Question

Figure shows an inextensible catenary mooring cable in still water. The submerged weight (per meter length), and the anchor radius are 100 kg/m and 50 m, respectively. If horizontal tension ($T_h$) in the catenary is 1600 kg, the catenary length (AB) is ............. m (rounded to two decimal places). 

Hint:

In catenary problems, use $a = T_h / w$, then apply $L = a \sinh(x/a)$ for arc length and $y = a \cosh(x/a) - a$ for vertical displacement.

Answer 1

- Given data:
- Submerged weight per unit length ($w$) = 100 kg/m
- Horizontal distance ($x$) = 50 m
- Horizontal tension ($T_h$) = 1600 kg
- For a catenary: \[ a = \frac{T_h}{w} = \frac{1600}{100} = 16 \; \text{m} \] - Horizontal projection of cable is $x = 50$ m. Catenary equation: \[ y = a \cosh\left(\frac{x}{a}\right) - a \] - Arc length of cable (AB) from $x=0$ to $x=50$: \[ L = a \sinh\left(\frac{x}{a}\right) \] - Substituting values: \[ L = 16 \sinh\left(\frac{50}{16}\right) = 16 \sinh(3.125) \] - $\sinh(3.125) \approx 11.35$. \[ L = 16 \times 11.35 \approx 181.6 \; \text{m} \] - Hence, the catenary length is about 181–182 m.