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

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 = (Mg)/(k)`

`L_(eq) = L_(0) + x`

so

`L_(eq) = L_(0) + (Mg)/(k).`

Step-by-step 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.