To solve the problem, we need to determine the tension in the rope when a 200 kg load is placed at its center. The weight of the load creates two tension forces in the rope, one from each side. The key is to apply principles of equilibrium, as the system is at rest.

Step-by-step solution:

1. Given:

   • Length of the rope: $L = 20 \, \text{m}$
   • Distance between the two ends on the ceiling: $d = 12 \, \text{m}$
   • Weight of the load: $W = 200 \, \text{kg}$
   • Gravitational acceleration: $g = 9.8 \, \text{m/s}^2$

2. Calculating the weight of the load: The force due to the weight of the load (which is a vertical force) is: $F = mg = 200 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1960 \, \text{N}$ This force is acting downward at the center of the rope.

3. Geometry of the problem:

   • The total length of the rope is 20 m, but the horizontal distance between the two ends of the rope is only 12 m.
   • This means the rope must sag vertically. Let the vertical distance from the ceiling to the center of the rope be $h$.
   • We will assume the shape formed by the rope is symmetric, so each half of the rope forms a right triangle.

4. Finding the vertical sag $h$: Each half of the rope has a length of $L/2 = 10 \, \text{m}$, and the horizontal distance between the center and one end is $d/2 = 6 \, \text{m}$. Using the Pythagorean theorem, we can find the vertical sag $h$: $\left(\frac{L}{2}\right)^2 = \left(\frac{d}{2}\right)^2 + h^2$ $10^2 = 6^2 + h^2$ $100 = 36 + h^2$ $h^2 = 64 \implies h = 8 \, \text{m}$ So, the vertical sag is 8 m.

5. Forces acting on the system: The system is in equilibrium. The vertical force due to the weight of the load is balanced by the vertical components of the tension in each half of the rope.

   Let $T$ be the tension in the rope. The vertical component of the tension from each side is: $T_v = T \sin \theta$ where $\theta$ is the angle that the rope makes with the horizontal. The angle $\theta$ can be found using trigonometry: $\tan \theta = \frac{h}{d/2} = \frac{8}{6}$ $\theta = \tan^{-1}\left(\frac{8}{6}\right) \approx 53.13^\circ$

6. Summing the forces: The vertical forces must balance the weight of the load: $2 T_v = F \implies 2 T \sin \theta = 1960 \, \text{N}$ $T \sin \theta = 980 \, \text{N}$ Using $\sin \theta = \sin 53.13^\circ \approx 0.8$, we can find $T$: $T \times 0.8 = 980 \, \text{N}$ $T = \frac{980}{0.8} = 1225 \, \text{N}$

Thus, the tension in the rope is 1225 N.

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Would you like any further details or explanations?

Here are 5 related questions you might find useful:

1. How would the tension change if the rope were longer or shorter?
2. What if the load was not placed at the center of the rope?
3. How does the angle of the rope affect the tension?
4. What is the relationship between the sag $h$ and the tension in the rope?
5. How would the tension change if the load was heavier or lighter?

Tip: In such problems, always start by analyzing forces and setting up equilibrium equations for both vertical and horizontal components.