Let's break down the problem step by step: ### **Given Data** - Force in member \( A \): \( P = 10 \) kN (at 40° left of vertical) - Force in member \( B \): \( Q = 20 \) kN (at 20° right of vertical) We need to find **the magnitude and direction of the resultant force** applied to the bracket. --- ## **Step 1: Resolve the Forces into Components** Assume upward as the positive y-direction and rightward as the positive x-direction. ### **Force A (10 kN at 40° left of vertical)** - \( F_{Ax} = -10 \sin(40^\circ) \) (negative because it's to the left) - \( F_{Ay} = -10 \cos(40^\circ) \) (negative because both forces are downward/compression) ### **Force B (20 kN at 20° right of vertical)** - \( F_{Bx} = 20 \sin(20^\circ) \) (positive because it's to the right) - \( F_{By} = -20 \cos(20^\circ) \) (negative because it's downward/compression) --- ## **Step 2: Calculate Each Component** ### **Force A:** - \( \sin(40^\circ) \approx 0.6428 \) - \( \cos(40^\circ) \approx 0.7660 \) - \( F_{Ax} = -10 \times 0.6428 = -6.428 \) kN - \( F_{Ay} = -10 \times 0.7660 = -7.660 \) kN ### **Force B:** - \( \sin(20^\circ) \approx 0.3420 \) - \( \cos(20^\circ) \approx 0.9397 \) - \( F_{Bx} = 20 \times 0.3420 = 6.840 \) kN - \( F_{By} = -20 \times 0.9397 = -18.794 \) kN --- ## **Step 3: Add the Components** ### **Resultant X-component:** \[ R_x = F_{Ax} + F_{Bx} = -6.428 + 6.840 = 0.412 \text{ kN} \] ### **Resultant Y-component:** \[ R_y = F_{Ay} + F_{By} = -7.660 + (-18.794) = -26.454 \text{ kN} \] --- ## **Step 4: Magnitude of the Resultant** \[ R = \sqrt{R_x^2 + R_y^2} \] \[ R = \sqrt{(0.412)^2 + (-26.454)^2} \] \[ R = \sqrt{0.170 + 699.738} \] \[ R = \sqrt{699.908} \approx 26.47 \text{ kN} \] --- ## **Step 5: Direction of the Resultant (Angle from the Positive X-axis)** \[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \] \[ \theta = \tan^{-1}\left(\frac{-26.454}{0.412}\right) \] \[ \theta = \tan^{-1}(-64.2) \] This is a very steep negative angle, almost straight down: \[ \theta \approx -89.1^\circ \] So, the resultant is almost vertically downward. --- ## **Final Answers** **Magnitude:** \[ \boxed{26.47 \text{ kN}} \] **Direction:** \[ \boxed{89.1^\circ \text{ below the positive x-axis (almost straight down)}} \]