Let's solve the problem step by step: ### **Given:** - \( P = 10 \) kN (Force in member \( A \)) - \( Q = 20 \) kN (Force in member \( B \)) - Angle of member \( A \) with the vertical = \( 40^\circ \) - Angle of member \( B \) with the vertical = \( 20^\circ \) ### **Step 1: Resolve Forces into Components** Let’s assume upwards and rightwards are positive directions. #### **Force in Member \( A \):** - Magnitude: \( P = 10 \) kN - Angle with vertical: \( 40^\circ \) left of vertical Components: - \( P_x = -10 \sin 40^\circ \) (negative, as it is to the left) - \( P_y = -10 \cos 40^\circ \) (negative, as it is downward) #### **Force in Member \( B \):** - Magnitude: \( Q = 20 \) kN - Angle with vertical: \( 20^\circ \) right of vertical Components: - \( Q_x = 20 \sin 20^\circ \) (positive, as it is to the right) - \( Q_y = -20 \cos 20^\circ \) (negative, as it is downward) --- ### **Step 2: Calculate Components** #### **Member A:** - \( \sin 40^\circ \approx 0.6428 \) - \( \cos 40^\circ \approx 0.7660 \) \[ P_x = -10 \times 0.6428 = -6.428 \text{ kN} \] \[ P_y = -10 \times 0.7660 = -7.660 \text{ kN} \] #### **Member B:** - \( \sin 20^\circ \approx 0.3420 \) - \( \cos 20^\circ \approx 0.9397 \) \[ Q_x = 20 \times 0.3420 = 6.840 \text{ kN} \] \[ Q_y = -20 \times 0.9397 = -18.794 \text{ kN} \] --- ### **Step 3: Sum Components for Resultant** #### **X Components:** \[ R_x = P_x + Q_x = (-6.428) + (6.840) = 0.412 \text{ kN} \] #### **Y Components:** \[ R_y = P_y + Q_y = (-7.660) + (-18.794) = -26.454 \text{ kN} \] --- ### **Step 4: Magnitude of Resultant** \[ R = \sqrt{R_x^2 + R_y^2} \] \[ R = \sqrt{(0.412)^2 + (-26.454)^2} \] \[ R = \sqrt{0.170 + 699.734} \] \[ R = \sqrt{699.904} \approx 26.47 \text{ kN} \] --- ### **Step 5: Direction of Resultant** \[ \tan \theta = \frac{R_x}{|R_y|} \] \[ \theta = \arctan\left(\frac{0.412}{26.454}\right) \] \[ \theta = \arctan(0.01558) \approx 0.893^\circ \] Since \( R_x \) is positive and \( R_y \) is negative, the resultant points almost directly downward, \( 0.89^\circ \) to the right of the negative y-axis (downward). --- ## **Final Answers:** ### **Magnitude:** \[ \boxed{26.47\ \text{kN}} \] ### **Direction:** \[ \boxed{0.89^\circ} \text{ to the right of downward vertical} \]

**1. Introduction** In analyzing forces acting on structural members and their connections, it is essential to understand how forces are resolved into components, how to determine the resultant force, and how to interpret these forces in terms of magnitude and direction. When two members exert forces on a common point, such as a bracket, their forces can be represented as vectors with magnitudes and directions. To analyze the combined effect, the forces are broken down into their horizontal and vertical components. Using trigonometry, these components can be calculated based on the known angles and magnitudes. The resultant force is then found by vector addition, involving summing the components and calculating the magnitude and direction of the resultant vector. **Explanation:** In structural analysis, forces acting along members are typically resolved into their components to simplify the analysis of their combined effects at joints or brackets. The principles of vector addition and trigonometry are fundamental in determining the magnitude and direction of the resultant force, which is critical for assessing the load transferred to the support or connection point. --- **2. Identifying the Correct Option Among the Choices** Given the problem involves calculating the forces exerted by members A and B on a bracket, and their resultant force, the correct option should reflect the accurate magnitude and direction derived from vector addition. **Justification:** The force in member A (10 kN) at an angle of 40°, and the force in member B (20 kN) at an angle of 20°, are resolved into their horizontal and vertical components. Summing these components yields the