Ativity 2 Torces Two forces act on an object. The first is a horizontal 50 N force. second is a 40 N force acting at an angle to the hosizontal. Use a graphical method drawn to a scale of \( 1 \mathrm{~mm}: 1 \mathrm{~N} \) to find resultant if the 40 N force is inclined at: \( 30^{\circ} \) to the horizontal \( 60^{\circ} \) to the horizontal.

To determine the resultant force when two forces act on an object using a graphical method, we'll break down the process step-by-step. Although we can't draw the vectors here, we'll use numerical calculations to simulate the graphical approach based on the given scale of \(1 \, \text{mm} : 1 \, \text{N}\). ### Given: 1. **Force \( \mathbf{F}_1 \):** 50 N (horizontal) 2. **Force \( \mathbf{F}_2 \):** 40 N at an angle to the horizontal - **Case 1:** \(30^\circ\) to the horizontal - **Case 2:** \(60^\circ\) to the horizontal ### Objective: Find the magnitude of the **resultant force \( \mathbf{R} \)** for both cases. ### Method: We'll use the **Law of Cosines** to calculate the magnitude of the resultant force when two vectors are at a certain angle to each other. \[ R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos(\theta)} \] where: - \( R \) = magnitude of the resultant force - \( F_1 \) and \( F_2 \) = magnitudes of the two forces - \( \theta \) = angle between the two forces ### Calculations: #### **Case 1: \(40 \, \text{N}\) at \(30^\circ\) to the horizontal** 1. **Identify the Angle Between Forces:** - \( \theta = 30^\circ \) 2. **Apply the Law of Cosines:** \[ R = \sqrt{50^2 + 40^2 + 2 \times 50 \times 40 \times \cos(30^\circ)} \] \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660 \] \[ R = \sqrt{2500 + 1600 + 4000 \times 0.8660} \] \[ R = \sqrt{2500 + 1600 + 3464} \] \[ R = \sqrt{7564} \approx 87 \, \text{N} \] #### **Case 2: \(40 \, \text{N}\) at \(60^\circ\) to the horizontal** 1. **Identify the Angle Between Forces:** - \( \theta = 60^\circ \) 2. **Apply the Law of Cosines:** \[ R = \sqrt{50^2 + 40^2 + 2 \times 50 \times 40 \times \cos(60^\circ)} \] \[ \cos(60^\circ) = 0.5 \] \[ R = \sqrt{2500 + 1600 + 4000 \times 0.5} \] \[ R = \sqrt{2500 + 1600 + 2000} \] \[ R = \sqrt{6100} \approx 78.1 \, \text{N} \] ### Summary of Results: - **When the 40 N force is inclined at \(30^\circ\):** - **Resultant Force \( R \approx 87 \, \text{N} \)** - **When the 40 N force is inclined at \(60^\circ\):** - **Resultant Force \( R \approx 78.1 \, \text{N} \)** ### Graphical Interpretation: If you were to draw this graphically: 1. **Scale:** \(1 \, \text{mm} = 1 \, \text{N}\) 2. **Draw \( \mathbf{F}_1 \):** Draw a horizontal line 50 mm long. 3. **Draw \( \mathbf{F}_2 \):** From the end of \( \mathbf{F}_1 \), draw a line at the specified angle (30° or 60°) 40 mm long. 4. **Resultant \( \mathbf{R} \):** Draw a line from the origin (start of \( \mathbf{F}_1 \)) to the end of \( \mathbf{F}_2 \). The length of this line represents the magnitude of the resultant force. Using this method with accurate measurements will give you the resultant forces calculated above.