A flexible chain of weight W hangs between two fixed positions A and B which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is θ. What is the tension in the chain at the midpoint?

(A) (W/2) sec θ
(B) (W/2) tan θ
(C) (W/2) cot θ
(D) None

To solve this problem, we need to consider the situation where a flexible chain is suspended between two fixed points, and we are asked to determine the tension at the midpoint of the chain.
Key Information:
• The weight of the chain is WW.
• The chain hangs between two fixed points AA and BB, both at the same horizontal level.
• The inclination of the chain with the horizontal at both the points of support is θ\theta.
• We are asked to determine the tension at the midpoint of the chain.
Concepts Involved:
1. Shape of the Chain: The chain forms a curve that is typically a catenary curve when the chain is flexible and under the influence of gravity. However, for small inclinations or approximations, we assume that the chain forms a shape where the tension is approximately constant along its length.
2. Force Balance and Tension: At any point along the chain, there are two forces acting on it:
o The vertical component of the tension that supports the weight of the chain.
o The horizontal component of the tension that keeps the chain in equilibrium.
Since the points AA and BB are at the same height, the horizontal components of tension at both points are equal. The tension at any point in the chain will vary along the length, but for simplicity, we focus on the midpoint.
At the Midpoint of the Chain:
At the midpoint of the chain, the chain is symmetrically distributed. Therefore, the weight of the chain is evenly distributed on either side of the midpoint, each supporting half of the total weight WW. The tension at the midpoint can be derived using the equilibrium conditions for the forces at that point.
• The vertical component of the tension at the midpoint supports half of the weight, i.e., W2\frac{W}{2}.
• The horizontal component of the tension at the midpoint remains constant throughout the chain, but it does not affect the vertical equilibrium at the midpoint.
Given the angle of inclination θ\theta at the points of support, we can use trigonometry to find the relationship between the tension and the weight. The tension at the midpoint can be expressed as:
T=W2cos⁡θT = \frac{W}{2 \cos \theta}
This is because the vertical component of the tension at the midpoint should balance half of the weight of the chain, and the angle θ\theta at the supports affects the vertical component of the tension.
Answer:
The tension at the midpoint of the chain is:
T=W2sec⁡θT = \frac{W}{2} \sec \theta
Thus, the correct answer is (A) W2sec⁡θ\frac{W}{2} \sec \theta.