[Physics] Does it take more force to press two springs in series

newtonian-mechanicsspring

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?

Best Answer

I will answer your first and second question in order.

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.

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