I have recently started studying Common Catenary (rope with uniform mass hanging between two points not on the same vertical line) .
While deriving the equation of the common catenary , we assume that the bottom most part of the string is perfectly horizontal and therefore calculate the equations by assuming that the bottom most part of the string's tension(say T0) is only horizontal.
It amused me when i was solving a problem in which a particle of mass m was placed at the vertex of the string . After placing the mass at the vertex , the tension at the bottom-most point of the string was not perfectly horizontal and now the string was inclined at some angle at the vertex .
My doubt therefore is that how can we prove that the tension at the lowest point of the string (when there is no weight present at the bottom of the string) is perfectly horizontal ?
I am asking this to the math community as i wanted some dervations to analyse the case.Thanks !!
Best Answer
It is our own physics assumption in force modeling.
There are no concentrated forces kept at the vertex or at any other point on the cable. All the weight $ W=\int q ds $ that acts comes from the the cable uniformly distributed along its length.
To find the shape of the cable we make a calculation on the basis of existence of a constant minimum tension $H \text{ = your} \;T_0$ at cable bottom point of horizontal tangency at $V$.
An assumption is made that the tension $T$ increases from minimum $H $ and increases as the arc length from bottom point V can be represented by three forces of a right triangle.
We assumed from physics of static equilibrium that
$$ T^2= H^2+ W_{cable}^2 \text { or} \quad T^2= H^2 +{\int q ds} ^2$$
That is, we made an a priori underlying starting physics assumption that at point $V$ it is given:
$$T=H$$
and so far it is not a consequence of any mathematical calculation.
The calculation is carried out thus:
$$ \tan \phi =\dfrac{\int q ds}{H} ;\text{ Let} \dfrac{H}{q}= c \tag 1$$
Differentiate w.r.t $x$ on which it is primed,
$$ \sec ^2 \phi \dfrac{d\phi}{ds}\cdot \dfrac{ds}{dx}=\dfrac{\sec \phi}{c}, \tag2 $$
$$\text{because} \cos \phi= \dfrac{dy}{ds} \text{ and } \dfrac{d\phi} {ds}= \dfrac{y''}{({1+y^{'2}})^{3/2}} ,\tag3 $$
it leads to the ode of catenary
$$\dfrac{y^{''}}{\sqrt{1+y^{'2}}}=\dfrac{1}{c}\tag4 $$
Let $$y'= \sinh \dfrac{x}{c} \tag5 $$
RHS of (4) then reduces to $1/c$ okay.
Integrating the above with IC $ x=0, y=c$ we can define shape of catenary:
$$ y=c \cosh\frac{x}{c} \tag6 $$
At any point including V ( where it seems indeterminate but is actually a constant)
$$H= \dfrac{W }{\tan \phi}. \tag7$$