1. **State the problem:** A body of mass 8 kg is suspended by two strings Mx and Nx attached to the ceiling. The strings make angles of 55° and 65° with the downward vertical. We need to find the tension in each string. 2. **Known values:** - Mass, $m = 8$ kg - Gravitational acceleration, $g = 10$ m/s$^2$ - Angles with downward vertical: $\theta_M = 55^\circ$, $\theta_N = 65^\circ$ 3. **Calculate the weight of the body:** $$ W = mg = 8 \times 10 = 80 \text{ N} $$ 4. **Set up the equilibrium equations:** Since the body is at rest, the sum of forces in both horizontal and vertical directions must be zero. - Let tensions be $T_M$ and $T_N$ in strings Mx and Nx respectively. - Resolve tensions into vertical and horizontal components: - Vertical components: $T_M \cos 55^\circ$ and $T_N \cos 65^\circ$ - Horizontal components: $T_M \sin 55^\circ$ and $T_N \sin 65^\circ$ 5. **Vertical equilibrium:** $$ T_M \cos 55^\circ + T_N \cos 65^\circ = 80 $$ 6. **Horizontal equilibrium:** The horizontal components must balance each other: $$ T_M \sin 55^\circ = T_N \sin 65^\circ $$ 7. **Express $T_N$ in terms of $T_M$ from horizontal equilibrium:** $$ T_N = T_M \frac{\sin 55^\circ}{\sin 65^\circ} $$ 8. **Substitute $T_N$ into vertical equilibrium:** $$ T_M \cos 55^\circ + T_M \frac{\sin 55^\circ}{\sin 65^\circ} \cos 65^\circ = 80 $$ 9. **Simplify:** Calculate the trigonometric values: - $\cos 55^\circ \approx 0.5736$ - $\sin 55^\circ \approx 0.8192$ - $\cos 65^\circ \approx 0.4226$ - $\sin 65^\circ \approx 0.9063$ Substitute: $$ T_M (0.5736) + T_M \times \frac{0.8192}{0.9063} \times 0.4226 = 80 $$ Calculate the fraction: $$ \frac{0.8192}{0.9063} \times 0.4226 \approx 0.3819 $$ So: $$ T_M (0.5736 + 0.3819) = 80 $$ $$ T_M (0.9555) = 80 $$ 10. **Solve for $T_M$:** $$ T_M = \frac{80}{0.9555} \approx 83.7 $$ 11. **Find $T_N$:** $$ T_N = 83.7 \times \frac{0.8192}{0.9063} \approx 83.7 \times 0.9045 = 75.7 $$ 12. **Final answer:** - Tension in string Mx, $T_M \approx 84$ N - Tension in string Nx, $T_N \approx 76$ N