# Does it take more force to press two springs in series?

Let's say I have a spring that takes force $$F_1$$ to fully compress it from a relaxed state with spring constant $$k_1$$ and total displacement from relaxed state $$x_1$$, which I believe is related through Hooke's Law $$F_1 = k_1 x_1$$.

Now say I have a second spring with $$F_2 = k_2 x_2$$.

Then if I put one spring on the other in series to make some combined string (label it $$3$$), is the force required to compress the two-spring system equal to $$F_{3} = \frac{k_1 k_2}{k_1 + k_2} (x_1 + x_2)$$? If so, is this more or less force required compared to either springs $$1$$ or $$2$$?

I tried comparing and it seems like $$3$$ takes less force than $$1$$ if $$1$$ takes more force to compress by itself than $$2$$, and $$3$$ takes less force than $$2$$ if $$2$$ takes more force to compress by itself than $$1$$. Is this right?

• The force required to compress the two spring system how much? $x_1$, $x_2$ or $x_1 + x_2$? – JMac Oct 8 at 19:13
• @JMac I am interested in full compression, if the first spring compresses fully by $x_1$ by itself, and a different/second spring compresses fully by $x_2$ by itself, then I assume the system in series can compress fully over a distance $x_1 + x_2$ correct? – user709833 Oct 8 at 19:17

I think your analysis is fairly correct, I'll try to explain why it makes sense, and what falls a bit short.

When you have springs in series, the force applied acts on each spring. For the case where you have two identical springs ($$k_1 = k_2$$, $$x_1 = x_2$$) in series, you can see that to get the displacement $$x_1 + x_2$$, $$F_1 = F_2 = F_3$$. This is because since the force acts on each spring in series, to compress both by that amount, you can just apply that same force.

When the two springs are different, it doesn't work out that nicely. Your analysis shows that the force $$F_3$$ will always in between the two $$F_1$$ and $$F_2$$ which makes sense, at least mathematically.

The issue is with the wording fully compressed. When dealing with springs that cannot compress beyond their maximum value (i.e. $$x_1$$ is as far as spring 1 will compress), then the simple math doesn't actually work.

In such situations, the force would always be the force required to compress the stiffest spring to it's maximum value. That is because applying more force to a fully compressed spring does not compress it any further.

In the case where $$k_2 > k_1$$, for example, when you apply $$F_1$$, the spring with $$k_1$$ compresses fully, and the other spring only compresses some. If you add more force, spring 1 doesn't compress any more; but the equation you were using doesn't account for that. Instead, you are analyzing the situation where $$F_3$$ is applied ($$F_1 < F_3 < F_2$$), and the math would say that spring 1 is compressed more than $$x_1$$ when $$F_3$$ is applied (and spring 2 is compressed less than $$x_2$$; but the total compression is still $$x_1 + x_2$$) but if $$x_1$$ is the maximum compression, that math obviously doesn't represent reality.

Once you recognize that spring 1 can't actually compress more than $$x_1$$, and spring 2 can't compress more than $$x_2$$ you should be able to see that therefore, to displace both springs in series as much as $$x_1 + x_2$$, you actually need to apply the force required for the stiffer spring; not the in-between force $$F_3$$.

Sorry this got long winded, I just really like springs.

• So are you saying that $F_3 = \max(F_1, F_2)$? – user709833 Oct 8 at 20:21
• @user709833 Exactly. It should make sense too, since the force applied is the force acting on each spring, and you know that to compress the stiffer spring fully, you need to apply that max force. Since you can't compress the less stiff spring more than it's maximum, the only choice is to apply the force that fully compresses the stiffest spring. – JMac Oct 8 at 20:26
• I assume this also doesn't matter with the order either? If the stiff spring $2$ is in top position in the system, compressing the whole system will take $F_2$, and likewise if that stiff spring is on the bottom of the system and the more flexible spring is on top? – user709833 Oct 8 at 20:28
• @user709833 Yup, exactly. Hopefully you can picture some springs in your head to see how that makes sense (but I'm probably just the weird one for doing that). – JMac Oct 8 at 20:31

To compress two springs by a distant of $$\Delta$$x each will require more force than to compress one spring by $$\Delta$$x. But to compress both springs by a total of $$\Delta x$$ will probably be easier than compressing one spring by $$\Delta x$$

Since the force required to compress a spring depends on x, the further you compress it the more force is required.

