1. **State the problem:** We have a right triangle with base $2.3$ m and height $1.2$ m, fixed support on the left, pinned support on the right, a horizontal force of $6$ kN to the left at the fixed support, and a distributed load of $5$ kN/m acting downward along the hypotenuse. We need to find the reaction forces at the supports. 2. **Calculate the length of the hypotenuse:** $$\text{hypotenuse} = \sqrt{2.3^2 + 1.2^2} = \sqrt{5.29 + 1.44} = \sqrt{6.73} \approx 2.594 \text{ m}$$ 3. **Calculate the total distributed load:** $$w = 5 \text{ kN/m} \times 2.594 \text{ m} = 12.97 \text{ kN}$$ This load acts downward along the hypotenuse. 4. **Determine the location of the resultant distributed load:** The resultant force acts at the midpoint of the hypotenuse, so at $1.297$ m from the left end along the hypotenuse. 5. **Resolve the distributed load into horizontal and vertical components:** The angle $\theta$ of the hypotenuse with the horizontal is: $$\theta = \arctan\left(\frac{1.2}{2.3}\right) \approx 27.3^\circ$$ Horizontal component: $$F_x = 12.97 \times \cos(27.3^\circ) \approx 12.97 \times 0.889 = 11.53 \text{ kN}$$ Vertical component: $$F_y = 12.97 \times \sin(27.3^\circ) \approx 12.97 \times 0.458 = 5.94 \text{ kN}$$ 6. **Set up equilibrium equations:** Let $R_{Ax}$ and $R_{Ay}$ be the horizontal and vertical reactions at the fixed support (left), and $R_{By}$ be the vertical reaction at the pinned support (right). The pinned support cannot resist horizontal force. - Sum of horizontal forces: $$\sum F_x = 0 \Rightarrow R_{Ax} - 6 - 11.53 = 0 \Rightarrow R_{Ax} = 17.53 \text{ kN (to the right)}$$ - Sum of vertical forces: $$\sum F_y = 0 \Rightarrow R_{Ay} + R_{By} - 5.94 = 0 \Rightarrow R_{Ay} + R_{By} = 5.94$$ - Sum of moments about the left support (taking counterclockwise as positive): The moment arm for $R_{By}$ is the base length $2.3$ m. The moment arm for the distributed load components must be projected onto horizontal and vertical directions. Moment due to distributed load about left support: The vertical component $5.94$ kN acts at horizontal distance $1.297 \times \cos(27.3^\circ) = 1.297 \times 0.889 = 1.153$ m. The horizontal component $11.53$ kN acts at vertical distance $1.297 \times \sin(27.3^\circ) = 1.297 \times 0.458 = 0.594$ m. Moments: $$M = R_{By} \times 1.2 - 5.94 \times 1.153 - 11.53 \times 0.594 = 0$$ Calculate: $$R_{By} \times 1.2 = 5.94 \times 1.153 + 11.53 \times 0.594 = 6.85 + 6.85 = 13.7$$ So: $$R_{By} = \frac{13.7}{1.2} = 11.42 \text{ kN}$$ 7. **Find $R_{Ay}$:** $$R_{Ay} = 5.94 - 11.42 = -5.48 \text{ kN}$$ The negative sign means $R_{Ay}$ acts downward. **Final reaction forces:** - Horizontal reaction at left support: $R_{Ax} = 17.53$ kN (right) - Vertical reaction at left support: $R_{Ay} = -5.48$ kN (downward) - Vertical reaction at right support: $R_{By} = 11.42$ kN (upward)