To find the resultant force, we need to resolve each force into its x and y components, sum these components, and then calculate the magnitude and direction of the resultant vector. Step 1: Resolve each force into its x and y components. Force 1: $F_1 = 6\,N$ (acting vertically upwards, along the positive y-axis) $F_{1x} = 6 \cos(90^\circ) = 0\,N$ $F_{1y} = 6 \sin(90^\circ) = 6\,N$ Force 2: $F_2 = 10\,N$ (acting at $30^\circ$ above the positive x-axis) $F_{2x} = 10 \cos(30^\circ) = 10 \left(\frac{\sqrt{3}}{2}\right) = 5\sqrt{3}\,N$ $F_{2y} = 10 \sin(30^\circ) = 10 \left(\frac{1}{2}\right) = 5\,N$ Force 3: $F_3 = 3\,N$ (acting horizontally, along the positive x-axis) $F_{3x} = 3 \cos(0^\circ) = 3\,N$ $F_{3y} = 3 \sin(0^\circ) = 0\,N$ Force 4: $F_4 = 3\,N$ (acting at $60^\circ$ below the negative x-axis) This force is in the third quadrant. Both x and y components will be negative. $F_{4x} = -3 \cos(60^\circ) = -3 \left(\frac{1}{2}\right) = -1.5\,N$ $F_{4y} = -3 \sin(60^\circ) = -3 \left(\frac{\sqrt{3}}{2}\right) = -1.5\sqrt{3}\,N$ Step 2: Sum the x-components to find the total x-component ($R_x$). $$R_x = F_{1x} + F_{2x} + F_{3x} + F_{4x}$$ $$R_x = 0 + 5\sqrt{3} + 3 - 1.5$$ $$R_x = 5\sqrt{3} + 1.5\,N$$ Numerically: $R_x \approx 5(1.732) + 1.5 = 8.66 + 1.5 = 10.16\,N$ Step 3: Sum the y-components to find the total y-component ($R_y$). $$R_y = F_{1y} + F_{2y} + F_{3y} + F_{4y}$$ $$R_y = 6 + 5 + 0 - 1.5\sqrt{3}$$ $$R_y = 11 - 1.5\sqrt{3}\,N$$ Numerically: $R_y \approx 11 - 1.5(1.732) = 11 - 2.598 = 8.402\,N$ **Step 4: Calculate the magnitude of the resultant