Worked Exercise — Ball Between Two Parallel Springs

Problem

A ball of mass $m = 20\;\text{kg}$ is squeezed between two vertical walls, placed $10\;\text{m}$ apart, and connected to each one by a horizontal spring. The springs have zero natural length, $\ell_0^{(1)} = \ell_0^{(2)} = 0$ (they are ideal elastics “at rest at zero”), with constants $k_1 = 100\;\text{N/m}$ and $k_2 = 200\;\text{N/m}$. At what distance $\ell_1$ from the left-hand wall is the ball at equilibrium?

Solution

Step 1 — Forces on the ball. Let $\ell_1$ be the length of the left spring and $\ell_2$ that of the right one. With zero natural length, the horizontal forces are $F_1 = k_1\ell_1$ (leftward) and $F_2 = k_2\ell_2$ (rightward).

Step 2 — Translational equilibrium. The sum of the horizontal forces must vanish:

$-k_1\ell_1 + k_2\ell_2 = 0 \;\Rightarrow\; 200\,\ell_2 = 100\,\ell_1 \;\Rightarrow\; \ell_1 = 2\,\ell_2$

Step 3 — The geometric equation. One equation with two unknowns is not enough: the third relation comes from the geometry of the diagram, since the sum of the two spring lengths is the distance between the walls:

$\ell_1 + \ell_2 = 10\;\text{m}$

Step 4 — Solving. Substituting $\ell_1 = 2\ell_2$:

$3\,\ell_2 = 10 \;\Rightarrow\; \ell_2 = 3{,}33\;\text{m}, \qquad \ell_1 = 2\ell_2 = 6{,}66\;\text{m}$

The requested distance from the left-hand wall is therefore

$\ev{\ell_1 \approx 6{,}66\;\text{m}}$

Teaching note: the problem could not be solved without the diagram, which provides the geometric constraint.

Links

Topics: Statica ed equilibrio Concepts: Equilibrio statico · Forza elastica e legge di Hooke Skills: Scomposizione vettoriale Objects: Molla