Question about springs in series

In summary, the problem involves a weight and two identical springs in series. The weight exerts a force on the top spring, causing it to compress by 5 inches. This same force is also exerted on the bottom spring, causing it to also compress by 5 inches. Therefore, both springs in series are compressed a total distance of 10 inches. This is because the force is balanced between the two springs in equilibrium.
  • #1
Ovan
3
0
Please help with this question, and if possible, please explain why.

If one spring is compressed 5 inches by a 10 pound weight. How much is two identical springs in series compressed by the same weight (total compressed distance)?
 
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  • #2
I'll try to ask you questions to lead you to the correct answer.

First, the way I understand the problem you have 4 objects, the weight, top spring, bottom spring, ground. Assuming that nothing is accelerating, what forces act on each object?
 
  • #3
The force of the weight.

And am I correct to understand that this force acts equally on each spring?
Which would mean that both springs (in series) are compressed the same distance as the individual spring is. Which in return would mean that the springs in series would be compressed a total distance of 10 inches(twice as much as the single spring).

Is this correct?
 
  • #4
If one spring with a certain compression C exerts force F, then what force would be produced by two springs, each with compression C?
 
  • #5
Ovan said:
The force of the weight.

And am I correct to understand that this force acts equally on each spring?
Which would mean that both springs (in series) are compressed the same distance as the individual spring is. Which in return would mean that the springs in series would be compressed a total distance of 10 inches(twice as much as the single spring).

Is this correct?

It is basically correct.

The weight acts on the top spring. In equilibrium, there is no acceleration, so that force is balanced by the force of the top spring on the weight.For that, the spring has to be shortened by ΔL=F/k. (F is the external force, the weight of the object)

The tension is the same in the whole spring, so the top spring acts with force F at the top of the bottom spring. The bottom spring balances that force in equilibrium, so it exerts F force on the top spring. For that it has to be shortened by ΔL=F/k.

ehild
 

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  • #6
Thank you for your answer and explanation, much appreciated!
 

FAQ: Question about springs in series

How do springs behave when they are connected in series?

When springs are connected in series, they behave as a single, longer spring with a combined spring constant. This means that the total force required to stretch the springs will be equal to the sum of the individual forces needed to stretch each spring.

How does the spring constant change when springs are connected in series?

The spring constant of a series of springs is equal to the inverse of the sum of the reciprocals of each individual spring's constant. In other words, the spring constant decreases when springs are connected in series.

Is there a limit to the number of springs that can be connected in series?

Technically, there is no limit to the number of springs that can be connected in series. However, as more springs are added, the overall spring constant decreases and the system becomes more difficult to stretch.

How does the stretching distance change when springs are connected in series?

The stretching distance of a series of springs is equal to the force applied divided by the combined spring constant. This means that the stretching distance increases when springs are connected in series, as the spring constant decreases.

Can springs in series be used to create a stronger spring?

No, connecting springs in series does not create a stronger spring. While it may increase the overall force required to stretch the springs, the maximum stretch distance and potential energy of the system will remain the same. To create a stronger spring, the individual springs must be combined in parallel.

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