As part of a new project I’m working on, I need to replace a spring with other means of force application, for that I need to measure how much force a certain spring applies. I devised a plan.
Hooke’s law dictates that the force applied by a spring is proportionate to its change in length (either stretched or compressed) relatively to its idle length.
F=-kX , where F is the force applied, k is the spring’s stiffness constant and X is the difference in length (the “-” sign indicates that the force is applied in the opposite direction of the change in length).
This makes our life very easy. All we need to do is apply a known amount of force to the spring, measure the difference in length and calculate the the k constant.
Conveniently, I had a 2kg scale weight just lying around (we know that gravity pulls the weight with approx. 19.6N of force), a threaded hook, vice grips and a caliper.
First, I needed to know the idle length of the spring, for that I measured it’s length while not applying any pressure on the spring (important). Next, I put the spring on a hook’s stem between a nut and a washer (for bigger diameter springs use a washer on either side of the spring) and hung the 2Kg weight from it. Notice that the threads are covered with a smooth plastic sleeve to prevent friction from messing with the measurements.
Now I could measure the length of the spring while applying about 2Kg of force (total accuracy is not important), compare it to the idle length and calculate the spring’s constant. After that, I could measure the spring’s length in working configuration and measure the amount of force it applies.
These are my measurements:
Idle length: 2.49cm
Length compressed by 2Kg: 1.53cm
Length in working configuration: 1.37cm
From the above measurements I can derive k: 19.6=k*(0.0249-0.0153) => k=2041.66N/m. And finally, find the force of the spring in working configuration: F=2041.66*(0.0249-0.0137)=22.87N (~2.33Kg).
I have two springs so if I want to replace them, I need to apply about 45.74N or 4.66Kg.
