## Step 1: **Free-Body Diagram and Tension Analysis** Let’s analyze the forces at point \(A\): - The weight of block \(D\): \(W_D = 4g\) - The weight of block \(E\): \(W_E = 6g\) - Tension in \(AB\): \(T_1\) - Tension in \(AC\): \(T_2\) - The downward force at \(A\): \(F\) **Horizontal and Vertical Components:** - For equilibrium at \(A\): \[ \sum F_x = \implies T_1 \cos\theta_1 = T_2 \cos\theta_2 \] \[ \sum F_y = \implies T_1 \sin\theta_1 + T_2 \sin\theta_2 = F \] **Geometry:** - \(AB\) and \(AC\) have horizontal lengths: - \(AB = 6-x = 4\) - \(AC = x = 2\) - Both have vertical distance \(s\). So, \[ \cos\theta_1 = \frac{4}{\sqrt{4^2 + s^2}} \] \[ \sin\theta_1 = \frac{s}{\sqrt{4^2 + s^2}} \] \[ \cos\theta_2 = \frac{2}{\sqrt{2^2 + s^2}} \] \[ \sin\theta_2 = \frac{s}{\sqrt{2^2 + s^2}} \] **Tension values:** - \(T_1 = W_D = 4g\) - \(T_2 = W_E = 6g\) **End of Step 1 Explanation:** We have resolved the geometry and set up equations for equilibrium at point \(A\), relating the sag \(s\) to the weights and positions of the blocks. --- ## Step 2: **Set Up the Equilibrium and Solve for \(s\)** **Horizontal equilibrium:** \[ T_1 \cos\theta_1 = T_2 \cos\theta_2 \] \[ 4g \frac{4}{\sqrt{16 + s^2}} = 6g \frac{2}{\sqrt{4 + s^2}} \] \(g\) cancels: \[ 4 \cdot \frac{4}{\sqrt{16 + s^2}} = 6 \cdot \frac{2}{\sqrt{4 + s^2}} \] \[ \frac{16}{\sqrt{16 + s^2}} = \frac{12}{\sqrt{4 + s^2}} \] Cross-multiplied: \[ 16\sqrt{4 + s^2} = 12\sqrt{16 + s^2} \] Divide both sides by 4: \[ 4\sqrt{4 + s^2} = 3\sqrt{16 + s^2} \] Square both sides: \[ 16(4 + s^2) = 9(16 + s^2) \] \[ 64 + 16s^2 = 144 + 9s^2 \] \[ 16s^2 - 9s^2 = 144 - 64 \] \[ 7s^2 = 80 \] \[ s^2 = \frac{80}{7} \] \[ s = \sqrt{\frac{80}{7}} \] --- ### **Final Answer:** \[ \boxed{s = \sqrt{\frac{80}{7}}} \] --- ### **End of Step 2 Explanation:** We substituted the known values, simplified, and solved for the sag \(s\) using the conditions for equilibrium.