Spring Compressed on Parabola

The slope of the function $y=x^2$ is $2x$ , so that at $x=\dfrac{1}{2}$ , the slope is $1$ , and the distance between the beads is $1$ m initially.

Each $1$ kg bead has a downward force of $10$ N, so initially the force compressing the spring is also $10$ N where the slope is $1$ .

(If the two beads are pushing towards each other with vector forces of $F, -F$ , then the compression force is $F$ , not $2F$ )

If $x$ is the distance from the $y$ axis to one end of the spring, then $2x$ is the total length of the compressed spring, and the force required to hold it at that length instead of $1$ meter is, using Hooke's Law where the spring constant is $5$ N/m

$F_{spring}=5(1-2x)$ N

The horizontal force exerted towards the $y$ axis by a $1$ kg bead on the parabola is (see Note below about this)

$F_{bead}=10(2x)$ N

Solving $F_{spring}=F_{bead}\;$ gets us $x=\dfrac{1}{6}=0.1666....$
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Note: $\;$ Given a tangent line of slope $s=\dfrac{y}{x}$ , if a bead travels down along that tangent, then the work (energy) it does pushing in the $x$ direction is the same as the work (energy) gravity does on the bead in the $y$ direction, using the Conservation of Energy law. $\;\;$ Work (energy) equals Force times Distance, so that we have $F_x x=F_y y$ , $\;$ or $F_x=F_y \dfrac{y}{x} = F_y s$ . $\;\;$ Here, in this problem, $F_y = 10$ N, and $s = 2x$ . $\;\;$ What could be confusing is that the term $s=2x$ looks like distance, but, here in this problem, it's a ratio, not distance.
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Addendum: An alternative way (per Bart Meijer) of solving this would be to look at this using work done on the spring and change in gravitational energy of the beads, which are, respectively

$W= \dfrac{1}{2}kx^2 = \dfrac{1}{2} \cdot 5 \cdot (1-2x)^2)$

$PE=mgh=2 \cdot 10 \cdot ((\dfrac{1}{2})^2-x^2)$

For equilibrum, the equation to solve would be

$\dfrac{d}{dx}( \dfrac{1}{2} \cdot 5 \cdot (1-2x)^2) = \dfrac{d}{dx}(2 \cdot 10 \cdot ((\dfrac{1}{2})^2-x^2)$

which yields the same result, $x=\dfrac{1}{6}$

A bead rests when the sum of the spring force $F$ and the gravitational force $G$ is perpendicular to the wire. Sorry, I'm not drawing here - please do or imagine :)

Let $\alpha$ be the angle between the tangential at the resting point and axis x. The slope of the tangential $tg(\alpha)$ is the first derivative of the function $f(x)=x^2$ .

$tg(\alpha)=2x$

from similar triangles:

$tg(\alpha)=\frac{F}{G}$ , where $F=kl$ and $G=mg$ , ( $l$ being the distance the spring is compressed: $l=L-2x$ )

so

$2x=\frac{k(L-2x)}{mg}$

$2x=\frac{5(1-2x)}{10}$

$x=\frac{1}{6}$

The slope of the parabola at any point $(x,x^{2})$ is $2x.$ Let's focus on the right half of the picture.

The force of gravity at any point is $10N$ straight down. We can break this into components. One is perpendicular to the wire the other parallel to (along down) the wire. We only need the latter, which is $10\sin(\tan^{-1}(2x))$ (This can simplify but I plan to solve by graphing, so I'll leave it like this.)

I didn't know a spring's force was proportional to the distance from rest, but anyway, this means the force from the compressed spring is $10(.5-x)N$ straight to the right. We can break this into components. One is perpendicular to the wire the other parallel to (along up) the wire. We only need the latter, which is $10(.5-x)\cos(\tan^{-1}(2x))$

To be in equilibrium, graph these and find the point of intersection: $(.16666666,3.1622777)$

So the solution is $x=0.1666667$

The spring is compressed.

tana = 2x

F = N.sina

P = N.cosa

=> F/P = tana <=> k.(l-2x) = mg.2x

<=> 5(1-2x) = 1.10.2x

Solve the EQN => x = 0.1666667

In my opinion it's easier to think of such problems in terms of energy rather than force.

Here there are two potential energy one from gravity ( $mgh$ ) and one from the spring ( $\tfrac{1}{2} k s^2$ ). Let $y_b$ be the height of the beads and $x_b$ be the $x$ -coordinate of the right most bead. Then $mgh = 20 y_b$ and $\tfrac{1}{2} k s^2 = \tfrac{1}{2} 5 (1 -2 x_b)^2$ . So the total energy is $E = 20 y_b + \tfrac{1}{2} 5 (1 -2 x_b)^2$ . Using that $y_b = x_b^2$ one gets $E = 20 x_b^2 + \tfrac{1}{2} 5 (1 -2 x_b)^2$ .

The system is stable when the derivative of energy is 0: $0 = 40 x_b -10 (1-2x_b)$ . Solving for $x_b$ yields $\tfrac{1}{6}$ .