I'm trying to determine the shape of the cable of a suspension bridge, where it is assumed that the total weight of the load, the cable and the roadway per horizontal distance is constant.
In my attempt, I first considered an infinitesimal element $dx$. Assuming the element is stationary, all the forces affecting it must add up to zero (Newton's second law). Thus we get the following expressions for the vertical and horizontal forces
$\begin{cases} T_1sin(\theta+d\theta)=T_2sin(\theta)+W \\ T_1cos(\theta+d\theta)=T_2cos(\theta) \end{cases} \tag{1}\label{1}$
where $T_1$ is the tension force pulling $dx$ upwards and to the right, $T_2$ is the tension force pulling $dx$ downwards and to the left. $W$ is the load that $dx$ experiences and can also be rewritten as $W=\rho dx$ ($\rho$ is the load per horizontal distance). The diagram of the forces is shown below.
Combining the equations from $\eqref{1}$, I get
$$ tan(\theta+d\theta)=tan(\theta)+\frac{W}{T_2cos(\theta)}, \tag{2}\label{2}$$
$$ \frac{tan(\theta+d\theta)-tan(\theta)}{d\theta}=\frac{W}{T_2cos(\theta)d\theta}. \tag{3}\label{3}$$
At this point, I'm using the definition of the derivative to rewrite $\eqref{3}$ as
$$ \frac{d}{d\theta}tan(\theta)=\lim_{\theta \to 0} \frac{\rho}{T_2cos(\theta)} \frac{dx}{d\theta}, $$
$$ sec^2(\theta) = \lim_{\theta \to 0} \frac{\rho}{T_2cos(\theta)} \frac{dx}{d\theta}, $$
but this doesn't get me anywhere. I then try a different tactic and rewrite $\eqref{2}$ as
$$ y'(x+dx)-y'(x)=\frac{W}{T_2cos(\theta)}, $$
$$ \frac{y'(x+dx)-y'(x)}{dx}=\frac{\rho dx}{T_2cos(\theta)dx}, $$
$$ y''(x)=\frac{\rho}{T_2cos(\theta)}. \tag{4}\label{4}$$
I cannot integrate $\eqref{4}$ as there is an unclear dependence between $\theta$ and $x$. Alas, I'm stuck.
How can I get the shape of $y(x)$ using the method I outlined? I know that it's possible to take a special case $\theta=0$ (the vertex of the parabola) and the answer is immediate. But I would like to see a derivation for an arbitrary point $x$, not involving the vertex.
Best Answer
HINTS:
Two variable tensions in the cable are not suitable to fix up a differential equation /relation because we need to compute many steps at many points.
Only one variable cable tension $T$ should be involved, along with an external variable cable self weight $W$ .
The triangle of forces including three forces should be drawn. I have used same length color and direction for representing force vectors. Horizontal force inside the cable $H$ is unknown at present.
$$ \tan \theta = \frac{W}{H}= \frac{\rho x}{ H}= \frac{dy}{dx} $$
Integrate with BC at bottom of parabola shape cable $ (x=0,y=0) $
$$ y = \frac{\rho x^2}{2H}; $$