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A uniform rope of length $l = 12 metres$ is suspended between two fixed nails $A$ and $B$ that are in the same horizontal level. The rope makes an angle $θ = 37 °$ with the horizontal at the nails. Find the radius of curvature of the rope at the lowest point in metres.

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answer is 8.

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Detailed Solution

Let the tension at highest point be F_o and at lowest point be F.

Then $F_{o} C o s θ = F$ and $F_{o} \sin θ = \frac{λ l g}{2}$
Therefore $F = \frac{λ l g}{2} \cot θ$ and if F is tension at lowest point and R is radius of curvature then

$F d θ = λ R d θ g$ .

Calculating we get R = 8 m.

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