Let's solve the problem step by step. --- ## **Step 1: Draw the Free-Body Diagram** The 20-kg mass is suspended by three cords: - \( T_1 \) (horizontal, attached to the left wall) - \( T_2 \) (makes a 60° angle above the horizontal, attached to the ceiling) - \( T_3 \) (vertical, supporting the mass) --- ## **Step 2: Write the Force Equations** The mass is in equilibrium, so the sum of forces in both x and y directions must be zero. Let: - \( T_1 \): tension in the horizontal cord (left) - \( T_2 \): tension in the cord at 60° above the horizontal (right) - \( T_3 \): tension in the vertical cord (downward; supports the weight) - \( m = 20\,\text{kg} \) - \( g = 9.8\,\text{m/s}^2 \) **Vertical (y-direction):** \[ T_2 \sin 60^\circ + T_1 \sin 90^\circ = T_3 \] But \(T_1\) is horizontal, so \(T_1 \sin 90^\circ = T_1 \cdot = \). So, \[ T_2 \sin 60^\circ = T_3 \] **Horizontal (x-direction):** \[ T_1 = T_2 \cos 60^\circ \] **Weight:** \[ T_3 = mg = 20 \times 9.8 = 196\,\text{N} \] --- ## **Step 3: Substitute and Solve** ### **1. Calculate \( T_2 \)** \[ T_2 \sin 60^\circ = T_3 \] \[ T_2 = \frac{T_3}{\sin 60^\circ} \] \[ \sin 60^\circ = \frac{\sqrt{3}}{2} \approx .866 \] \[ T_2 = \frac{196}{.866} \approx 226.4\,\text{N} \] --- ### **2. Calculate \( T_1 \)** \[ T_1 = T_2 \cos 60^\circ \] \[ \cos 60^\circ = .5 \] \[ T_1 = 226.4 \times .5 = 113.2\,\text{N} \] --- ### **3. Tension in the vertical cord (\( T_3 \))** \[ T_3 = 196\,\text{N} \] --- ## **Step 4: Final Answers** \[ \boxed{ \begin{aligned} T_1 &= 113.2\,\text{N} \\ T_2 &= 226.4\,\text{N} \\ T_3 &= 196\,\text{N} \end{aligned} } \] --- ### **Summary Table** | Cord | Tension (N) | |------|-------------| | \(T_1\) | 113.2 | | \(T_2\) | 226.4 | | \(T_3\) | 196 | If you need a diagram or further breakdown, let me know!