Re: Why is only 1 force taken into account when there are actually 2?

The question: Assume an upright spring with a weight G on top of it. According to a statics FBD there are two forces acting upon the spring: G (weight) downwards and F(the earth's support upwards). But the displacement from equilibrium is only G/-k not G+F/-k why?

Hello Kalle,

I did a lot of head scratching on this one! I couldn’t really find a good explanation, other than the fact that in a static situation, or a ‘force couple’, only the applied force enters the equations, not the reaction force. It seems to me that it must be accepted by convention to solve the problem.

The force ‘F’ is called a ‘reaction’ force. Notice that its direction is opposite (and equal) to force ‘G’**. In any static application of a force to an object, there must be some type of reaction force equal and opposite to the applied force. If there was not a reaction force, then the applied force would cause the object to accelerate. The reaction force is not an additional applied force, and is not added to the compression calculation.

In your example, I think that gravity, and the earth’s upwards support may be making this hard for you to visualize as to why F does not enter the equation. Consider the following example: Take your same spring and attach one end to a vertical wall. Now, press the spring towards the wall using the same amount of force as weight G. The spring will compress to the same height as before, indicating that the total forces applied are exactly the same**. It should be clear in this example that there is only one force being applied to the spring, yet there will be the same reaction force at the wall. That is, the wall will “push back” on the spring with a force equal to G. And again, the displacement will only be G/ -k.

** (In reality, the reaction force F would be somewhat larger than G because it would include the weight of the spring itself. However, in most learning examples we assume that the weight of the spring is zero, for simplicity).

Another good analogy is a tug-of-war contest with an elastic rope: Side ‘A’ pulls with force X. If the rope is at a standstill, then side ‘B’ must also be pulling with force X (or –X). Now tie one end of the rope to a tree, send the A team home, and don’t tell the B team. In either situation, the tension on the rope is X.

I hope that this helps you.

Best Regards,

Jay Shapiro

[note added by MadSci Admin: In addition, consider that in most spring-compression situations, this one included, we take the compression with one end of the spring fixed, so that the compression distance is relative to the fixed end. An alternative approach would be to consider the compression relative to the center of the spring, in which case you actually can include the reaction force. However, in that case each half of the spring is considered separately, and you know, I assume, that a spring that is half as long as the original will have a spring constant that is twice as big! (see this previous answer in our archives) So each of G and F will compress the spring only x/2, but since the compression is happening from each end (towards the center), the total compression of one end relative to the other end is again x!]