Let's break the problem into two steps as requested: --- ## **Step 1: Find the angle θ of the cable with the horizontal** ### **Explanation** - The tension graph \( T \) vs. \( y/L \) shows that at \( y/L = 1 \), the tension \( T = T_S = 550\,\text{N} \). - The force \( F_a \) creates a counterclockwise torque about the hinge, balanced by the torque from the cable tension. - For equilibrium at \( y = L \), sum of torques about the hinge: \[ F_a \cdot L = T \cdot L \sin\theta \] At \( y=L \), \( T = 550\,\text{N} \). - But \( F_a \) is not directly given. However, we can use the graph for \( F_h \) (horizontal force from hinge) at \( y/L = \), which is \( F_h = 200\,\text{N} \). #### **From the diagram and question:** - At \( y/L = \), the hinge must provide all the horizontal support, so: \[ T\cos\theta = F_h \] At \( y/L= \): \[ T = \implies F_h = 200\,\text{N} \] But at \( y/L=1 \), \( F_h = \) (from the graph). - At \( y/L=1 \): The entire horizontal force is taken by the cable, so: \[ T\cos\theta = F_a \] At \( y/L=1 \), \( F_h = \), \( T = 550\,\text{N} \), so all horizontal support is from cable. - Setting \( F_a = T\cos\theta \): But we need another equation. Recall at \( y/L= \), \( F_h = 200\,\text{N} \), and at this point \( T = \). #### **From the equations:** - For \( y/L = 1 \): \[ F_a = T\cos\theta \] But we don't know \( F_a \). #### **Let’s check the torque balance at \( y/L=1 \):** - At \( y=L \): \[ F_a \cdot L = T \cdot L \sin\theta \implies F_a = T\sin\theta \] - But above we had \( F_a = T\cos\theta \). Since the force is horizontal, it's the horizontal component of the tension that balances \( F_a \). So, by geometry, the cable pulls at an angle θ, and its horizontal component must equal \( F_a \): \[ T \cos\theta = F_a \] And vertical components balance the weight, but that's not directly needed here. #### **Combine with torque:** \[ F_a \cdot L = T \cdot L \sin\theta \implies F_a = T\sin\theta \] Thus, from above, \[ T \cos\theta = T \sin\theta \implies \cos\theta = \sin\theta \implies \theta = 45^\circ \] --- ### **Step 1 Answer** #### **(a) Angle θ** \[ \boxed{45\ ^\circ} \] **Explanation:** The balance of torques and the horizontal force leads to the conclusion that the cable is at a \(45^\circ\) angle with the horizontal. --- ## **Step 2: Find the magnitude of \(\vec{F}_a\)** ### **Explanation** - From above, at \( y = L \), \( F_a = T \cos\theta \). - We know \( T = 550\,\text{N} \), and \( \theta = 45^\circ \). \[ F_a = T \cos\theta = 550\,\text{N} \cdot \cos(45^\circ) = 550\,\text{N} \cdot \frac{1}{\sqrt{2}} \approx 389\,\text{N} \] --- ### **Step 2 Answer** #### **(b) Magnitude of \(\vec{F}_a\)** \[ \boxed{389\ \text{N}} \] **Explanation:** At \( y/L = 1 \), the tension in the cable is \(550\) N, and with \( \theta = 45^\circ \), the horizontal force applied is \(389\) N. --- ## **Final Answers** | Part | Value | Units | |------|-------|-------| | (a) | 45 | degrees | | (b) | 389 | N |