• Each spring has a different displacement, $x_1$ for one spring, $x_2$ for another, and then combined $x_1 + x_2$ – user709833 Oct 8 at 18:57
• thats true if k's are different. – jmh Oct 8 at 19:02

If you are pressing the two-spring system by the same amount as the one-spring system, then it takes less force. What you are asking, mathematically, is if $$\frac {k_1 k_2}{k_1 + k_2} \leq k_1$$, always. The answer is yes, since this is the same as $$\frac{k_1}{\frac{k_1}{k_2} + 1}$$; since $$k_1$$ and $$k_2$$ are both positive, it should hopefully be clear why this is less than $$k_1$$.

• I don't understand why these answers are assuming a different context than the one I stated in the post. Am I being unclear anywhere? One spring compresses $x_1$ by itself, another compresses $x_2$ by itself. When hooked up together I assume that means the resulting system compresses by $x_1 + x_2$. – user709833 Oct 8 at 18:59
• Yes, you are being unclear. If we don't make these assumptions, then the question is ill-posed, because it doesn't make sense to ask which situation takes more force if you don't specify information about the new length. – Danny Oct 8 at 19:27
• When I say "compress by some amount" I mean fully compress, like if you put a spring on a table upright and pressed down on it until it squishes to some minimum height. If we know spring $1$ and $2$ individually I want to know something about what happens when you compress the system fully in series. – user709833 Oct 8 at 19:30
• It still depends on what $x_2$ is. The maximum compression you're fully describing is a property of each spring, and is not something you get from analyzing the situation with only Hooke's Law. – Danny Oct 8 at 19:44

So you are in a situation like this:

Split the springs into their own free body diagrams

and notice that they share the same force $$F$$ and that the combined displacement is the sum of each displacement

$$\delta = \delta_1 + \delta_2 = \frac{F}{k_1} + \frac{F}{k_2}$$

The above is solve for the force to get the combined stiffness

$$F = \frac{1}{\frac{1}{k_1} + \frac{1}{k_2}} \delta$$

Notice that $$\frac{1}{\frac{1}{k_1} + \frac{1}{k_2}} < \mathrm{min}(k_1,\,k_2)$$ so it takes less force to compress the same as any individual spring.

If I put one spring on the other in series to make some combined string (label it $$3$$), is the force required to compress the two-spring system equal to $$F_{3} = \frac{k_1 k_2}{k_1 + k_2} (x_1 + x_2)$$?

This is already practically answered by ja72 but here is a more efficient way.

Since the tension in the springs are the same we have $$k_1x_1=k_2x_2$$ The total extension of the block is $$x_1+x_2$$, thus by using $$F=k_{\text{eff}}x_{\text{tot}}$$ we have $$k_{\text{eff}}=\frac{F}{x_{\text{tot}}}=\frac{F}{x_1+x_2}=\frac{1}{x_1/F+x_2/F}=\frac{1}{1/k_1+1/k_2}=\frac{k_1k_2}{k_1+k_2}$$ To find the force displaced we simply use Hooke’s law ($$F=kx$$) and for a arbitrary distance $$\Delta{x}$$ the force applied will be $$F=\frac{k_1k_2}{k_1+k_2}\Delta{x}$$, since you are asking for it to be fully compressed then you would indeed get $$F=\frac{k_1 k_2}{k_1 + k_2} (x_1 + x_2)$$

Which springs take the most amount of force to compress?

Basically the question you are asking is if $$F=\frac{k_1 k_2}{k_1 + k_2} (x_1 + x_2)$$ is greater than $$k_1 x_1$$ or $$k_2 x_2$$. Well this simply depends on what $$k_1$$ and $$k_2$$ are! You can see this if we make the expressions $$\frac{k_1 k_2}{k_1 + k_2} (x_1 + x_2)=k_1x_1$$ $$\frac{k_1 k_2}{k_1 + k_2} (x_1 + x_2)=k_2x_2$$ there is clearly some $$k_1$$ and $$k_2$$ value where the forces will be equivalent (I will not do the math as it is too much work but I urge you to try it).

Again some of the other answers may contain the information I have put out but I just want to be precise in answering the question.