# Hooke's Law simple question

Just got introduced to Hooke's law

Simple Q: Three light springs, each of natural length l and with spring constant k, are arranged vertically between two points a distance 3l apart. One end of the first spring is fixed to a point on the ceiling. A weight of mass m is connected between the other end of the first spring and the upper end of the second spring. Another weight of mass is attached between the second and third springs. The lower end of the third spring is attached to a point on the floor, directly below the original point on the ceiling. Find the equilibrium positions of the weights.

Solution:

Let the upper weight be a distance $l + x_1$ below the ceiling and let the second weight be a distance $2l + x_2$ below the ceiling. Let us consider the forces acting on the upper weight:

$mg + k(x_2 -x_1) = kx_1$

Second weight:

$mg = k(x_2 -x_1) + kx_2$

before I go on with the solution I'm already confused.

Where did they get $x_2 - x_1$ from? $T = k\delta l$ from hooke's law, but $\delta l$ is surely $x_1$ or $x_2$ where did they get $x_2 -x_1$ from?

## 1 Answer

Force on the upper weight consists of there components:

1. Gravity $mg$
2. Tension of upper spring $F=-k(l-(l+x_1))=-k x_1$
3. Tension of middle spring $F=k(\lambda-l)=k(x_2-x_1)$ where $\lambda=(2l+x_2)-(l+x_1)$ is actual length of middle spring.

Second weight is done similarly.

To understand the problem better you'd better make a drawing of the system and forces acting on each component.