Let's solve each part step-by-step, with explanations after each step. --- ## **Part A: Find \( F_3 \) in terms of \( F_1 \)** ### **Step 1: Setting up the Equilibrium Condition** Since the three forces produce a zero resultant force (\(\vec{F}_R = \)), their vector sum must be zero: \[ \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \] Given: - \( F_2 = \frac{1}{2} F_1 \) - \( F_1 \) is along the x-axis - \( F_2 \) is at 90° from \( F_1 \) (so along y-axis) - \( F_3 \) makes an angle \(\theta\) with the x-axis (direction unknown yet) Let’s write the components: - \( \vec{F}_1 = F_1 \hat{i} \) - \( \vec{F}_2 = \frac{1}{2}F_1 \hat{j} \) - \( \vec{F}_3 = F_3(\cos\theta \hat{i} + \sin\theta \hat{j}) \) Sum the x and y components: \[ F_1 + F_3 \cos\theta = \quad \text{(x-component)} \] \[ \frac{1}{2}F_1 + F_3 \sin\theta = \quad \text{(y-component)} \] **Explanation:** We break each force into components and set up the equilibrium equations along both axes. --- ### **Step 2: Solving for \( F_3 \) in terms of \( F_1 \)** From the x-equation: \[ F_1 + F_3 \cos\theta = \implies F_3 \cos\theta = -F_1 \] From the y-equation: \[ \frac{1}{2}F_1 + F_3 \sin\theta = \implies F_3 \sin\theta = -\frac{1}{2}F_1 \] Now, square and add both equations: \[ (F_3 \cos\theta)^2 + (F_3 \sin\theta)^2 = F_1^2 + \left(\frac{1}{2}F_1\right)^2 \] \[ F_3^2 (\cos^2\theta + \sin^2\theta) = F_1^2 + \frac{1}{4}F_1^2 \] \[ F_3^2 (1) = \frac{5}{4} F_1^2 \] \[ F_3 = \frac{\sqrt{5}}{2} F_1 \] **Explanation:** By squaring and adding, we use the Pythagorean identity (\(\sin^2\theta + \cos^2\theta = 1\)) to eliminate \(\theta\) and solve for \(F_3\). --- ### **Final Answer for Part A:** \[ \boxed{F_3 = \frac{\sqrt{5}}{2} F_1} \] --- ## **Part B: Find the Angle \( \theta \)** ### **Step 1: Use Component Equations to Get \( \theta \)** From above: \[ F_3 \cos\theta = -F_1 \\ F_3 \sin\theta = -\frac{1}{2}F_1 \] Divide the second equation by the first: \[ \frac{F_3 \sin\theta}{F_3 \cos\theta} = \frac{-\frac{1}{2}F_1}{-F_1} \] \[ \tan\theta = \frac{1}{2} \] **Explanation:** Divide y by x component equations to get \(\tan\theta\), eliminating \(F_3\) and \(F_1\). --- ### **Step 2: Solve for \( \theta \)** \[ \theta = \arctan\left(\frac{1}{2}\right) \] \[ \theta \approx 26.6^\circ \] **Explanation:** Use the arctangent function to find the angle. --- ### **Final Answer for Part B:** \[ \boxed{\theta = 26.6^\circ} \] --- ## **Summary of Answers** - **Part A:** \( F_3 = \frac{\sqrt{5}}{2} F_1 \) - **Part B:** \( \theta = 26.6^\circ \)