Tension in a hanging chain

Consider a very small part of the chain with length $dl$ and mass $dm=\frac{Mdl}{L}$ as in this image:

image

We know that $T_x$ do not change along the chain and $dT_y=gdm$ . Therefore, at the end points, $T_{y0}=\frac{Mg}{2}$ .

Since $n=\frac{\sqrt{T_x^2+T_{y0}^2}}{T_x}$ , so $T_x=\frac{T_{y0}}{\sqrt{n^2-1}}=\frac{Mg}{2\sqrt{n^2-1}}$

We have $\tan \theta=\frac{T_y}{T_x}$ , so $\frac{d\theta}{\cos^2 \theta}=\frac{dT_y}{T_x}=\frac{gMdl}{LT_x}$ .

$\frac{d\theta}{\cos \theta}=\frac{gMdl \cos \theta}{LT_x}=\frac{gMdx}{LT_x}$ .

$\int\limits_0^{\theta_0} \frac{d\theta}{\cos \theta}=\int\limits_0^{\frac{d}{2}} \frac{gMdx}{LT_x}$ .

$ln(\tan \theta_0 + \frac{1}{\cos \theta_0})=\frac{gMd}{2LT_x}$

(with $\cos \theta_0=\frac{T_x}{T_0}=\frac{1}{n}$ )

$d=\frac{2LT_x}{gM} ln(\tan \theta_0 + \frac{1}{\cos \theta_0})=\frac{ln(\sqrt{n^2-1}+n)}{\sqrt{n^2-1}} L =1.14(m)$ .

My solution assumes knowledge of the equations and properties of a catenary curve. In general, the shape of any hanging rope is known as a catenary. You can see the equations of a catenary here , and the proof here .

Note that the tension at the lowest point is completely horizontal. Let $T_0$ be the magnitude of the tension at the lowest point. Since the rope is stationary, the horizontal component of tension must be fixed throughout the rope. This shows that the horizontal component of tension at the points of suspension must also be $T_0$ . Let $\lambda$ be the mass per unit length of the catenary, and let $g$ be the acceleration due to gravity. The vertical component of tension at the point of suspension is simply half the weight of the rope, i.e $\frac{\lambda g L}{2}$ . The total tension on the points of suspension is given by: $T= \sqrt{T_0^2 + \left ( \frac{\lambda g L}{2} \right ) ^2}$ From the problem statement, we have $T= n \times T_0$ , so: $nT_0= \sqrt{T_0^2 + \left ( \frac{\lambda g L}{2} \right ) ^2}$ $\implies n^2 T_0^2 = T_0^2 + \left ( \frac{\lambda g L}{2} \right ) ^2$ $\implies \frac{\lambda g L}{2} = T_0\sqrt{n^2-1}$ $\implies \frac{T_0}{\lambda g}= \frac{L}{2\sqrt{n^2-1}}$ The equation of the catenary will then be: $y= \left ( \frac{T_0}{\lambda g} \right ) \times \cosh \left ( \frac{x}{\frac{T_0}{\lambda g}} \right )$ $\implies y= \left ( \frac{L}{2\sqrt{n^2-1}} \right ) \times \cosh \left ( \frac{x}{\frac{L}{2\sqrt{n^2-1}}} \right )$ Let the distance between the points of suspension be $d$ . Note that in our equation, we take the point directly $1$ unit below the vertex of the catenary as the origin. The length of the catenary from $x=0$ to $x= \pm \frac{d}{2}$ is given by: $L= 2 \times \left ( \frac{T_0}{\lambda g} \right ) \times \sinh \left ( \frac{\frac{d}{2}}{\frac{T_0}{\lambda g}} \right )$ $\implies L= 2 \times \left ( \frac{L}{2\sqrt{n^2-1}} \right ) \times \sinh \left ( \frac{\frac{d}{2}}{\frac{L}{2\sqrt{n^2-1}}} \right )$ $\implies \frac{1}{n^2-1} \times \sinh \left ( \frac{\frac{d}{2}}{\frac{L}{2\sqrt{n^2-1}}} \right ) = 1$ $\implies \sinh \left ( \frac{d \times \sqrt{n^2-1}}{L} \right ) = n^2 - 1$ $\implies \frac{d \times \sqrt{n^2-1}}{L} = \sinh^{-1} (n^2-1)$ $\implies d= \frac{\sinh^{-1}(n^2-1) \times L}{\sqrt{n^2-1}}$ Plugging the numerical values from the question gives $d \approx \boxed{1.14 \text{ m}}$ .

Let $\lambda$ be the mass per unit length and $T_{x}$ be the horizontal tension in the spring.

At uppermost point, $T_{y} = \frac{3 \lambda g}{2}$ , $T_{y}^2 + T_{x}^2 = 49 T_{x}^2 \Rightarrow$ $\frac{\lambda g}{T_{x}} = \frac{8}{3} \sqrt{3} = \frac{8}{\sqrt{3}}$

In such a curve if we define $x = 0 , y=0$ at lower most point,

$\frac{dy}{dx} = sinh(\frac{\lambda g}{T_{x}}x)$

$dl = \sqrt{dx^2 + dy^2} = dx\sqrt{1 + (\frac{dy}{dx})^2} = \sqrt{1 + sinh^2(\frac{8}{\sqrt{3}}x)}$

$\int_{0}^{\frac{d}{2}} \sqrt{1 + sinh^2(\frac{8}{\sqrt{3}}x)}dx = \frac{3}{2}$

Solve to get $d = \frac{\sqrt{3}}{2}log(2 + \sqrt{3}) = \fbox{1.14 m}$

We have: $T\text{cos }\!\!~\!\!\text{ }\theta ={{T}_{0}}~$ $\cos \theta =\frac{1}{7}$ $\theta \approx 1.4274$

We also have the famous result:

$s=a\tan \theta =a\sinh \frac{x}{a}$

$\therefore a=\frac{1.5}{\tan 1.4274}\approx 0.2165$

Thus,

$\tan \theta =\sinh \frac{x}{0.2165}$

Solve for x,

$x\approx 0.5703$

But this is only half the width. Hence, the result is 1.14 m