110 N \( 25^{\circ} \) North of West 80 N \( 58^{\circ} \) East of North 65 N \( 18^{\circ} \) East of South 30 N \( 52^{\circ} \) North of West

To determine the **resultant force** from the four given forces, we'll follow these steps: 1. **Resolve Each Force into Its Horizontal (East-West) and Vertical (North-South) Components** 2. **Sum All the Horizontal Components and All the Vertical Components Separately** 3. **Calculate the Magnitude and Direction of the Resultant Force** Let's proceed step-by-step. --- ### **1. Resolving Each Force into Components** We'll assume the following coordinate system: - **Positive X-axis:** East - **Negative X-axis:** West - **Positive Y-axis:** North - **Negative Y-axis:** South #### **Force 1: 110 N at 25° North of West** - **Direction Interpretation:** Northwest quadrant.  - **Components:** - \( F_{1x} = -110 \times \cos(25°) \)  (East-West component; negative because it's West) - \( F_{1y} = 110 \times \sin(25°) \)   (North-South component) #### **Force 2: 80 N at 58° East of North** - **Direction Interpretation:** Northeast quadrant.  - **Components:** - \( F_{2x} = 80 \times \sin(58°) \)     (East-West component; positive because it's East) - \( F_{2y} = 80 \times \cos(58°) \)     (North-South component) #### **Force 3: 65 N at 18° East of South** - **Direction Interpretation:** Southeast quadrant.  - **Components:** - \( F_{3x} = 65 \times \sin(18°) \)     (East-West component; positive because it's East) - \( F_{3y} = -65 \times \cos(18°) \)   (North-South component; negative because it's South) #### **Force 4: 30 N at 52° North of West** - **Direction Interpretation:** Northwest quadrant.  - **Components:** - \( F_{4x} = -30 \times \cos(52°) \)   (East-West component; negative because it's West) - \( F_{4y} = 30 \times \sin(52°) \)   (North-South component) --- ### **2. Calculating Each Component Numerically** Let's compute the numerical values using a calculator: - **Force 1:** - \( F_{1x} = -110 \times \cos(25°) \approx -110 \times 0.9063 \approx -99.69 \, \text{N} \) - \( F_{1y} = 110 \times \sin(25°) \approx 110 \times 0.4226 \approx 46.49 \, \text{N} \) - **Force 2:** - \( F_{2x} = 80 \times \sin(58°) \approx 80 \times 0.8480 \approx 67.84 \, \text{N} \) - \( F_{2y} = 80 \times \cos(58°) \approx 80 \times 0.5299 \approx 42.39 \, \text{N} \) - **Force 3:** - \( F_{3x} = 65 \times \sin(18°) \approx 65 \times 0.3090 \approx 20.09 \, \text{N} \) - \( F_{3y} = -65 \times \cos(18°) \approx -65 \times 0.9511 \approx -61.82 \, \text{N} \) - **Force 4:** - \( F_{4x} = -30 \times \cos(52°) \approx -30 \times 0.6157 \approx -18.47 \, \text{N} \) - \( F_{4y} = 30 \times \sin(52°) \approx 30 \times 0.7880 \approx 23.64 \, \text{N} \) --- ### **3. Summing the Components** - **Total Horizontal Component (X-axis):** \[ F_{\text{total\,x}} = F_{1x} + F_{2x} + F_{3x} + F_{4x} = (-99.69) + 67.84 + 20.09 + (-18.47) \approx -30.23 \, \text{N} \] - **Total Vertical Component (Y-axis):** \[ F_{\text{total\,y}} = F_{1y} + F_{2y} + F_{3y} + F_{4y} = 46.49 + 42.39 + (-61.82) + 23.64 \approx 50.70 \, \text{N} \] --- ### **4. Determining the Magnitude and Direction of the Resultant Force** - **Magnitude (\( F_{\text{resultant}} \)):** \[ F_{\text{resultant}} = \sqrt{F_{\text{total\,x}}^2 + F_{\text{total\,y}}^2} = \sqrt{(-30.23)^2 + (50.70)^2} \approx \sqrt{912.44 + 2570.49} \approx \sqrt{3482.93} \approx 59.02 \, \text{N} \] - **Direction (\( \theta \)):** The angle is measured from the positive X-axis (East) towards the positive Y-axis (North). \[ \theta = \arctan\left(\frac{F_{\text{total\,y}}}{F_{\text{total\,x}}}\right) = \arctan\left(\frac{50.70}{-30.23}\right) \approx \arctan(-1.679) \approx -59.03° \] Since the X-component is negative and Y-component is positive, the resultant force is in the **Northwest quadrant**. To express the direction correctly: \[ \theta_{\text{North of West}} = 90° - |\theta| \approx 90° - 59.03° \approx 30.97° \] **Therefore,** the angle is approximately **31° North of West**. --- ### **5. Final Resultant Force** - **Magnitude:** Approximately **59.0 N** - **Direction:** **31° North of West** --- ### **Summary** By decomposing each of the four forces into their respective horizontal and vertical components, summing these components, and then calculating the resultant's magnitude and direction, we've determined that the combined force is approximately **59.0 N** directed **31° North of West**.