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.
Which springs take the most amount of force to compress?
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.