Forces \( F_{1}\left(10 \mathrm{~N}, 090^{\circ}\right), \mathrm{F}_{2}\left(20 \mathrm{~N}, 210^{\circ}\right) \) and \( \mathrm{F}_{3}\left(4 \mathrm{~N}, 330^{\circ}\right) \) act on a particle. Find, correct to one decimal place, the magnitude of the resultant force.

To find the magnitude of the resultant force acting on the particle, we'll break down each force into its Cartesian components (x and y), sum these components, and then calculate the magnitude of the resulting vector. ### Given Forces: 1. \( F_{1} = 10 \text{ N at } 90^{\circ} \) 2. \( F_{2} = 20 \text{ N at } 210^{\circ} \) 3. \( F_{3} = 4 \text{ N at } 330^{\circ} \) ### Step 1: Break Down Each Force into Components **Force \( F_{1} \):** - \( F_{1x} = 10 \cos(90^{\circ}) = 0 \) - \( F_{1y} = 10 \sin(90^{\circ}) = 10 \text{ N} \) **Force \( F_{2} \):** - \( F_{2x} = 20 \cos(210^{\circ}) = 20 \times (-\frac{\sqrt{3}}{2}) \approx -17.32 \text{ N} \) - \( F_{2y} = 20 \sin(210^{\circ}) = 20 \times (-0.5) = -10 \text{ N} \) **Force \( F_{3} \):** - \( F_{3x} = 4 \cos(330^{\circ}) = 4 \times \frac{\sqrt{3}}{2} \approx 3.46 \text{ N} \) - \( F_{3y} = 4 \sin(330^{\circ}) = 4 \times (-0.5) = -2 \text{ N} \) ### Step 2: Sum of Components - **Total \( x \)-component:** \[ F_{\text{total}_x} = F_{1x} + F_{2x} + F_{3x} = 0 - 17.32 + 3.46 \approx -13.86 \text{ N} \] - **Total \( y \)-component:** \[ F_{\text{total}_y} = F_{1y} + F_{2y} + F_{3y} = 10 - 10 - 2 = -2 \text{ N} \] ### Step 3: Calculate the Magnitude of the Resultant Force \[ |\mathbf{F}_{\text{resultant}}| = \sqrt{(F_{\text{total}_x})^2 + (F_{\text{total}_y})^2} = \sqrt{(-13.86)^2 + (-2)^2} \approx \sqrt{191.88 + 4} \approx \sqrt{195.88} \approx 14.0 \text{ N} \] ### **Final Answer:** The magnitude of the resultant force is **14.0  N**.