Forces acting on rope I recently asked my physics teacher this question, and he could not give me a clear answer:
How gravity acts on a streched rope?
Consider example like this one below: if we strech a rope of mass m between points distance d apart, how low will the rope hang? And what will be the tension on the rope? Is rope a perfect parabola in this scenario? I just find it extremely quaint and I couldn't find the answer anywhere.

 A: The shape of a freely hanging massive rope in gravity is a catenary. Parabola comes as a  natural guess to early learners. Even Galileo once thought it to be a parabola.
Tension is varying at every points but calculation of tension at end points can be done by force balance equations.
$$Derivation:$$
Consider a freely hanging massive rope with uniform linear mass density $ {\lambda}$. Let $x$ and  $y$ axes represent horizontal and vertical co - ordinate axes. Point$(0,0)$ be the lowermost point i.e. the point where $\frac{dy}{dx} = 0$

Observe : By horizontal force balance horizontal component of tension is same everywhere. Assign it a variable $T_H$ which can be calculated easily by vertical force balance at the end points  and torque balance equation about center of mass (You will need this equation in calculation of $T_H$ in asymmetrical case.)

Between any arbitrary $x$ and $x+dx$:
$$T_H \left(\frac{dy}{dx}_{x+dx} - \frac{dy}{dx}_x\right) = {\lambda}dsg$$
Since, $$ds^2= dx^2+dy^2$$
$$or , ds = dx \sqrt{1+\left(\frac{dy}{dx}\right)^2}$$
The above given equation becomes -
$$T_H \left(\frac{dy}{dx}_{x+dx} -\frac{dy}{dx}_x\right) = {\lambda}gdx\sqrt{1+\left(\frac{dy}{dx}\right)^2}$$
$$or,  \frac{d^2y}{dx^2}=\frac{{\lambda}g}{T_H}\sqrt{1+\left(\frac{dy}{dx}\right)^2}$$
Now to further progress from here , making use of $u$
substitution , put $\frac{dy}{dx} = u$.
$$\frac{du}{dx} = \frac{{\lambda}g}{T_H} \sqrt{1+u^2}$$
The solution of this differential equation is -
$$u = \sinh \left(\frac{{\lambda}g}{T_H}x\right)$$
and thus applying proper boundary conditions-
$$y=\frac{T_H}{{\lambda}g} \cosh \left(\frac{{\lambda}g}{T_H}x\right) - \frac{T_H}{{\lambda}g}$$
See , the given equation is for symmetrical case . By making the situation asymmetric there will only be difference in boundary conditions.
A: The shape is the catenary.
To derive it take a small section of the rope and balance the forces. The shape of the rope follows the curve $y = y(x)$.

Here the segment with length ${\rm d}s = \sqrt{{\rm d}x^2 + {\rm d}y^2} = (\sqrt{1 + y'^2}) {\rm d}x$ has weight per length of $w$ and thus the total vertical external force applied is $w {\rm d}s$.
The segment is pulled to the left by tension with horizontal component $H$, and to the right by the horizontal component $H+{\rm d}H$. But since there are not external forces along the horizontal, it means that $${\rm d}H = 0 \tag{1}$$, or that the horizontal component of the tension is always constant along the rope.
Similarly, the left side is pulled down by the cable by the vertical component $V$ and pulled up by $V+{\rm d}V$. The balance of forces in the vertical direction is $(V + {\rm d}V) - V - w{\rm d}s =0$ or $$ {\rm d}V = w {\rm d}s \tag{2}$$
The tension is also always tangent to the rope which means that at any point
$$ (\text{slope}) = y' = \tfrac{V}{H} \tag{3} $$
Next re-arrange the above as $V = H y'$ and take the derivative with respect to $x$ on both sides
$$ \require{cancel} \tfrac{ {\rm d}V}{{\rm d}x} = H y'' + \cancel{\tfrac{ {\rm d}H}{{\rm d}x}} y' $$ or
$$ w \tfrac{ {\rm d}s}{{\rm d}x} = w \sqrt{1+y'} = H y'' \tag{4}$$
The solution to the above is given by an equation of the form $$y(x) = y_0 + a \left( \cosh \left( \frac{x-x_0}{a} \right) -1 \right) \tag{5}$$
The $\cosh$ function is what gives it the catenary shape. I leave it up to the reader to evaluate $y'$ and $y''$ and prove that (5) solves (4).
Note that the point $(x_0,\,y_0)$ is the lowest point on the curve, and the parameter $a$ is called the catenary constant (with length units) describes the radius of curvature of the rope at the lowest point.
A: It is interesting to note that the questioner’s intuition about the curve being a parabola is correct if the suspended weight varies linearly with $x$ rather than with the curve length $s$. A similar analysis to those in other answers then gives
$\displaystyle \frac{d^2y}{dx^2}= \text{constant}$
A practical example of this is a suspension bridge where the weight of the cables is much smaller than the weight of the horizontal suspended roadway.
A: You are looking for this word: catenary
Deriving this from Newton's laws is not trivial. Fortunately, there's a derivation on Wikipedia.
