Let's solve step **(a):** _Determine the reactions at A and B._ --- ## **Step 1: Calculate the Support Reactions at A and B** The frame is pin-jointed at **A** and has a roller at **B**. Loads at **C**: 20 kN (vertical, downward), 10 kN (horizontal, right). Let's denote: - \( A_x, A_y \): reaction components at A (pin—both directions) - \( B_y \): reaction at B (roller—vertical only) ### **Sum of Forces in X-direction:** \[ \Sigma F_x = : \quad A_x - 10 = \implies A_x = 10\,\text{kN} \] **Explanation:** Only A can provide horizontal reaction. The applied horizontal load is 10 kN to the right, so \(A_x\) must be 10 kN to the right. --- ## **Step 2: Sum the Forces in Y-direction and Take Moments** ### **Sum of Forces in Y-direction:** \[ \Sigma F_y = : \quad A_y + B_y - 20 = \] \[ A_y + B_y = 20\,\text{kN} \] ### **Sum of Moments about A (Clockwise Positive):** Take moments about point A: Distances: - **C** is 3 m horizontally from A, 4 m vertically (so, use geometry for distances). - **B** is 4 m horizontally from A. But from diagram: - **Horizontal distance from A to C:** 3 m - **Vertical distance from A to C:** 4 m **Vertical load at C:** Horizontal distance from A to C is 3 m. \[ \Sigma M_A = : \quad (B_y \times 4) - (20 \times 3) = \] \[ 4B_y - 60 = \implies B_y = 15\,\text{kN} \] Now substitute \(B_y\) into earlier equation: \[ A_y + 15 = 20 \implies A_y = 5\,\text{kN} \] --- ## **Explanation:** - **Step 1:** Found \(A_x\) by balancing horizontal forces. - **Step 2:** Used moment equilibrium about A to find \(B_y\), then vertical force balance to find \(A_y\). --- ## **Final Answer: Reactions at A and B** \[ \boxed{ \begin{aligned} A_x &= 10\,\text{kN} \quad \text{(right)} \\ A_y &= 5\,\text{kN} \quad \text{(up)} \\ B_y &= 15\,\text{kN} \quad \text{(up)} \\ \end{aligned} } \]

# Solution to the Problem Let's break down the solution into two parts, focusing on the reactions at supports A and B of the frame. ## Step 1: Calculate the Horizontal Reaction at A ### **Sum of Forces in X-direction:** \[ \Sigma F_x = 0: \quad A_x - 10 = 0 \implies A_x = 10\,\text{kN} \] **Explanation:** The only horizontal force is the 10 kN load applied to the right. Since there are no other horizontal forces acting on the frame, the horizontal reaction at A must equal this load to maintain equilibrium. --- ## Step 2: Calculate the Vertical Reactions at A and B ### **Sum of Forces in Y-direction:** \[ \Sigma F_y = 0: \quad A_y + B_y - 20 = 0 \implies A_y + B_y = 20\,\text{kN} \] ### **Sum of Moments about A (Clockwise Positive):** Consider the moments around point A: - The moment due to the vertical load at C (20 kN) is \(20 \times 3 = 60\,\text{kN \cdot m}\) (clockwise). - The moment due to the reaction at B is \(B_y \times 4\,\text{m}\) (counterclockwise). Setting the sum of moments around A to zero: \[ \Sigma M_A = 0: \quad (B_y \times 4) - (20 \times 3) = 0 \] \[ 4B_y - 60 = 0 \implies B_y = 15\,\text{kN} \] ### **Substitute \(B_y\) into the Vertical Force Equation:** Using the previously found equation for vertical forces: \[ A_y + 15 = 20 \implies A_y = 5\,\text{kN} \] --- ## Final Answer: Reactions at A and B The support reactions can be summarized as follows: \[ \begin{aligned} A_x &= 10\,\text{kN} \quad \text{(horizontal, right)} \\ A_y &= 5\,\text{kN} \quad \text{(vertical, up)} \\ B_y &= 15\,\text{kN} \quad \text{(vertical, up)} \\ \end{aligned} \] Thus, the reactions at the supports are: - **At A:** \(A_x = 10\,\text{kN}\) (right), \(A_y = 5\,\text{kN}\) (up) - **At B:** \(B_y = 15\,\text{kN}\) (up) \[ \boxed{ \begin{aligned} A_x &= 10\,\text{kN} \\ A_y &= 5\,\text{kN} \\ B_y &= 15\,\text{kN} \\ \end{aligned} } \]