1. **Problem Statement:** Determine analytically the magnitude and direction of the resultant force from four forces acting at a point: (i) 10 N pull N 30°E (ii) 12.5 N push S 45°W (iii) 5 N push N 60°E (iv) 15 N push S 60°E 2. **Understanding the problem:** We will resolve each force into its horizontal (x) and vertical (y) components using trigonometry. Then sum all x-components and y-components separately to find the resultant vector components. 3. **Coordinate system and angle conventions:** - North is positive y-direction. - East is positive x-direction. - Angles are measured from North or South towards East or West as given. 4. **Resolving each force into components:** (i) 10 N pull N 30°E - Angle from North towards East: 30° - $F_{x1} = 10 \sin 30^\circ = 10 \times 0.5 = 5$ N (East, positive x) - $F_{y1} = 10 \cos 30^\circ = 10 \times 0.866 = 8.66$ N (North, positive y) (ii) 12.5 N push S 45°W - Push means force directed towards South 45° West - Angle from South towards West: 45° - $F_{x2} = -12.5 \sin 45^\circ = -12.5 \times 0.707 = -8.84$ N (West, negative x) - $F_{y2} = -12.5 \cos 45^\circ = -12.5 \times 0.707 = -8.84$ N (South, negative y) (iii) 5 N push N 60°E - Push means force directed towards North 60° East - Angle from North towards East: 60° - $F_{x3} = 5 \sin 60^\circ = 5 \times 0.866 = 4.33$ N (East, positive x) - $F_{y3} = 5 \cos 60^\circ = 5 \times 0.5 = 2.5$ N (North, positive y) (iv) 15 N push S 60°E - Push means force directed towards South 60° East - Angle from South towards East: 60° - $F_{x4} = 15 \sin 60^\circ = 15 \times 0.866 = 12.99$ N (East, positive x) - $F_{y4} = -15 \cos 60^\circ = -15 \times 0.5 = -7.5$ N (South, negative y) 5. **Sum of components:** - $R_x = F_{x1} + F_{x2} + F_{x3} + F_{x4} = 5 - 8.84 + 4.33 + 12.99 = 13.48$ N - $R_y = F_{y1} + F_{y2} + F_{y3} + F_{y4} = 8.66 - 8.84 + 2.5 - 7.5 = -5.18$ N 6. **Magnitude of resultant force:** $$ R = \sqrt{R_x^2 + R_y^2} = \sqrt{13.48^2 + (-5.18)^2} = \sqrt{181.74 + 26.83} = \sqrt{208.57} = 14.45 \text{ N} $$ 7. **Direction of resultant force:** - Angle $\theta$ measured from East towards North (counterclockwise) is: $$ \theta = \tan^{-1} \left( \frac{R_y}{R_x} \right) = \tan^{-1} \left( \frac{-5.18}{13.48} \right) = -21.1^\circ $$ - Negative angle means below the positive x-axis (East), so direction is 21.1° South of East. **Final answer:** - Magnitude = 14.45 N - Direction = 21.1° South of East