1. **Problem Statement:** We have three forces acting on an object at its center of mass. We need to find: a) The resultant horizontal force. b) The resultant vertical force. c) The magnitude and orientation of the resultant total force. 2. **Given Data:** - Force 1 (purple): 500 N, directed up and left at 65° from the horizontal right line. - Force 2 (blue): 300 N, directed up and right at 35° from the vertical line. - Force 3 (red): 400 N, directed right at 25° from the horizontal right line. 3. **Approach:** - Resolve each force into horizontal ($F_x$) and vertical ($F_y$) components. - Sum all horizontal components to get resultant horizontal force $R_x$. - Sum all vertical components to get resultant vertical force $R_y$. - Use Pythagoras theorem to find magnitude $R = \sqrt{R_x^2 + R_y^2}$. - Find orientation angle $\theta = \tan^{-1}(\frac{R_y}{R_x})$ relative to horizontal. 4. **Resolving Forces:** - Force 1 (500 N, 65° from horizontal left/up): - Horizontal component: $F_{1x} = -500 \cos(65^\circ)$ (left is negative) - Vertical component: $F_{1y} = 500 \sin(65^\circ)$ (up is positive) - Force 2 (300 N, 35° from vertical up/right): - Since angle is from vertical, horizontal component: $F_{2x} = 300 \sin(35^\circ)$ (right positive) - Vertical component: $F_{2y} = 300 \cos(35^\circ)$ (up positive) - Force 3 (400 N, 25° from horizontal right): - Horizontal component: $F_{3x} = 400 \cos(25^\circ)$ (right positive) - Vertical component: $F_{3y} = 400 \sin(25^\circ)$ (up positive) 5. **Calculate components:** - $F_{1x} = -500 \times \cos(65^\circ) = -500 \times 0.4226 = -211.3$ - $F_{1y} = 500 \times \sin(65^\circ) = 500 \times 0.9063 = 453.1$ - $F_{2x} = 300 \times \sin(35^\circ) = 300 \times 0.574 = 172.2$ - $F_{2y} = 300 \times \cos(35^\circ) = 300 \times 0.8192 = 245.8$ - $F_{3x} = 400 \times \cos(25^\circ) = 400 \times 0.9063 = 362.5$ - $F_{3y} = 400 \times \sin(25^\circ) = 400 \times 0.4226 = 169.0$ 6. **Sum components:** - Resultant horizontal force: $$R_x = F_{1x} + F_{2x} + F_{3x} = -211.3 + 172.2 + 362.5 = 323.4$$ - Resultant vertical force: $$R_y = F_{1y} + F_{2y} + F_{3y} = 453.1 + 245.8 + 169.0 = 867.9$$ 7. **Magnitude of resultant force:** $$R = \sqrt{R_x^2 + R_y^2} = \sqrt{323.4^2 + 867.9^2} = \sqrt{104588 + 753243} = \sqrt{857831} = 926.1$$ 8. **Orientation angle relative to horizontal:** $$\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{867.9}{323.4}\right) = \tan^{-1}(2.684) = 69.5^\circ$$ **Final answers:** a) Resultant horizontal force = 323.4 N (right) b) Resultant vertical force = 867.9 N (up) c) Magnitude of resultant force = 926.1 N Orientation = 69.5° above the horizontal to the right