resultant force's magnitude and direction. The calculated resultant magnitude (~26.47 kN) and the angle (~0.89° to the right of downward) match the detailed calculations. Therefore, the option corresponding to these results is correct. --- **3. Step-by-Step Calculation** **Step 1: Resolve forces into components** - Force \( P = 10\, \text{kN} \) in member A at \( 40^\circ \) from vertical: - Horizontal component: \[ P_x = -10 \times \sin 40^\circ \approx -10 \times 0.6428 = -6.43\, \text{kN} \] (Negative because it points to the left) - Vertical component: \[ P_y = -10 \times \cos 40^\circ \approx -10 \times 0.7660 = -7.66\, \text{kN} \] (Negative because it points downward) - Force \( Q = 20\, \text{kN} \) in member B at \( 20^\circ \) from vertical: - Horizontal component: \[ Q_x = 20 \times \sin 20^\circ \approx 20 \times 0.3420 = 6.84\, \text{kN} \] - Vertical component: \[ Q_y = -20 \times \cos 20^\circ \approx -20 \times 0.9397 = -18.79\, \text{kN} \] **Explanation:** Resolving the forces into components using sine and cosine functions allows the use of vector addition to find the resultant force at the bracket, considering both magnitude and direction. --- **Step 2: Sum components to find the resultant** - Horizontal component: \[ R_x = P_x + Q_x = -6.43 + 6.84 = 0.41\, \text{kN} \] - Vertical component: \[ R_y = P_y + Q_y = -7.66 + (-18.79) = -26.45\, \text{kN} \] **Explanation:** Adding the components component-wise yields the total horizontal and vertical forces acting on the bracket. --- **Step 3: Calculate magnitude of the resultant force** \[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{(0.41)^2 + (-26.45)^2} \] \[ R = \sqrt{0.1681 + 700.7} = \sqrt{700.8681} \approx 26.47\, \text{kN} \] **Explanation:** The magnitude of the resultant vector is obtained via the Pythagorean theorem, combining horizontal and vertical components. --- **Step 4: Find the direction of the resultant** \[ \theta = \arctan \left( \frac{R_x}{|R_y|} \right) = \arctan \left( \frac{0.41}{26.45} \right) \approx \arctan(0.0155) \approx 0.89^\circ \] Since \( R_x \) is positive and \( R_y \) negative, the resultant points slightly to the right of downward. **Explanation:** The angle indicates the direction of the resultant force relative to the vertical. --- **Final Results:** - **Magnitude:** approximately **26.47 kN** - **Direction:** approximately **0.89° to the right of downward** --- **4. Analysis of Why Other Options Are Incorrect** Suppose alternative options present different magnitudes or directions: - **Options with significantly lower magnitudes** (e.g., around 15-20 kN): These neglect the combined effects of both force components, underestimating the resultant magnitude. - **Options with larger magnitudes** (e.g., over 30 kN): These overestimate the resultant by possibly miscalculating the components or ignoring the vector nature of forces. - **Options indicating a large angular deviation (e.g., 30° or more):** Such angles suggest a dominant horizontal component, which isn't supported by the calculations given the force magnitudes and angles. **Explanation:** The calculated components and the resulting magnitude and direction are consistent with the principles of vector addition. Any option deviating significantly from these values does not align with the mathematical analysis or the physical configuration. --- **Conclusion:** The correct solution involves resolving the forces into components, summing these components vectorially, and calculating the resultant's magnitude and direction. The precise calculations yield a resultant force of approximately **26.47 kN** directed about **0.89° to the right of downward**. This matches the expected physical behavior and confirms the correctness of the selected option. **Final Answer:** \[ \boxed{ \text{Magnitude of resultant} \approx 26.47\, \text{kN} \\ \text{Direction} \approx 0.89^\circ \text{ to the right of downward} } \]