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A uniform massive string of mass $m$ and length $l$ is fixed between two rigid supports. Angles made by the tangents on the string at points of suspension are shown in the figure. The point C is the lowermost point of the string. Then choose the correct statements: $\left(\text{Use} \cos(37^{\circ})=\dfrac45\right)$

(A) Length of the string left of the point C is $\dfrac{16l}{25}$

(B)Tension in the string at point C is $\dfrac{12mg}{5}$

(C)Radius of curvature of the string at point C is $\dfrac{12l}{25}$

(D)Tension in the string at point B is $\dfrac{3mg}{5}$

I found the tensions in the left (call it $T_1$) and right parts (call it $T_2$) by simply using the equations:

$\begin{align}T_1 &\cos(37^{\circ}) + T_2\cos(53^{\circ})=mg \hspace{50pt} \text{(for vertical equilibroum)} \\ &T_1\sin(37^{\circ})=T_2\sin(53^{\circ}) \hspace{50pt} \text{(for horizontal equilibrium)}\end{align}$

So that $T_1=\dfrac{4mg}{5}$ and $T_2=\dfrac{3mg}{5}$ and that indeed is the correct answer.

For finding the radius of curvature and length, I need to form a differential equation, but that is where I am stuck. Ideally, the heights of the strings are equal in a catenary and one could use the equation $y=a\cosh\left(\dfrac{x}{a}\right)$, but how do we solve this case?

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You are on the correct track. If you place a coordinate system on the lowest point then

fig

$$ y(x) = a \left( \cosh \left( \tfrac{x}{a} \right)-1 \right) $$

is the shape of the catenary with $$ a= \frac{H}{w} $$

Here $H$ is the horizontal component of tension, and $w$ is the weight per unit length. $$ w = \tfrac{m g}{\ell} $$

The radius of curvature $R$ is found by equating

$$ \tfrac{x^2}{2 R} = \mathrm{taylor}( y(x), x=0, \mathrm{order}=2) = \tfrac{x^2}{2 a} $$

Or $R=a$.

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  • $\begingroup$ Thank you sir, for the answer. I am solving this type question for the first time, so I have very little knowledge about the catenary equation. Can you tell why or how $a=\dfrac{H}{w}$, I have seen this equation on wikipedia and on many other sites, but they just mention $a$ as some constant, don't define what. So how do we know that $a=\dfrac{H}{w}$? and does $a=\dfrac{H}{w}$ for all catenary equations? $\endgroup$
    – V.G
    Commented Dec 10, 2020 at 8:43
  • $\begingroup$ look at this answer where the equations are developed, and notice than in $$ H \frac{{\rm d}^2 y}{{\rm d}x^2} = \frac{m g}{\ell} \sqrt{1+\left(\frac{{\rm d}y}{{\rm d}x}\right)^2} $$ the factor $H/m g$ comes out naturally if you divide both sides of the equation with $m g/ \ell$. The equation to be solved is then $$ a \frac{{\rm d}^2 y}{{\rm d}x^2} = \sqrt{1+\left(\frac{{\rm d}y}{{\rm d}x}\right)^2}$$ where $w = m g / \ell$ and $a = H/w$ $\endgroup$
    – John Alexiou
    Commented Dec 10, 2020 at 22:36

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