Let's solve the problem **step by step**: --- ## **Given:** - Mass, \( = 20 \) kg - Angle with the horizontal, \( \theta = 60^\circ \) - Gravity, \( g = 9.8\, \text{m/s}^2 \) --- ## **Step 1: Draw Free-Body Diagram** Let: - \( T_1 \): Tension in the horizontal cord (left) - \( T_2 \): Tension in the cord at 60° to the horizontal (right) - \( T_3 \): Tension in the vertical cord (down), which equals the weight of the mass --- ## **Step 2: Write Force Equilibrium Equations** The mass is in **equilibrium**, so the **sum of forces in both x and y directions must be zero**. ### **Vertical Direction (y-axis):** \[ T_2 \sin(60^\circ) + T_1 \cdot = T_3 \] But from the diagram, \(T_1\) is horizontal, so it doesn’t contribute to the y-direction. \[ T_2 \sin(60^\circ) = T_3 \] ### **Horizontal Direction (x-axis):** \[ T_2 \cos(60^\circ) = T_1 \] --- ## **Step 3: Calculate \( T_3 \)** \[ T_3 = \text{Weight} = mg = 20 \times 9.8 = \boxed{196\, \text{N}} \] --- ## **Step 4: Solve for \( T_2 \)** From vertical equilibrium: \[ T_2 \sin(60^\circ) = 196 \] \[ T_2 = \frac{196}{\sin(60^\circ)} \] \[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx .866 \] \[ T_2 = \frac{196}{.866} \approx \boxed{226.4\, \text{N}} \] --- ## **Step 5: Solve for \( T_1 \)** From horizontal equilibrium: \[ T_1 = T_2 \cos(60^\circ) \] \[ \cos(60^\circ) = .5 \] \[ T_1 = 226.4 \times .5 = \boxed{113.2\, \text{N}} \] --- ## **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 |