Supporting a Parabola

Firstly, since the entire system is horizontally symmetrical we only need to consider the vertical component of the force applied.

The slope of the curve $y=x^2$ is $2x$ for any value of x, and so if $\theta$ is the angle of inclination then $\tan\theta=2x$ . Therefore, from the diagram, the vertical component of the force applied to an infinitesimal line segment with x coordinate $x_1$ in the normal direction is $\alpha\cos\theta=\frac{\alpha}{\sqrt{1+4x_1^2}}$ .

Now, consider the infinitesimal line segment to have a horizontal width of $dx$ and a vertical height of $dy$ .

By the pythagorean theorem, $\begin{aligned}(dl)^2&=(dx)^2+(dy)^2\\ dl&=\sqrt{(dx)^2+(dy)^2}\\ dl&=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\\ dl&=\sqrt{1+(2x)^2}dx\\ dl&=\sqrt{1+4x^2}dx\end{aligned}$

Now, to find the total force applied in the vertical direction (which will be the total force overall, since horizontal components on either side cancel out), we can integrate the formula for vertical force with respect to length $l$ : $\begin{aligned}F&=\int_{x=-1}^{x=1}\frac{\alpha}{\sqrt{1+4x_1^2}}dl\\ &=\int_{-1}^1\frac{\alpha}{\sqrt{1+4x_1^2}}\sqrt{1+4x^2}dx\\ &=2\int_0^1\alpha dx\\ &=2\big[\alpha x\big]_0^1\\ &=2\alpha\end{aligned}$ However, this total upward force must equal the weight of the wire, which is $1\text{kg}\times10\text{m/s}^2=10$ , and so $2\alpha=10, \therefore \boxed{\alpha=5}$