As discussed in Halliday and Resnick’s “Fundamentals of Physcis,” Hooke’s law states that the formula relating the force a spring exerts, as a function of its displacement from its equilibrium length, is force F = -kx. x here is a measure of the displacement of the free end of the spring from its unloaded, unstressed position. k is a proportionality constant called the "stiffness," and is specific to each spring. The minus sign is in front because the force that the spring exerts is a “returning” force, meaning that it opposes the direction of displacement x, in an effort to return the spring to its unloaded position. The spring equation usually holds for displacement x in both directions--both stretching and compressing displacement--although there can be exceptions. If you don’t know k for a specific spring, you can calibrate your spring using a weight of known mass.
The formula is accurate only up to a point. For large x it will not be accurate. Displacement x in both directions will not necessarily effect the same magnitude of restoring force, e.g. if the turns of the spring's coil are tightly bound in the relaxed, equilibrium position.
Determine the position of the free end of the spring, if hanging loosely--its other end affixed to something solid like a wall.
Decide what displacement x from the equilibrium position you want to know the spring force, measuring it in meters.
Multiply x by -k to find the force that the spring exerts to try to return to its equilibrium position. If x is in meters and k is in kilograms per second-squared, then force F is in Newtons, the SI unit for force.
If you don’t know k, proceed to the next step to determine it.
Find the spring’s proportionality constant k by hanging a weight of known mass m, preferably in kilograms, from the free end of the spring, after positioning it vertically. From the resulting displacement, you can determine k by the relation k = -mg/x, where g is the gravitational acceleration constant 9.80m/s^2, where the caret ^ indicates exponentiation.
For example, if the spring displaces x = 5 centimeters under a load of 5 kilograms, then k = - 5kg x 9.80 m/s^2 / (-0.05m) = 980 kg/s^2. So then you can subsequently solve for its restoring force F when the displacement x is, say, 10cm, as follows F = (-980 kg/s^2)(0.10m) = -9.8 Newtons.
- "Fundamentals of Physics"; David Halliday and Robert Resnick; 1992
- Georgia State U.: Elastic Potential Energy
- spring image by Edsweb from Fotolia.com