Question

Suppose a ball with mass M hangs vertically from a spring with stiffness k and relaxed length L0. At what length Leq will the ball encounter equilibrium?

Answer 1

Answer:

At equilibrium, the sum of vertical forces are equal to zero. Let's analyze the downward and upward components of the forces.

Downward forces: $F_{down} = Mg$ (Gravity)($g$ is the gravitational constant.)

Upward forces: $F_{up} = kx$ (Spring)

Equilibrium condition: $Mg - kx = 0$.

$x = \frac{Mg}{k}$

$L_{eq} = L_{0} + x$

so

$L_{eq} = L_{0} + \frac{Mg}{k}.$

Explanation:

If there is no masses hanging from the spring, the length of the spring is equal to $L_{0}$. When there is a mass M hanging, the spring will be stretched by an amount of $x$. This distance can be found by using the equilibrium condition, that the sum of the forces is equal to zero.