum F_y = 0 = A_y + B \sin 30^\circ - 60$$ - Sum of moments about A: $$\sum M_A = 0 = -60 \times 0.5 + B \sin 30^\circ \times 1 - 90$$ 4. **Calculate $B \sin 30^\circ$ from moment equation:** $$0 = -30 + B \sin 30^\circ \times 1 - 90$$ $$B \sin 30^\circ = 120$$ Since $\sin 30^\circ = 0.5$, $$B \times 0.5 = 120$$ $$B = \frac{120}{0.5} = 240\ \text{N}$$ 5. **Calculate components of B:** $$B \cos 30^\circ = 240 \times \cos 30^\circ = 240 \times \frac{\sqrt{3}}{2} = 240 \times 0.866 = 207.85\ \text{N}$$ $$B \sin 30^\circ = 240 \times 0.5 = 120\ \text{N}$$ 6. **Use force equilibrium to find $A_x$ and $A_y$:** - Horizontal: $$0 = A_x + 207.85$$ $$A_x = -207.85\ \text{N}$$ (to the left) - Vertical: $$0 = A_y + 120 - 60$$ $$A_y = -60\ \text{N}$$ (downward) 7. **Interpretation:** The negative signs indicate the directions opposite to assumed positive directions. **Final answers:** $$A_x = -207.85\ \text{N}$$ (left) $$A_y = -60\ \text{N}$$ (down)