Hanging cable

This problem can be defeated by looking at the torque about the lowest point on the cable.

As we move along the cable, the mass that must be supported by the vertical component of the tension is constantly changing. The horizontal component however is constant. Pick any segment of the cable and calculate the net force on it in the horizontal dimension, with any two endpoints $p_1$ and $p_2$ we must have $T_x(p_1) = T_x(p_2)$ (or else the cable would accelerate in either direction, but alas, it is stationary).

Therefore, it must be that the tension at the lowest point in the wire (which is purely horizontal) must be equal to the horizontal component of the tension at any other point.

We have three torques to consider, the vertical component of the tension at the support, which is equal to half the weight of the wire and acts through the perpendicular distance $l/2$ from the origin (point L in the diagram), the horizontal component of the tension at the support and acts through the perpendicular distance $h$ to the origin, and the weight of the wire which acts at the average perpendicular distance $l/4$ origin. This is diagrammed below:

Fig 1.

Balancing the torques, we have

$\begin{aligned} \lambda\frac{l^2}{8}g + T_xh &= \lambda\frac{l^2}{4}g \\ T_xh &= \lambda\frac{l^2}{8}g \\ T_x &= \lambda\frac{l^2}{8h}g \end{aligned}$

Plugging in the values from the problem, we find that the tension at the low point is $3500\mbox{ N}$ .

It is well-known that a hanging rope assumes the shape of a catenary, with equation

$y=a\cosh\frac{x}{a}$

where $a=\frac{T}{\lambda g}$ . For $x=0$ , we have the height $a$ . For $x=\frac{d}{2}$ , we have the height $a\cosh\frac{d}{2a}$ . But, the problem also tells us that $a\cosh\frac{d}{2a}-a=h$ . So, one can solve for $a$ in the equation

$a(\cosh\frac{50}{a}-1)=0.7$

and plug in $T=a\lambda g=(1785.83094m)(0.2kg/m)(9.81m/s^2)\approx\boxed{3500N}$

The free-hanging cable has a catenary curve of the form:

$y = a \cosh { \left( \dfrac {x} {a} \right) }$

where the y-axis pass through the lowest point, and

• $a = \dfrac {T_0} {\lambda g}$
• $T_0 =$ the tension of the cable at the lowest point
• $\lambda = 0.2$ kg/m, the mass per unit length of the cable
• $g = 9.8$ m/s $^2$

Therefore, $y(50) - y(0) = a \cosh {\left( \dfrac {50} {a} \right) } - a = h = 0.7$

By numerical method, it is found that $T_0 = \boxed {3500} N$

This is a great problem and I would thank Brilliant for this. I am sure that most of the others must have done this assuming the curve to be a parabola and the radius twice the focus but that is an approximation...

Since the curve will be that of a Catenary passing through (50 , a+0.7) we put these in the equation of catenary and get a .Since we need to find the tension(T) at the lowest point which is equal to the constant horizontal force H=a w (where W is the weight per unit length i.e. 0.2*9.8).

Thus answer is 1785.83 0.2 9.8 = 3500.2268