When an object of weight $W$ is suspended from the center of a massless string that is attached at both ends to the ceiling, with the rope making an angle $\theta$ with the ceiling at both attachment points, the tension in the string can be found as follows:

Step 1: Analyze the forces on the object

• The object is in equilibrium, so the sum of the forces acting on it must be zero.
• The forces acting on the object are:
  1. The weight of the object, $W$, acting vertically downward.
  2. The tension, $T$, in the string, which has two components:
     • A vertical component, $T \sin(\theta)$, balancing the weight of the object.
     • A horizontal component, $T \cos(\theta)$, which balances the tension in the other part of the string.

Step 2: Resolve the forces

Since the string is massless and the object is in equilibrium, the tension must be the same throughout the string. The vertical component of the tension should balance the weight $W$:

$2 T \sin(\theta) = W$

The factor of 2 comes from the fact that the string is split into two equal parts, each contributing a vertical component of tension $T \sin(\theta)$.

Step 3: Solve for the tension $T$

$T = \frac{W}{2 \sin(\theta)}$

Final Answer:

The tension in the string is:

$T = \frac{W}{2 \sin(\theta)}$

This equation tells you that the tension in each half of the string depends on the weight of the object and the angle $\theta$ between the rope and the ceiling.

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Would you like further details or explanations on this? Here are a few related questions to consider:

1. How does the angle $\theta$ affect the tension in the string?
2. What happens if the angle $\theta$ is 0 or 90 degrees?
3. How does the tension change if the object is not at the center of the string?
4. What would happen if the string was not massless?
5. How would the tension change if there were more than one object hanging?

Tip: In problems involving equilibrium, always remember that the sum of forces in both the horizontal and vertical directions must be zero!