I have an interesting question. I tried to analyze the equivalent spring constant of a multiple spring system.

When multiple springs are in parallel and have the same spring constant, all the springs compress/extend by the same distance when a force is applied. Hence, the equivalent spring constant is simply the applied force divided by the distance from equilibrium. $$F_{spring} = k\Delta x$$ for each individual spring. To continue, most sources mention that the sum of the forces exerted on each spring is the overall exerted force on the system.

However, I do not fully understand this. Why is it not possible for the overall force to simply act on all the springs undiminished?

In a similar vein, why is the force exerted on one spring in a series system transmitted undiminished to all the other springs?

Lastly, I was thinking about what would happen if the spring constants of the springs were not the same. For the system in parallel, if we placed a block on the ends of the spring before we started, the block would tilt if a force were to be exerted, due to the uneven spring constants. How would I hence go about thinking about such a system?

  • $\begingroup$ I believe I have an insight into this after thinking for a bit - for the springs in parallel, they exert a force on the block pushing them. If each spring pushed the block with the magnitude of the force that the block exerted on them, the block would have an overall net force in the other direction, which does not make sense. However, if the net force on the block is always balanced, does it mean it cannot accelerate? $\endgroup$ Feb 8 at 10:43
  • $\begingroup$ I was writing my answer while you were writing your comment. Our arguments are essentially the same. But – thinking about your last sentence – the blocks/bars don't have to be in equilibrium: if we start to the springs (or to stretch them more) then the bars are accelerating and the force we apply to each bar is somewhat greater in magnitude than the sum of the forces applied by the springs. $\endgroup$ Feb 8 at 11:15
  • $\begingroup$ @PhilipWood Does this have to do with the fact that in actuality, the bars have mass and can be accelerated? (i.e. I apply a force to the bar and the bar applies the same force on me, but it transmits less of that force to the springs, because it can have a net force to induce acceleration.) $\endgroup$ Feb 8 at 23:07
  • $\begingroup$ Yes. You have explained it very nicely. $\endgroup$ Feb 9 at 9:00

2 Answers 2


"Why is it not possible for the overall force to simply act on all the springs [in parallel] undiminished?"

Suppose that there are N springs and that they are vertical and run between two light horizontal bars to which they are attached at each end. To apply tension to then system we apply equal and opposite vertical forces, $±\mathbf F$ to the middle of the two bars. In equilibrium the net force on each bar must be zero, so each spring must exert $\mp \frac{\mathbf F}N$ on the bar. If the force exerted by each spring were $-\mathbf F$, the system could never be in equilibrium. [I'm neglecting gravitational effects. If you don't like this, turn the system through 90° so the springs and bars are in a horizontal plane.]

Try applying a similar argument for the springs in series.


We can discuss this problem using only few principles, like equilibrium, action/reaction, and degrees of freedom and constraints of the systems (along with the level of detail/abstraction). IMHO, this is the most important thing to understand. Once understood, everything else is algebra.

In what follows, I'll assume that a spring keeps each direction while we deal with the motion of points along a desired direction.

Point system with no dimension

At this level of description, the system has only 1 degree of freedom, i.e. the motion of the point along an axis. Parallel springs "are connected" on the very same point, since all the dimensions are negligible. Thus, we only need to care about the translation equilibrium, providing

\begin{equation}\begin{aligned} F^e & = (k_1 + k_2) \Delta x = k_{//}^{equiv} \Delta x && \text{parallel} \\ \Delta x & = \left(\frac{1}{k_1}+\frac{1}{k_2} \right) F^e = \frac{1}{k^{equiv}_{series}} F^e && \text{series} \end{aligned}\end{equation}

Finite dimensional system

With a finite dimensional system rigid system, connected to the ground with springs, we need to deal with both translation and rotation. The system may also be constrained to remove rotation. Let's study a system that is free to rotate, and another system with constraints setting rotation to zero.

System 1. Free rotation. We need to write equilibrium for both translation and rotation of the system, subject to external force and spring reactions.

enter image description here

\begin{equation}\begin{aligned} F^e & = k_1 (x+H\theta) + k_2 x && \text{translation} \\ 0 & = F^e (H-h) -k_2 x H && \text{rotation} \\ \end{aligned}\end{equation}

providing \begin{equation}\begin{aligned} x & = \left( 1- \frac{h}{H} \right) \frac{F}{k_2} \\ \theta & = \left[ \frac{1}{k_1}\frac{h}{H} - \frac{1}{k_2}\left(1-\frac{h}{H} \right) \right] \frac{F}{H} \end{aligned}\end{equation}

from which you can evaluate the positions of all the points of the rigid body: as an example, the displacement of the upper spring is $x_1 = x + H \theta$, and the center $x_C = x + \frac{H}{2} \theta$, or the point of application of force $x_F = x + \frac{h} \theta$

\begin{equation}\begin{aligned} x_1 & = \left( \frac{1}{k_1}\frac{h}{H} \right) F \\ x_C & = \frac{1}{2} \left( \frac{1}{k_1}\frac{h}{H} + \frac{1}{k_2}\left(1-\frac{h}{H} \right) \right) F \\ x_F & = \dots \\ \end{aligned}\end{equation}

System 2. Constrained rotation. We need to write equilibrium for both translation and rotation, subject to external force, spring reactions and constraint reaction.

enter image description here

\begin{equation}\begin{aligned} F^e & = \left( k_1 + k_2 \right) \Delta x = k^{equiv}_{//} \Delta x && \text{translation} \\ M & = \left[ h + H \frac{k_1}{k^{equiv}_{//}} \right] F && \text{rotation} \end{aligned} \end{equation}

The moment provided by the constraint $M$ guarantees the rotation equilibrium.


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